A 548569 QA 154 B72a 1837 PT. A QA 154 B722 1837 UNIVERSITY OF MICHIGAN 3 9015 06363 2130 QA Harte 154 B720 1837 Ch p IM } M Bonnycastle John A KEY ! *} 1 * 3 LAST NEW YORK EDITION * BONNYCASTLE'S ALGEBRA; AND ALSO ADAPTED TO THE FORMER AMERICAN AND LATEST LONDON EDITIONS OF THAT WORK: ¿ TO THE or } CORRECT SOLUTIONS 1 } CONTAINING TO { ALL THE QUESTIONS. THE THE WHOLE RENDERED AS PLAIN AS THE PRESENT STATE OF THE SCIENCE WILL ADMIT, BY JAMES RYAN, AUTHOR OF AN ELEMENTARY TREATISE ON ALGÉBRA, THEORETICAL AND PRACTICAL THE NEW AMERICAN GRAMMAR OF ASTRONOMY; &c. > NEW YORK: W. E. DEAN, PRINTER AND PUBLISHER, 2 ANN-STREET, 1837. } 1 1 1 03 · ༨ } ENTERED, According to the Act of Congress, in the year 1825, by JAMES RYAN, 1 In the Clerk's Office of the District Court of the Southern District of NEW YORK. M ķ { Galé Pro, M. M. Kosau 4-12- 1 + CONTENTS. Page PRACTICAL EXAMPLES for computing the numeral Values of va- rious Algebraic Expressions, or Combinations of Letters 1 2 Addition Subtraction Multiplication Division Algebraic Fractions Involution Evolution W • Irrational Quantities, or Surds Arithmetical Proportion and Progression Geometrical Proportion and Progression Resolution of Simple Equations Quadratic Equations Questions producing Quadratic Equations Of Cubic Équations Solution of Cubic Equations Solution of Cubic Equations by Converging Scries Of Biquadratic Equations Resolution of Equations by Approximation Of Approximation by Position Exponential Equations Binomial Theorem • · • Indeterminate Analysis Diophantine Analysis Summation and Interpolation of Infinite Series Logarithms Multiplication by Logarithms Division by Logarithms Rule of Three by Logarithms Involution by Logarithms Evolution by Logarithms. Miscellaneous Examples Miscellaneous Questions Application of Algebra to Geometry * * • • • • * • • • 1 • • ► • • • • 19 31 33 36 54 55 ib. 86 92 106 . 108 113 * • ་ · • • • • • • 134 137 . 140 W 148 • • 3 4 9 . 160 . 185 . 203 ib. 204 ib. . 205 206 • • 124 131 W . 207 . 209 240 16071 A KEY bhavana Can THE LAST NEW-YORK EDITION Practical Examples for computing the numeral values of various Algebraic Expressions, or combinations of letters. } 14 14 ΤΟ BONNY CASTLE'S ALGEBRA. REQUIRED the numeral values of the following quan- tities; supposing a=6, b=5, c=4, d=1, and e=0. 1. 2a2+3bc-5d=72+60—5—127. a 5. 2 xda+2a²e= C 2. 5a2b-10ab2c-900-1500+0=-600. 3. 7a²+b-cxd+e=252+5-4+0=253. 4. 5√ab + b²-2abe² 5/30 +2560-0—— 7.613871. OF I 5+6+1 16 12 16 6 1 4. KEVIN 1 6. 21/c+2√(2a+b−d)=6+2/16=14. 7. a √ (a² + b²) + 3bc √ (a²—b²)=6 √61 +60 √11: 245.8589862. 1 A 2 8. 3a²b+3/(c² + √[2ac+c²)]=540+ (16+ √64) =540+ (16+8)=540+3/24-542.8844991. 3 26 +c √56+3 √c+d_10+4 √25+3√4+1 16 9. 3a-c 2a+c 1. +0=2, or 4' } 18-4 3 1 S 2 1 44 • 3 2 [ plakat شار 2 1 B } Ex. 1. 1+16 a - z b za Ex. 6. ! Ex. 2. 5x-3a +6 +7 -3x-4a+26-9 2x-7a+36 2 Ex. 3. 2a+3b-4c- 9 Ex. 4. 3a+26-5 5a-3b+2c-10 a+56-c 6a-2c+3 10a76-3c-2 CL 7a } ADDITION. EXAMPLES FOR PRACTICE. MPLES } Ex. 5. x3+ax² + bx+2 x³+cx² + dx-1 Ex. 8. -2c-19 2x³+(a+c) x²+(b+d)x+1 6xy-12x2 3xy- 4x2 -2xy + 4x2 -3xy+ 4x2 4xy - 8002 1 1 Ex. 7. 4ax+3x-130 3αx+5x²+9x² 1/21/18 7xy-4x²+90. -6x²+x²+40 7ax+8x²+7xy 2a2 3ab2b³ —3a² -2a²+ a³ +363-5c3 1005ab-263+4c3 16a²+20ab-bc -80.· 13a2+22ab+36³+a³—c³+20-bc / 3 } : Ex. 9. t Ex. 1. } + 5a 7√bc b a Œ 8a 7c² 12 √bc ö+ a OC Ex. 10. 13a f b 3c2 +. Za 3 a - 3 b +6 Ex. 3. 3a+ b + c−2d -8b-8c+2d 4c2 a } # 8 +3a+9c-4d Ex. 5. 20αx 5 √x+3α 4ax+5√x a чора 16ax-10√x+4a e²+10 4bc3a² 6bc-5a² + 2e2 -15 -9bc-4a² —10e²+21 bc-6a² 9e²+16 5√bc 00 SUBTRACTION. EXAMPLES FOR PRACTICE. gagn +6 (ab + x) d Add __.9 (ab+x) -9( -) d -3 (ab + x) d Ex. 2. 3x-2a-b+7 4x+a-3b+8 x-3a+26-1 ·3a+2b−1 Ex. 4. 13x²-2ax+ 962 5x2-7ax- .b2 ! 1 1 1 · Ex. 3. x²+xy+y² x² - xy + y² T ¸x²+x³y + x² y² — x³y — x² y² —xy³ +x²y²+xy³+ys 001 Ex. 4. 3x²-2xy +5 x²+2xy-3 1 * Ex. 5. 2a2-3ax+4x² 5a26ax- 2x² 3x²-2x³y+5x2 +x²y 2 I +6x³y—4x²y²+10xy -9x26xy-15 3x²+4x³y-4x²y²+16xy-4x²-15 10a¹-15a³x+20a²x² da !. Ex. 6. 5x3+4αx²+3a²x+a³ 2x²-3ax +a². 1 5 * 4a²x²+6ax³- 8004 10a¹-27a³x+34a²x²-18ax³ — 8x¹ 10006 7ax¹ +yk -12a³x+18a²x²-24ax³ S ļ 1 10x 8αx1 + 6a²x³ +2α³x² -15ax²-12a²x³- 9α³x²-3α*x +5a²x²+4a³x³+3a²x+a³ a²x³-3α³ x² { } 第 ​as • 6 ** Ex. 7. 3x+2x²²+3y³ 2x³-3x²y²+5y³ 6x+4x³y²+6x³y³ -9xy²-6x¹¹— 9x2y5 +15x³y³+10x²y³+15y¤ 6x6 — 5׳y²—6x¹y¹+21x³y³+x²y5+15y° Ex. 8. x³-ax² + bx-c x²-dx+e xax¹ + bx³- cx² 1 dx¹+adx³ —dbx²+dcx +ex³-αex²+bex 1 x5 ~ (a+d) x¹+(b+ad+e)x³—(c+da+ae)x²+ (db+be)x-ec. a³+a2b+ab² +a²b+ab²+b3 Ex. 9. 1st. Multiply the factors I. and II. a²+ab+b² a+b 1 J ec a³ a² bab² a²bab²-b3 _a³-2a2b+2ab2b3 a³+2a²b+2ab²+b³ Next multiply the factors III. and IV. a²-ab+b² a-b 1 3 £1 1 by It now remains to multiply the first product I. II. by the second product III. IV. a³+2a2b+2ab2+b³ a³-2a2b+2ab2-b3 a®+2ab+2a¹b2+ a³b³ ~2a5b-4a4b2—4a³b³ —2b2b4 +2a¹b²+4a³b³+4a²b¹+2ab5 a²+ab ασ -66 2d. Change the order of the question, that is, multiply the factors I. and III., then II. and IV. together. Then, a²+ab+b² a² — ab+b² a+b a-b a² S I -ab-b2 ¦ b2 23 a³b³ — 2a2b42ab5-b6 a² + a¹·b²+a²b4 K a³ — a²b+ab² +a²b―ab²+b3 a²+a³b+a²b² at +bk Then multiply the products I. III. and II. IV. +a2b2 a² + a ²¿ 2 + b ₁ a2b2 ~ a ª b² a ² b¹¹ 76 +63 - a³b — a²b² — ab³ ασ -66 which is the product. 3d. Again, multiply the I. factor by the IV., and next, the II. by the III. a² —ab+b² a+b M +a²b²+ab³+b4 next, a²+ab+b² a-b a³+a²b+ab² Qu · a² b — ab² — b3 -b3 I Jakobsdal f 8 ન + It remains to multiply the product I. IV. and II. III. a³+b3 a³-b3 ab+a³bs -a3b3f6 aG -bo as in the foregoing cases. / It will now be proper to illustrate this example by a numerical application. I *** Suppose a 3, and b=2. We shall have a+b=5, and a b=1: further, a²=9, ab=6, and 624. Therefore, a²+ab+b²=19, and a2ab+b27. So that the pro- duct required is that of 5x19x1x7=665. 1 a6 G Now, aº 729, and 66-64; consequently the product is aº — bộ—665, as we have already seen. Ex. 10. a³+3a²x+3αx²+x³ a³-3a²x+3ax2 003 a®+3αx+3α¹x²+a³x³ ! 1 -3α⁰x-9α¹²x²-9a³x³-3x²x² ==~ -3a4x2 Ma +3a¹x²+9α³x³+9a²x²+3ax5 Ex. 11. a*-|- a²c² + c² a² - c² p a¤ 3 a® + a¹c² + a²c4 achacha cô a³x³-3α²x²-3ax +3a²x ... Co } 1 * 111 I ха 006 i 1 1 1 9 } } Ex. 12. a² + b² + c² - ab-ac-bc a+b+c Ex. 2. -18axy -8ax a³+ab² + ac²—a²b—a²c-abc +a²b+b³+bc2 — ab² abc-b²c a²c + b²c+c³ — abc — ac²—bc2 . +b3 as a2x4 Ex. 1. 16x28x, or Kata * CASE I. When the divisor and dividend are both simple quantities. 16x2 12a²x² 80c 8α2x Poganjk * DIVISION. 15ay23ay, or 15ay÷—3ay, or + -3abc+c3 Ex. 4. 12a2b² — 3a²b, or 15ay 3ay¹ 1 21a²c²¿ª³ ————7αc²x¹³, or — Ta Ex. 5. -15a²x²-5ax², or dat =21xy. 4 Ex. 3. 201 30+ + + = = = = = = -3]; and azt÷3 -a 5 5212. 2x ; and -15ay2 3ay B =5y Aptitudeng 21a²c²a 12a2b2 3a2b }—1—5y³. Mang 7αc²xt 15a2x2 5ax² C T 3ax* 3x 2 5y; and ax ax [ -4b; and * 3a; and \ W ! 10 Ex. 6. — 17x-α³c÷ 17 ac² 17x ac H 5 5 ≈3√xy. Ex. 1. Here Ex. 2. Akemes +17. Ex. 4. Ex. 3. Here } Ex. 5. 2x+1. Here Here Here When the divisor is a simple quantity, and the dividend a compound one. Ex. 6. Here Ex. 7. Here Ex. 8. Here I Typ 5x3a²c, or · 蕊 ​1 1 .5 CASE II. ; and 24√xy÷8√xy, or 3x²+6x²+3ax-15x 3x 3abc+12abx - 9 a²b 17x13c 5x3a² 3ab 40x363+60a2b2-17ab ·ab 15a²bc-12acx²+5ad² 5ас ·= x²+2x+α-5. 6 b c d z+4b z d² — Qb²x² 2bz 14a2-7ab21ax-28a 7a -20ab60ab3 - 12a²b² -4ab Sp =c+4x-3a. 1 pettand 20ax³-15αx²+10x+5α 5a 24xy 8√x27 Mumbai Jalg ·40a2b2-60ab =3ab23x²+ dz C =4x³+3x²+ 3cd+2d2-bz. =2ab+3x-4. 5-1562-3ab, 1 f } AXC CASE III. When the divisor and dividend are both compound quantities, 00 ax)α 002 ე3 004 (1+-+ +++, &c. Am a² a3 a4 23 ac K 202 a 002 * 202 a 拿 ​20180 203. a2 203 a² 81 1 003 a² a². 1 004 a3 Serdangelog ི, ི་ a3 11 201 @3 205 EXAMPLES FOR PRACTICE. ая Ex. 1. a-x)a²—2ax+x²(a—x ax 1 ax7x² ax+x² } -1 t î 1 12 I 1 Ex. 2. x-α)x³-3x²+3a²x-a³ (x²-2ax+a² 203 ax2 1 1 ·2ax²+3a²x -2ax2+2a2x a²x-a³ a²x-a³ Ex. 3. a+x)a³+5a²x+5ax²+x³(a²+4ax+x² ¿ a³+ a²x 4a²x+5αx² 4a²x+5ax² ax²+x³ ax²+x³ Ex. 4. y-8)2y³-19y²+26y-16(2y2—3y+2 2y316y2 digital 3y²+26y 3y²+24y ·004+1 •201203 2y-16 2y-16 Ex. 5. x+1)x+1(x¹x³+x² −x+1 x05+004 } 203 fo +1 ·x3 + x² x²+1 -202 C XC x+1 x+1 13 1 Again. x-1)x®-1(x+x+x³ +∞²+x+1 5 06-005 ევ5 " /22/22/22/22/ 48x³-72ax² * ! } Ex. 6. 2x-3a)48x3-76ax²-64a²x+105a³(24x²-2ax-35a² -4ax²-64a²x, -4ax²+6a²x -70a²x+105a³ →70a²x+105a³ Ex. 7. 2x²+3x-1)4x9x2+6x-1(2x²-3x+1 4x+6x³-2x² x-1 x-1 Almanna 8 8 6x37x²+6x -6x³-9x²+3x B.2 2x²+3x-1 2x²+3x-1 ་ " C 1 14 : Ex. 8. x²-ax+a²)x¹—a²x²+2a³x-a¹(x²+ax-a² -α x³ + a²x² I 3 1 01 } 201 ax³-2a²x²-f-2α³x ax3. a²x² + a³x Ex. 9. 3x-6)6x96 (2x3+4x²+8x+16 6x-12x3 → 1 ·α*x+x5 α¹x — a³ x² Šių -α²x²+a³x-a4 -α²x²+a³x-20% 12x³-96 12x³-24x2 Į Again. a+x)a³ +x5 (α-a²³x+a²x²-ax³+x+ aε+a⭑x 2422 -96 24x²-48x a³x²+x5 a²x²+a2x3 48x-96 48x-96 ·a²x³-+x5 a²x³-ax⭑ ax²+x6 ax4x6 · 15 ? Ex. 10. 2x+3)32x+243(16x¹-24x3+36x2-54x+81 i A } 32x5+48x4 -48x¹+243 -48x72x3 1 6 — ცენ pr Again. x-α)x6 ~α6 (x5+ax¹+a²x³ + a³x²+a*x+a³ a ax5-a6 ax5-a²x4 72x3+243 72x3108x² I J 108x2+243 -108x2-162x a²x²*—aε a²x²-a³x³ 1 a³x³-a6 a³x³-a¹x² 162x+243 162x+243 a²x²-a6 a4x²-a5x abx - ав абх a³x-a6 ασ } ¿ 6 16 ľ Ex. 11. b—y)b*—3y¹ (b³+b²y+by²+y³— b—2 Byt б-у b4-b³y Again. a42a³b -2a³b+4a²b ·2a³b-4a²b² ; P b³y-3y* b³y — b 2 y 2 Į b²y² — 3y+ b2y2 — by s a+2b)a¹+4a²b+8b4(a³-2a²b+4ab4ab2-(8b2+86³) ** +4a²b+4a²b 2 +4a²b+8a b²2 p +4a²b² 8ab2 +4a²b² + 8ab³ by³-3y+ by³— y¹ ཀ -8ab³- 8ab³ ·8ab3-16 b3 . / 2y4 remainder. į -8ab316638b4 -8ab2 — 16b4 +1663+2464 17 - f Ex. 12. x+a)x²+px+q(x+(p—a). x³-ax² Ex. 13. x²+ax (p-a)x+q (p—a)x+pa—a² -pa+a²+q pa² OC pa² + paz -pa 7 Žyg C (a-p) x²+qx (a—p) x² - (a²-ap)x XC pa² ac + (a²-ap+q)x Again, x-a)x³ —px²+qx−r(x²+(a−p)x+(a²¬ap+q)+&c. -3x+9x²-10x3 -3x+6x2 3x3 P pa 20 +a² + q храз 202 &c. &c. храг pa² x2 (a²ap+q)x-a(a²—ap+q) 3x27x3+5x4 3x²-6x³-3x² 1-2x+x2)1-5x+10x210x³+5x¹-x(1--3x+3x²-x³ 1-2x+ x2 +a³-a²p+aq &c. &c. ·x32x4x5 -x³+2x4x5 +&c. 1 1 1 ! 18 J I Ex. 14. a2-2ab+262) a++4b¹ (a²+2ab+262. a-2a³b+2a2b2 ! 1 + a5-2α¹x+ P Ex. 15. a² — 2ax+x²)α5—5a¹x+10a³x²-10a²x²+5αx¹—x³(a³ a³x2 photog 2a³b-2a2b2+464 2a³b-4a²b²+4ab3 Į i 3a+x+ 9a³x²2—10a²x³ -3a+x+6α³x². 3a²x3 1 Ex. 16. a²+ab √2+b²)a+b+ + pada 2a2b2-4ab34b4 2a2b2-4ab³+464 Momen G 3α³x²- 7a²x²+5ax 3a3x2 6a2x3+3axt # bang [3a2x+3αx2-x3 a²x²+2ax¹ — a²x³+2ax¹ — 2015. J (a²-ab √2+b² a¹+a³b √2+ a282 -a³b√2- a²b² -a³b√/2-2a²b² — ab³ √2 a²b²+ab³ √2+b^~ a²b²+ab³ √2+b¹ · J { 19 } CASE I. To find the greatest common measure of the terms of a , fraction. Ex. 4. Here x³ —α³) x¹ — a¹(x x²- a³x ALGEBRAIC FRACTIONS. } a³x-a¹ Divide by a³, then x-a)x³-a³ (x²+ax+a² x³-ax² ax²- a³ ax² α²x Dividing the remainder by x Therefore xa is the greatest common measure sought, Ex. 5. Here a³ — a²x—ax²+x³)αª¹ — x²± A a²x — a³ a²x-a³ (a a¹ — a³x — a²x² + ax³ a³x + a²x² — ax³ — x¹ a³ + a²x — ax² — x³)a³ — a²x—ax²+∞¹³ ( 1 ˆ a³ + a² x-ax² Ax²-x3 a²x-x3 Divide by x; a²x²)—a²+x²( — 1 -a²+x², Call -2a2x+2x3 Divide by 2x; -a²+x²)a³+a²x— ax² — x³ ( — α. a³ — ax² } Therefore a2x2 is the greatest common measure sought. It frequently happens that the common measure of quantities of this kind is better found by resolving both 1 · 20 numerator and denominator into their component factors, in which it will be useful to remember that the difference of any two even powers is divisible both by the difference and sum of their roots; and that the difference of two odd powers is divisible by the difference of their roots, and their sum by the sum of their roots: thus, in the last ex- ample, (a²+x²)(a² — x²) a³ — a²x — ax²+203 (a²x²)α- ( a² — x²) x where it is obvious that a²x² is a common measure of both terms; and dividing by it, the fraction reduces to a²+x² which is in its lowest terms; and consequently a a¹ at _x4 Add OC JF a²x² is the greatest common measure, as was found by the former rule. And the same method may be advantageously used in other examples of this kind. Ex. 6. Here a²x²+a¹) x²+ax³-a³x-a¹(1 x²+a²x²+a¹ Divide ax³ — a²x² — a³x — 2α¹ 2a4 by a: x³-ax²-a²x-2a³)x¹+a²x²+a¹(x+a x¹ — ax³ — a²x² - 2α³x ax³-2α²x²+2a³x+a* a²x². ax³ } Padang Divide by 3a², and we have x²+ax+a²)x³-ax²-a²x-2a³(x-2a x³+ax²+a²x 1 p 3a²x²+3α³x+3a¹ > a³x-2a⭑ -2ax²-2α²x-2a3 -2ax²-2α²x-2a3 Therefore ²+ax+a² is the greatest common measure sought. (% 21 Ex. 7. Multiplying the denominator by 7, we have 7a²-23ab+6b²)35a³-126a²b+77ab2—42b³(5a 35a3-115a2b+30ab2 Multiply by -11a2b47ab2-42b3 77 -77a2b+329ab2-294b3(-116 -77a2b+253ab2 K 76ab2-22863 Dividing by. 7662, and we have a-3b)7a2-23ab+6b2(7a-2b 7a2-21ab 2ab+6b2 2ab+6b2 Therefore a—36 is the greatest common measure sought. x³+ax²+ bx²-2a2x+bax-2ba² Ex. 8. x²-bx+2ax-2ab x³+ax² + bx²-2a²x+bax-2ba² 204 Here I divide the nu- (x+2α)x(x-b) merator by x+2a, and the quotient comes out exactly; therefore x+2a is the greatest_common measure sought. Ex. 9. Ma 3-ax2-8a2x+6a3)x¹-3ax³-8a²x²+18a3x-8α¹(x-2α ax³-8α²x²+ 6α³x -2ax³+12a³x-8a* −2ax³+2a²x²+16a³x-12a* -2a²x²- 4a³x+4α¹ 6663 Dividing by 2a2, and we have x3+2ax²-2a²x Ketundegag x²+2ax-2α²)x3 ax²-8a2x+6a3(x-3α 3ax²-6α²x+6a³ 3ax² -6a²x+6α³ Therefore 2+2ax-2a2 is the greatest common mea- sure sought. C 22 Ex. 10. Here, dividing the numerator by 5a³, and the denominator by b, we have a²+2ab+b²) a³+2ab+2ab2+b³ (a a³+2a²b+ab² 1 { * ab2b3 Dividing by b², and we have a+b)a²+2ab+b²(a+b a²+ab Therefore a+b is the greatest common measure sought. Ex. 12. Here the numerator being the least com- pounded, and b rising therein to a single dimension only, I divide the same into the parts 6a5-4a3c2, and 15a¹b- 10a²bc2, which, by inspection, appear to be equal to 2a³(3a²-2c²,) and 5a²b(3a²—3c².) Therefore, 3a² —2c². is a divisor to both the parts, and likewise to the whole, expressed by (3a² —2c²) × (2a³+5a2b); so that one of these two factors, if the fraction given can be reduced to lower terms, must also measure the denominator; but the former will be found to succeed: thus, 3a²-2c²)9a³b-27 a²bc-6abc²+18bc³(3ab-9bc 9a3b- 6abc2 ab+b2 ab+b² Ex. 4. Here- -27 a²bc+18bc³ -27a2bc18bc3 Therefore 3a²-2c2 is the greatest common measure sought. 201 i CASE II. To reduce fractions to their lowest terms. (x² + a²)(x² — a²) ___ x²+a²· ·at x³ (x² - α²) 003 x5-α2x3 a²x3 ენ by dividing both terms by x²². ↓ t ľ Ma • • 1 23 t Ex. 5. 6a²+7ax-3x²)6a²+11ax+3x²(1 ~6a²+ 7ax-3x² 4ax+6x² Divide by 2x; 2a+3x)6a²+7ax-3x²(3a-x 6a²+9ax tion sought. 1 Therefore 2a+3x); { Ex. 6. Here 6a² + 7ax 6a²+11ax+3x² -2ax-3x² -2ax-3x² Mult. by 1 Kadhalang 2x³-16x-6 2(x³-8x-3) 2 3x³-24x-9 -3(x³ — 8x—3) Ex. 7. Here, multiplying the numerator by 5, and we have 15x-2x+10x2-x+2)45x5+10x3+20x2-5x+4(3x 胄 ​5 Mult. by ✩ 3x2 3a- 00 3a+x 45x56x130x³-3x²+6x 15x¹-2x²+10x²-x+2)30x¹-100x³+115x²-55x+25(+2 6x420x323x²-11x+5 Watkykjy the frac- 30x¹-4x³+20x²-2x -96x395x²-53x+21 Multiply the last divisor by 32, and we shall have -96x³+95x²-53x+21)480x¹-64x³+320x²-32x+64(−5x 32 480x¹-475x³+265x²-105x +4 411x355x²+ 73x+64 -96x395x2-53x+21)13152x+1760x2+2336x+2048(--137 13152x³ 13015x²+7261x—2877 14775x2-4925x+4925 24 Dividing the latter by 4925, it becomes 3x-x+1; which by another operation, exactly divides -96x3+ 95x2-53x+21; and therefore is the common measure; 3x3 + x²+1 and the reduced fraction is Batte Ex. 8. Here, the denominator being the least com- pounded, and d rising therein to a single dimension only, I divide the same into the parts 4a2d-4acd, and —2ac² +2c; which, by inspection, appear to be equal to 4ad(a-c), and -2c2(a-c). Therefore a-c is a divisor to both the parts, and likewise to the whole, expressed by (4ad-2c²)×(a-c); so that one of these two factors, if the fraction given can be reduced to lower terms, must also measure the numerator; but the latter will be found to succeed thus, a —- c) a² d² ——— c² d² — a²c² + c¹{ad² + cd² — ac² —ç³ a² d² -,ac d² Į a—c). the Ans. acd² — c²d² acd² - c² d² -a²c² + c¹ -a²c²+ac³ * A Therefore, a-c is the greatest common measure; a² d² — c² d² — a²c² + c¹ 4a³d — 4acd — 2 ac²+203(: Ex. 3. Here 1. beberkang -ac³+c¹· -ac³+c¹ Ex. 4. Here 5a the fraction required. i 1 CASE III. To reduce a mixed quantity to an improper fraction. 2x 1xa-2x a -2x Ans. ad2cd2 ac² - c³ c3 S 4ad-2c2 a a ɑ 3x-b 5a² - (3 x — b) a a 1 1 5a²-3x+b a • 25 { Ex. 5. Here x Ex. 6. Here 5+- Ex. 7. Here 1- Sa 2α-x+1 a 1 5x+10x2-x+3 5x Ans. 1 30- E 1 J Ex. 8. Here 1+2x- a+x² 2ax-(a+x²) ___2ax-a 2a 2a 2x-7 15x+2x-7 3x x—a—1) ___ a−x+a+1 Samant a a Ex. 5. Here a Therefore a+x+ a 3x -1 pap Ex. 4. a-x)a²+x²(a+x a² ax 5x 10x²+4x+3 5x ax Lanmananana To reduce an improper fraction to a whole or mixed quantity. Ex. 2. Here, (ax—x³)÷x—a—x² Ans. 2a Ex. 3. Here (ab—2a²)÷ab=(b—2a)÷b—1— b 睿 ​Jordaną 2x2 x³-y³ My Gandaman x-Y CASE IV. ax+x² 202 a 00 Mag x-3 5x(1+2x)-(x-3) 5x Ans. بعد 2a 17x-7 3x ·004 C2 . 2x2 is the mixed number sought. x²+xy+y², as is readily found by division. 3 Ex. 6. Here (10x²-5x+3)-5x=2x-1+5x Ans. 26 1 CASE V. To reduce fractions to other equivalent ones having a common denominator. Ex. 2. Here by the rule, 2xxc=2cx new numerators. 1 $ } b xa= axc= 2xc ab Hence and are the fractions sought. bc vac Ex. 3. Here aXc a = ab 1 ww U ac (a+b)b=ab+b2 } b xc = b c common denominator. ab+b2 ac Hence and bc are 'the fractions sought. bc Ex. 4. Here 3xx 3c 9cx ac common denominator. A 2b×2a-4ab new numerators. dx2ax 3c=6acd) 1 × 3c × 2a=6ac 6acd common denominator. 9cx 4ab Hence Dagd and are the fractions required. 6ac' 6ac' 6ac Ex. 5. Here the three fractions, when reduced, are 3 2x 5a+4x and 5 4' 3' Therefore, 3x3x5=45 2xX4×5=40x (5a+4x)X4X3=60α+48x Hence fractions required. 7a2-7xa 6xα-6x² 14a-14x' 14-14x' new numerators. ¿ 4×3×5=60 common denominator. 60a +48x 60 P 45 40x The fractions therefore are and 60' 60' Ex. 6. Here a×7×( x)=7α². 7xa 3x X2X(a-x =6xα- 6x2 2x7x(a+x)=14a+14x ) 2X7X(a-x)=14a-14x com. denom. are the new numerators. and ... 14a+14x 14a-14x J VA numerators. 1 1 1 27 Ex. 5. Here CASE VI. To add fractional quantities together. 2x 5x 14x 25x 39x Ex. 4. Here + + 5 17 35 35 35 3x x 15x 2ax 15x+2ax + 10a 10a 10a + Ex. 6. 27x-14 35 · OC XC XC 12x 8x 60 13x Here+-+ -+ + 2 3 4 24 24 24 4.00 Ex. 7. Here + 7 6a+ Ans. 9x+ + 2a 5 I 5 p 1 Ans. 12 x-2_20x, 7(x-2)_20x+7x-14 35 Ex. 8. This may be written 2a+3a+a 2x 8x 18x 40x =6a+ 5 9 45 45 Ex. 9. Here the fractions are By the rule, 3xx (a−x)× a=3a²x-3ax2 a X5 X a =5a2 (a-x)x(a-x)5 } • + 3a²x-3ax²+5x2 =2a+2+ 5a²-5ax Ex. 10. This is the same as 9x+ 5x²+10x 6x-9 15x 15x 5x²-16x+9 15x " big g 9x+ Ans. 35 Ans. =6a- M at 22x 45 nume- rators. =5a²-10ax+5x2 Their sum =10a²-10ax+3a²x-3ax²+5x2, and 5× (a—x)×a=5a²—5ɑx common denominator. 10a2-10ax+3a²x-3ax²+5x² Hence the sum is 2a+ 5a²-5ax Ans. Ans. 2x 5 Ans. 3x a a and ? 5'ɑ-x a 12 ⭑-2 3 5x²-10x-6x+9 15.x } a XC 8x 9 2x-3 5x 1 9x 28 # Ex. 11. Here 5x+ Ex. 4. Here 15y- 8ax+3αx²+603 =5x+ =5x+ 12x3 12x2 We have not in all these examples followed exactly the pro- cess described in the rule, at least not so as to exhibit the opera- tion, both in order to save room, and to indicate to the student a more concise way of setting down his work; and the same will be observed in the following case. CASE VII. To subtract one fractional quantity from another. Ex. 3. Here 12x 3x 60x 21x 39x 7 5 35 35 35 1+2y 120y-(1+2y) 118y-1 4.x =x+ Ans. 35 8 8 8 Hence Ex. 5. Here axx (b+c)=abx+acx ax×(b−c)=abx—acx 2a, a +2,x + 3x² 4.00 Sala -2bx+2ba 2bc Therefore Difference 2acx (b−c)x(b+c)=b2-c2 common denominator. 2acx b2_c2 is the difference sought. Ex. 6. Here x COC 2bc P 8ax 3αx²+6x3 =5x+ 12x3 12x3 8a+ 3αx+6x2 XC M a S + XC (x+ 26 Co с 2ba-2bx 2bc a Ex. 7. Here again we have a+ α-x a+x + a+x a+x a 00 Whence (a+x) x (a+x)=a²+2ax+x² (α-x)x(α-x)=a²-2ax+00² S (a—x) Sum =2a2+2x² (a×x)×(a−x)=a²x² common denominator. 2a2+2x² is the difference required. a2002 Ans. ac numerators. x- a с (a- —∞)—–—– 1 1 XC 26 a + x a-XC numerators. 1 " K Sta 29 L C7X0 Ex. 8. This is the same, when properly arranged, as 42x+147, 40x-48 x+ ax x+ + 168. 168 2x+7.5x- 6 + 8 21 82x+99 168 CLIC =ax-x+ 168ax-86x+99 86x-99 the answer required. 168 168 Ex. 9. Here subtracting the first from the second, we 11x-10 3x-5 26x-10 3x-5 182x-70 7 105 have x+- 15 7 168ax-168x+82x+99 168 45x-75 137x+5 x+ the answer sought. 105 105 Ex. 10. First to find the difference of the fractions a- OC @+a and a(a+x) a(a-x)' a(a-x)x(α-x)=a³-2a2x+ax²¿ } a(a + x)x(a+x)=a³+2a³x+ax² } Difference -4a²x (d) and consequently a- 15 32x+5 105 Ex. 5. Here a(a+x) xa(a-x)-at-a2x2 common denominator. Hence the second fraction subtracted from the first is a(a−x)=a¹ -4a²x - 4x a¹ — a²x² a² - x² 4x is the difference sought. a²-x²² Ex. 4 Here* X CASE VIII. To multiply fractional quantities together. 3x 5x Ans. 2 3b 5x2 26 numerators. 3x² 5a · 2x-3x² X 5 2α 2x 4002 Ex. 6. Here X X 3 7 These points are placed here to denote such factors as cancel each other. a a+x Ans. t 1 8x3α 21a+21x Ans. 1 + 30 મ 1 Ex. 7. Here X X } 2x 3ab 5ac =15ax Ans. a с 26 Ex. 8. By reducing these to improper fractions, they be- 2a² + bx 3a²x-b 6a+x+3a2x2b-2a2b-b2x X a aoc f > come a²x It is, however, frequently more simple to multiply quantities of this kind together as in common multiplica- tion, thus: } ' bx a 2a+- 3a- b Coc 6a²+3bx grap 26 f2. a² b2 6a²+36x a2¹ is equivalent to the preceding fraction. Ex. 9. Here OC 26 a 3x x+1 x-1 3x³-3x X 1 2a X Jaya Ex. 10. The third fraction a+ a² m² a² fr X the product, which a+b2a2+2ab ax a² a-x a- XX a² So that we have X a-b ax + x² ax Now observing what has been before stated relative to the factors a²x²=(a+x)(a-x), and a2-b2=(a+b)(a—b), also ax+x²=(a+x); our product may be written in the form Ans. (a+x)(a−x)(a+b)(a−b)× a², (a+b)(a+x)x(a-x) Which, by cancelling such factors in the numerator and denominator as are like, becomes ** } 31 ' L (a_b) a² à³_a²b ac 00 Ex. 6. Here Ex. 7. '. Ex. 8. Ex. 9. CASE IX. To divide one fractional quantity by another. 7x 3 3 7x X 7x2 Ex. 5. Here Ans. X 5 3 15 5 OC 4x² 5x 1 { 7 x+1 6 Here Here X t Janga ac X X 1. 00 00 I 3 2x 5 pa This may be written -x (2α+x)x C X 4x2 1 4x X 7 ^ 500 35 x+1 4.00. x-b x(x+b) } the answer. 5 2a+x Ans. 03 - 203 OC c²+cx+x² Where it is to be observed that c-x is a divisor of c3-x³ as stated in Case I. ·)₁ = + Ex. 10. These two fractions are readily resolved into the following factors: (x²+52) (x² — b²) (x-b)(x-b) XC 23 Ex. 3. (—2x²¹³)³ —— 3 Ans. Ans. INVOLUTION.' RULE I. (x²+b²)(x²-b²) 00² +62 x(x²-b2) OC 34a8a 5468 Ex. 1. (2a2)=23a0=8a6 Ans. (2a2x)=2a8x116a8a Ans. Ex. 2. 8 27x6y⁹ Ans. Ans. 33/9. 3a²x 818ax¹ Ex. 4. 562 62578* Ex. 5. In the preceding page of the introduction we have -An. 1 32 3 (a+x)³ a³+3a²x+3αx²+x38 (α + x)" a+x ܫ a¹+3a³x+3a²x² +ax³ + a³x+3a²x²+3ɑx³+x¹ a¹+4a³x+6a²x²+4ax³+x Ans. Again, in the preceding page, by writing a for x, and y for a, we have 1 (a—y)³ — a³—3a²y+3ay²——y³ (a—y)²=a²—Qay +y² INVOLUTION. 1 3 a5—3a¹y+3a³y² —a²y³ —2a¹y +6a³y²—6a²y³-+Qay¹ + a³y²—3a²y³+3ay¹—y³ (ay)5a55ay+10a³y²-10a³y³+5ay-y5. 1 } CI ax³+ RULE II. Ex. 3. Although the rule prescribes the finding the coefficients separately, it must not be understood as ab- solutely necessary, being merely stated in those terms. for the sake of perspicuity generally the whole opera- tion is performed in one line, thus, 4.3 4.} {α + x)¹ — a¹+4a³x+· -a²x²-|-· 6.2 .3 2 4 Or performing the divisions and multiplications (a+x)¹=a¹+4a³x+6a²x²+4αx³+x¹. 1 } -001. And in the same manner, we have (α-x)5a5-5a¹x+10a³x²-10a2x35ap+- 25. Ex. 4. Here (a+x)6: a6+6α$x+15a²+20a³ ³+15a²x²+6ax²+a6, and (a—y)"= ¿—7y+21a y²—35@y³+35á³y* --21aªy5-+-7ays → 6 ! 33 Ex. 5. Here (2+x)= 25+5.2¹x+10.23x²+10.22x3+5.2x²+x5 Or establishing the powers of 2= 32+80x+80x²+40x3+10x+x5 In the other part of this example the quantity is a tri- nomial, which it is better to put under a binomial form, thus: Ex. 3. Ex. 4. Here 5ax2. { (a−bx)+c } ³=(a−bx)³+3(a−bx)²c+3(a−bx)c²+c³ Then involving the several powers of a-bx we have (a-bx)³-a³-3a2bx+3ab²x²-b3x3 3c(a-bx)2-3a²c-6acbx+3cb²x² 3c2(a-bx)=3c2a-3c2bx C3—C3 Whence by addition (a-bx+c)³- a³+3a²c+3c³a+c³-3a²bx-6acbx-3c2bx+3ab²x²+ 3cb²x² -63x3 Ans. EVOLUTION. V Ex. 5. Here Ex. 6. -Zax² 3 CASE I. To find the root of a simple quantity. Here √4a²x6=√4√ a² √x6=2ax³ Ans. -125a³x6-3-1253/a33/x6➖➖ Here Ex. 8. Here 3 Ex. 7. Here 3 V 4 256a¹x=1/2564/ a¹÷/ x8=4ax². 2a2 √ √α¹ 4a1 J 2α 9x²y² v9vx² √ y² 3xy 8a3 3/8/a3 125x6/1253x6 -32a5x10/32/ α55/ x¹º 5x** 10 243 bertang 기 ​Jennant 243 } D } ་ 34 2a2+2ax) +2ax 2a²+4ax+xc² Ex. 3. x¹ 201 To extract the square root of a compound quantity. Ex. 2. a¹+4a³x+6a²x²+4ax³+x¹(a²+2ax+x² Ans. a¹ 2x²-x) - 00 2x²-2x+1) 4x³-x) } XC 4x³-2x+3) Ex. 5. 4006 x42x³-3x²-} x + 1 (x²-x+1 Ans. 16 200 EVOLUTION. 2002-200+30 1 x² - 1 x + 16 Ex. 4. 4x6-4x+12x²+x²-6x+9(2x³-x+3 Ans. CASE II. 4a³x+6a²x² 4a³x+4a²x² 3x -2003 + 3x² -2x3 + x2 Automat 2x³+2x²) +2x² 2x³+4x²+3x) 2x34x²+6x-4 2a²x²+4ax³ + x1. 2a²x²+4ax³ +001 1 4x-12x30c² ·4.004 +02 12x³-6x+9 12x³-6x+9 x+4x+10x+20x³+25x²+24x+16(x³- 4x510x4 4x5+ 4x4 6x+20x+25x2 6x+12x3 92 [2x²+3x+4 8x³+16x²+24x+16 8x3+16x²+24x+16 ^3 I ! புரம் Ex. 6. b 2a+ 2a b 2a 2+1- 2a+- {2+}) 2+1-1) b 8 4 C a 4a2 a²+b(a+ a2 1 +b+ +b Melangk b2 b2 8a3)- 4a2 b2 4a2 Ex. 7. 1+1(1+1~}+7~&c. 1 1 1+1 drą, al 35 b2 4a2 { b 2a 1-1+2 11 + 1/1/14 62 8a³ + &c. 63 64 + 8a164a6 63 74 8a4 64a6 4a)-12ax 4a212ax +90² 1-3/4 EVOLUTION. Ex. 3. 4a²-12ax+9x2(2a-3x Ans. CASE III. To find any root of a compound quantity. [ 1 'દ્ 36 | * [z Ex. 4. a²+2ab+2ac+b²+2bc+c²(a+b+c a² 2a) 2ab · (a+b)²=a²+2ab+b² 2a) 2ac (a+b+c)²=a²+2ab+2ac+b²+2bc+c² Therefore a+b+c is the root sought. Ex. 5. x-6x5+15x120x³+15x²-6x+1(x²---- 006 3x4)-6x6 წენ I (x²-2x)³x6-6x+12x18x3 3x1) 1 16a4 32a³) -96α³x 3x4 (x²-2x+1)³=x-6x+15x4-20x315x²-6x+1 32x5 80x180x1 12x3 Brigad Ans. x²-2x+1 Ex. 6. 16a1-96a³x+216a²x²-216ax³+81x¹(2a-3x 1 4 (2α-3x)¹—16aª—96a³x+216a²x²-216ax³+81x¹ Ex. 7. 32x580x80x³-40x²+10x-1(2x-1 [2x+1 (2x-1)5—32x³-80x¹+80x³-40x²+10x-1 } IRRATIONAL QUANTITIES, OR SURDS. CASE I. To reduce a rational quantity to the form of a surd. Ex. 3. Here. (5)2=25, therefore √25 Ans. 1 { 1 37 Ex. 4. Here (3x)3-27x3, therefore 3/27x3 Ans. Ex. 5. Here (-2a)*=16a4, therefore 16a¹ Ans. Ex. 6. Here (a²)5-a¹º, therefore a¹0 Ans. And √(a+b)²=a+b+2√/ab..√(a+b+2 √ab) Ans. 10 5 3/ } α Again, (√2_ =3/ 2a 3 3 Ans. Ex. 2. Here a 1 2a 4a² 4a Ex. 3. Here a Also, b√a Va Note to the above Case. Here 5√/6=√25× √/6=√/150 Ans. Ans. 13 1 2 6 Ex. 1. 1 a Ex. 2. Here√5a=√25× √5a= √1/ Ex. 3. Here 72a 108 1 • 2 3*6 1 • J. 9 1 1 4a2 i || CASE II. To reduce quantities of different indices to others that shall have a given index. 1 -IQ HIM a² b2a 8a3 = √ 27 × √ =3/ 1 8 2 1 X X 6 IN G11 V. 1 6 P 1 a a √ 62 f2 =3, 4a 11 K Ans. Ans. 9 4a2 = 3/ 1 Hence (53) and (62) or 125; and 36 are the an- swers sought. 1st index. 2, 2d index. 1 8 X =4, 1st index. 72a3 108a2 1 1 1 8 48412, 2d index. D 2 * * 38 ! Therefore (24)* and (42)*=16* and 163. Whence 168 and 168 the answer sought. 2 1 2 Ex. 4. Here 1 1 1 m Therefore (a) and (2) are the quantities sought. Ex. 5. Here 4 1 1 2*4=2x Adam becomes and mn Therefore (a³) and (61)š Ex. 2. Here and X 1 1 1 6 I 4 1-8, 1st index. 1 3 →IA 2 6 2 2 1 2 6 36-314, 2d index. J X=3, Note to the above Case. 4 and 1/2 = 1/2 4 2, 2d index. Ex. 3. Here 26 3-6 Therefore (a³) and (a²)³t Ans. " n' m n mn 1 and Hence (41) and (53) Ans. 1 1 3 3, 1st index. are the answers. 1 4 1 Ex. 4. Here again = and 3 12 4 12 pagtaas and 1 2 Therefore (a1)12 and (63)17 Ans. 1 1 Ex. 5. Here reduced to a common denominator 3 12 Therefore (am) and (b) Ana, 1 3 ကရေ 1 1 { · 1 39 To reduce surds to their most simple forms. Ex. 3. Here 125=√(25x5)=55 Ans. Ans. Ex. 5. 1 Ex. 6. Ex. 7. Ex. 8. Ex. 4. Here √294=√(49×6)=7√√/6 Here/56/(8x7)=23/7 Ans. Here 1923 (64 x 3)=4/3 Ans. Here 7/80-7 (16x5)=28/5 Ans. Here 93/ 81-93/ (273)=273/ 3 Ans. Ex. 9. Here, reducing the radical, we have 5 5X6 V 1 36 6 5 √30 3 Therefore ✓ Ex. 10. 3/ 3 5 10100 Ex. 6. Here Also 6 CASE III. 3/ 3 1 121 3 16 12, the answer. Ex. 11. Here √.98a²x= √(49a² × 2x)=7a √2x Ans Ex. 12. Here √(x³-a²x²)=√ { x²(x-a²)} = x √ (x-a²). X = √30 6 121 3X4 1 64 4 3 ✓ 1 242 12: hence CASE IV. √30 Ans. 4. 3 16 To add surd quantities together. Ex. 5. First ✓72=√(36×2)=6√2 And √128=√(64×2)=8√2 Ans. 14√2 180=√(36x5)=6√5 405 (81x5)=9√5 Ans. 15√5 Ex. 7. First 33/40=33/ (8x5)=6/5 And 1353/ (27×5)=33/ 5 Ans. 93/5 + • ! 40 } Ex. 8. Here 4/ 5443/(27×2)=123/2 And 53 128-53 (64×2)=203/2 Ans. 323/2 Ex. 9. Here 9√243 9√( 81×3)= 81 √3 And 10/363=10/(121×3)=110√3 Ans. 191√3 Ex. 10. By first reducing the fractional surds, we have 2 6 1 √3 = √ = = 3/3 √6 And 27 50 Ex. 11. Here k 2 Hence 3√3× √6=√/6 27 50 3 And 70=7x10V6=10V6 7X 1 =3 V Commenta 54 1 100 10 √(9×6)= 1 2100 1 Ans. 1 =21/2 1 2 .3 21 41 And √3264-42 1 Hence 123 +3=62+32=632 Ans. =31√6 · 10 } 1 Ex. 12. Here a²b=√(a²xb)=='a√\/b } ×b) 2 1 1 3 10 ò√6 2 And √4bx+ + = = = √ (4x+ × b) == 3 a² √/b 3 Ans. (+3x²)√/b ↓ 1 1 41 i CASE V. To find the difference of surd quantities. Ex. 1. Here 2√50=2√(25 × 2)=10√/2 And √18= √( 9×2) = 3/2 Ex. 2. Here And Difference 7/2 320=3/ (64 × 5)=43/ 5 3 40 (8×5)=23/5 } P Ex. 3. Here. V And √ Difference 23.5 3 15 1 √ 5 √/15 25 5 5 15 1 81 27 1 Difference Ex. 4. Here 2√ } = 2 √ ? = √/2 √/8 √8= √(4x2)=21/2 And Difference √2 Ex. 5. Here 33/3279 And / 72=√(8×9)=23/9 Difference 9 18 1 Ex. 6. Here√27 = 3/2 3 K 9 32 V15 4. 45 V 15 Difference 18.1 V-V-16 64 18 18 1 1218 42 " 1 1 Ex. 7. Here √/80a*x =√(16a+ x5x)=4a² √5x And √20a2x3√( 4a²x² × 5x)=2ax√5x t Ex. 8. Here 8 And 23 1 To multiply surd quantities together. Ex. 5. Mult. 5/8 By 3/5 Prod. a³b=83/ (a³× b)=8a/b ab 23 (a©×b)=2a²/b Difference (8a-2a²) 3 / b Product Ex. 6. Mult. 18 3 By 53/4 Ex. 7. CASE VI. 15/40151/(4x10)=-30/10 Ans. Difference (4a² 2ax)√5x Mult. √6 By Ex. 8. Mult. By 53/72=53/ (8x9)=103/9 2 15V9 1 Prod. √/54 = 30 √(9×6) = 10 V 1 30V 10V6 18 5/20 Prod. Ex. 9. Mult. 2 √3 By 135 /360=√(36×10)=15√/10 27/15 Answer. 1 Ex. 10. Here 724a³ × 120 a 289 241 2 X 4 2 харха 43 1 1 27 1 69649 8 Ex. 11. Mult. 4+2√/2 By 2√2 ¹¹×a²º±=8706¼µµ Τ } + A Ex. 5. Here 8-4-4 the answer. 1 Ex. 12. Mult. (a+b)=(a+b)mn 1 By (a+b)m=(a+b)mn 8+4√/2 -4√/2~4 *5 CASE VII. To divide one surd quantity by another. 61/54 3√ 2 Ex. 6. Here 472 23/18 Therefore Product =(a+b) Ex. 7. Here 5- 1 글 ​1 And √1135 V/1/5 = = √ 4 3 Ex. 8. First =2√27=2√/(9x3)=6/3 Ans. 69 1 80 × √3 == =23/4, which will not reduce lower, 3 2 23 3 5 42÷22 5 27 m 69 mta mn 2 33 n 17 4 Ans. X 4 And √² ÷√3 = √ √ ( x ) == X 72V 3. 11 m IQ m 2 3 81 5 23 24 12 .8 S 69 1 Pagpakas 9 √3 √/3 Answer. 165 84 4X2 2 "31 11 √/2 44 F 165 2 Hence 84 X 3 1/2=55 42 Ex. 9. Here 41+2× And √a÷V ab=a³÷a rahma²+==== 63 1 2 Ex. 10. Here 3213 a ; therefore 7) the answer. 16 And Va Therefore a the answer required. 275 648 // Consequently Ex. 12. Here Ex. 3. Here Ex. 11. Here 9-4 8 11 a 1 a2 Ex. 4. Here a 3 9 75 11 825 424 1 1 - Mag And anam amn amn 825 424 :00 √2 Answer. 3 162 4 648 X 5 55 275 ÷a³=að²—3—a} 2 9 3 27 8 16 1 1 a 8 m X 53 √20+√12 √5+√3 Note to the above Case. n M M-n amn the answer. a the answer. m. M ! the answer. mn a 1/4, or 2, the answer. } | many 45 * 4 卯 ​Ex. 5. Here Ex. 6. Here a(a²x²) + ↓ 1 1 a+x By / 1 (α-+x)¹ ÷ Square Ex. 7. Multiply √x+3y By √x+3√y Square Mult. by CASE VIII. 画 ​1 To involve or raise surd quantities to any power. Ex. 3. Here (33/3)²=(3 × 3³)²—9.33—93/9 Ans. Ex. 4. Here (17/21)3=173√(212×21)=173×21 √/21=103173/21 Ans. 1 1 Ex. 5. Here (†√/6)'=(¿×6³)ª = 1 × 6² = -/- 62 64 62 1 Therefore the Ans. 36 Ex. 6. Multiply 3+2√5 3+25 9+6/5 +6√5+4√25 9+12√/5+20=29+12√/5 20 :(a+x) a x+3√xy +3√xy+9y E (a²-x²) +6√xy+9y x+3√ y 1 7 1 Ans. Ans. x√x+6x√y+ dy √x +3x√y +18y √ x + 2 7 y √ y Cube xx+9x√y+27y√x+27y√y L 1 1 1 46 Ex. 8. Here Mult. √3-√2 By √3-√2 rs 1 1 Ex 3. Ex. 4. را Square = 5- 2√/6 Square =5-2√/6 4th power =49. 3 3-√6. -√6+2. V To find the roots of surd quantities. Here 25-10/6 -10/6+24 0024 a 49-20/6 the answer. CASE IX. 1 8 Here—2—a³× a) = 23a²/ a Ans. 3 27 16 3 23 23 × 2 -/-/ 81a0=3a 6 ૩૯ a 103(102+10)=10/10 Ans. 8 1 Q2 10 2x-2√/a) -4x√a+4a - 4x √a +4α Ex. 8. Here a+2√ab+b(√a+√b Ans. · 2 √a+√b) 2√ab+b 2√ab+b 1 Ans. Ex. 5. Here a 3.3 Ex. 6. Here a√3 =(-)×(3)³*=Gª Consequently ◊ G√3)=V ()}=(3)² = (³)}=} {√³•• За 1 √3α. 9 Ex. 7. Here x²-4x √α+4α(x-2 √ď Ans. } 1 3 1 $9.5 ་ ་ 1 1 J 3 CASE X. To transform a binomial or residual surd into a general surd. · Ex. 4. First (3—√✓/ 5)—9—6 √ 5+5=1,4—6√ 5, therefore✓ (14-65) Ans. Ex. 5. Here (√ 2—2 √ 6)²—2—4√ 12+24=26~- 412, therefore (26-83) Ans. 1 47 Ex. 6. Here (4-7)2=16-8/7+7=23-8√7 Hence (23-87) Ans. Ex. 7. In examples involving cube root radicals, it is useful to know the following form of the cube of a bino- mial: viz. (a+b)³=a³±b³+3ab(a±b) Hence (23/3-33/9)3-24-243-183/27(23/3-3 /9)=-219-54(23/3-33/9) Consequently (−219-54(23/3—33/9)) is the general surd required. น Hence 3 ก ·To extract the square root of a binomial surd. ✓ } CASE XI. V Ex. 3. Here [+}√ (a²—b)] *=√ [3+3 √ (36 -20)]=√(3+2); and √[a-√ (a²—b)]= √ [3-3. ✓ (36-20)]=√(3-2) (3+2)±√(3-2)=√5±1 Ans. Ex. 4. Here ✓ [} a+ ¦ √ (a²—6)]=√ [3³+}√ (529–448)] — ✓ (323 323 + 33 ) S ✓ [ √ α- √ (a² — 6)]=√ [²¾³−}√ (529–448)]= a ✓ (33³ - 3/3) Hence (23+3)± √ (3³ ~})=4±√7 Ans. * IL It may here be observed, that b denotes the quantity under the second radical after its coefficient has been introduced. present example, b=20, because 2 V 5V 20. Thus in the I t 48 ¿ 1 > Ex. 5. Here √ [}a + } √ (a² —b)]=√ [18+√ (1296-1100)]= ✓ (18+7); and 33 15 √(2+2) and ✓ [ } a — } ✓ (a² —b)] = √ [18—3√ (1296–1100)]= ✓ (18-7) Hence✓ (18+7)±√ (18—7)=5±√11 Ans. Ex. 6. Here 1 √ [§a+{√ (a²—b)]=√ [3³+{√(1089—864)]= 2 √ [}a — √ (a²—6)]= √ [³ ³ —√ √ (1089—864)]= ✓ (33-15) 2 33, } Ex. 9. Here 1 = √( − } + }) =√1=1; 3) 33 Hence√(+5)+(-15)=√24±3 Ans. Ex. 7. Here√ (a+}√ (à²+b))=√ (3+}√ (1+48)) = √(3 + 3) = √4=2; 2 And√ (a-√ (a²+b))=√(}~{√ (1+48))= ✓ (3-3)=√ -3. Whence (1+ −48)=2+ √ -3 Ans. Ex. 8. Here (a+}√ (a²+b))=√ (}+{√ (9+16)) = √(} + { ) = √4=2; 1 And (a-(a²+b))=√ (3—{√ (9+16))= ✓ (3-5)=√-i. 1. Whence (3±√ −16)=2±√1 Ans. 1 (a+}√ (a²+b)) = √(−√ + √ √ (1+8)) And√ (a-√(a²+b)) = √ ( − }~ {√(1+8))=' ·√. ( − 1 − 3 ) = √ −2. 2 Whence✓(-1√-8)=1±√2 Ans. Ex. 10. Here√(}a+}√ (a²—b))=√ (}a²+}√ (a* -4a²x²+4x¹)) = √(a²x²); $ ¿ 1 49 2 * And ✓ (}a—1√ (a² —b))=√ (}a² ——✓ (a¹ — 4a²x²+ 4x4))=x. Whence √(a²+2x √ (a² — x²))=x+ √(a²x²) Ans. Ex. 11. Let 12-24 be reduced to a general surd, and it becomes √(+36+242). Hence 6+2 √2=a; and +36 +24√2=b; therefore ( a + 1 √(a² —b)) = √(3+ √2+√(44+24/2-36-24√/2)) =√(3+√/2 + 3 / √8)=√(3+2√2)*=1+√/2. Again, √(√(a² —b))=√(3+√2−3√(44+24 √/2-36-24√/2))=√(3+√2−}√8)=√(3+√2— · St √/2)=√/3. Whence √(6+2√2 −√/12-√/24)=1+√2−√/3 Ans. CASE XII. To find any root of a binomial surd. Ex. 1. We have A²—B² — 250 = 5 × 5 × 5 × 2 ; ..QX5³×2=n³, whence Q=4, and n=5X2. Then 3/ {(A+B)× √Q},or³/ {(68+√(4374)×2)}=}/ 268 +r7, the nearest integer. A√Q 684-136/1, 10 n 10 2 + 7 + 7 7+7 S y 25 2/1 2 and the radical part√ 1=s and 49+10 59 14 14 4-t, the nearest integer. And ts=4, √ (t²s²—n)=√ (16—10)=√ 6, and 2/ Q=/ Q=√/ 4 =2; and so the root to be tried is whose cube, upon trial, I find to be 68-4374. 4-√6 //2 Ex. 2. We have A2-B2-175-121-54=2×3× 3×3. QX3³×2=n3; whence n=3x2=6, and Q=4 T | ܀ p * The square root of this expression is found as in the foregoing ex- amples. E 2 50 (because 33x2x4=3³× 2³n³). Then [(A+B) 3 3 × √ Q] = [(13+11) x2]=48+=r=4. A√ Q= ✓ 175x√4=2√ 175=10/7, and the radical part √ 6 ~ + 2 2 4 + 4 2 t=1 2√7 28271 in the nearest integer. And ts=√7,√ [(t²s² —n)] = √ (7-6)=√1=1. 20% Q ts+ √ [(t²s² —n)] ___√7+1 =4=2; whence Ans. 20/ Q /2 √7=s, and' 3 Ex. 3. We have A2-B2-1; QX1=n³, whence Q=1, and n=1. Then [(A+B)x√Q]=√10+= r=2 nearly. AQ=27, and hence the radical part 2 + 1/2 5 2√7 10 -1)=√ ✓ 7=s, and 1 r+2 r 25 17 ✓ (t²s² —n)=√ ts+ √ (t²s² —n) ___√7+ √3 20/ Q M 4 3 ✓ 4. 1 f 102 w Ans. And ts= Therefore 2 5 5 Ex. 4. We have A2-B2-3; Qx3=n"; hence Q= 34, and n=3. Then (A+B) + Q} = // {√5046 // √ Q}={/ +√(5043) × √ 81}={(71+71) ×x9r 4 nearly. ✓ y *The rule in the Introduction, which was first given by Newton in the UNIVERSAL ARITHMETIC, fails when texactly, as it would be in the above example, if the calculation was performed accurately. I pro- posed this particular example at the Mathematical Club: "Required to know, if the cube root of 2 V7+3 V 3 can be found by the rule given by Newton, page 139, Universal Arithmetic, for extracting any root of a bi- nomial surd; and, if not, to show where that rule fails, and what altera- tion is to be made in it, so as to obtain the root ?" Robert Adrain, LL. D. Professor of Mathematics, &c. Columbia College, has since ably inves- tigated the subject, and he found the rule not only to fail in this, but in a great variety of other examples; and has also discovered the rule to be defective. See "The Exercises of the Mathematical Club of the City of New-York." } 51 A√ Q= √ 5046 × √ 81261√ 6, and hence the radi- it n J cal part 6s and 1t in the nearest in- 2s 26 teger, and ts=√ 6,√ (t² s² ~n) = √(6—3)=√3 and 2/Q=20/81=9. And therefore ts+√ (t²s²—n) 22/Q √6+ √ 3 /9 Ex. 5. Here a=45, and b=1682; whence (a² —b) =(2025-1682)=7, and n³-3(3/ (a²-b))n=90, or n³-21n=90, whence it readily appears from inspection, that n6. Whence 3 3 4+ 4 3 / (45+√/1682)=1+±√(36 — 4√/343)=3+ √√/8= 3 3+√/2 (45—√1682)—§ — 1√(36 — 4√/343)=3 — ±√/8= 3-√/2. 3 Ex. 6. Here a=9, and b=80; whence (a²—b)= 3/ (81—80) — 1, and n³-3(3/ (a²—b))n—18, or n³-3n= 18. Whence it readily appears, by making trials of the di- visors of 18, which are 1, 2, 3, &c. forn, that n=3; whence 3/ (9+√80)=3+1√ {9−43 (81—80)=3+}√ 5, (9/80)=-3√(9-43/(81-80)3-5. 3 the Ans. 3 P Ex. 7. Here a=20, and b=32368; whence (a² +6) 3/ (400+32368) and n³-3(/(a²+b))n=40, or n³- 96n40; when it readily appears, by inspection, that n =10; whence {20+32368}=√100-43/(400+32368)} ? 1 =5+√-7, {20√3236810-1√ {100—43/ (400+32368) 5--7. Ex. 8. Here a 35, and b=28566; whence (a²+b) =3(1225+28566)=31, aud n³-3(/(a²+b)n)=70, or n³-93n=70; when it readily appears, by inspection, that n=10; whence * 1 4 # −6, 27, 100 27) 27 (35+√28566)=√(100-4×31)=5+ (35-28566)=√(100-4x31)-5--6. Ex. 9. Here 81+√-2700=27×(3+ √100; then the cube root of 3+ √ can be now more easily found. We have a 3, and b=100; whence 3 (9+100) 6. = 3, and n³ — 3 { ³/ (a² + b) { n=n³—7,n=6; when it readi · } 37 ly appears that n=-2; whence / (3+ √ −100)=-1 + = √(4-4×3)=-1+3-3, we shall have by multi- plying by 3, (which is the cube root of 27,) -3+2 √ - 3; and in like manner, (3--100)=-1-3√3, which, multiplied by 3, as before, becomes-3-2-3. 3 3 27 K 3 27 CASE XIII. To find such a multiplier, or multipliers, as will make any binomial surd rational. I Ex. 5. Given surd Multiplier Product [ Ex. 6. Given surd Multiplier Product Ex. 7. Given surd Multiplier Product } 52 16 1 Balag √/5-√x √5+ √x 5- √a+√b Va-vb 00 as required. 7 Ex. 8. · Given surd 1-3/ 2a ་ a- b at√b a - √ b a2b the answer. as required. J Square of the terms 1+ 3/4a² Product with sign changed +/2a Therefore 1+ 2a+ 4a² multiplier. Mult. by 1-V/2a ? —- 1+2a+3/4a² -2a-4a²-3/8a3 Product 1-8a3-1-2a as required. F 1 53 Ex. 9. Given surd 3/3-1/2 3 Ex. 10. Given surd a+b 3 Square of the terms 3/9+13/4 Product with sign changed +6 Whence/9+6+134 is the multiplier required. n-3m,2m 'b' &c. ་ Here m- and n=3, and a Ex. 4. Here √42-√/18 4. m b + a 3-3 32 protadament 4 a b¹ = √ / a° — √ / (a©b³)+ √ /_ a³b6−√ / b⁹.¸* Pogledaj 3-3 a 4 Ans. Ex. 5. Here OC CASE XIV. a 8 } 3-6 3 a b ¹ + a 474 a-vo a-vo Ex. 6. Here a + √b a + √b which is the answer sought. 'N- m G To reduce a fraction whose denominator is either a simple or a compound surd, to another that shall have a rational denominator. √/6 X X 3 + √ ¤¯¯¯³+ √ œ ˆˆ˜ 3 — √ √ x 00 a M V6 √7+ √ √7+√3XV7−√3_ √7-√3 3 N a 9 a-vo LJjW | 'b' 2m 3-√x_3x-x√x a 9-x Ans. √b a²+b-2a√//b a²-b Ple Ex. 7. Here, by the preceding rule, the multiplier for the denominator is 49+ 35+ 25. 3 10. 10 4935+25 Whence X 3/7-3/5-77-3/5/49+/35+25 *Hero, by taking n=4, and putting a³ and 63 for a and b in the formu- la, we shall obtain the same result. > 54 HOW Stan } 10(3/49+3/35+25) 7-5 the answer required. Ex. 8. Here 54935 +25)= 2/3 3/3 X 3' 3 3/9+3/10 3/9+ 10 3/81—3/90+3/100/33/81—3/90+3/100) 3 81-90/100 33/9-33/ 10+¾/300 1 19 Ex. 9. Here SCA 4 4 4-4/5 √4 + √ √ 4 + 25^4-√5 /4-1/5 X 4 4 4(1/4-1/5)_4(1/4-1/5)2+3/5 2/4-3/5 X 2-2/5 2+2/5 4(√/ 4—1/ 5)(2+√5) —4(√ 5−√/ 4)(2+√√/5)= 4-5 4(−√/10−2√/2+(2+√/5) ×√.5) Ans. ARITHMETICAL PROPORTION AND PRO- GRESSION. 101 2 Ans. {2+100×2}× =202 X bart passa Ex. 3. Here the formula S=(a+1)× becomes s (1+1000) x 500-500500 Ans. Ex. 4. Here the formula s 9+10, 101 n 2. 2 {2a+(n-1)d becomes N d =101210201. Ex. 5. Here s=(a+1)=(1+24)×12=300 Ans. Ex. 6. Here the formula la+(n-1)d, becomes =2+(365-1)2=2+728-730 Ans. `1 Ex. 7. Here s={2a~(n−1)d'}} = {20—20× 4 55 l 40 21 X -140 Ans. 3 2 3 2 Ex. 8. Here the first term is 1+1=2, and the last 100+100 200, the number of terms 100. x21 = (20-20) x 21 == Therefore (a+1)=(2+200)50=10100 yards, or 5 miles 1300 yards, the Answer. I GEOMETRICAL PROPORTION AND PRO-. GRESSION. Ex. 3. Here the 1st term a-1, the ratio r≈2, the number of terms n=20; =1X-553-20X1 1-()8 28-1 28 gr for SATILAN 1.X 5s. 3 d. 1 pn I Whence the formula s=a(—— 22011048575 Ans. Ex. 4. Here a—1, r, and n=8, whence sa( / } becomes 1× 28-1 255 127 27 220-1 2-1 or sum required. Ex. 5. Here a=1, r=1, and n=10; whence s=a rm1 1-(1)10 310-1 310_1 59048 I Jap 310 × 39 X 239366 X 9841 =1× 19683 p pm-1 1 the answer, 128 128 =1 T gue Ex. 6. Here a=1,r=2, n=32. Whence s=(a r 2321 2-1 ལྤཀ] 3232 — 1—4294967295 farthings—44739247. EQUATIONS. RESOLUTIONS OF SIMPLE EQUATIONS. CASE I, 1: Here 2x+3x+17, by transposing gives 2x-x 17-3-14. Whence a14 Ans. 5 56 $ 2. Here 5x-9=4x+7, by transposing, gives 5x-4x =7+9=16. Whence a=16 Ans. 3. Here x+9-2-4, by transposing, gives x=4+2 -9=-3 Ans. 4. Here 9x-8=8x-5, by transposing, gives 9x-8x -8-5-3. Whence x- 3 Ans. 5. Here 7x+8-3-6x+4, by transposing, gives 7x-- 6x=4+3—8--1. Whence x=-1 Ans. -2 the answer. CASE II. · 3x OC + 1. Here 16x+2=34, by transposing, gives 16x=34 -2-32, and by division x- -2 Ans. 32 16 2. Here 4x-8=-3x+13, by transposing, gives 4x+ 3x=13+8, or 7x-21, and, by division, x= 21 77 36 3 Ans 3. Here 10x-19=7x+17, by transposition, 10x-7x =17+19, or 3x-36, and by division, x= -12 Ans. 3 4. Here 8x-3+9=7x+9+27, by transposition 8a +7x=27+9-9+3, or 15x-30, and, by division, x= K I 30 15 5. Here 3ax-3ab-12d, by transposition, 3ax=12d 12d+3ab +3ab, and, by division, x= 4d =b+ Ans. За a CASE III. S " 1. Here +24; then multiplying by 4, (which is a multiple of 4 and 2,) 6x=x+96, or 6x-x=96, or 2 4° 96 5x96, and, by division, x= 19 Ans. 5 1 57 have at 2. Here++62; then, multiplying by 3, we 3x 3x + 186, and multiplying by 5, gives 5x+ 5 2 15x 3x+930; and this, multiplied by 2, gives 10x+6x 2 p +15x=1860, or 31x1860, and, by division, x =60 Ans. 3. Here x-3 -3 x Wallplate x+19 + 20- 2 3 ; then, multiplying by 2 6, (which is a multiple of 2, 3, and 2,) 3x-9+2x=120 -3x-57, and, by transposition and division, x- 72 8 1 x+3 x+1 x+2 4. Here + ; then, multiplying 2 3 4 by 12, we get 6x+6+4x+8=192-3x-9, or 13a- 169; and, by division, x 169 -13 Ans. 5. Here bx b, x+a+ C 2bcx abc+cb² + d C 13. x+a a 2x a+b + + b d pag C a 2bx, ab+b² + d. a S -16- STA { 1860 31 a and, by multiplying by a and d succes- sively, we have adcx+a²cd+abdx=2bcdx+a²bc+acb², or, by transposition, adcx+abdx-2bcdx-a²bc+acb2- a²bc+acb² — a²cd ade+abd-2bcd a²cd, and, by division, x= =9 Ans. ; then, multiplying by , multiply by c, cx+ca+bx= Ans. CASE IV. 1. Here 2x+3=9; then, 2x=9-3=6 by trans- F } 1 58 position, and 4x=36 by squaring, or 9 Ans. 2. Here √(x+1)-2-3; then √(x+1)=3+2=5, or +1-25 by squaring, and consequently x=25-1= 24 Ans. 3 3. Here (3x+4)+3=6; then /(3x+4)=6-3= 3, or 3x+4=27 by cubing, and 3x=27-4-23, or x= 23 7 the answer. 3 4. Here √(4+x)=4-√x; then 4+x-16-8√/x+ x by squaring, and, by transposition, 8/x=12, or 2√x 9 =3; and 4x=9 by squaring, or x=-2 Ans. 4 „ 5. Here √(4a²+x²)=†/ (4b¹+x¹); then, by squaring 4a²+x²=√(4b¹+x¹), and squaring again, 16a¹+8α²x² + x446¹+x4, by transposition and division 2- and consequently a=√( I C S bt-4a⭑ 2a2 36 4 -) Ans. root of both sides x+1= a² 4 #་ CASE V. 1. Here 9x²-6-30; then 9x2=30+6=36, or x² 36 9 2. Here 3+9-36; then a³-36-9-27, or x=3/27 -3 the answer. 3. Here x²+x+1= I -4, and consequently x=√4=2 Ans. α square root of both sides, x+b, or x= ~ 1 bt-4a¹ 2a2 81 ; then, by extracting the square 4 1 4}, or x=4-1=4 Ans. 4. Here.x²+ax+b2; then, by extracting the a IQ pampalakad. Ans. 5 59 5. Here x²+14x+49=121, then, by extracting the square root of both sides of the equation x+7=11, or x =11-7=4 Ans. } 2. Here 10-x: x * 1. Here 3cdx=20abc, consequently, x= 2 རཱུལ. : a:: 5bc: cd; then 3x × cd=a×5bc, or 20abc 20ab Ans. 3cd 3d CASE VI. S =2 Ans. 10 ext. and means, and 3x-10, or x=- 3 Ans. 3 3. Here 8+8x: 4x :: 8: 2; then 16+16x=32x by mult. ext. and means, and 32x-16x-16, or 16x=16; 16 therefore x= 16 1 Ans. } 3: 1, then 10-x-2x by mult. 4. Here x 6—x x: 2:4; then 4x-12-2x by mult, ext. and means, and 4x+2x=12, or 6x=12; therefore 12 6 • 5. Here 4x: a : 9√x: 9; then 36x=9a√x by mult. ext. and means, or 4x-avx, and by squaring 16x²=a²x ; Ans. a2 therefore, by division, x= 16 EXAMPLES FOR PRACTICE. Ex. 1. Here 3x-2+24-31, by transposing Gives 3x=31+2−24—9; Whence x=3 Ans. Ex. 2. Here 4-9y=14-11y; or 11y-9y-14-4, or 2y=10; Whence y5 Ans. 60 { * f Ex. 3. Here x+18=3x-5, or 3x-x-18+5, or 2x=23; whence x- .11. Ac Ex. 4. Here x++ Here x+2+3 1 Ex. 7. 00 X Ex. 6. Mult. + * } Ex. 5. Multiply the given equation by 2, and we have 4xx+2=10x-4; whence 10x+x-4x=4+2, 1 Mult. by 6. 6x+3x+2x=66, or 11x=66'; Whence x6. Mult. 8 or x- OC 1 or 7x=6; whence x- x+3 1 77 2 3 4 10 30x+20x-15x-42, or 35x=42, 42 6 1 Ans. 35 5 Ex. 9. Here x+a: J Į 11 મ Kad 23 2 3 připrav 4. 13x-45; whence x= 6 7 { by 60, gives 6x+18+4x=48-3x+15, or 45 6 13=313 Ans. Ex. 8. Here 2+√3x= √(4+5x) being squared, Gives 4+4/3x+3x=4+5x; x-5 002 x + a' Whence 4/3x=5x-3x-2x Squaring 48x4x2, or dividing by 4x, we have x=12. by 12, gives or x²+2ax+a²x²; Whence 2axa², or 2x: P K a, or x= I a la 2 61 Ex. 10. Here √x+√(a+x)=√√(a+x)' √(a + x) Gives (ax+2)+a+x=2a, or √(ax+x²)=a-x; Hence by squaring ax+x²-a²-2ax+x², a Conseq. 3axa², or 3x=a, or x= 2=1080 Ans. 3 1 1 C Ex. 11. The equation 3 2 Mult. by 12 gives 3ax-3b+4a=6bx—4bx+4a; Whence 3ax-2bx=3b, or x(3a-2b)=3b, a²x². Therefore Ex. 14. Here 36 Consequently x=3a-26 4 Ex. 12. Here √(a²+x²)=√/ (b¹+x1), by squaring Gives a²+x²=√(b+x4), squaring again Gives a¹+2a²x² + x² = b²+x¹; f4 ax-b 4- J Whence 2a²x²-b¹—a¹, or x— M 2a2 Ex. 13. Here (a+x)+√(a−x)=√ax, by squaring Gives 2a+2√(a² — x²)=ax ; do -2a Whence √(a²x²): ; squaring again, 2 OC a + a 1+x a 2a a²x²-4a²x+4a² 4 a + 1. XC ax+a+ax bæ bx-a 3 1-x2 mult. by F 2 ·4x² — a²x²—4a²x, or dividing by x, a²x -4a², or -4.x 4a² a²+4° 1 at 2a 1x2 b, or =b, becomes by reduction = b; A } #4 } } W + Ex. 15. By squaring a+x=√{a²+∞ √(b²+x²) { We have a²+2ax+x²=a²+x√(b²+x²) Transposing 2ax+x²=x√(b²+00²) Divide by x, 2a+x = √(6²+00²) By squaring 4a2+4ax+x²=b2+x²; 62 Whence 2a-b-bx², or bx²-b-2a B-2a Therefore a√2a b J Whence 4ax-b²-4a², or x= મ 3/ The answer required. Ex. 16. Multiplying the given equation by 2, we have √(x²+3a²)—√(x²´—3a²)=2x√ a. By squaring 22-2√(x*-9a4)=4ax2, or √(x-9α¹)=2ax²-x2 62 Squaring again a¹-9a¹-4a²x²-4ax¹+x², or 9a3 (4a-4a²)x49a¹, or x=/ 4-4a Ex. 17. Here √(a+x)+√(a-x)=b, by squaring 2a+2√ (a² — x²)= b², or √(a² — x²) — — b² — a ; Whence a²x²-1b₁b² a + a² P b And x=√(b²a—16¹)=√(4a—b²). 3 43-2a b b32a 36 4a Ex. 18. Given equation (a+x)+3(ax)=b By cubing both sides after the form of Ex. 7. Case x. Surds, we have > ; a 2a+33 (a²—x²) {³/ (a+x)+† (a−x) } =b³, 3 But since (a+x) + (a−x)=b, this becomes. 2a+3√(a²x²)xbb³, or 33/ (a²x²) = $ (a²x²); or 1 whence 1 ¿ n ¡ 1 a² - 203 1 Ex. 20. Here √+1 OC V. Gives ļ grantkat Therefore x=√{ Ex. 19. Here a+ √x=√ax, which divided by x Va Gives +1=√α, or Voc -1+√a, or √a-1; Or * 63 (b³ — 2a)³ 2763 Va Whence √(a−1)√x=√a, or √x=√(a−1) Therefore x= J V. b3-2α 36 a² ( p Va √x Patentkammtan b3—2ά. 36 a (va-1)² .00 1 x + 1,, x 1 + +2√ x-1 ∞+1 2x²+2 x²-1 2x²+2+2x²-2 = x²a² — a²'; Whence 4x² — a ²x² a Therefore x- √(a² —4) Ex. 21. This equation transposed, is x+1 za, by squaring 002 1 x² 1 Gdje dag 1 3 a² +2a2; or by multiplying a², or (a² —4)x² = a² the answer. √(a²+ax)+√(a²-ax)=a, squaring 2a² + 2√(a¹ — a2x2)=a2; whence 4. Japan √(a¹ — a²x²) — — a²; squaring again, aª— a²x²=1a¹, or a²x²—¾a¹ ; Whence a=√3a²=÷√√/3. 2 a4x2 Ex. 22. Here √ (a² — x²) + x √ (a² — 1) = a²√/ (1 — x²) then, by transposition, (a²x²) a² √(1-x²) — x √ √(a² = (a2-1); squaring, a²x²-at-a¹x²- 2xa²√(1-∞²) (a ²-1) + a²x²-x2. Again, by transposition, and division, 2x √(1 − x²) (a² — 1) — a² — a²x² + x² — 1 — ( 1 − x²) (a² — 1 ); by squaring, 42(1 — x²)(a² — 1)=(1 — x²)² (a² — 1)³, M S - 64 and 4x2 (1 x²)(a² - 1), dividing the whole by (1-x²) (a²-1): therefore by actually multiplying, 4x2 — a² — a²x²x²-1, or 3x²+a²x²-a2-1; whence a²= 2 = a² — 1 *^* a³ +3 a² - 1 ; and consequently, x=√( ) Ans. a² +3 Ex. 23. Here √(x+a)=c−√(x+b); squaring, x+ac² 2c√(x+b)+x+b, and transposing c²+b-a 2c√(x+b)=c²+b—a, or √(x+a)= 2c c² + b —α 2 —b, ~) 2¢ whence x+b= the answer. 1 Patayang Ex. 24. Here V • Gives c² +b-x. 20 Whence b a+x Ifa b a+x + a b a + x +v. squaring, p C + )², and x= 00 + √ C C a-x a 20 Ja =1/ 4bc a² — x² 0, or abbx+ac+cx "a2-002, Therefore ab-bx+ac+cx=0', And bx-cx ab+ac Consequently x= ab+ac b-c RULE I. T =0, a 4bc a²x²⁹ b + c b-c EXAMPLES FOR PRACTICE. √ 4bc a²x² a2 22 ? ; 1 by The resolution of simple equations, containing two unknown quantities. · 1. Here 4x+y=34, and 4y+x-16, then from 1st,x= " 65 34—1, and from 2d, x=16—4y; therefore, by equality, 4. 34-Y 4. 16y—y—64—34, or 15y=30; whence 16-4y=16-8-8. 00 2. Here 2x+3y=16, and 3x-2y=11; then, from 1st, 16-3y and from 2d, x= therefore by equality, 2 11+2y_16-3y and conseq. 22+4y=48-9y, or 13y 3 16-4y, and, consequently 34-y=64-16y, or 30 y= 2, and x 1-5 5. 2 26 —48—22—26; whence y=13 11+2y 3 2x 3y 9 + 5 1 20' 1 =2, and x= 3 11+4 15 3 3. Here 3x 24 61 and + 4 5 120 the 1st by 20, and the 2d by 20; we shall have 8x+15y =9, and 15x-8y=10; whence, from the former equa tion, x= ; there- and from the latter x= J 101-8y 9-15y 15 8 9-15y .8 fore, by equality, 225y; hence 161y=533, or y=533÷161=}; also x= 9-15y 9-5 = } 8 8 ; then, mult. 101-87 15 or 813-64y=135 – Ma 4. Here x+2y=a, and x-2y=b; then, from, the 1st, x=2a-4y; and from the 2d, x=2b+4y; hence 26 +4y=2a-4y, or 8y=2a-2b; whence y-a-1b, and x=2b+4y=2b+a−b=a+b. 1 66 1 1 I 5. Here 2+32=8 8, 2y 3 2y 3 _ 21 3 18- -16- 36 3 1st, x=18- 22 3 Pal 6. Here+2= 2y and from the 2d, x=3+32; hence 3+3. 2 12 3 -12. 73 or 18+9y=96—4y; whence y=13 =6, and x=16 > 3' 8 lm 3 12. 6 1, then from the 1st, x=16 and from the 2d, x= or 4y=54-2y; whence y= " } 30 2x 3y 7. Here x+y=80, and 3 9y -80-y, and from the 2d, x= 8 9y+8y=640; whence 30-371-17-421-7 9, 9, and xy:: 4:3; then, from the 4 Ay 640 y = 17 =37 -3 54 6 1.1 17' ; .hence 1 L M ; hence, POČ r 16 9y 8 4y 3 9, and x= $ ; then, from the 1st, x perdendekat Ay 3 and x=80—y 14 80-y, or 8. Here y-6=, and x=y+6; then, from the 1st, THERE ARE 1 } x-2y-12, and therefore, by equality, 2y-12=y+6; whence y18, and xy+6=18+6=24. 67 } 440 11 or y RULE II. 1. Here 2+7y=99, and 2+7x=51; then, from the 1st, x=693-49y; which value, being substituted for x, Y in the second, gives +7(693-49y)=51, or 2400y= 33600; whence y=14, and x-693-49y=693-686-7. x + y 00 2. Here re-12=2+8, and if 8 P JAUTOD 4 5 3' then, from the second, x= EXAMPLES FOR PRACTICE. A y= =40; whence x= 18y+2100 47 ing substituted for a, in the first gives Y +8, or, clearing of fractions and transposing, 11y=440,' 4 } 2y-x 4 +27; ; which value, be- 9y+1050 47 C 12- 18y+2100 720+2100 47 47 -60. 3. Here x+y=s and x²-y=d; then, from the first, x=s—y, or x²=s²-2sy+y²; which value being substi- tuted for x, in the 2d, gives s2-2sy+y2-y³d, or y= s2-d s² -d s² + d { 25 ed for x, in the second, we shall have and xs-y=S— 28 2$ 4. Here 5x-3y=150, and 10x+15y=825; then, 150+39 from the 1st, x= 5 1 ; which value being substitut- 1500+30y 5 + 15y 525 =825, or 21y=525; whence y= =25, and a=== 21 150+3y 150+75 =45. 5 5 1 * 6.8 T 5. Here x+y=16, and x: y :: 3 : 1; then, from the 2d, x=3y, which value of x, being substituted in the 1st, we shall have 3y+y=16, or y=4, and consequently a =12. 6. Here x+12=12, and y+2=9; then, from the 2d, x=18-2y; which value, being substituted in the 1st, we shall have 18—2y+2=12; hence y=4, and x= =10. $ 2 7. Here x: y': : 3:2, and x²-y2-20; then, from 3y the 1st, x= or x²= x2 9y2 which value being substitut- " 2 4 ed in the 2d, gives { √16=4; and x= 8. Here 9y2 4 3y 12 2 12: =2+13, ac 2 galler 1 2 +27; then, from the 2d, x- ; -y2-20, or y216; hence y=== =6. x+8 4 660-27y_y being substituted in the 1st, we shall have 4 +25; hence, by transposing and reducing 28y=560,or 560 y= =20, and x= 660—27y__660-540 60. 28 ૭ X and +++16= RULE III. } 5 660-27y, which value, 2 2 EXAMPLES FOR PRACTICE. 2x-y 4 h Ex. 1. Here y+6. +6y=21, and 23 3 then, from the 1st, x+24y-76, and from the 2d, y+15x =63. Multiply the 1st by 15, and we have 15x+360y 1140, then subtract the 2d equation from this, and we 4. 5x a 69 1077. shall have 359y=1077, or y=359 =3, and from the first, x=76—24ý=76—72—4. Ex. 2. Here 3x+7y=79, and 2y=9+; then, from the 2d, 4y-x=18. Multiply this equation by 3, and we have 12y-3x-54, to which add the 1st, and we 133 shall have 19y=133, or y= =7, and, hence from the 19 2d equation x=4y-18-28-18-10. 1 Ex. 3. Here 30x+40y=270, and 50x+30y=340; then, multiplying the 1st by 5, and the 3d by 3, we shall have 150x+200y=1350, and 150x+-90y=1020, then, by subtracting the latter from the former, gives 110y= 330 270-40y 330, or y=110 =3, and, from the 1st, x= 30 270-120 5. 30 : 1 Ex. 4. Here 3x-3y=2x+2y, and x+y: ay 3:5; then, from the 1st, x=5y, and, from the 2d, 5x+5y= 3xy, or by substituting the value of x, found from the bst, we shall have 25y+5y=15y2, and dividing by y and 30 transposing y==2, and x=5y=10. Ex. 5. Here x2y+xy-30, and 3+y=35; then, adding 3 times the 1st equation to the 2d, gives x³-3x²y; +3xy²+y³=125, and by extracting the cube root x+y =5, and, from the 1st, (x+y)xy=30; hence by substi- tution 5xy=30, or xy=6, and x²+2xy+y²-25, from which subtract 4 times the last equation, and we shall have x2-2xy+y2-1, and by evol. x-y-1, adding this equation to x+y=5, and we shall have 2x-6, or x=3; whence y=2. Ꮐ * 70 18 1 3x-5y 2x+y 2 5 C a+b 2 Ex. 6. Here + ; then, from the first, 27y-11x=30, and from the y 3 с Vab } 2d, 9x-2y=96. Multiplying the 1st of these equations. by 2, and the second by 27, we shall have 54y-22x= 60, and 243x-54y=2592; then, by adding the two last. 2652 equations together, we have 221x=2652, or x= 221 12, and 9x-2y=96; hence 2y=9x-96-108-96- 12; hence y= 6. 12 2 ( Ex. 7. In the first of the two equations, (x+y): a: (x-y): b, and x²-y2c; ..b(x+y)=a(x—y), or (b+a)y=(a—b) x ; 202 or y ; x²(4ab)=c(a+b)2; whence x2 — (a-b)2 (a+b)2 а-в Whence y= x, and y² a+b Substitute this for y2 in the 2d equation, and we have x2. (a-b)2 (a+b)2 -x2 x²(a+b)² - (ab)²x²=c(a+b)², or x² { (a+b)² — (a—b)² } =c(a+b)², or Stefan =c, or y² = x² P α. I 1 Mult. 1st by d, Mult. 2d by a, By subt. M C 3, and 8 -b C y= V. the answer. ab 2 Ex. 8. Given ax+by=c * c(a+b)² 4ab dx+ey=f$ dax+dby=dc dax+ay af dby-aey=dc-af x-2y 4 Sad Poprad 1 c(a+b)2 c(a-b)2, 4ab 4ab C K or markomge 1 Kurkkuj 71 } - dc-af af-dc ae ae-ab Whence y=db Mult. 1st by e, eax+eby=ec Mult. 2d by b, bdx+eby=bf By subtraction, eax-bdx-ec-bf; ec-bf bf-ce Whence x ca - bd bd-ea Ex. 9. Given x+y=a, and x²—y²—b; Here divide x² — y² —b b -y==; 1 P by x+y=a; and we have x- b Whence by add. 2x=a+7, or x= b And by subt. 2ya - a Ex. 10. Here x²+xy=a y²+xy=b By add. x²+2xy+y²=a+b, or a²+b 2a a² - b or y= 2a x(x+y)=α, or x√(a+b)=a y(x+y)=b, or y√(a+b)=b; (x+y)²=a+b, or x+y=√(a+b). Now the two proposed equations may be put under the form Whence by division, x= ľ t a √(a+b) The Resolution of Simple Equations, containing three or more unknown Quantities. PRACTICAL EXAMPLES. 1 b ; and y=√(a+b) Ex. 3. Given x+y+ z= 53 -x+2y+3x=105 x+3y+4x=134 Subtract 1st from 2d, y+22=52 Subtract 2d from 3d, y+z=29 } • I J 72 1 Now, subtracting the latter from the preceding one, we have z=23. Also from the last, y=29-z=29—23=6, And from the first, x-53-y-2-53-29-24, That is, x=24, y=6, and z=23. Given x+y+} z=32 3 x + 2y +32=15 1 x + y + z = 12 Multiplying the first by 6, and the second and third by 60, we have 1 Ex. 4. 6x + 3y+ 2z=192 20x+15y+12%=900 15x+12y+10x=720 Again, multiply the first by 10, the second by 3, and the third by 4, give 60x+30y+20z=1920 60x+45y+36z=2700 ·60x+48y+40x=2880 Subtracting now the first of these, from the 2d, and the second from the 3d, we have 15y+16z=780 3y+ 4z=180 Mult. the latter by 5, 15y+20z 900 Subtract 15y+16z=780 } + 4z=120, or z=30 192-3y-2z 6 180-4z =20, and x= But y= 3 Therefore x12, y=20, and z=30. Ex. 5. Given 7x+5y+2x= 79 8x+7y+9x=122 =12 x+4y+52 55 Multiply the first by 8, the second by 7, and the third by 56, and we have 73 لو 56x+224y+280z=3080 Subtract the first from the second, and the second from the third, and we obtain 1 56x+ 40y+ 16z= 632 56x+ 49y+ 63z= 854 9y+ 47% 222) 175y+217z=2226 ( Multiply the first of these by 175, and the second by 9, and we have # 纂 ​1 1 1575y+8225z=38850 1575y+1953z=20034 By subtraction, 222-47z But y= -9, and x=- 9 That is, x=4, y=9, and z= 6272z-18816, or z=3 55-4y-5z 1 Ex. 6. Given x+y=a x +z = b y + z = c By addition, 2x+2y+27=a+b+c, or x+x+ z = a + b + } c; From which, subtracting each of the three given equa- tions respectively, we have c 10+10+10 Ja-3b+} c Z a + 3 b — c The values sought. z = − a + b + y= XC * Ex. 7. Given 2+3+62, ++=47, and 4 5 ac Y Z +=+ 38. 4 5. 6 By clearing of fractions, these become 6x+4y+3x=744, 20x+15y+12z=2820, and 15x+12y+10%=2280, G 2 1 74 Multiplying the first by 20, the second by 6, and the third by 8, we shall have. 120x+80y+60z=14880 120x+90y+72z=16920 120x+96y+80%=18240. And, subtracting the first of these equations from the second, and the second from the third, there will arise 10y+12%=2040, and 6y+8x=-1320. Or, multiplying the first of these two equations by 3, and the second by 5, we shall have 30y+36z=6120, and 30y+40x=6600. Whence, by subtracting the former of these from the latter, we have 4z480, or s= =120. And, consequently, by substitution and reduction, x= 24, and y=60. Z Ex. 8. Given +y=x+100, y-2x=2z-100, and x+100=3x+3y; then, by transposing, 2+y-x=100, y-2x-2x=-100, and 2-3x-3y=-100. And, by adding twice the first to the second, and subtracting the third from the first, we shall have 3y-4x=100, and 2x` +4y=200; multiplying the second of these two equa- tions by 2, and adding the result to the first, we have, 11y=500, or y=45. And, consequently, by substi- tution and reduction, x=9, and 63 Ex. 9. Given x+y+z7, 2x-3y+3%, and 5x+ 5z=3y+19; then, by adding the first to the second, and 3 times the first to the third, and transposing the terms of the second and third, we shall have 3x-22-10, and 8x +8x=40. Or, dividing the latter of these two equations by 4, we have 2x+2x=10, and, adding this to the for- mer, 5x20, or x=4, and, consequently, z=1, and y=2. 7 Ex. 10. Given 3x+5y-4x-25, 5x-2y+32=46, and 3y+52-x=62; then, adding 3 times the third to the first, and 5 times the third to the second, we shall have 14y+11=211, and 13y+28x=356. A * 75 Or, multiplying the first of these two equations by 13, and the second by 14, there will result, 182y+143% 2743, and 182y+392z=4984. And, subtracting the former from the latter, 2241 2492=2241, or z= 9, 249 And, consequently, by substitution and reduction, y=8, and x=7. Ex. 11. Given x+y+z=13, x+y+u=17, x+z+ u=18, and y+z+u=21; then, assume S=x+y+z+ u, and the above equations will be transformed into the following ones, } S-u=13, S-2=17, S-y=18, and S-x=21. Add all these equations together, and we have 4S-x -y-z-u—69, that is, 4S-(x+y+z+u)=69. But x+y+z+u=S, therefore, 4S-S-69, or 3S= 69, and S=23, and, consequently, by substituting for S its value in the four transformed equations, we shall have 23-u-13, 23-z=17, 23-y=18, and 23-x=21. And, consequently, x=2, y=5, z=6, and u=10. 1 Ex. 1. Let one of the parts =x, Then the other will be And by the question 15-x=x, or 15-x, 7 60-4x=3x, or 7x-60; whence x= 60=84, the one part; and 15-x=63, the other. Ex. 2. x Let the value of the purse be a; then the mo- ney =7x; and by the question 7x+x=20, or 8x —20, or x=2º=2s. 6d. the value of the purse, and, consequently, 17s. 6d. the money contained in it. Ex. 3. 1 SIMPLE EQUATIONS. Let the number of sheep' =x; then, by the question, x+x+x+7=500; or 2x-492; mult. by 2, 5x=985; Whence, 95-197, the number sought. x= 76 } I Ex. 4. Let the length of the post; then, by the question, x+x+10=x; multiply by 12, 3x+ 4x+120 12x; 0 Transposing, 5x=120, or x=13°24' feet, the answer required. 1 Ex. 5. Lethe number of guineas a; then, x-x -72, or multiply by 20, 20x-5x-3x= 1440; whence 12x=1440, or x= 1440—120 12 guineas. Ex. 6. Let B's share, then, by the question, A's share B's share 3x, 2x, I Consequently, x+2x+3x=300, or 6x=300; whence x= -300-501. B's share, 2x=100%. a's share, and 3x1507. c's share. 6 Ex. 7. Let the age of the wife at the time of thẹ marriage, and that of the husband 3x; Then, after 15 years, their ages will be x+15, and 3x+15; and, by the question, 3x+15= 2(x+15); or 3x+15=2x+30; Whence, by transposing, x=15, the age of the wife; and 3x=45, the age of the husband. Ex. 8. Let the number sought =x; then, by the 2(x-5) question, x-5 will be the remainder, and 3 2 40; whence, 2x-10=120, or 2x=130, and consequently, x=130-65, the number sought. Ex. 9. Let the less number of voters =x, then the greater number =x+120, and, by the question, x+x+120-1296; Whence, 2x=1296-120-1176; consequently, x= 588, for one candidate, and x+120-708, for the other. Ex. 10. Let a represent the age of c, then 3x is the age of B, and 6x the age of 4. Now by the question, x+3x+6x=10x-140; whence 1 I 1 may reg x=414, c's age, 3x=42, B's age, and 6x=84, A's age. M Ex. 11. Let the equal sum laid out by each be x then a leaves off with x+126, and в with x- 87; and by the question, x+126-2(x-87); or x+126=2x- 174; Whence x= 3007. the first stock of each. Ex. 12. Let the price of the harness price of the horse 2x, and the price of the chaise and by the question, x+2x+6x=60l. or 9x=60; 60 Whence 61. 13s. 4d. the value of the harness; OC= x 9 2x=131. 6s. 8d. the horse; and 6x=407. the chaise. K spla Ex. 13. Let x denote the number of beggars; then by the question, 3x-8 was the number of pence he had about him; x; then the 6x; 1 Which from the other part of the question may also be denoted by 2x+3; Whence 3x-8=2x+3, or, by transposing, x=11, the number of beggars. Ex. 14. Let x denote the value of the livery; then x+8 is the whole amount of his hire for the year, or for 12 months; 7x+56 Hence, as 12 7:: x+8: 12 the hire for 7 months; but by the question the servant received a+ 7x+56 22; whence =x+23, or 7x+56=12x+32, 12 W And by transposition, 5x-24, or x=24-441.-41. 16s. In the preceding examples only one unknown quantity has been employed, but it will be more convenient in many of the following questions to use two or more un- known letters, according to the nature of the question.. Ex. 15. Here let x denote the son's share, and y the daughter's; Then the value of their shares will obviously have to 箐 ​1 י 78 1 each other the same ratio as half a crown to a shilling ; that is, as 5 to 2. Hence, then, we have xy:: 5 5:2 And x+y=560 $ 5y From the first, 2x=5y, or x= 2 5y Whence from the second, 2+y=560, 7 or 5y+2y=7y=1120: whence y=1120-1601. the daughter's share, and x= 4007. the son's 2 5y share. Ex. 16. Here it may be observed, that every number consisting of two digits is equal to 10 times the digit in the tens place plus that in the units. If therefore a be put for the former, and y for the lat- ter, the number itself will be denoted by 10x+y; and the number with the digits Inverted by 10y+xc; Hence by question, 10x+y+18=10y+a; Whence the second equation gives 9x-9y=-18, or x-y=-2, or x=y—2; Which substituted in the 1st, gives 10(y-2)+y= 4(y-2)+4y. And this, by multiplication and transposition, becomes 10y+y-4y-4y=20-8, or 3y=12; Whence y=4, and x=y-2=2; Therefore the number sought is 24. Ex. 17. Let a represent the equal income of each; then by the question, A's yearly expenditure is 4.00 5' and that of B, 4.x Categ { -5 +50; In four years, therefore, в spends 16x 5 ,} + 200; which 1 ་ } f 79 exceeds his income in the same time (viz. 4x) by 100; hence we have the equation 16x 5 +200=4x+100, or 16x+1000=20x+500; 500 Whence 4x=500, or x= 1257. the income sought. 4 Ex. 18. Let the number of persons in company be x, and the number of shillings each paid y; then xy will be the whole reckoning. Now had there'been three persons more in company, viz. (x+3), each would have paid (y-1) shillings; whence we have P (x+3)(y-1)=xy; And from the other conditions of the question, (x-2)(y+1)=xy; K Whence, from actual multiplication, these become xy+3y-x-3=xy xy-2y+x-2=xy 1 And consequently by addition we have 2xy+y-5-2xy, ૭ { M Or, cancelling the 2xy on both sides, y—5=0, or y= 5, the number of shillings each paid. And by subtracting the second equation from the first, 5y-2x-1=0; 5y-1 Whence 2x-5y-1, or x= ber of persons in company. Ex. 19. Leta denote the money he had about him ; then by borrowing x and spending one shilling, he had left 2x-1. 1 Also, at the second tavern, after borrowing 2x-1, he had 4x-2; but spending one shilling, he had left 4x-3. At the third tavern, he borrowed 4x-3, and then had 8x-6; and after spending one shilling, he had left 8a-7. 24 2 1 =12, 12, the num- P 1. } T } 80 { At the fourth tavern, borrowing 8x-7, he had 162–› 14; but after spending another shilling, he had left 16x -15; which by the question is equal to nothing. Whence 16x-15=0, or x=150s. 111d. the money he had at first. 16 Ex. 20. Let x and y denote the two parts; then by the question, x+ y=75, and 3x-7y=15 Mult. the first by 3, Subtract the. 2d, " they And we have 210 10 10y=210 Whence y= -21 the least number, And x=75-y=54 the greatest number. Ex. 21. Let x the whole quantity of the mixture; +25 was the quantity of spirits, and x-5 the quantity of water. then Which altogether made the whole x; therefore by addition, x+x+20=x; or by mult. by 6, 3x+2x+120=6x; f 3x+3y=225 3x-7y= 15 vited Whence, by transposing, x=120, and conseq. x+25=85 gallons of spirits, x535 gallons of water. Ex. 22. Let x- y= Then 21x the number of guineas, and the number of moidores ; the shillings paid in guineas, And 27y the shillings paid in moidores. Now, the whole number of pieces used being 100, and the number of shillings paid being 2400, we have x+y= 100 21x+27y=2400 $ Multiply the first by 27, and we shall have 27x+27y=2700, The second 21x+27y=2400; 1 { ↓ 1 81 Hence, by subtraction, 6x=300, or x=50, the number of guineas; and, consequently, y=100—x, or y=50, the number of moidores. 1 Ex. 23. Let x be the number of days; then, 14x miles will be travelled by one, and 16x miles by the other; Hence, 30x-197, or x=1976d. 133h. the time re- x: 30 quired. Ex. 24. Let x denote the weight of the body; then, x+9 is the weight of the head, and, by the question, x=3x+9+9, or x=18; Whence, x=36, the weight of the body, x+9=27, the weight of the head, and 9, the weight of the tail; consequently, 36+27+9=72lbs. the weight of the fish. Another Solution. Let 2x the weight of the body; then, 9+x: x the weight of the tail. ..9+9+x=2x; W by transposition, x=18; .. the fish weighed 36+27+9=72lbs. x x Ex. 25. Let and represent the three parts re- 2'3' quired, and the three latter conditions of the question will be answered; For, the first multiplied by 2, the second by 3, and the third by 4, will obviously be all equal to x, and therefore, equal to each other. Hence, then, it only remains to fulfil the equation, 88 4' 20+10+1=10, 4 Or, multiplying by 12, to clear it of fractions, 6x+4x+3x=120; Whence, 13x=120, or x=130. H 1 1 1 2 } { } 120 2X13 are the parts sought. Ex. 26. Let 2x, 3x, and 4x, be the three parts; then, it is obvious that the first, of the second, and of the third are equal to cach other; 3 Wherefore, there only remains the equation Whence, 9x=36, or x=36 2x+3x+4x=36; = 36=4, Consequently, 2x=8, 3x=12, and 4x=16, are the parts required. Ex. 27. Let the value of the first horse be x, and of the second y; then, by the question,' or, by transposition, x+50=2y y+50=3x 2y-x=50 −y+3x=50 $ { 8 120 Therefore, =413' 3x13 い ​CO 82 1 1 3 \ 1 " 131 and Multiply the latter by 2, and we shall have −2y+6x=100; to which adding 2y-x= 50, we have I 5x=150, or x=30%. the value of the first horse; and, from the second equation, y=3x-50-401. the value of the second. 120 4 X 13 Ex. 28. Letxa's money, and y=n's. Then, when n gives a 5s. the latter will have x+5, and the former y-5; and if a gives в 5s. then A will have x-5, and в 3+5. Sy+5=2(x-5) (x+5=3(y-5) Now, by the question, Or, multiplying and transposing, 2a:— y=15& x+3y=20 Multiply the latter by 2, and we shall havo -2x+6y=40, to which adding 2xy=15, the sum gives 5y=55, or y=11, B's money, } ↓ # } 4 13. =2- 1 ! Fu 1 83 - 15+y And 2x-15+y, or x == 13, A's money. 2 1 Ex. 29. Let x and y be the two numbers; then, by the question, the difference is to the sum as 2 : 3, and the sum to the product as 3: 5;, that is, 현 ​1 15+11 2 x-y: x+y: xy :: 3 X 5 Which, by multiplying extremes and means, gives ·Y: x+y: :: 2:3 25y+5y=15y² Hence, dividing by y, and transposing, 15y=30; or y=2, one number, And x=5y=10, the other. " 3x-3y=2x+2y Į 5x+5y=3xy From the first, x=5y; which, substituted in the se- cond, gives per qu Ex. 30. Let x the number of shillings he had at first; then, by the question, he lost x, and, therefore, had 3x left, to which he won 3s. 3x+12 After this, he had 3x+3= 4 one-third, and had then two-thirds of it remaining; viz. 6x+24 12 6x+48 12 36x+288 A ; to which adding 2s. won, he had Fi Then losing one-seventh of this, he had six-sevenths of it, viz. left; which, by the question, was 12s. of which last he lost } 84 36x+288 Whence 12, or 36x+288-1008; 81 And, consequently, 36x=720, or x=20s. the money he had at the beginning. Ex. 31. Let x be the number of leaps the greyhound takes, and y the length of each; then, by the question, 3:4::a: fx, the number the hare takes, and • 3:2 y 3y, the length of each; ▼ 84 $ But the hare being 50 of her leaps before the grey- hound, he has to pass over +50 leaps of the hare. Now, each of these, viz. x and x+50, multiplied by their respective-lengths, will obviously be equal; viz. xy = ( x +50) 3 y, Or, dividing each side by y, and reducing, 3x=x+100, or 9x-8x=300; Whence, x=300, the number of leaps the greyhound takes before he catches the hare. Another Solution. Let 3x the number of leaps the greyhouud must take ; ..4x the number the hare takes in the same time, and 4x+50 the whole number she takes ; but 2 3 3x 4x+50; ..9x=8x+100; papp by transposition, -100. Hence, the greyhound must take 300 leaps. Ex. 32. Put w, x, y, and z, for four parts; and let m be that quantity to which each part becomes equal af- ter the operations expressed in the question are per- formed; ང་ Then we shall have w+x+y+z=90. Also w+2=m x-2=m 2y SMA Q & C 2 =m m Or w-m-2 พะ x= m + 2 m Y 2 z=2m P Whence adding together the four last equations, w+x+y+x=4m+5: 2 or substituting 90 for w+x+y+z from the first, 4m+90, or 9m 180; whence m-20; consequent- ly, w=m- m—2—18; x=m+2=22; y= :10; and z= 2 KAMA 2m 40, the parts required. Ex. 33. Let x, y, and z, be the three numbers; then m } by the question, x+y+z=324, x::: 5: 7, and x+z =2y, or x+z-2y=0; now, by subtracting the last equation from the first, we have 3y=324, or y=334- 108, and, consequently, by substitution and reduction, x=90, and z=126. 85 Ex. 34. Let a be the number of days the man would be in drinking it by himself; then will be the quantity he drinks in a day, and is the quantity the woman drinks in a day: ** Therefore the quantity both drink together is +3² 0. But by the question, 12(+)=1, or multiplying by 30x, 360+12x=30x; whence, by transposing, 18x= · 360, or x=20, the number of days sought. 00 Ex. 35. Let a be the number of men in the side of the less square, and a+1 the number in the side of the greater; then a² will be the whole number of men in the former, and (x+1)²x²+2x+1 in the latter; } 1 30 Whence x2+284, and x²+2x+1-25 will each ex- press the whole number of men; from which we have this equation, x x²+2x 24=x²+284, or 2x=284+24=308; Whence x=154; and consequently x²+284=24000, the whole number of men. Ex. 36. Let x, y, and z, be the number of days in which ▲, î, and c, respectively would finish the work; A, then 1 81 1 A will do part of it in one day, ac 1 1 B will do part, and c- 2 pig part. Y Then, by the question, we shall have 1 1 1 + У 12 pedag } + 1 → 86 As } t + Z 1 G I OC Qr by division, + Y 1 00 1 10 F And consequently by addition pla -18 - 31 720' 41 720' 49 720' 1 1 1 + 1 2 2 2 1 1 1 ft y 十六​十 ​Ma + y Z Z 12 From this, subtracting each of the three first equations, we have 1 or, z= 二十​二十 ​XC Y or, y= Z 9 1 1 or, x= 1 ·Ex. 2. Given x² 8 \ 121 Z 720 P Therefore x= 720 31 720 41 720. 49 121 10 360 QUADRATIC EQUATIONS. 17 23 days for c 31 EXAMPLES FOR PRACTICE. Ex. 1. Given x²-8x+10=19; By transp. x²-8x= 9; Whence x=4±√(16+9)=4±5, Therefore 9, or -1. 23 17 days for B 41 34 14 days for a. 49 Given x²-x-40-170; by transp. x²-x=210; 2 1 29 2± ? 841 Whence x=√(}+210)=}±√= 4 15, or -14. 1 1 87 Ex. 3. Given 3x²+2x-9=76; by transp. 2 85 whence 3x²+2x=85, or x²+3α= 1 85 -1 x = = = = = = = ± √( + 353 ) = = = = v 3 Therefore x=- .3 12-20 20- 1 M -XC Whence x= Therefore x 1 16 1 3 3 Ex. 4. Given ²+7=8; by transp. 5 or x2. 8' 3 =5, or 1 77 1 3+6=12 ; Ev 1 Ex. 5. Given√=22 2 or 256 9 2 10 3x=8 10. — ± √ √ + ) == ± √ Agg 5 6' 17 3 Mult. by 2; x 1 1 133. 1 √ x = ± √ ( + z ) = 3 + √ 3 1 √x=44; whence Therefore √x=(+3)=7, or -19 49 36 400 9 19 3 361 9 Conseq. x=72-49, or Ex. 6. Given x+√(5x+10)=8 By transposing √(5x+10)=8—x, By squaring 5x+10=64-16x+x², Where 21x=-54; therefore } } • t 88 1 ( 1 | * 1 1 Y 21 441 21 225. x = 2 ± √( 4 - 5 4 ) = 2 + √ ² 20 4 Ex. 7. Given (10+)*—(10+∞) ** That is, 18, or 3, the answer. a -2. Here, since the first index is double the second, the equation is a quadratic; therefore by the rule 1 # 21 15 £ 2 2 Ex. 10. (10+ a ) = ± √ (i + 2) = ± √ ₁₂ = 2; ] 1 1 9 2 êt 4 Ex. 8. Given 2x¹-x²+96=99; by transp. 3 2x¹x²-3, or x¹. 2 1 3 Whence a²= ±√(+2)= ±√25 16' Therefore x². 1 == 4. S 1 Whence (10+x)=2; (10+x)=4, 10+x=16, and x= =6. 6 4' } 1 4 1 K -x02 and x= √. 1 6 1 Ex. 9. Given x+20x³-10-59; by trans. x+20x3 69; whence a-10/(100+69)=-10+13=3, or -23; and x=3/3, or -3/ 23. } 4. 2 1 1 1 8 1 Whence = √(+=+25 an 3 ± √6. Given 3x²º—2x²+3=11; by transposition, 2 8 3x²n-2x²-8, or x2n ğən zxn=3 1 1 9 *This value of a does not answer the condition of the question; be- cquae, from the transposed equation, a must be less than 8, See my Elementary Treatise on Algebra, Theoretical and Practical. 89 f 6 "Therefore "--- -2, and x=2″· 3 Ex. 11. Given 51/ x-3√x=1}, or t 3√x-5x=-13; then 1 5 1 4 By division x zx Where x ± √ √36 1 5 25 15.3 Whence x 6 ± 6 ·•Conseq. x= < or ²= x2 4 Ex. 13. 11 4 13 =3- or 81' Ex. 12. Given√(3+2x²)=2+3x² 3' 9 9 x2 02² = = 1/3 ± √ / 16 + 16 4. 33 6 Given x√( Oc 1 or 3 4. 5 9 31 ) == 0 + √ 36 3 Mult. by and we have x√(3+2∞³)=1+x² 2' P pack Squaring, 3x2+2x¹= 3 9 Whence ¹+2x² ; therefore by the rule 16 9 6 16 -x): 1 81 } 1 1 +-x² + x²; 3 = 2 18 ¡±√ 16' √2, or = √(−3±3√/2). x= ↓ 1+02 √x x√(6—x²)=1+x2; or by squaring 6x2. ·x²-1+2x²+x04; Whence 2x-4x²-1, or x¹-2x² — — }; A ; mult. by √x, 1 + } • Į 4 1 Therefore 21+√(1-3)=1±√2, Consequently x=√/(1++√/2).' Ex. 14. Given√(1-x³)=x²; multiplying by ∞,, we have √(1-x³)—x³; or 1—x³ —x6, or x²+x³-1; whence also we have, 1 "That is, x2 Ex. 16. ax Į ,00 Ex. 15. Given a√(−1)=√(x²—b²). By squaring x² — x²-b², or 2x²—ax—b², Jag 90 -1 x3 x³ = − } ± √ (} + 1 ) = = = ± √5, or √ = ===(~+++√/5)+ a ·00 82 ེ? ارة a b2 a. a² = = √(1 + 2/2 ) = = = = = √ 16+2): 4. (8b²+a²). Given √(1+x—x²) — 2 (1 + x − x²) or x 1 -1 Here (1+x-x²) — 2(1+x- - 1/(1+x=-=-=²) = = = = 1/1) 18 Therefore I 2 1±√16.18=1±√1141 1 1 + 12 3' Kamerak 1 2 (1+x—x²)=1±√(√16 - 1/8 ) = Mega x= }} = √(}+} })=3+}√/14. 2 ΟΙ Consequently 1+x-x² =({})² = 1/1, or (-)² 9' 1 I 6' 36 From the first, we have x2 x, which gives x = 3 ± √ ( } + })=3±7√41; From the second, we have x2-x=3, which gives 35 } 1 / 91 1 " 1 1 Ex. 17. Given √(x Given √(x−)+√(1 -- ) + √(1 − = ) = x XC Mult. by √(x-1) —√(1 — 1), and we have J X ¤ — 1 = x √/ (x — — ) — ∞ √/ (1 — — ) — → - 1) —~~√(1–2); V 00 1- 1 · = = √(x − 1 ) — - OC 00 (x — — 1) — 2 √/ (x ac √(1 − 1), x =√(-)+(-); therefore, 20 − 1 ) + 1 = 0; 22/10 X = +x− 1 = 2 √(x − 1 ), -2√/(x− −), or, which is the same, XC OC divide by x, + 1 and from the first, +1=0; or, by extracting, √(x − 1)—1—0, or a Gardagklapky S 20 Consequently, x} = √(}+1)=}±√/5. St M 1 OC Ex. 18. xen Given a¹”—2x3"x"=6;* then this expres- sion is evidently to (2n) — (x²x²)=6. Put y=a2n-an; then, y-y=6, and y-y+1=6+1=25, by completing the square; y-==2}, by extracting the root. Whence, y=3; therefore, a"-"=3; and, completing the square, x-x+1=3]=, or a"-}= 13, by extracting the root; therefore, "+}√13, and x=V (+3√13), the Ans. 1, or x²-x=1, 1 *See Note, page 131, in the book, for the method of resolving this, and the equation in the following example into factors; or iny Treatise on Algebra. } } £ I 92 ——a; Ex. 19. Given x42x3 + x a Here, this equation may be expressed thus, (x2-x)² - (x²-x)=α; then,. (x² -— x ) ² — ( x² -—-—- x)+1=a+1, by completing the square;. x²-x-1)= ± √(a+1), by extracting the root; and a²- 002 x = ± √(a + 1). Again, x2-x+1=3±√(a+1), by completing the square; x-3=√[±√(a+1)], by ex- tracting the root; therefore, a±√ [±√(a+)], the → Ans. QUESTIONS PRODUCING QUADRATIC EQUATIONS. QUESTIONS FOR PRACTICE. CTICE. * Ex. 1. Let x and y represent the two parts; then we have x+y=40, and x²+y2=818, + I From the first, x=40-y, or x2-1600-80y+y2, By substitution, 1600-80y+2y2-818, or·· 2y2-80y=-782, or y²-40y=-391; Whence, y=20±√(400-391)=20+3=23, or 17, But x=40-y=17, or 23; Therefore, 17 and 23 are the parts sought. Ex. 2. Let a represent the number sought; Then, by the question, (10-x)x=21, or x²-10x=-21; Whence, x=5±√(25-21)=5+2=7, or 3. Ex. 3. Let x and y be the two parts; then, by the question, we have 35(24-2y)=y(24-y), or 840-70y=24y-y2, or y2-94y=-840';: x+y=24) 35(x-y)=xy S Here, x=24-y, which, substituted in the second equation, gives ↓ 93 1 Whence, y=47±√(472-840)=4737=10, or 84.* Consequently, x=24-y=24-10-14. Ex. 4. Let one part x; then, the other will be 20-x, And, by the question, 20(20-x)=x², or x²+20x=400, Therefore, x=-10+√(100+400)=-10+10√/5, And the other part =20-x=30—10√/5. Ex. 5. Let x and y represent the two parts; then, x+y=60, and xy: x²+y²:: 2 : 5, or 5xy x+y=60, and = x²+ y², 2 Squaring the first, x2+2xy+y2-3600, Subtract the second, a2-2xy+ y²=0, 7200 We have 44xy=3600, or xy= =800, 9 800 From this we have x= 800 Y y y—60, or y²÷60y= 800; √(900-800)=30+10=40, or 20, Consequently, x=60-40—20, or 60—20—40. In all quadratics of this kind, in which x may be changed for y, and y for x, in the original equations, with- out altering their form; the two values of one of the quantities may be taken for the values of the two quan- tities sought. first, gives Whence, y=30+ ; which, substituted in the # By squaring the second, Subt. it from twice the first, And we have Ex. 6. Here, in order to avoid radicals, let us as- sume a² and y² for the two parts; then, by the question, x²+y2146, and x-y=6. Which may now be solved the same as Ex. 1. Ano- ther method is as follows: ? T *This value of y does not answer the conditions of the question.. I x² -2xy + y² = 36 2x2 +2y2=292 x²+2xy + y²=—256 dang 94 Hence, by extracting But • x+y=16 x-y= 6 Whence, by addition, 2x=22, or x=11, and x²=121, And, by subtraction, 2y=10, or y=5, and y2=25. Ex. 7. Let x and y represent the two numbers; Then, by the question, And 4 y= 00 tion x+y=3 16xy 5 Whence Squaring the first, Subtract 4 times the second, And we have Whence, x-y=16; and since also By addition we have also 2x-36, or x=18 And by subtraction 2y= 4, or y I 2. But the more direct solution is as follows : 36 From the second equation x Y 36 the first gives+y=20, or y³—20y: Y 4 1 gra y Whence y=10±√(100—36)=108-18 or 2, Therefore 18 and 2 are the parts sought. Ex. 8. Let x and y be the two numbers, and conse- 1 quently and their reciprocals. Then by the ques- where the 2d becomes x+ and 1 223 1 + Y Apmak เ 16xy 4 igo 2 x+y=201 xy=36) 16 5 x²+2xy + y²=400 4xy =144 x²-2xy + y² 256 x+y=20 1 LO 3 which substituted in 20 or xy= 48 5 Consequently = ; 12y 5 4. gives 12+y=3, or 5+12y²=16y, or y²—' Totally 1 5 12 36 ; • which substituted in the first. 4 5 zY— 12 ; : 95 I 2 Whence y=√ Therefore and 1 4 9 5 are the quantities sought. 6 Ex. 9. Let x and y represent the two numbers. xy Then by the question, x-y=15, and 22=y³, The second equation gives xy=2y³, or x=2y²; Whence by substitution in the first, we have 15 2y²—y=15, or y²—}y; and hence 2 1 1 5 2)=3+1 = 2, or o 6 2' 6 6'. → 5 1 15 11 y = ± √ ( 6 + 5 ) = ± 4 -5 2 Consequently x=15+y=15+3=18. Here the two values of y are not those of x and y, be- cause y is made to represent the less number, and cannot, therefore, be changed for a without altering the condi- tions of the question. Ex. 10. Let x and y be the two numbers; then by the question, x-y=5, and x³-y³-1685, 3 3 By the 1st, x=5+y, or x3=125+75y+15y²+y³ Consequently 125 +75y+15y²+y³—y³=1685, That is, dividing by 15, y²+5y=104; 3; 675 25 Whence y= ±√(+104), or y=- 5 21 22 + 3/2 Therefore x=5+y=5+8=13; Consequently 8 and 13 are the numbers required. Ex. 11. Let x be the number of pieces, and Y the shillings that each piece cost; then, by the question, xy=675, and 48x=675+y. From the 1st, y: =8; ; whence by substitution we have ac A 96 1 } I 30 $ 675 48x=675+ or 48x2-675x=675, And 225 32 X 225 16 2252 322 675 675 15 Or x2 16. 225 225 255 + -)= + 16 32 32 45,, the shillings each cost. Ex. 12. Let x and y represent the two numbers; then, by the question, x(x+y)=77, or x²+xy=77, And y(xy)=12, or xy-y2-12. By subtraction we have x2+y2-65; whence y=√ (65-2); which by substitution in the first, gives 77-202 x²+x√(65—x²)=77, or √(65-x²)= And by squaring, 65—x² " ·±√ y = OC Whence a²= 219 2x¹-219x2-5929, or x 2192 16 kaļ Marke -OC = 4 The 225 { ±√( ; whence 1 772—154x²+x¹ x2 219 2 219 23 242 +7=-=603, or 4 4 土 ​4. —2964), or 196 4 15, OC 1 1 -49, xc² Consequently √60, or √49=7; and adopting the latter y=√(65-2)=√(65-49)=4; that is, 7 and 4 are the two numbers required. > x²-2964; 1 or 3 Y Ex. 13. Let a represent the number of sheep, and the shillings each cost, and consequently y+2, what they sold for; then, by the question, we have 1 1 ; the latter, And (x-15)(y+2)=1080 by multiplying, gives xy+2x-15y-30-1080, Or since xy=1200, we have 2x-15y 1200 ; and 15y=- ; which substituted Ꮖ Whence x- Now y= } in the latter, gives 2x. y+2 2y 11 斗 ​97 xy=1200 2 Juttgart 18000 —90, or 2x²-18000-90x, or x²+45x=9000; 45 45 195 + 2 2 the number of sheep, and 16s. the price each cost. Ex. 14. Let x less number, less number, and xy the greater; then xyxxx²y2-x², and x²y²+x²=x³y³ — x³, by the question; and, by division, y=y2-1, and y²+1=xy³ -XC. From the first, y²-y=1, and, by completing the square, y²−y+1=1+1=4; therefore y√5, by y²+1 extracting the root, or y=3+√5. Again, x= 5 y³-1 21 (5+ √ (5+√5), the Answer. XC 18000 452 ±√(~—+9000)= 1200 75 X -90. (because y²-y+1, and y³=y²+y=2y+1,)= 1 -2 1 1 1 + = √5, and 2y= √5x(+ √5)= 2 y. 2, 1+√5 } 75, Ex. 15. Let (m+n).and (m-n) represent the two numbers, ↓ Then, by the question, (m+n) —(m-n) =2n=8 (m+n)+(m-n)-14560 And Now (m+n)m¹ +4 m³n+6m² n² + 4mn³ + n¹ ( m ———— n ) 4 ———— m * — 4m³n + 6m² n² 4mn³n²; pl 12 ↓ 1 1 t • 1 1 1 Whence, by subtraction, 8m³n+8mn³—14560, Or by division, and substituting n=4, we have m³+16m 455 1 98 Mult. by m, m¹+16m²-455m=65×7m Add 49m² to both sides, and it becomes m¹+65m²-49m²+65×7m Ma Therefore by completing the square, 652 m²+65m²+ =(7m)²+65(7m)+ 4 65 2 65 2 Whence m²+ :7m+ or m²=7m, or m=7; Consequently m+n=7+4=11, one number, And mn=7-4, 3, the other. Ex. 16. Let x be the whole number of persons, and Y the number of shillings each would have had to pay; then, after two were gone, the number was only (x-2), and their reckoning y+10. Now, by the question, xy=175, and (x-2)(y+10)=175, ; therefore 5x- OC 652 4 From the latter, xy+10x-2y-20-175, or Since xy=175, we have 10x-2y=20, or 5x-y=10, 175 175 But y= 10, or 5x210x=175, OC or x²-2x=35; Whence x=1+√(1+35)=7, the number sought. Ex. 17. Let x be the number of persons at first, and y the shillings each would have received; then x+2 was. the number at last, and y-1 what each actually re- ceived; hence the following equations, xy=144, and (x+2)(y-1)=144, From the latter, xy+2y-x-2=144, or since xy=144 144 288 Zy-x=2; from the first, y= ; therefore -x=2, OC or x²+2x=288 ac f Ex. 18. Let Consequently, x=-1±√(288+1)=-1+17=16 xc² 1 Y 20¹k y2 في حرج 'bers in geometrical progression; then we have y 2 I xc2 73 x, y, and y² 0 -+ x + y +² XC y+ x2 Y 201 99 -+x²+ y²+ 85-b Make x+y=s, and xy=r; then will x²´+ y² — s² — 2r x², y² + OC 204 y2 хуя y2 x2 :(a—s)*—2r. 1 Here equations first and third are drawn from this con- sideration, that the sum of the squares of any two quan- tities is equal to the square of their sum minus the double rectangle of them. Whence by adding the first and third, we have 7 =15-a -a-s; Ans. represent any four num- +x² + y²+ s²+(a—s)²-4r=b,· And from the second, we have r³+y³=xy(a—s), or x³+y³=r(a—s). Also since a³+y³=(x+y)³—3xy(x+y), or =s³—3rs, We have r(as)=s³—3rs, or r= 1 y+ 2s² 2as+a² 202 1 '2s+a' Which value, substituted for r in the above, gives 453 s²+(a—s)² - =b, or P -b, or 2s+a 453 2s+a St :8; 1 ? A 1 b s²+-s a 4 Whence 4s3-4as2+2a2s+2as22a2s+a³-4s32sb tab, Or by reduction and transposition, 2as+2bs a³-ab, or 1 1 1 As 22th f V 2a Whence n²: 100 S Consequently s=- (b), 4a2 Where, by substituting a=15 and b=85, we ob- tain s=6; 1 b, :±√(・ 222 b2 -1 but r= =8. Hence then, x+y=6, and xy=8; from which are determined x=2, and y=4; and therefore the numbers sought will be 1, 2, 4, and 8. Ex. 19. Let m+n be one of the numbers, and m-n the other; then we shall have (m+n) +(m—n) =11, or 2m=11 (m+n)²+(m-n)5-17831-a. $3 216 2s+a 27 Now (mn)5m5+5m¹n+10m³n² + 10m²n³+5mn¹+n³, and (m-n)”—m5—5m¹n+10m³n² — 10m²n³+5mn*——-nº. Consequently by addition we have 10mn¹+20m³n²+2m5=a, or a-2ms n¹+2m²n² 10m Or substituting for m its value, found as above, in the second term, we have 121 2 -121 14641 ±√( + 4 16 -); Whence, by substituting for a and m, we derive n², or n=23; consequently, m+n=5}+11=7, and 5)-1=4, That is, the two numbers are 4 and 7. 1 + a² 2 a-2m³ 10m a-2m5 10m 1 I 101 Ex. 20. Let x-3y, x-y, x+y, and x+3y, be the four numbers; then, by the question, (x-3y)(x-y)(x+y)(x+3y)=176985-a, or (x²-9y2)(x²-y2)-a, or x10x²y²+9y¹=a; Or, since by the question, (x+3y)-(x+y)=2y= 4, or y=2, this becomes x¹—40x²=176841 چه Consequently, x²=20√(176841+400)=441; Hence, x=√441=21, and x-3y=15, x-y=19, x+y=23, and x+3y=27, the numbers sought. Ex: 21. Let x= the number of hours' march of the first detachment, and y the miles per hour; then, x+1 will be the hours of the second, and y- the miles per hour; Whence, by the question, we have xy-39, and (x+1)(y-1)=39, or xy-1x+y-1-39; Or, since xy=39, we have 1x+y=1=0, 4 or 4y-x=1. 39 39 y y Again, x= ; whence, 4y- -1, or 4y2-39-y, Or y²y; therefore, y=√(+32) = 20 = 31. Consequently, 31 and 3 miles per hour are their rates of marching. g Ex. 22. Let x and y represent the two numbers; then, by the question, x²+xy=140 y² —xy = 78 $ By addition, x²+y²=218, or y= √(218—x²). 140-x2 But, by the first equation, y= 140-x2 Q } XC ; → Whence and reducing 19600-280x2x218x2-x+, or 2x¹-498x2—19600; √(218-x²); or by squaring ? } * 1 - 1 1 102 Whence, by dividing by 2, we have x4249x²-9800 ; –9800 That is, x- 151 2 MADE 249 2. ±vi 2492 4 249 -9800), or x²== ± 2 viz, x²=200*, or 49; therefore by adopting the lat- 140-49 =13, 7 ter we shall have x=7, and y= the numbers sought. x3: Ex. 23. Let x³ greater, and y³= less; x³ — y³ — 135, and x-y=1, by the question. Dividing the former equation by the latter, 409 x²+ xy + y²=108, And x²-2xy+y2=48, squaring the 2d. Hence 3xy=36010, by subtracting: 6 And, from the second equation, x=ity; which va lue, being substituted for x in the above equation, and we have 3y × (1+y)=10, or y²+zy; whence by com- 7 10 7 12 pleting the square, and extracting the root, y= } 77 49 10. 23 1 1 ± √(144 + 3), or y=— 12+12=1; hence x=1+ 1 10 1=2; and, consequently, a³=15, and y³=277 Ex. 24. Let x-y, x, and x+y= the numbers; then, (xy)²+x²+(x+y)2=93, or 3x²+2y2-93, and 3(x-y) +4x+5(x+y)=66, or 12x+2y=66, by the question. From the second equation, y=33-6x, hence y²=1089 -396x+36x2, which value being substituted for y in the * This value of o² does not answer the condition of the question; be- cause we see from the first equation that 2 must be less than 140, Į 103 25x2-264x=-695; whence x= first, we shall have 3x²+2178-792x+72x²-93, or 132 一 ​25 139 25 1 1 ± √² 75 } 17424 695 (Song) 625 25 77 ±25 ; hence x= or x=5, and by taking * 132 or x 25 the latter value for x, we have y=33-6x=3; and, con- sequently, the numbers are 2, 5, and 8. Ex. 25. Let x, y, and z= the numbers; then, x-y : y−z:: x: 2; whence xz-zy=xy—zx, ox (x+z)y= 2xz. Bút x=191-y, and xz=4032, by the ques- tion; ✓ > Therefore (191-y)y=2x4032, or y2-191y=-8064; Hence y=63, by completing the square, &c; x+z=191—y—191–63128, and xz= From the square of the first, x²+2x+2=16384,, Take 1 times the second, Now xz=1032. 4xz =16128, x²-2xz+z² = And we shall have 256. Or, x-z-16, and x+z=128; whence, by addi- tion and subtraction, x=72, and z=56; And, consequently, 72, 63, and 56, are the numbers. Ex. 26. Let x denote the least number, and 2/ the common difference. Then the four numbers will be ex- pressed by x, x+y, x+2y, and x+3y; and the four spe- cified sums by x+2,x+y+4, x+2y+8, and r+3y+15. Whence, by the nature of geometrical proportionals, we have (x+2)x(x+2y+8)=(x+y+4), and (x+2)× (x+3y+15)=(x+y+4)×(x+2y+8): that is, y²+4y =2x, and 2y2+10y+2=5x, by multiplication and trans- position. Hence, 5y+20y=4y²+20y+4; therefore, y²=4, or y=2; whence a-6; and, consequently, the numbers are 6, 8, 10, and 12. Ex. 27. Let x and y the numbers; then, (x+y)× (x − y) = x²-y²=5, and (x+y2)x(x²-y³)=65, by the question. By dividing the latter equation" 104 1 == 1 we have x²+y2-13; whence, by addition and subtrac- tion, x=3, and y=2. greater, and y2 1 Ex. 28. Let x2 less; then,.x2- x=y²+y, and x²+y2-10, by the question. From the first, by adding to both sides, we have x2-x+1=y²+ y+, and, by extracting the square root, x-y+}, or xy+1; which value being substituted for x, in the se- cond equation, we have y²+2y+1+y²=10, or y²+y= 9 ; henco, y=-19, and xy+1=√19. 2 1 4 Padangly á And, consequently, a 5+1.9, and y2-5-19. Ex. 29. Let x=sum, and y= difference; then, x+y= greater, and x-y= less. Therefore, 2-y²+ 2x=61, and 2x²+2y2-2x=88, by the question; or, di- viding the second equation by 2, we have x²+y² 44; hence, by adding this to the first, we have 2x²+x =105, or x²+3x=105. Whence, a 1±√(√6 +¹³)=-1±√4=-} 297; hence, y=√(44-x2+x)=√(44-49+7)= 2 4. √/2. Kateg = gadget 2 5 Workpl ² x: Therefore, 7+1/2 and 7-2 are the numbers. Ex. 30. Let = greater, and y= less; then, (x-y) x(x²-y²)=576, and (x+y)x(x²+y2)=2336, by the question; that is, x³-x²y-xy²+y³=576, and x3+x²y +xy²+y³=2336; hence, 2x2y+2xý²=1760, by sub- traction; and, by adding this to the second equation, we have a3+3a2y+3xy2+y³-4096, or x+y=4096-16, by extracting the cube root;, but 2x²y +2xy²=2xy(x+y) 1760; then 32xy=1760, by substitution; or xy-1760 3.2 55. Predato Whence, the sum and product being given, (by ques- tion 7,) we have x 11, and y=5. paper P 1: " Ex. 31. Let x, y, and z= the numbers; then, x+y +z=20, xz=y², and x²+y²+z2140. From the first equation, x+z20-y, or x²-2x+2²-400-40y+y²; # 105 x= hence, x²+z²+y2—400—40y+y2, by substitution, or x²+x²+ y²=400-40y; therefore, 140-400—40y, 40y =400-140=260, or y=6}; and, consequently, 20-y+√(400-40y-3y²) 5 2 63+√3, and 16' 20-y-√(400-40y-3y²) :6-V3- 632 - √3 500. 2 16* Z 16 • + POSTE : Ex. 32. Let x greater, and y= less; then will xy= 320, and x³-y³: (x-y)³: 61: 1, by the ques- tion; hence, dividing by x-y, we have x²+xy+y²: (xy)2 61 1; from whence (by subtracting the con- sequents from their antecedents,) we have 3xy: (x—y)², :: 60: 1, or xy: (x−y)² : : 20 : 1; hence, xy=20(x- y)², or 20(x—y)²=320 by substitution, or (xy)- 16; and, consequently, x-y=4. Now, xy+4, which va- lue being substituted for x in the first equation, we have y²+4y=320, and, consequently, y=-2√324-2 18-16, or -20, and by taking the positive value of y, 320 we have x= =20. 16. NOTE. This can be done otherwise by putting a= = } sum, and y difference. Then x+y= greater, and -y= less, from whence the answer can be easily found. Ex. 33. Let x= the share of A, or the first term of the x 1 progression, and let the common ratio oras Itoy; then 'y' x+xy + xy²+ xy³-700, and xy³-x: xy² —xy :: m : n, by the question. From which proportion we have y³—1 (by dividing the whole by m—n+ √(m²—2mn—3n²) 2n K (y-1)x my, or y²+y+1= n ico 3 y-1). Hence, y is found = 25--7 4 24 MY 22 Podes Whence, x(= K 700 1+ y + y²+y³ Kapal dat from the first 106 1 27 X 700 equation) is given, =27+36+48+64 the four shares are 108, 144, 192, and 256 dollars. OF CUBIC EQUATIONS. To exterminate the second Term from a Cubic Equation, Ex. 3. Given equation x³-6x²-10, or x³-6x²—10—0. Here, x=x+2, Therefore, Therefore, Whence, we shall have z3-12≈—26=0, or z³-12%=26, as required. Ex. 4. Given equation y³-15y²+81y-243=0. Here, y=x+5, y=x+152+ 75x+125 -15y2: -15x2-150x-375 +8ly = +81x+405 -243 -243 BVB A ¿ m Therefore, -10 --- Whence, we shall have x³ ³+622+12+ 8 —6z² — 24z-24 -6x²= Apart Ex. 5. Given equation x3+ 4 p 3 +-x² 7 +÷x 8 9 16 00 { or x36x88, as required. 7 3 x² + 8x 1 Here, xy- 3 3 003 x³ —y³ — — y² + 16Y 3 3 +73² Categ x²-6x-880, K -10 -108. Therefore 4 zombat jerame 77 • 9 16 1 64 3 864 my 8y- 32 -9 16 :0. J } 107 Then, Then, Hence, we have y³ +169. Ex. 6. Given x¹+8x³-5x²+10x-4=0. Here, xy-2, x¹y+8y³+24y2—32y+16 8y³ — 48y²+96y-64 -5y2+20y-20 +10y-20 8003 -5x2. 10x= 4 Ja 11 3 or y³ +169 = 43 A Madag 3003- +3x²- 5x= -2= Bat Whence, we have y¹-29y²+94y—92—0. Ex, 7. Given x43x3+3x²-5x-2=0. Here, xy+e, x¹y¹+4ey³+6e²y²+4e³y + e¹ 3y39ey2-9e2y-3e3 3e³ + 3y²+6ey+3c² Mama.lang " 11 3 16-4 as required. 3 2 уз 2==000 0, Now, the value of e, by which the third term is taken away, is had by resolving the quadratic equation 6e²-9e +3=0; the roots of which are found 1, or; hence, by substituting y+1 for x, in the given equation, we find y¹+y³—4y—6—0, an equation wanting the third term. Ex. 8. Given 3x3-2x+1=0. SpĚSTÍ, 4 Y 5y-5e -2 Here, let x JadaG ' 1 JO thén, 3x²-2x+1 +1=0,or 3-2y²+y³=0; hence, y³-2y²+3=0; the roots of which are the reci- procals of the former. y ; 108 A 1 x=/; then, 6 Ex. 9. Given x4 y¹ y3 64 2.63+ { 1 • Therefore, Therefore, y¹—3y³+ 12y2162y+720, the equation required; the roots of which are six times those of the former. (gamma S 1 St 0-1= 3-1 -3-1 SOLUTION OF CUBIC EQUATIONS. 1 Ex. 1. Given x³-3x²-6x=8, to find x. Here xy-1, y³-3y²+3y-1 +3x² +372-6y+3 x³ -6x -6y+6 -8 8 } --- oc²: C S K 500- y2 3.62 Reduced equation y³—9y=0; Consequently y=0; and, dividing by y, we have y²= 9, or y+3, or 3; whence the three values of x are x=y-1 x=y FY-1 x=y-1= 1 3 + = x² - 2 x + 4 J 1 B 31+ 4.6 1 2 4 1 203 x³-x³-x² += 2 3 Ex. 2. Given x³+x²=500, to find x. First x-z-, Z2 2 3 % + J G 1 18 1 as required. 1 Jong Supercar 18 1 27 9 -500 =0. Here let Reduced equation 23-12-499350, or x³-3x=49925; posted here often, 17 : 1 109. Whence we have a Jamanand ===3/ ly by our formula b2 a³ b b2 ==V | ₁ + √(+ ) { + V 1 ½ -VG+)! — 27 } 49925 2 1 and b=4993; consequent- 27 + √ 3 (4992) 2 4 (49925)2 1 4 729)}, Which last expression, being reduced, gives 1 z=√ {249.9629+√(249.96292-729)}+ ⚫ becomes Gild 1 729) } +3 49925 مة الله Or z=499.9257+.0001-7.950, 3 √ {249,9629—√(249.96292-729)} e 1 -5 Consequently x-z-3-7.950-.333-7.617 Ans. Ex. 3. Given equation a³-3x²-5. First x-y+1, x³-y³+3y²+3y+1 Therefore, -3x²-372-6y-3 -5 Whence y³-3y-7-0, or y³-3y=7. Here a being equal to -3, and 6-7, we shall have, by the formula, -3/ 49 49 y = V { { +√(² + − 1)} + √ {{−√(4-1)) =√ 6.854102 // 2 +.145898; or y=1.899616+.5254852.425101: Hence, y+1=2.425101+1=3.42510Ï. K 2 110 Ex. 4. The given equation x3-6x=6, being in its proper reduced form, we have L 3 x=³/ {3+√(9—8)} + {3−√(9—8)}, or x=(3+1)+(3-1), or x=3/4+372, Answer. Therefore, 3 Ex. 5. The given equation a³+9x=6 is here already in its reduced form; whence we have immediately. x= {3+√(9+27)} + {3−√(9+27)}, or x=(3+6)+(3-6), or x=9+/-3, or x=9-3/3, Answer. Ex. 6. Given x3+2x2-23x-70. Let x=y-3; 8 x³—y³—2y²+zY— 27 8 9 Ską 1 +2x²= +2y²-3y+ -23x= -70 1460 27 4 J and consequently y=3 Į 73 Whence y³-3y —219%-1460=0, or z³-219z=1460. z=3/ { 730+ √[(730)2_(73)²] } +³/ { 730−√/[(730)²(73)²] 3 'or 46 -23y+3 -70 =0, assume { 2 z= 37 (730+379.319126)+3/ (730—379.319126; hence Stic z=3/(1109.319126)+(350.68087)=10.35188+7.05187, 2 17.40375 3 y=3; then, 1 =5.80125; I Whence xy-3-5.80125-.66666 5.13459 Ans. { } 23 111 Ex. 7. Given equation x³-17x2+54x=350, to find x. This equation, by exterminating the second term, be- comes 15 27 Whence, by the formula, we shall have 21 21 3 ==} 1 3 23-42-2= =407 21 21 3 √/ {20337 — √ [(20327)² - (143)³]}, V − — Or by reducing and simplifying the expression, z=3} {5502+√/[(5502)²—(127)³] } + {20337+√/[(203327)² —(143)³] } + -(14금​)]}+ 1 M {5502— V[(5502)² — (127)³] } ; 1 or z=3√ [5502+√/28223621]+[5502— √28223621]; 1 3 or z=√ [5502+5312.1776]+[5502— 5312.1776]; 1 or 2= z 1=3/31/ 3/10814.1776+189.8224-9.287391. 14.954068, Answer. · Consequently x=a+ Ex. 8. Given x³-6x=4. by a few trials, to be equal to 1 17 3 Honda Here x is readily found, 2, and therefore E > 2 foot { & 112 x+2)x36x-4(x²-2x-2 x3+2x2 ་ JAPA ·2x²-6x -200²-4x T Whence 2-2x-20, or x²-2x-2; the roots of which quadratic are 1+√3, and 1-√/3; and conse- quently -2, 1+√/3, and 1-√/3, are the roots of the proposed equation. Ex. 9. Given x³-5x+2x-12. Here x is readily found by a few trials, to be equal to 3, and therefore x-3)x³-5x²+2x+12(x²-2x-4 003 - 3x2 な ​-2x-4 -2x-4 Jakaya KAJA -2x²+2x -2x²+6x * -4x+12 -4x+12 T Whence x²-2x-40, or x²-2x=4; the roots of which quadratic are 1+√5, and 1-1/5; and conse- quently 3, 1+√5, and 1-√/5, are the three roots of the proposed equation. • * 113 1 1 SOLUTION OF CUBIC EQUATIONS, BY. CONVERGING SERIES. 26 Ex. 1. Given x³+9x=30. Here a=9, and b=30; 27 × 900 100 25 whence 2762 276²+4a³ 27 × 900+4×272 ¯¯112 112 28 60 8.11 25 X-B 10 3 // [2(27b2+4a³)][2(27 × 900+4×272)] (252) 53/(216×294)/ (36750) 126 21 Consequently, formula first, we shall have 1 2.5 25 X-A 6.9 28 12.15 28 14.17 25 X 18.21 28 20.23 25 X 24.27 28 26.29 25 X 30.33 28 32.35 25 x=F 36.39 28 C D E 38.41 25 X G 42.45 28 44.47 25 X-H 48.51 28 f ; and 3 1.0000000 (A) .1653439 (B) .0721898 (C) .0405828 (D) .0257221 (E) .0174913 (F) .0124581 (G) .0091693 (H) .0069160 (I) 1 50.53 25 XI 54.57 28 › 56.59 25 X-K 60.63 28 62.65 25 X-L 66.69 28 68.71 25. X 72.75 28 74.77 25 X-N 78.81 28 80.83 25 X 84.87 28 86.89 25 X P 90.93 28 M 92.95 25 X 96.99 28 98.101 25 X-R 102.105´´28′ O 104.107 25 X=S 108.111 28 x 25T 114.11728 110.113.25, 116.119 25. X-U 120.12328 122.125 25 X-V 126.129 28. 128.131 25 X-W 132.135´´28 134.137 25 X X 138.141 28 } 114 + 1 .0053164 (K). .0041481 (L) .0032775 (M) .0026163 (N) .0021068 (0) .0017091 (P) .0013950 (Q) .0011460 (R) .0009456 (S) .0007837 (T) .0006521 (U) .0005445 (V) | .0004561 (W) .0003832 (X) .0003228 (Y) 115 1 140.143 144.147 28 146.149 25 X 150.153 28 152.155 25 X A 156.159 28 x25x M 158.161 25 X B 162.165 28 164.167 25 X 168.171 28 -Z 170.173 25 X-D 174.177 28 176.179 25 X 180.183 28 182.185 25 186.189 K -E X F 28 3 Log. 1.3770034 Log. (36750) Còlog. 21 No. 2.18. Therefore x=2. 18. Ex. 2. Given x³-2x=5. Here a-2, and b=5. 5 Then 22/23/2√20; b 27X25-4×8 27 × 25 643 675' .0002726 (Z) .0002307 (A) .0001955 (B) .0001662, (C) .0001416 (D) .0001206 (E) .0001031 (F) .0000966 (G) 1.3770034 0.138934 1.521752 8.677780 0.338466 1 and 2762-4a3 2762 8 Consequently, by formula second, 116 3 we shall have x=+³/ 20(1 B- 11.14 643 17.20 643 X- C X 15.18675 21.24 675 29.32 643 35.38 643 X F X G- &c.) 33.36 675 39.42 675 NOTE. This series converges so slowly, that it re- quires to calculate more than 60 terms in order to ap- proximate to the answer; which is always the case when 4a³ is considerably less than 2762; then the nu- merator of the multiplier is nearly equal to the denomi- nator, as above, the numerator 643 is nearly equal to 675, and, therefore, the series must converge slowly; conse- quently, this method is not useful except when 4a³ is nearly equal to 2762; then the numerator is small in comparison with the denominator, and the answer is. found by summing a few terms of the series. Therefore Cardan's Rule, in Case 2, is more expeditious than this method, when 4a³ is considerably less than 2762; as in the example x³-2x=5. Here a= 2, and b5; then V. 25 8 → 10 102 b 62 b b2 a³ VIA x = V { + √ ( + − Z ) } + V {22 - √₁₂ 3 4 4 8 5 --g 9 643 {-√013) 2 643 5.8 643 X A 3.6 675 4 1252 + √(20 1643) + 108 +2 V: (2.5+2.44)+† (2.5 2.44)=3/ (4.94)+3.06=1.703+.391=2.094, Answer. The same method of reasoning is likewise applicable to Case 1. D } Septe 3 12; and X 9.12675 23.26 643. X E- 27.30 675 Kat KATE 25 5 2)} + √ { 5 - √(²)} = V {{ 27 4 a3 127)} Ex. 3. Given x³-3x-3. Here a=3, and 6—3; then 3 27b2-4a³ 27.9-4.27 b 3 23/ 21 13 = 21 21 = √12; 23/ p 2762 27.9 9-4 + - 117 1 I } Consequently, by formula 2, we shall have, ૭ 3.6 } 51.54 53.56 57.60 $59.62 63.66 65.68 1. 5 5.8 5 9.12 11.14 15.18 9 17.20 21.24 23.26 27.30 29.32 33.36. 35.38 5 39.42 41.44 45.48 47.50 10 10 10 X X-B 9 5 X-C 5 × E 10 5 × F X-G 5 X-H يرات من X-I X-K X-L bla bla 69.72 M 71.74 5 75.78 X-N Li 3 મ → 1 L { { p Synony 1.0000000 (A) -.0617283 (B) -.0127012 (C) -.0040246 (D) -.0015083 (E) —.0006186 (F) ~.0002684 (G) -.0001210 (H) -.0000561 (I) —.0000265 (K) -.0000127 (L) -.0000062 (M) -.0000030 (N) -.0000014 (0) MŠ, 12 1 118 1 1 ધ f 77.80 81.84 + X=0 Sum, Com. Log. .9189230, Log. 12, No. 2.1038, Therefore, x=2.1038, Answer. Then, 4. = 1; and 5 3 Ex. 4. Given x³-27x-36. 3 / (14400) 15 1 2.5 4 X=A 6.9 8.11 4. X-B 12.15 5 2762 362 4a³ —27b²¯¯¯ 4.273—27.362¯¯ 4.272362 14.17 4 -26 -12 3 // [2(4a³ — 2762)]† (272—324) 15 ¯¯¯ / ; Consequently, by formula 4, we shall have X=C 18.21 5 20.23 24.27 26.29 30.33 1 X-D X-E Here a 27, and b=36. 27 × 362 } } 3 -.0000007 (P) 1 -0.0810770 .9189230 -1.963279 0.359727 0.323006 ¿ Pagkat 4 324 405 Bund +1.0000000 (A) -.1481481 (B) +.0579423 (C) -.0276712 (D) +.0157145 (E) -.0095747 (F) } 119 + + + + + + of anfor 32.35 36.39 38.41 42.45 44.47 48.51 50.53 54.57 56.59 4. X-K 60.63 5 62.65 66.69 5 X-F → X÷G × H GLA GLA GLA XI X-L 68.71 4 72.75 74.77 4 X-M X-N 78.81 5 X 80.83 4 X-0 84.87 5 86.89 90.93 92.95 96.99 GLA X = Q 98.101 102.105 104.107 108.111 110.113 114.117 116.119 4 120.123 XR GIA G 4 X=S 4 X-T X÷U 다 ​I } I # : +.0061103 (G) -.0040295 (H) +.0027232 (1) = —.0018756 (K) +.0013113 (L) -.0009284 (M) +.0006639 (N) -.0004790 (0) +.0003489 (P) -.0002553 (Q) +.0001878 (R) -.0001388 (S) +.0001030 (T) —.0000833 (U) +.0000623 (V) ↓ 122.125 126.129 ว and 4 X-A G Again, ( Sum, Log, .8919368, Log. 3/14400, Colog. 15, No. 1.446, .1603503 Therefore, 1.446 is one of the negative roots, or va- lues of x. and X÷V + 4a³ - 2762 1349900 26996027* J 2762 4a³-27625° 4a³ - 2762 2762 4. J 4 120 5.8 4. 11.14 4 X-B+ 9.12 5 15.18 4.32, the other two roots of the given equation. Ex. 5. Given x³-48x-200. First x-y+16, and x³—y³+48y²+ 768y+ 4096 -48x²-48y2-1536y-12288 Then +200 + 200 Reduced equation y3-768y-7992-0, or y³-768y= 7992. r Here a=768, and b≈7992; whence 23/ 23/3996; 2 4×27-27× 362. 4 1 2 1349900 3.6^26996027 -.0000467 (W) +.8919368 -1.9503308 1.3861208 8.8239087 =9 Hence, +(10935)× {1+ ** 4.7683-27.79922 27.79922 Bús 10935; 2 3.6 X÷C *&c.}=5.7657, or - Consequently, by formula 3, we shall have 86393600 1724545728 1.0000000 (A) +.0055563 (B) 121 + + 15.8 1349900 9.12 26996027 11.14 1349900 X 15.18 26996027 3/845 1 Then 3 2 X Therefore x-31.91+16=47.91 Ans. Ex. 6. Given x³-22x-24. Here a=22, and b=24. -48 [8.10648-54.576] ; and K +1.0054563 .0023632 Log. 1.0054563 Log. 23 (3996), or (31968) 1.5015718 3 No. 31.91, 1.5089350 12 123/ (845)2 845 J ·26 †/ [2(4a³ — 27.242)] 1 486 845 A 2.5 6.9 8.11 X B C X 486 12.15 845 14.17 486 X 18.21 845 20.23 486 X -D 24.27 845 26.29 486 X E 30.33 845 15552 486 15552 42592-15552 27040845 Hence by formula 4, we shall have, } B. J 3 L2 123/ (714025) 845 -.0001030 (C) +.0000030 (D) • … p Al 2762 4a³-2762 1.0000000 (A) -.1065088 (B) +.0299485 (C) -.0108452 (D) +.0044273 (E) -.0019393 (F) } JA + + Kampuni dig + + + 1 1 32.35 486 X F 36.39 845 38.41 486 845 486 H Ka +.0008897 (G) -.0004218 (H) +.0002049 (I) -.0000960 (K) +.0000483 (L) —.0000245 (M) +.0000126 (N) -.0000068 (0) +.0000035 (P) -:0000018 (Q) +.9156906 -1.9617485 1.0791812 1.9512378 17.0731433 .0653108 Therefore, one of the negative roots or values of x is -1.162=-r. l X 42.45 44.47 48.51 50.53 54.57 56.59 486 60.63 845 62.65 486 X÷L 66.69 845 68.71 486 X 72.75 845 74.77 486 X 78.81 845 80.83 486 X 84.87 845 86.89 486 X P 90.93 845 * G X 845 486 XI 845 X # K M 1 N Log. .9156906 Log: 12 Log. / (714025) Colog. 845 No. 1.162 122. Kat V J } + 1 123 + and + + + + Again, ( + 4a³-2762 2762 4a³-2762845' 1 486 3.6 X 845 5.8 486. X 9.12 845 11.14 486 X 15.18 845 17.20 486 X 21.24 845 23.26 486. X 27.30 845 29.32 486 X 33.36 845 35.38 486 X 39.42 845 41.44 486 X H 45.48 845 47.50 486 X 51.54 845 I 53.56 486 X 57.60 845 59.62 486 X-L 63.66 845 65.68 486 X 69.72 845 ૭ J :)=√( 4. 486 Hence, A B C D E F G K 1 -M 42592-15552. 4 -15552)=6760, "" 1.0000000 (A) +.0639053 (B) -.0136130 (C) +.0044657 (D) -.0017327 (E) +.0007357 (F) —.0003305 (G) +.0001543 (H) -.0000742 (I) +.0000364 (K) -.0000180 (L) +.0000091 (M) -.0000046 (N) +1.0535355 1 1 t Log. 1.0535355 Log. / 6760 No. 4.581 # Therefore, до ཆུ།༽ 2 Last number 124 } Result Or, +5.162 -4.000 And, consequently, +5.162, -1.162, and -4, are the three roots; but the root is only required, and this is the affirmative root 5.162. .0226491 .6383244 .6609735 +.581 +4.581. of BIQUADRATIC EQUATIONS. Ex. 1. Here the given equation, viz. f 1 z ³ — ( ——— b² + d) z: = b² + d) z = 1.2 108 x¹-55x²-30x+504=0, being of the proper form for solution, we have by the se- cond rule, b=55, c——30, and d=504. The cubic, or reduced equation, therefore, viz. 1 A 1 b² + c² = 1 bd becomes 23-756-7811107, Where the root of this cubic being found 31.66666, or 313, our two quadratics are «²+(√/2 { 313+55 } )∞=-(313-55)+√ {(313-552-504} 6 55 55 x²—(√/2{313+} )x=-(313—~~)~ √/ { (313-5)²-504 } 125 I or x²+10x-21; and x²-10x=-24. The first gives x=5/(25-21), or x=3, and 7. And the second, x=-5√(25-24), or x=-6, and ·4; That is, the four roots are 3, 7, 4, and 6. -; whence, **—2^—22³ + 23 2² — 201 22. x = 2 - 3/1/ Ex. 2. Given equation 4+2x²-7x²-8x=-12. First, in order to exterminate the second term, we have 1 -7002 -8x= +12 +2x³= +2x³-3x²+5%- or 2¹=882 1-8 289 16 1 12 = + 22 m ~7z2+7x- Whence, 224 ± √( pag 3 25 9 or 4' 4' and =±√ 2 Į TOGG 1 1 Reduced equation 24-82+14- 140=0. 16 -8202-14 10: 16 HAN+ -8x+4 +12 ايه من ايه 1 +₂,0 4 7 4 or 25 Consequently, z=± √ + = + ₂ or 17 8 225) = = = = = = = = = 士 ​16 4 -5 2 pag colâ 2' t f 126 And therefore xz- has the four following values, viz. 4 Ex. 3. 1 Whence OC 8 ys' y³ — ( OC 23 OC Jurid 11 - 100 12 10/2 5 ♡ IQ 3 2 T ૭ 10 t 5 I -3 ૭ +14002- + 4:00 8 004 80c3- HIQ 1 1[Q 4 10. 2² + { √ √ ² (= = = = = = =) } 2 1 2 11 1 41Q J 2 Given x48x314x²+4x-8 to find x. First x=x+2; x¹=2¹+82³+24x²+32x+16 1 4 + 2²={√26- {√² (= = = = 10 ) } = 3 ) } ==== 2: 3 11 2 Kangar Ka 212 2 2 Reduced equation 24-10z²-42+8=0, From which we obtain the following cubic, -+8)y=- -6 pag ૭ 4. Chantalang Peter 3 -823-482296z—64 +1422+56%+56 4x+ 8 8 y³ —163y=19 The root of which cubic is y=-1.3333, or 3 -1000 16.80 +=+ 8 3' 108 Therefore the two quadratic equations are 16 ) + √ √ √ √ 4 100 Startal 3 6 11 27 -4 10 -3 -2. 4 10 3 Or z²+22=2, and z²-2%=4; 3-)- 1 6 VI 3 4 10 4. -) 2_8 { 28 -5)²=-8} 127 Whence the four roots, or values of z, are z=−1+√3, and z=-1-√3, z=+1+5, and 2+1-√/5; And since xz+2, we have the four following values of x, viz. x=3+√/5, x=3—√/5, x=1+3, x=1-3, as required. The student will observe here, that as the quantities under the latter two radicals, in the above quadratics, arise from the square of a negative quantity, their roots must be taken negative, and not affirmative. Ex. 4. Here the given equation, viz. x¹17x²-20x-6=0, is already in the proper form for solution, having b=- 17, c -20, and d—6; whence our cubic is -173 202 17 × 6 -6)z = 108 3 P 172 ≈3 ( 12 + 8 53. or, z³ — 18- 31223 29- 108' The root of which cubic is z=2.3333, or s 2=2/1/1; Whence our quadratics are p > 7 17 17 x² + { √2 (3 + 1/7 ) } x = − ( − 17 ) + √ { ( 177 )² + 6 } v 3 6 7 17 7 17 7 17 00-10216+10=-15--√15-풍​+이 ​} Or x²+4x-2, and x²-4x=3, Therefore the four roots, or values of x, are x=−2+√2, x = −2 — √2; -2 x=+2+√7, x+2-7, as required. 1 z Ex. 5. Given x¹ -3x² 3x²-4x-3=0. Here,b 3=0. Here, b = −3,c= -4, and d--3: whence, by substituting the values in the equation 23(b²+d)e=gb³+1c2bd, and simplify- ing the results, we shall have 23+2+11=0; the root of which last equation, as found by trial, or by one of the former rules, is--r. Whence our quadratics are 128 x²+ {√2(−}+1)}∞——(−3 −3) + √ or x²+x=3; and x²- } p From the first of which x—— 1 second = Ex. 6. √~3.. 27x3: Given -19x3+132x²-302x+2000. Here a=- 19, b132, c302, and d=200. Then, by substituting these values, in the cubic equa- tion, (Rule 1,) we shall have, by simplifying the results, 23—2173z=1362; whence, by Cardan's rule, z=8.6112 Fr. Hence, by substituting this value of r in the two quadratic equations, (Rule 1,) and reducing the results, we have the following quadratics: 22- 5.08724z=-3.46262 22-13.9127542--57.75978 Whence the four roots of the given equation are deter- 'mined by solving those two quadratics. NOTE. The rule given by Hutton, in his Mathematics, for discovering whether the roots of a biquadratic equation are real or imaginary, fails. See Hutt. Math. vol. ii. page 269, first New-York edition. The error in Hutton's rule will appear evident from the above given equation. The same error is likewise committed by Maclaurin. Ex. 7. Given equation x-27x+162x²+356x-1200—0- 27 First x=x+ 4 2187 19683 xx¹¹+27%³+ -z²+ p 1 +162x²- +356x -1200 {√2(−3+1)} x = − -(-골​-용​)-vi (-곪​-골​) 2+3), or x²-x=-1: √13, and from the } x = − (−− } — ³) + √ { ( − 1 −1)²+3 } Segment) Gr -2723 podatak 531441 2+ 8 16 256 2187 59049 531441 4. 16 2 -Z +162222187%+ 64 59049 8 +356z+2403 -1200 K 曩 ​129 i ratics z²+ {√[2(r~}b)x]}=−(r+}b)+√ { (r+¿b)² —d } z² = {√ [2 ( r ~ { b) x] } = ~(r+3b) — √ { (r + } b) ² — α } gives the four following values of z, viz. 2 - —d} z=−4.19392; z=-9.25000 2= 6.90306; z= 6.54086; But xz6.25; therefore the values of x are, x= 2.05608; x=- 3.00000; a=13.15306; x= 14.79086. my my Reduced equation 24-1113²+82 +2356- :0, 256 From which we draw the following cubic, viz. y³+1322154y=75539412, And the root of this equation substituted in our quad- ~ 1 # Ex. 8. Given x-12x²+12x-3=0, to find x. Here the equation, in its proper form for solution, being ——12, c=12, and d=-3; our cubic is 1728 144 23-(12-3)=- 12, 108 8 or z³—9% -10, Where we have, from inspection, z=2; whence the following quadratics: x²+{√/2(2+4)}x=−(2−2)+√ { (2—2)²+3} x²- {√2(2+4)} x=-(2-2)−√ {(2-2)²+3} Or ²+/12.x √3, x²√12.x=√3; Whence a=√12±√(3+√3), x=+ √ 121√(3-√3); Or the four values of x in numbers are, ·3.907378; x=.443277; 00 Xc 2.858083; *=.606018. Yang magtangg Ex. 9. Given a¹-36x²+72x-36-0. Hereb 36, 72, and Z-36; FOR M Toggl 130 8 Whence, by substituting these values, in the equation, ≈ ³ - ( + b² + d) x =}b³+}c²-}bd, and this reduced be- e 108 comes z3-72z: 216, where it is evident, by inspec- tion, that z=6, and this value substituted for r, in the two quadratics according to Rule 2, we shall have, after 'sim- plifying the results, a2+6x=6, and x²-6x=-6: from the first of which x= 33.8729836; and from the se- cond +31.7320506. Therefore, the four roots of the given equation are, .8729836, 1.2679494, 4.7320506, and -6.8729836. } Ex. 10. Given ¹-12x+47x2-72x+36=0. To take away the second term, let x=y+3; then the reduced equation becomes y4-7y2-6y=0. Here y=0, or 0 is evidently one of the roots of this last equation; whence it is reduced to y³-7y-6=0: the roots of which as found by a few trials, or by one of the former rules, are 1, 2, and 3: and, consequently, the four roots of the given equation are 1, 2, 3, and 6. X Mag Dom Ex. 11. Given x1+24x³-114x²-24x+1=0. Here, a=24, b=—114, c=- —24, and d=1; then by substituting these values in the cubic equation, Rule I., and reducing the result, we have 23-1228z=-16272. Let z2y; then y³-307y=-2034. Now the root of this is found by a fow trials, or by one of the former rules, to be 9; hence z=2y=18=r. P Jag K Again, by substituting this value of r in the two quad- ratics according to Rule I., we shall have, after reducing the results, x²+28x=1, and x²-4x=1; from the first -14/197, and from the second x=25. | Ka Ex. 12. Given ¹-6x³-582-114x-11-0. Here a ·6, b——58, c—— -114, and d—-11; then by sub- stituting these values in the cubic equation, Rule I, we have, by reducing the results, 3—98}z=−167135. Now, the root of this cubic being found, by one of the former rules for cubic equations, and substituted for r in the two quadratic equations, according to Rule I., we ľ 131 shall have the four roots of the equation. Or, the an- swer, in the form in which it is given in the introduc- tion, is found thus : Let k=(ac-d)=182,l=c²+d(a²-b)=2512; and then A3-A2+kА-31—A³+29A2+182A - 1256= 0: and from this equation, by trial, A will be found —4; aA 1 C and then B=(2A+}a²—b)²= √75=5√/3, C= 2B 3/3. 10 4 And x=(Ba) ±(1² a²+}aB²±C — A}} = {√3 + 3/2 ± √ (17 ± 23 23), the roots required. See Simpson's Algebra, page 150. RESOLUTION OF EQUATIONS, BY APPROXIMATION. therefore x=4.1+2; Then x²-16.81+8.2z+2² Ex. 2. Given equation 2+20x=100. Here a few trials show the root to be nearly 4.1; let 20x=82. +20z } = 100; } Therefore rejecting z², we have 98.81+28.2x=100; 1.19 28.2 4x Moderat .042; And consequently x=4.1+8=4.1+.042=4.142; Whence by repeating the operation, and assuming x= 4.142+z, four other figures may be obtained, which gives the value of x=4.1421356, as required. Judg or z Ex. 3. Given x39x²+4x=80. By trials we find the root nearly 2; assuming, there- fore, x=2+z, we have x²=8+12+6x²+-33 9x236+362+9z² 8+ 4z { p =80; 132 } Whence, by rejecting the second and third powers of z, we have 28 5, 52 And therefore x=2.5 the root, nearly. Assume now x=2.5+z, and we have 15.625+18.75% +7.522+23 52+52% 80; or z= 9x256.25 +45.z +92² 4x=10. +4z Whence rejecting the higher powers of %, 81.87567.752-80, 1.875 67.75 -.0276, Where x 4.5-.0276-2.472, nearly; and, if again we were to assume x=2.472+z, we should get four other decimals, which would give x=2.4721359. Appl or z= 1 Remark. In the preceding examples we have involved every power of a completely; but it is obvious, that as the higher powers of z are always rejected, it is unne- cessary to carry the expansions beyond the second term, as shown in the following solution: Ex. 4. Given equation ¹-38x³+210x²+538x+289. =0. +210x2 +538x +289 38x³-1026000-102600z—&c. 189000+ 12600z+&c. 16140+ 538x+&c. 289 Whence 10571+1853820, or z= Here, since the second term is equal to all the other terms of the equation, it is obvious that a is less than 38; and by a few trials we find it to be nearly 30; let there- fore x=30+z; then reserving only the, first two terms of each expansion, we have x04 810000+108000z+&c. HOME Ja Magd BČAN Matlaging Pa M 1 | 80; =0. 10571 18538 :.5,- 1 T ! 133 Assume, therefore, x=30.5+ And x 30.5, nearly. z, and we shall have (rejecting the decimals as incon- siderable) 001- 38x3 +210x2 +538x +289 M 865365+113490z -1078159-107816z 195352+ 12810z 16409+ 538z 289 Whence, -744+ 19022z=0, 205 6x1 10x3 112x2 -207x=- +110 744 .039; whence x=30.53, nearly ; 19022 Or z= And, by assuming again x=30.53+z, a still nearer ap- proximation may be obtained. But in equations of this kind, in which the coefficients are very large with re- gard to the root, the approximations are always very slow, adding no more than one new figure at every operation ; whereas, when the coefficients are small with respect to the root, each operation doubles the number of figures. last obtained. Ex. 5. Given x+6x-10x3-112x2-207x+110= 0, to find x. Here, x=4, nearly; assume, therefore, x=4+z; then, 1024+1280z+&c. 15361536x+&c. 4802-&c. 896z-&c. 828-207-&c. 640 -1792 110 C Si :0; 590 1233 } 0, And -590+1233z=0, or z=- :.47; Whence, x=4.47, nearly. And, by assuming x=4.47 +z, another approximation may be obtained, till at last. we find x=4.46410161; but it would be useless to go through the entire operation in this place. M 2 1 134 } h OF APPROXIMATION BY POSITION. EXAMPLES FOR PRACTICE. Ex. 1. Given x10x²+5x=2600. ► Here, it is soon discovered that x is a little more than 11; let us assume, therefore, x=11.0, and x=11.1 ; Then, by the rule, 11.1 1367.631 1232.1 55.5 2655.231 2655.231 2596. I Then, jer ta 00 203 +10x2 = 50 Battleg Results Therefore, 11.1 11.0 Bygging 32= ; 59.231 Or .1 :: 4 : .00673; Whence, x=11.0+.00673-11.00673, Ans. Remark. In this example, one of our suppositions ap- proached so near the truth, that we have been enabled to obtain seven places of figures true, in a single operation ; which is a degree of approximation very seldom acquired with so little labour. Ex. 2. Given 2x4-16x340x²-30x+1=0. Here, a is nearly 1; assume, therefore, 2 T -128- Gold OC S 11.0 jykazani 2x4 -16x3 +160+40x240 60= -30x 30 + 1=+ 1 =+ 1 + 5 Results 3 1331 1210 55 Making 2596 Ką 2600 2596 Battag [ 2 16 Kateg * [ } 135 1 a +1 +5 Therefore, 2 3 1 Or 8 • 1 :: 3 : .3 Whence, x=1.3, nearly; assume, therefore, 1.3 OC 1.2 *Then, 5.7122 -35,152 +67.6 -39. + 1. ++ +.1602 -.9008 8 Then, 32768 8192 1536 256 40 42792 74727 42792 Or 31935 .1602 Results T Therefore, 1.3 1.2 MAN : Or 1.0610 : .9008: .085 Hence, x=1.2+.085-1.285, nearly; and, by repeat ing the operation, we have the still nearer approximation 1.284724. 2x4+ 4.1472 -16x327.6480 +40x²+57.6000, -30x36.0000 =+ 1 + 1.0000 .9008 £ Ex. 3. Given x+2x+3x³+4x²+5x=54321. Here, we find the value of x to be between 8 and 9; assume, therefore, 1 SE DE x ევ5 2x1 3a3 4x2 5x 0 3 .1 Results Therefore, 9 8 1 .0 -.9008 9 59049 13122 2187 324 45 74727 1 54321 42792 :: : 11529 1 : .3 J 136 Whence, x 8.3, nearly. And, by assuming x=8.3 and 8.4, another approximation will be obtained; but the successive corrections are very small, and will occupy more room than can be devoted to a single example. Ex. 4. Given (7x+4x²)+ √(20x210x)=28. Assuming here x= 4 and x-5, we have 4. 00 5 Jeg t 8 =(7x²+4x²) 9.91596 16.7332=√(20x²-10x)=21.21320 Results 24.7332 31.12916 24.73320 די ' 113.2895 5 4 Therefore, Javljan 115.7578 113.2895 Or 6.39596 : 1 :: 3.2668 : .51; Whence, we shall have x=4.51, nearly. And, by repeating the operation, x=4.510661. Ex. 5. Given 1442-(x²+20)}+{196x2 — { = (x2+24)2114. Assuming x-7, and 8, we have ✯ 8 XC 28 24.7332 Magn 47.9061144x²-(x²-20)²=46.4758 65.3834=√196x² — (x²+24)² } =69.2820 115.7578 31.12916 Results Therefore, 8 7 { Or 2.4683 * 1 :: 114 113.2895 .7105 : 2; Whence, a 7.2, nearly. Assuming, now, x=7.2, which is found too great: let us therefore take a 7.1 and 7.2, and we have 11 1 TAS J 113.8805 7.1 114.4002 113.8805 47.9791144x²-(x²+20)}=47.9999 65.9014=√{196x²-(x²+24)²}=66.4003 Log. 4.8 Mult. by 137 00 3.38196 3.26995 Results Therefore, 7.2 7.1 .1195 .023; Or .5197 : .1 : :: Whence, x=7.1+.0237.123, nearly. And if, for a new operation, there be taken x=7.123 and 7.124, we shall have x=7.123883. And a further assumption of this kind will about double the number of decimals. EXPONENTIAL EQUATIONS. Ex. 2. Given x200, to find x. Here, we soon find that x is nearly 5; let us, Ꮳ fore, assume x=4.8, and x=4.9. Then, the log. 2000-3.3010300; and Log. 4.9 0.6901961 0.6812412 4.8 4.9 54499296 27249648 3.26995776 : 7.2 114.4002 Therefore, 4.9 4.8 114 113.8805 there- 62117649 27607844 3.38196089 3.30103 3.26995 Or .11201 .03108 .027; .1 : Whence, x=4.8+.027-4.827, nearly. And repeating the operation, by assuming x=4.827, and x=4.828, we shall obtain x= -4.82782263. 1 1 1 138 Ex. 3. Here, (6x)-96; and our formula must there- fore be xx(log. 6+log. x)=log. 96. Where log. 96=1.9822712, And x is obviously nearly 2; assume, therefore, x= 1.8, and 1.9; then, x Log. 6. 0.77815 Log. 1.8 0.25527 1.03342 1.8 2.00811 1.86015 iw ? " 826736 103342 1.860156 Results Therefore, 1.9 1.8 >* • Log. 8.6 0.9344935 8.6 Log. 6.0.77815 Log. 1.9 0.27875 } .14796 : .1 :: .12212 : .0826; Whence, x=1.8+.08261.8826, nearly. Which five figures may be doubled by another opera- tion, making x=1.8826432. 1.98227 1.86015 Ex. 4. Given equation 123456789. Here, after a few trials, or from inspection in a table of powers, we find x is between 8 and 9, but nearer the latter than the foriner. Assume, therefore, x=8.6, and x=8.7. 1.05690 1.9 951210 105690 2.008110 Then, log. 123456789-8.0915149, 56069610 74759480 8.03664410 Results Log. 8.7 0.9395193 8.7 65766351 75161544 8.17381791 139 8.17381 8.03664. I Therefore, 8.7 8.6 .13717 . .1 : :: .05487 : .04; Whence, x 8.6+.04-8.64, nearly. And, repeating the operation, by assuming a equal to 8.64 and 8.641, x is found 8.6400268. 1 Ex. 5. Given x = (2 x — x ) = . • This will be more convenient under the form 00t 2 (x* — x) x = 2x A p Now, in order to find a first approximate assumption, may be observed that 2x must be greater than a", or 2 greater than -1. it The same will also be obvious if we put the equation (20¹² — 6) under the form +2²-1=2. x²=2.4647, and 2.2391 1.8226 As 4165 20 Since, then, a is less than 2, but nearly equal to that number, let us assume a 1.8; then, x=2.8806, and K (x*—x), XC Also, x-1 0.6387 1.6004 1st result 2.2391 And when x=1.7, then, (x-x) 0.3728 OC Also, 1.4498 2d result 1.8226 : 8.09151 8.03664 1.8 1.7 Var .1 1 OFERTE Therefore, by the rule, 2.0000 1.8226 .1774 1 好 ​: .04; 1 1 140 1 Whence, x=1.7+.04-1.74, nearly. plakang.2) And repeating the operation with the assumption a 1.74 and x1.75, we find x1.747933. m pr m Also, by another assumption, a still nearer approximate value of x may be determined; and so on, to any degree of accuracy required. n M P BINOMIAL THEOREM. Ex. 5. Here, the proposed binomial being (1+1), we have P=1, Q Q====== 1 1 m or (1+1)³, L spalvos ju 1, 1 Whence, -AQ J 1 N WE ARE AN INTE 3n Sta M. N 2n m-2n 1+ 1 + 1/12 m -3n G 1 =1 ·BQ : X ·CQ 1 3=1=A, 1 1 | A 1-3 4. 1-4 6 X 2.4 1-6 1.3 1 -1.3.5 X X E, 4n 8 2.4.6 1 2.4.6.8 Where the law of continuation is sufficiently obvious; 譬 ​and therefore we have (1+1)=√2= 1.3.5.7 1.3.5 2.4 2.4.6 2.4.6.8 2.4.6.8.10 1 1.3 + •DQ= X ! IN Q 1 1 X IQ = 1 B, X 1 1 2.4 1 ! 1 Kangad, dapat 1 X 1 1.3 2.4.6 -- π 1 Integ a &c. 1 1 Ex. 6. Here, we have to convert (8-1), or (23-1) into a series. Make 2-a;, then it becomes (a³-1); Where Pa³, Q==, and -1 Prop m n 1 3' 141 m pï =(a³)ñ =(a³)³ m -m =α = A, 1 a 1 -1 -AQ 12 = 3 x x = 1/3 = 3/2 == 0 X Port B, 3а2 N 2n a B m -2n 3n 2 Vallaval Kapt Berita 22 Įkelkl m n B m 1.2.5 3.6.9a8 Therefore, ³ (a³ —1)= 3/ 1 1.2 1.2.5 1.2.5.9 &c. 3.a2 3.6a5 3.6.9a8 3.6.9.12a¹¹ Or, substituting 2 and its powers for a, we have 3 (8-1)=3/7= 3. BQ= in n 2n A m m x==(a²)}" =(a°)}=a=a, pñ 3 ∙CQ a ·AQ X-X 1 1 3.22 3.6.24 Bababa 3n 2n 1-3x-12x = 10 X Kontaktné pamat 6 X a3 1-6 1 ጎጂ 1 1 Ex. 7. Here again (243—3)5 — (35 — 3) 5 — ( a5 —— 3) 5, — — M 3 1 m by writing a=3; also pa³, q and whence, 12 5 ·BQ= 1 Blyth 1.2 X 9 3.6.a5 ∙CQ: as 1-5 10 P X Whence, 1-10 P 3 -3 5at per may may mga adv 1.5 3.6.9.27 X 5a¹ -1.2 3.6.q5 Buttertag Katana pakalpojuma sagata 1 a3 B, 3 X a5 Ç, -4.32 -4.9.33 X 15 5.10.a a5 5.10.15a¹ Whence, by writing 3 for a, and cancelling the like powers of 3, we have / (243-3)=5/ 210— 4 4.9.14 5.10.15.20.316 X Paglalagh 1.5.9 3.6.9.12.2¹0 4.9 5.38 5.10.37 5.10.15.311 N Tatting P by Jogg -4.32 5,10.9 3 as D, Saga } A > } &c. D. &c. • 1 મ m Sp m n m- pn-an-a pr m + m n m Ex. 8. Here, p=a, Q= ∙AQ· M 2n m~2n 3n m-3n 4n S 2 a m Palakka n -BQ = Magkakata AQ n an 1 X 2n m-2n {1+ -DQ: CQ= 113 1 X -BQ= a} {1 ± =A, ~~ =a³3=s, A, 1-2 4 1-4 OC. 20 X ==== b 6 1-6 8 4 +00 a X X ±x 2a X X 1-3 6 1- 6 J ±b a X OC + 2a α a² X 2.4a2 Ex. 9. Here, we have Pa, Q: Whence, - 202 31/1 -a 142 ±3x3 2.4.6a3 3a ±b ± 2.4a2-2.4.6a3 Alp 3a ±:00 a ±x a X Whence, (xa)²= 002 3x3 b 1 a a² X ±± aux Ex. 10. Here, P=a, Q I X =B, > 土 ​3a 3.6a² 3.6.9a³ a to а and B, + x +3x3 a 2.4.6a³ a ·002 2.4a2 Ha b m n 3 therefore we have (a±±6)³ 262 2.563 2 +00 -3.5x4 a 2.4.6.8a⭑ || 1/2 a =α, ·262 CQ= 3n X 9 3.6a2 From which the law of continuation is obvious; and ; therefore, -a 3.5x 2.4.6.8a1, &c.} ±0 α -262 1 Q3 3.6a² £2.563 1 3.6.9a3 and =D, and :C, a α = E; 777 1 22 > ጎ 4' ·a =D, 2.5.764 3.6.9.1201±, &c.} m 1 1 M 143 171 janga la m Pnanɑ— ጎ m~ N ·AQ= n m 2n Mak at { 1 α m n m m m pn S S dang ·BQ= ·N 1 Ela Pag m 17 4 ↓ ɑ¤—^, 1 X n m-2n 3n CQ= 1 4 p 2 -A Q = 3 x ·BQ = Padded 8 S X 8 Galerapping Sp the present the multiplier a a ³ ³ 2 13-1 a 4a -b 4a X Therefore, b 362 &c.} 4a 4.8a2 In the last three examples, we have repeated the frac- tional root of the first term in every line; but it is ob- vious that this is not necessary, as we may leave it out of every term; only remembering to introduce it at last as a general multiplier of the series. -b 1 4 3n 12 4.8a2 α Where again the law of the series is discovered; and we have (a—b)¹. -aª X Or, we might have put our examples under a different form, as in the following instance: Ex. 11. Here, the proposed quantity may be put un- der the form a³×(1+); and, therefore, omitting for -1=A, 1 X 200 a -a ·362 1 X XC P=1, Q==, and a JAPA =B, 3a x 2-3 2x X X 6 3a a 2-6 2x2 9 3.6a² ·b a a* X B, X 3.763 3.7.1164 4.8.12a3 4.8.12.16a4* C 9 4.8a2 -b -2x² 3.6a2 362 1 XC ทาน 2 n a -a =C, -3.763 4.8.12a3 GALA 3 we have- ; whence, a 2.4x3 3.6.9a³ } =D, D, } 144 m -3n 4n pag M Apply n 1 2x 3 x 11 + 3 a Valent Pr 1n m Sp a³× {1+ AQ m- n Which, by cancelling the multiplier 2, in the numera- tor and denominator, becomes 2x 3a Making J ·DQ= 3n m Kak ገኒ m 2n KAŽDÉ 2.4x3 X 12 3.6.9a2 } Whence, (a+x)³², or a³(1+2)³ = α 2012 2.4x3 + 2.4.7x4 За 3.6a2 3.6.93 3.6.9.12a++, &c. } Ex. 12. Here, P=1, @=∞, 1 2-9 •BQ= 15 2 1 5X1 plakimetatud CQ= Part Calen 2-5 10 2-10 15 > 1=^, X 202 9a2 XC -2x X 2x 5 5.10 Ex. 13. Here, 2x 5 5 + N Whence, we shall have X P=α, a, q X 5.10 1 2.3x2 (ax) 81 by 4203 4.7204 92a3 a AC --x a ሃኪ -2.4.7x⭑ 3.6.9.12a B, } X > x, and 1 9².12a¹‡, &c. } OC 1 -2.3x² 5.10 Therefore (1- X 2.3x2 2.3.8x3 2.3.8.13x4 5.10.15 5.10.15.20 and m Madaga m Make N E; Q Q 5.10.15 C, 2.3.8x3 =(a+x); therefore 1 In 1 ,, D, paddl $ &c. m -'an a =A, Whence we have pn And by omitting this factor in the subsequent opera- tions, 1 145 } m n m. •AQ=· p A 2n m-2n 3n m -3n 4n m n ·BQ = Suddenl G m -AQI n m pran a 2 -CQ n m DQ= 2n m- 2n ( 11/17/14+2348 1F Ex. 14. Here, 3n in 3n An 2.4.6a3 Whence, introducing the general factor a 1 we have therefore, P=a, Q= -1 } 1 1 -BQ: X whatsa S CQ med -+-x а 2a 1-2 -2 Fx ±x +3x² X X 2a 4 α 2.4a2 •DQ= 6 1-6 8 X Fx 1-4 +3002 ±x X The X 2.4a2 a My X F 2a 2.4a² 2.4.6a3 a 8 a 1-3 ±x、 Fx 3a X 3.5x3x X a (ax): +x a (a + x) * 3.5x3 3.5.7x + 1 =B, jangga Tjera Fx X За and a ±x a 1 = a( a + x) =²; 3 m n 6 1-6 4x2 +x X 9 X 3.6a2 a 1--9 74.7x3 X 12 3.6.9a3 7 X C, 3.5x3 2.4.6a3 +3.5.7x4 2.4.6.8a¹ 2.4.6.8a* A, (and, omitting this term,) Apparateterig +4x² 3.6a2 D, F4.7x3 3.6.9a3 +x a or 1 꿈​. Whence, 3 E, F, &c. } +4.7.10x 3.6.9.12a¹ 1 a ; Whence, introducing our two general factors ax= 1 N 2 146 a EL EX m we have a(ax) Ex. 15. Here, 事 ​00 4x2 1午​十 ​F 3a 3.6a² -AQ: n M-N m In m mately, japan 2n m -2n Paglaya 3n m-3n 4n S 10 -BQ= 5 1 5 CQ тро 1 1 -DQ= 00 P=1, Q=1 and I' Ex. 16. 2nd) S OC X-X ī -1-5 10 1=A, ४ 4.7x3 3.6.9a³ 1 1 (1+x) 5 X 15 -1-15 20 1-5+ 5.10 } 5 stand X a 2 8 x0 X 5 1 m n Whence, 11 I d -1-10 6x2 00 -6.1103 X X1= 5.10.15 5.10 Lament B, Therefore, I + 4.7.10x4 3.6.9.124 F, &c. { 3.6.9.12a¹ } - 3/3/13 =(1+x) ; consequently, OC == +6x2 5.10 D, ∞ +6.11.16x4 X 5.10.15 1 5.10.15.20 1 6.11x3 6x2 6.11x3 6.11.16x+ 1 5.10.15. 5.10.15.20 1 x Here, (+)³_(a+á)* − (a+œ)² S 1 (1+x) 5 + 2 (α-x)² (a-x)· =(a+x) (a² — x²) — ' — a+¤; we have r=a², Q==, and J a2 ¡ (a+∞) 3 a+x (a+x) = (a² = x²) 2 omitting, till the expansion is effected, the leading factor -002 [ m N Š &c. Whence, plakateg 2 433 1. E, 147 And, consequently, m m -1 2 pn — (a²)ñ — (a²) ☎ — a m N m -AQ: -n 2 ·BQ = 2n m-2n 3n m -3n 4n 1 X-X a CQ= -DQ= Bra 1-2 4 1 Plakaty 6 J 1 -+ a Mult. by a+x 202 a² 1 x2 X 2a3 002 2a3 ་ 4 3x1 X X 1 x2 2a3 a 2.4a5 M -x2 a2 1-6 3.5a6 X 8 2.4.6a7 X A, , •202 Wedd a2 X 3x1 2.4a5 3.5x6 2.4.6a7 002 a2 कर Answer, 1+-+2a2+2a3 + Or (a²x²) 2 = 3x4 3.5x6 + + 2.4a5 2.4.6a¹ 3004 3.500€ x2 1-+- + + 2a² 2.4a1 2.4.6a6+, &c. 3.5.7x8 2.4.6.8a9 +, &c. Xx3 3x5 *3.5x7 +-+ + + α 2a3 2.4a³ 2.4.6a7 2.3 3x1 3x5 x 2x (a + 2) = (1 + 2 x ) + α (g) X a B X :D, +, &c. 2.4a4 2.4a5 : E, +2.40s+, &c. Or, the expression in this example may be reduced to a more simple form by taking But, in order to have the same answer as above, the terms of the result must be divided by a x, and its powers. 148 1 INDETERMINATE ANALYSIS. C 8y-16 3 Ex. 1. Given equation 3x=8y-16. 8y-16 Here x= 2y-1 =2y-5+ 3 3 Let 2y-1 3 XC PROBLEM I. p; then 2y-1=3p; or 2y=3p+1; 3p+1 Whence y p+1 which latter will obvi- 2 2 =P += ously be a whole number, provided p be taken any odd number. Assuming p=3, we have y= 3p+1 -5, and x 2. EXAMPLES FOR PRACTICE. Whence x= / :8, which are the least values sought. Ex. 2. Given equation 14x=5y+7. Here it may be observed, that since 14x and 7 are both divisible by 7, 5y must be so likewise; consequently y must be divisible by 7. Let therefore y=72, and we have 5y=352; which substitute for 5y, gives the equation 14x=35x+7, or 2x=5z+1; 52 +1 2+1 =2x+ 2 2 Therefore =p, or z=2p—1; whence, 2+1 2 10p-4 2 ; 2 5z1 5p-2; and y=7%=14p-7. 2 Hence the general values of x and y are x=5p-2, and y=14p-7, where p may be assumed 1, or any inte- ger whatever; the former value, viz. p=1, gives x=3, and y=7, the least values. А -wh. P 149 Ex. 3. Given equation 27x=1600-16y. Here again it will be observed, that since both terms on the right-hand side are divisible by 16, the left-hand member, viz, 27x, must be so likewise; which cannot be except x itself be divisible by 16; make therefore x= 16%, and our equation becomes 27.16z=1600-16y; or 272=100-y; Whence y -100-27%; and x=16%, where z may be assumed at pleasure, providing 27% is less than 100. If z=1, then x=16, and y=73, z=2, then x=32, and y=46, z=3, then x=48, and y=19, Which are the only three answers. Ex. 4. Let 7 and 11y be the two parts required, then we have 7x+11y=100. Whence x= Make now 7d-2 3 obtain d3b-1. Therefore y 1 100-lly 7 2+3y 7 =2d-1+1. Again, make 3 Kateg =14-2y + 2 + 3y 7 d; and we have 3y=7d-2, or y= =b, and we 7d-2 216-7-2 .3 3 100-11y 100-776+33 7 7 9 And x= =19-116; Where it is obvious that b cannot be taken greater than making therefore b=1, we have x=19-116-8, and y=76—3=4; therefore 8×7-56 is one part, and 4× 11-44, the other. Ex. 5. Given equation 9x+13y=2000; Whence x= d+1 3 I -76-3, 2000—13y_222—3+ 2-4y—wh. 9 J 150 1 Make now 2-4y 9 2-9p 4 y= Here making Hence 2-p 4 2-9p =p, or 4y=2-9p, or 2-P_wh. 4 Whence making And y= 254-11x 5 -2p+ 2000-13y 2000-117q+52 9 9 P Stat =q, we have p-2—4q ; 2-18+36q -9g-4, 4 and x= -228-13q, Where q may be assumed at pleasure, providing only that 13q be less than 228; it may therefore be an inte- ger from 1 to 17; which latter therefore denotes the number of possible solutions. If q=1, then y=9q-4-5, and x=228-13q=215 If q=2, then y=14, and x=202 If q=3, then y=23, and x=189 &c. &c. &c. Ex. 6. Given equation 11x+5y=254. Here y= =51-2x- 1+x 5 Matatanda 1+x 5 254-11x 254-55p+11 5 5 { =p, we have x=5p-1, =53-11p, Where must be assumed less than ܕܒ p any number from 1 to 4. Ifp=1, then x=5p-1-4, and y=53-11p=42 p=2, then x 9 ; and y=31 p=3, then x=14; and y=20 p=4, then x= 19; and y= 9. wh; 53 11 ; that is Kattank Ex. 7. Given equation 17x+19y+21z=400. In questions of this kind, in which only the number of 151 solutions is sought, the answer is more readily obtained from the following rule: Let ax+by=c, be any proposed indeterminate equa- tion, and find the value of p and q in the equation, ap— bq=1; then the number of possible solutions of the equation ax+by=c, is equal to the difference between cps ca * the integral parts of the fractions b a In our proposed equation, by transposing 21z, we have 17x+19y=400-21z; and by giving 2 the several va- lues 1, 2, 3, 4, &c. we obtain the following set of equa- tions; p being 9, and q=8; Equations. 17x+19y=379; I 17x+19y=358; 17x+19y=337; 17x+19y=316; 17x+19y=295; 17x+19y=274; 17x+19y=253; 17x+19y=232; 17x+19y=211; No. of Solutions. 9.379 8.379 19 *9.358 19 17 9.337 8.337 17 19 9.316 8.316 19 9.295 8.295 17 19 9.274 17 8.358 19 9.232 19 17 9.253 8.253 19 9.211 19 17 $8.274 1 * See Barlow's Theory of Numbers. 17 8.232 17 8.211 17 1 ↓ Ta Kottakka Vapo V Ka 1 =1 1 1 :1 1 0 0 :0 M 152 ? 1 1 17x+19y=190; 17x4-19y=169; 17x+19y=148; 17x+19y=127; 17x+19y=108; 17x+19y= 85; 17x+19y 64; 17x+19y= 43; 17x+19y= 22; Apple 1 *9.190 8.190 19 17 9.169 8.169 19 9.148 19 9.127 19 9.108 1 19 9.85 19 9.64 19 9.43 19 9.22 7 3.202 77 3.191 17 8.148 77 3.180 77 17 8.127 1 17 8.108 17 8.85 17 8.64 17 8.43 17 8.22 17 5 2.202 S 5 2.191 -0 5 2.180 5 1 -10 Ans. 19 Total number of solutions Given the equation 5x+7y+11z=224. value, from 1 to 19, which equations; also in the equa- 3 and 7-2. 2. Hence No. of Solutions. 3.213 2.213 Ex. 8. Here z may have any gives the following set of tion 5p-7q=1, we have Equations. 5x+7y=213; 5x+7y=202; 5x+7y=191; 5x+7y=180; Y * When any of tho left-hand fractions are exactly equal to an integer, tho quotient must be diminishod by an unit. =1 1 1 0 :0 :6 ! mig alay Mald -5 př 1 153 ད་ י + A 5x+7y=169; 5x+7y=158; 5x+7y=147; 5x+7y=136; 5x+7y=125; 5x+7y=114; 5x+7y=103; { 5x+7y= 92; 5x+7y= 81 5x+7y=70; 5x+7y= 59; 5x+7y= 48; 5x+7y= 37; 5x+7y=26; 5x+7y= 15; 1 J 3.169 2.169 y 5 3.158 2.158 77 3.147 77 5 3.136 2.136 7 3.125 7 3.114 7 3.103 77 3.92 7 3.81 7 3.70 7 3.59 7 3.48 7 3.37 17 3.26 5 2.147 17 3.15 7 5 2.125 5 2.114 5 2.103 5 2.92 5 2.81 5 2.70 5 2.59 5 2.48 5 2.37 5 2.26 5 2.15 5 -4 K :4 Linking :4 Stand 3 3 2 Q =2 PROJE ته 1 1 =1 Total number of solutions =59 Ans. O 154 1 Ex. 9. Let x be the number of half guineas, and y the number of half crowns; then the number of six- pences in each of the former being 21, and in each of the latter 5; also the number of sixpences being 800, we have the following equation: 21x+5y=800; Here the equation 21p-5b-1, gives p1, and 6=4; 1.800 4.800 Whence by the rule 5 21 =7, Answer. Where, as in the preceding examples, the first quo- tient is distinguished by unity being exactly integral. This rule, which is taken from Barlow's Theory of Numbers, is much shorter than that which depends upon an actual determination of the several solutions: the latter is omitted here as presenting no difficulty to the student who has attended to the preceding solutions. Ex. 10. Let x represent the number of guineas I have to give, and y the number of louis-d'ors I am to re- ceive; then by the question, } 21x-17y=1; Hence 17y=21x-1, or y=x+ Make 4x-1 17 } or x=1p+ ཡ MAY =p; then 4x=17p+1, P+1 p+1 4. And consequently x=4p+? +1 4 Where, if we make 4x-1 17 my —wh. b; then p46-1, wh. =17p-4; 4x-1=21p-5; I Whence also we have y=x+ 17 Where p may be taken any number at pleasure; if we take p=1, then x=13, and y=16, which are the least numbers, viz. I must give 13 guineas, and receive 16 louis. Val } 155 } Ex. 11. Let x, y, and z, be the number of gallons of each sort respectively; then by the question x + y + 2=1000 12x+15y+182=17(x+y+z) By transposing the terms of the latter equation we have 5x-2y+z=0 x+y+z=1000 But By subtraction, 6x+3y=1000 Where it is obvious the answer cannot be obtained in in- tegers, because the first side of the equation is divisible by 3, and the other is not. We may therefore assume x or y at pleasure; taking for a the value given in the an- 1 swer of the Introduction, viz. x=111 1000 – 6×1111 __ 333 111- 9' W – x = 111+; y = 9 3 9' 77 question t and z=5x+2y=777; 1 1 ry That is, 111 at 12s.; 111 at 15s.; and 777 at 18s. Ex. 3. Let x Jogg INDETERMINATE ANALYSIS. OC 2 Apply 6 Let x the number sought; then by the } x - 3 13 20 -2. and PROBLEM II. Ga we have Berlinden whole numbers. p; then x=6p+2. Make 6 Substitute this value of x in the second equation, and we have 6p2-36p-1 13 S 13 -wh. 14 156 ↑ Make * Ex. 4. question 6p-1 · 13 • pag Make 20 -5 -7 x-5 7 Let now 2+1 =r; then q6r−1; and p= 6 13r-2; and consequently x=6(13r-2)+2=78r- 10; where r may be taken at pleasure. If r=1, then x=68. Let x be the number sought; then by the whole numbers, 7p+3 9 Let now q; then and } 00 2 9 p: Make =p, or x=7p+5· Substitute this value for x in the second equation, and we have 29-3 17 7p+5-2 7p+3 9 9 139+1 6 Again, let there be taken 7(2s-1)+3 2 r-1 2 =29 +2 + 1 6 7s q; and we have 'p= 29-3 wh. 77. =r, and we shall have q= +1+2+1 wh. 2 1 p wh. เ 13(6r-1)+1 6 99-3 7 -wh. 2, and p= =9+ 7r+3 રે =s, or r=2s-1; p Then 4= 9s-3; whence we have x=7(9s-3)+5=63s-16; where s may be taken at pleasure. If s=1, then x=47 and if s=2, x—110, &c. 3r 1 9(7s-2)-3 my 157 , → Ex. 5. Here by the question we have x-16 X-27 and 39 56 00 -16 S Make , Make =p, or x=39p+16. 39 This, substituted in the second equation, gives 39p+16-27 39p-11 wh. 56 56 17p+11 56 Make Then q= 2r-1 5 Let therefore Then Also make 177-6 5 q, or p= s+1 2 Consequently r W OC x-5, x ry 17(5-2)-6 5. s; then r= 59+6 17 8 whole numbers; =p- 56q-11 17 wh. K 5q+6 17 =3r-1+ 5s+1 2 and 5(2t-1)+1 2 17p+11 - 56 =r; t, and we have s=2t−1 ; =3q1+ 2C 8 9 02 2r-1 -wh. 5 =25+ s +1 2 5q+6 17 =17t-8; p= 56t-27; and x=39(56t-27)+16-2184t-1037; Where t may be assumed at pleasure. When t=1, then x=1147, the least number agreeing with these conditions. =5t-2, and q= Ex. 6. Here putting a for the number sought, we must have wh. 56(17t-8)—11 17 all whole numbers: 158 x-5 From the first, by putting =p, we obtain x=7p 7 +5; and this substituted in the other two, gives 7p-2 and whole numbers: 8 7p-2 8 9+2 Make 1 Let now 7p-3 9 g; then 7(8r-2)-3 56ŕ-17 9 Let therefore p= p= r, and we have q=7r—2 ; 8(7r-2)+2 Whence again p= =8r-2; 7 Now, substituting this value of p in the third equation, we shall have 89+2 7 9 2r+1 9 or 6r-2+ 2r+1 9 s-1 Kompletn Where 2 5-1 And let =t, and s=2t+1. 2 From this we readily draw the following values: 9(2t+1)-1 s=2t+1; r = 9t+4, 2 q=7(9t+4)-2=63t+26, 8(63t+26)+2 7 wh. :9+ = a whole number; =s, or r= 9s-1 2 PATR is likewise wh. 9+2 7 K 72t+30, and ▸ wh. 48+ x=7(721+30)+5=504t+215, Where t may be assumed at pleasure. If t=0, then x=215; If t1, then x=719; s-1. 2 1 If t=2, then x=1223; If t=3, then x=1727, &c. 159 > 鼻 ​: Ex. 7. Let x be the number sought; then, by the question, x x x x x x OC 9' 8' 7'6' 5' 4'3' Now, first, if ! and - 00 9 XC must all be whole numbers. 2' 00 and be whole numbers, 8 must necessarily be so. We have, therefore, in the present case,, only to find x x x OC 9'8' 7'5 whole numbers, which must have place, if x be made equal to 2520, the product of all these denominators. Ex. 8. Here, if we put x for the number, the condi- tions are, that and 00 00 OC 6' 4' 3 X-5 7 and x x x 00 00 2' 3'4' 5'6' must all be integers, or whole numbers. But the first five of these fractions, when brought to a common denominator, are 30x 20x 15x 12x MOVIE 10x and , 60 60 '60 '60 60 Whence, as any multiple of a whole number is a whole 20 whole number, we have only to make and numbers. 60 Or, as in the preceding examples, these conditions may be reduced, by observing that if x x ▬▬▬▬▬▬ 2'5' 6 x-5 7 W OC and are whole } OC numbers, and must be so likewise; XC 4. 3 And the least value of x, which answers the three for- mer conditions, is x=2×5×6=60; x, therefore, must be some multiple of 60. 160 Let then x= whole number. Now 3p-2 7 Therefore 60p, and the last condition is that 60p-5 my sequently p= Make Again, let 2—1 3 EQ₂ then 3p-2 77 7(3r+1)+2 3 gr =9p-1 Where r may be assumed Wwwg p= S K 3p-2 7 whole number. Belgra =r, and we have q=3r+1; and con- If r=0, then p=3, and x=60p=180. DIOPHANTINE ANALYSIS. 79+2 3 I QUESTIONS FOR PRACTICE. Ex. 1. Since x+1, and x-1, are to be both squares, let xy-1, then x+1=y², which fulfils the first con- dition; and therefore it only remains to make y2—2=0. 1 =2q+1+2 =¹ 3 9 — Let y²-2=(y-1)=y2-2y+1, and we shall have 2y=3, or y=3, or y²=2; Consequently y²-1={-1=4, the number sought. Ex. 2. Here x+4, and x-+-7, must become squares. Let x-4-p², or x=p2-4, then x+7=p2+3=0. Let p²+3=(p+9)²=p²+2pq + q²; 3-92 Whence p 27 60p-5 7 7r+3. 0, or any integer whatever. and, consequently, x= , 9–22q²+q* : 4q2 and if we take 7, we find x 4r252 which we may substitute for r and s any integer numbers whatever. Let s be taken 2, and r1, we have 'a 13, whence x-4=2, and x+7=209. THE SULLINGS pla 121 16 9¸§¹¹ — 22r² §² fer got: " in 1 1 161 Ex. 3. Here 10+x, and 10-∞, must become squares. This might be done by the method that has just been em- ployed; but let us explain another mode of proceeding. It is evident that (10+x)x(10-x)=100-x², must likewise become a square; now its first term being already a square, let 100-x²-(10—px)²=100-20px 官 ​+p2x2; therefore x=- 20p p²+1 Now it is only necessary to make one of these formulæ =, because it necessarily follows that the other will be 10p²+20p+10 10(p²+2p+1) a square; now 10+x= p²+1 p²+1 10 10p²+10 or ; but p + I' (p²+1) 2 and since p²+2p+1=0, the denominator, hence 10p2+10=; and here it is necessary to find a case in which that takes place. When p3, then 10p²+10=90+10=100=□. Let p=3+q, and we shall have 100+60q+10g2O. Let the root of this be 10+qt, and we shall have 100+60g by which 60-20t +1092-100+20qt+q22, whence q= t2-10' 20p means we shall determine p-3+q, and x= Let p²+1 t=3, then q=0, and p=3; therefore x=6, and, conse- quently, 10+x=16, and 10-x=4.* Ex. 4. Here x²+1, and x+1, are to be both squares. Let x+1=p2, or x-p2-1; then x2+1=p¹-2p²+2 ; which last formula is of such a nature as not to admit of a solution, unless we already know a satisfac- tory case; but this is a square when p=1; therefore let p=1+q, and we shall have a²+1=1+49²+4q³+q¹ which may become a square several ways. Let 1+2q² the side of this square; then 1+4q2+4q3+q₁ = 1 + • ghly , *See, for a more general solution of this and several of the following questions, Euler's Algobra, vol. ii. p. 174. } 162 1 鲅 ​4q²+4q4; therefore q, and p=3; consequently x=4º. Ex. 5. Let x2, y2, and z2, be the numbers sought. Then x²+z²=0, y²+z²=0, and x²+y²=0. x02 002 Or, 2+1=0, 22 And, by putting But a2 را 22 2 s4+2s²+1 452 - 32 22 - and Y ___ r² — 1 2. 2r XC Z 28 ร +1=0, and of z2 $2 1 X.33 4r²(s²-1)²+4s²(r² — 1)² 22 002 22 + 1 y2 and +1 JAN 22 which are both evidently squares; and therefore it only remains to make 4r2 x² y² 22×32 22 " a square. s² = 1 ) 2 + ( 25 483 44 and 2r ; or a 1 r2 :)²= 2 Whence s-24, and r=-22. · And Ja pod Y Z 2r pt+2p²+1 1 to a square number. 4.1.25.2. Or, r²(s²-1)²+s²(r2 — 1)²= r²(s+1)2 × (s-1)²+s² (r+1)²× (r−1)² to a square number. And, by mak- ing r-1s+1, or r=s+2, we shall have (s+2)²x (s+1)2 × (s-1)²+s2x(s+-3)2 × (s+1)2 to a square number. Or, (s+2)2 × (s-1)²+s2x(s+3)=2s¹+8s³ +6s²-4s+4 to a square number. Jabatan tang Let, now, the root of this last square be assumed 5²-s+2: then, 2s+8s36s2—4s+4 = (5/s² —s+2)² =2581-535s2+s²-4s+4; or 2s+8s3=25s1 - 553; or, 2s+8=25s-5. y 2 z2 (s² —1)², (r² — 1)² + 4.52 4.r2 we shall have " Vektorat = 575z 48 ? 183z and y= 44 In order, therefore, to have the answer in whole num- bers, let z= 528, and we shall have x -6325, and -5796. And consequently, (528)2, (5796)2, and (6325)2, are the squares required. y= 00 $21 2 25 Jobgarag 575 48 163 Ex. 6. Let the numbers sought be x and y ; then x² +y=0, and y²+x=. And, if r-x be assumed for the side of the first square x²+y, we shall have x²+y= r²-2rx+x², or y=r2-2rx. Therefore, by reduction, p2 - Y Again, if y+s be assumed for рад - у 2rx=r²-y, or x= 2r the side of the second square, we shall have y²+: 2r =2sy+s², =(y+s)²=y²+2sy+s2. Whence also } or r²-y=4rsy+2rs². Whence, by transposition and 2r²s+s² r²-2rs2 4rs+1' and x= ; p² -y division, we have y= 27 4rs+1 where r and s may be taken at pleasure, provided r be greater than 2s2. But to find the answer in the book, according to Eu- ler's method, let x²+y=(p-x)²; and, at the same time, the other y²+x=(q-y)2, and we shall thus obtain the two following equations, y+-2px=p², and x+2qy=q², · and y= from which we easily deduce x= -Z2 berkarat p² - Y 2r 2pq²-p² 4pq-1 Let p 1, and q=3, we shall then have a- 77 , whence we derive a²+y=80+15=288=(137)³, and y²+x+25=100=(10). ΤΟ 9 40 I 49 3 64 Ex. 7. Let a², y², and 22, be the numbers required; then, by the nature of harmonical proportion, x² -y² : y² 2x²z2 a²: 2²; hence y²: x² 22 x²+233 (x²+22) a square; therefore, since the nume- 2qp² — q² 4 pq - 1 4 in which p and q are indeterminatę. 2 ਝੰਡ, and 3 20' Y x222 X M (x + 2)² + ( = = = = = 2 2 rator is already a square, (~1~2)²+(~7~)² -)2 must be a 164 / x+z square; we have, therefore, to find two squares and (~72)², such, that their sum may be a square. This 2 x + x Ꮗ . S will be accomplished by taking 2rs, and p2. 2 2 2* s2; for then yr²+s², as appears from note, page 228 x+z XC -z in the Introduction. 2rs r2: 2 Ka +s²; and z G the root of 2 =2rs+r²-s²; then, y, or Therefore, 2rs-p²+s², and 2rs+²-s2, will be the p² + s² roots of three squares in harmonical proportion. Or, if each number be multiplied by ²+s², we shall have 2r³s +2rs³ —r¹+s¹,6r2s2-4-s1, and 2r3s+2rs³+r—s¹, for the three roots, where r and s may be taken at pleasure. 3 J Ifr=1, and s=2; then x=35, y=7, and 2=5; and the three squares are 352, 72, and 52, or 1225, 49, and 25. Ex. 8. Let x³, y³, and z³, be three cubes, such, that their sum x³+y³+z³, may be equal to a cube —v³; then, by transposing one of the terms, we have x3+ y³ — v³. 23. In order to satisfy the conditions of this equation, put x=u+w, and y=u-w; then we have ³+y³= 2u(u²+3w²). Also, put v=r+s, and 2-r-s, and then we have vz=2s(s²+3r2); whence 2u(u²+3w²)= 2s(s²-3r²), or u(u²+3w²)=s(s²+3r²). 3 Assume u mt+3np, and w=nt-mp; then u(u²+ 3w²)=(mt+3mp) (m²+3n²)(12+3p2). Also, assume s= at+3cp, and r=ct-ap; then s(s²+3r²)=(at+3cp)(a² +3c²)(1²+3p²); whence (mt+3np)(m²+3n²)(1²+3p²)= (at + 3cp)(a² + 3c²)(1² + 3p²); or dividing by t²+3p2, (mt + 3np) (m² + 3n2)=(at + 3cp) (a² + 3c2), and t 3c(a²+3c²)—3n(m²+3n²) P. m(m² + 3n²)—a(a²+3c²) م All Clot x+z 2 2x222 x²+x²' 62 52 1 Hence, x= + 1 x- Dagblad 2 will be 54 2 > ·) 2 Ma 6x2g2 — pt — gd r² + s² 165 1 Or, if p be taken m(m²+3n²)-a(a²+3c2), then t= 3c(a²+3c²) — 3n(m²+3n²); where a, c, m, and n, may be any numbers taken at pleasure. If we suppose n=0, and a=c, then 12c³, and p=m³ —4c³; hence u=12mc³, w =4mc³ —m¹, s=3cm³, and r= =16c¹-cm³ ; therefore x= u+w=16mc³ — m¹, y=u—w—8mc³+m³, z=r--s=16c4 -4cm³, and v=r+s=16cª+12cm³. Now, if c=m=1, then x=15, y=9, z=12, and v=18; or dividing by 3, x=5, y=3, z=4, and v=6; so that 33+43 +53—6³. Again, if we suppose a=0,. and c1, we shall find · 3n — (m² + 3n³)+m, y=−3n+(m² +3n²)—m, z= (3n-m) (m² + 3n²)+1, and v=-(3n+m) (m²+3n²)+1; where, if m -1, and n=1, then x=- ·20, y=14, z= -7, so that we have 143+17373-203. Ex. 9. Let x be one of the parts, then 100-2 will be the other part, which is also a square number. As- sume the side of this second square =2x-10,* then will 100 x²(2x-10)²=4x² - 40x+100; and, conse- quently, by reduction, x=8, and 2x-10=6. There- fore 64 and 36 are the parts required. Wedding 17, and v Ex. 10. Let x and y denote the numbers sought; then, by the question, x2 y2-x-y, and x²+y²=0. The first equation, divided by x-y, gives x+y=1; hence x=1-y, and x2+y21-2y+2y=、a square. Assume 1-ry for its side: then 1-2y+2y2-1-2ry+ 24-2 r2-2r r²y²; whence y=42_Q' and x=1-y X J 1 Sedan * ✓ 722 r may be taken at pleasure, provided it be greater than 2. If r=3, then the answer. 4. 3 and x= y= ** 77 7 Spraggadget ; where Job * lf x-10 had been made the side of the second square, in the above solution of this question, instead of 2x-10, the equation would have been ∞2—20x+ 100=100-2; in which case, the sido of the first square, would have been found =10, and -10, or the side of the second square =0; for which reason the substitution a-10 was avoided; but 3x-10, 4a-10, or any other quantity of the same kind, would have succeeded as well as the former, though the results wouid have been less simple. This question is done generally, in example 4, preceding questions for prac- ice in the Introduction. Р 166 Ex. 11. Here, if we call x and y the two numbers, we have to find x+xy=0 y+xy=0 Now if we assume x-m², and y=n², these become m² + m²n² = m²(1+n²) n² + n² m² — n² ( 1 + m²) which must be both squares; that is, we must have 1+ n², and 1++m², both squares. Assume n2+1=(n+r)²=n²+2rn+r²; Then, from this we have n= 5.2 manner we have m ❤ 1 1 And we have x= greater than 1. 28 p • 1-72 2r Consequently x= 2r r and s may be assumed at pleasure. 9 If r=2, and s=3, then x= and y= 16 9* Which numbers answer the conditions of the question; but they are both square numbers, which is not a neces- sary condition. The question may, therefore, be otherwise resolved, as follows: Let x and x-1,be the two numbers; then we have to x+x(x-1)=x², and find both squares. only remains to find x²-1 a square. x²-1(x-r)²x²-2rx+r², Make r² + 1 2r -p2 1-$2 -)³, and y=(-)²; where -)2. 25 ; and x-1: 5 If r3, then x= 3 swer the conditions of the question. M and in the same (x-1)+x(x-1)=x²-1, The first of which is so, and therefore it 2 3' 16 ; where may be any number } F 1 which numbers an- 167 1 Ex. 12. Let x², y², and z² represent the required squares; and we shall have to solve the equation x²+z²=2y²; In order to which, let x=m+n, and z=m-n, then x²+z² = 2m²+2n² = 2y²; In which case, it only remains to find m²+n²=y². J Let m—p²-q², and n=2pq, and we have at once m² + n² = ( p² — q²)²+4p²q²= ( p²+ q²)² — y². Hence the following general results, viz. x=p²—q²+2pq, and x²-( p²-q²+2pq)² y= p²+q² x² = (p²+q²)² z= p² -q²-2pq z ² — ( p² — q² — 2 pq)² Where P and 9 may be assumed at pleasure. If p=2, and q=1, then x=7, y=5, and z=- Consequently, the squares are 49, 25, and 1. 2 น 1 Ex. 13. Let x²-y, x², and x²+y, be the three numbers in arithmetical progression; Then, we have to find a2, x²+y, and 2-y, rational squares; or x²+y=m², and x²-y=n²; Assume y=2rx+r², then we have x²+y=x²+2rx+r²=(x+r)², And therefore it only remains to find 002 x²- 2rx-p³ —n². Ad Assume x²-2rx — r² — (x — m)² = x² - 2mx+m², S Aft m²+r2 And we shall have x 2m-2r' " 1 ; Where m and r may be taken at pleasure. If m=5, and r=4; then, x=4, and 27 2-1681. Also, y=2rx+r²=41×4+16=180; whence, the three numbers will be 30, 210, and 3901. And if these numbers be multiplied by any square number, the same conditions will obviously obtain : Hence, multiplying by 4, we shall have 120, 840, and 1560; which are the numbers given in the answer in the Introd. / • { 168 And if these last be again multiplied by 4, we shall have 482, 3362, 6242, which are all integral, and equally answer the conditions of the question. ず ​The question may be otherwise solved as follows: Let x, y, and z, be the numbers, and assume x+y=m² x+z = n² y + z = r² Then we shall have F X m² + n² ——— p²) = 3/31 ( y = 1} ( m² — n² + r²) z = {} ( — m² + n² +r²) Where, since the numbers are in arithmetical progres- sion, we have And consequently Where x+z=2y; or n²=m²—n²+p²: Therefore we have only to find m²+r²=2n²; that is, three square numbers in arithmetical progression : Whence, from Example 12, we shall have m² = ( p² — q² — 2pq)² n² = ( p² + q² ) ² r² = ( p² — q² + 2pq)² Ex. 14. L then by the question, 1 x = ( p² + q²)² - 4pq(p² —q²) y = {} (p² + q²)² 2 2 z = ( p² + q ²)² + 4pq( p² —q²) P and q may be taken at pleasure. Let x, y, and z, be the numbers sought; x²+y+x=m² y²+x+z=q² z²+x+y=r² Assume x2+y+z = (x+n)²= x²+2nx+²: 2 Then we shall have x- A y + z- n² Angelegenhed 2n This value of x, substituted in our second and third equations, gives f • 169 الدو 1 v² + y + z = n ² + z = p² = -ge 2n ·+y= r²; 22+ Hence, assume y²+ And 22+ Y y + z-n² 2n y + z = N2 2n y+z-n² 2n -+y=(zs)²=22-28z+s², +z=(y—p)²=y²—2py+p², And we shall now obtain From the 1st. 2: From the 2d. n² —y—4pny+2np² 1+2n n² — y — 2ny+2ns² 1+4ns n²-y-4pny+2np² n²-y-2ny+2ns² 1+2n 1+ 4ns And consequently by reduction, 4nsp²+2n²s+p² — n² — s² — 2ns² 2p+2s+8nps-2-2n 16 8 Where n, p, and s, may be assumed at pleasure. If n=1, p=4, and s=11, then y=1, z=, and x= which are the answers given in the Introduction. Ex. 15. Assume 8x and 15m for the two numbers; then, by the question, (8x)2+(15x) is to be a square. But (8x)²+(15x)²=289x²=(17x)2; therefore, a may be any number at pleasure. If x=72, we have 576 and 1080 for the numbers sought. Whence, V i } Ex. 16. Let x and y be the numbers sought; then, by the question, we have to solve the equations. x²+xy, or x(x+y)=m² y²+xy, or y(x+y)=n²; P 2 1 1 170 ༈ Assume xp, and y-q², and these become p²(p²+q²)==m² q²(p² + q²)=n² Which conditions will be fulfilled, if we find p²+q²= } a square. Now, we have already seen (Example 12) that for this purpose we have only to assume pr²-s², and q=2rs Therefore, ap²=(r²-s²)² And y=q²=4r2s2 Where r and s may be assumed at pleasure. If r=2, and s=1, then x=9, and y=16. These, however, are square numbers, which is not a necessary condition. Now, it is obvious that any any multiple whatever of the same numbers 'will equally answer the purposes re- quired in the question. We shall have, therefore, a more general solution by taking x=t(r²—s²)², and y=4tr2s2; where r, s, and t, may be assumed at pleasure. Ex. 17. Let x and y represent the two numbers; then, by the question, x²+ y² — 1 — m² x²-y²-1=n² Assume xy+1; then, these two equations reduce to (y+1)2+ y²-12y²+2y=m² (y+1)² — y² — 1= 2y=n² Where, putting the first under the form 2y(y+1)=m², and substituting 2y=n², this becomes 'n²(y+1)=m², Consequently, y+1 must be a square; let, therefore, y+1=p2; then, since we have found 2y=n2, we have, by subtraction y=n²-p²+1 And, consequently, an²-p2+2 Where n and p may be assumed at pleasure. If n=4, and p=3, then y=8, and x-9, the numbers required. K 171 Ex. 18. Without attending to the particular numbers in the question, let us endeavour to resolve any given square number into two other square numbers. For this purpose, let a² represent the given square that is to be so resolved, and put a2 and y2 for the required squares. Then we have to satisfy the equation a² = x² + y², or a² — y² = x². ་ In order to which, let us assume px a+y= q qx 2α- a-y= p From which we have, by addition and subtraction, K px, qx __( p² + q²) x + q px 1 OC: P Pq qx___ ( p² - q²) x 3 2y= Whence, by multiplication and division, q P pq 2pqa p² +9² (p² - q²)a p²+p2 21 y= Where the indeterminates p and q may be assumed at pleasure. But it is obvious that 2pq and p²+q², as also p²—q² and p²+q², being incommensurate, the values of x and must be fractional, except a be divisible by p²+q². Y That is, unless a be divisible by the sum of two squares; and then the question will admit of as many integral answers as a has divisors of this form. 1 ;' Now, by the question, a=65; and 65 is divisible by 5- 2212, by 13=32+22, and by 65-72+42-8²+12. Hence, we may assume p=2, and q=1; or p=3, or p=3, and q=2; or p=7, and q=4; or p=8, and q=1; F 1 172 1 # Which give the four following solutions, 652—16² + 632-562+332602+25252²+392. These being the only integral answers the question ad- mits of. Ex. 19. Let x², y, and be the numbers required, and put the given number 19-a; then, by the question, уз 202, x²+a, y+a, and 2+a, are to be all squares, or Assume x²+a(r—x)²= r²—2rx+x², and y+a=r²; 2-2 a Then, we have x= x²+a— m² y + a = n²· y2 002 +a=v²; ac Wherefore, ==2r, and+a=42+a, which must be 雌 ​a square. Let, therefore, 4r2+a=(2r+s)²=4r²+4rs+s²; And, consequently, y =( a In which case, we shall have r— α· $2 iC2 } 5 ( 36 ) 2 2r Galatasa 19-1 4. 5 Kwa } 4.S Where s may be taken at pleasure, provided be greater than ɑ. 1 and y=r²-a; y—r² 5 )2 But, by the question, a=19; if, therefore, we assume 5 y a-52 19-1 s=1, we have y: )2–19 4' x 2 and x=1+9=36 4. Hence, the three numbers are 25 5 y2 and 22 1296''Y plate Sp 4' M Ab₂ S2 4s › Splende debaj 25 a ·S2 48 92-81. -)2 -9, * 173 The question may be otherwise answered, thus : Let 4x2, and x2a, be two of the numbers; then (x² -α)² 4x2 will be the third number; and, by the question, 4:00²+a =m² x²-a+an²' (x² -α)² a 4002 -+a=(x²+a)²=v2. Here, the second and third equations being squares, it only remains to find the first 4x2+a=m². Assume 4x²+a=(2x+s)²=4x²+4sx+s²; a -$2 ; 45 Whence, x= sure, provided only that s² be less than a. Ex. 20. Let x and y be the two numbers we have to find; then, x²+ y²+xy=m²; Assume x²+xy+y²=(x+r)²=x²+2rx+r², y2g2 And we shall have x 2r-y Wherer and y may be taken at pleasure, provided y be greater than r, but less than 2r. If y=3, and r2, then x=5- y=5, and r=3, then x=16 y=7, and r=5, then x=8. | ggg d where s may be taken at plea- + Ex. 21. If x, y, and z are taken for the three num- bers, we have to find x²+zy=0 y²+xz= z²+xy=0, Or, dividing by x2, we have, in that case, zy 1+ 202 Z 2/22/2 x2 z2 x2 + 28 Da + 20 Y II 11 > 1 1 + 174 ac Y Or if, in order to simplify the expression, we put m, and Z Z n, the above will become mn +1: 22 m² +n=0=s² n²+m=0=1². From the first, mn=r2-1; assume, m=r+1 n=r-1,' And there now remains to find w2 3-2w (r+1)²+(r−1)=s² (r−1)²+(r+1)=t², or p2+3r=s2 p2r+2=12; Assume r²+3r=(r+w)²=r²+2rw+w²; Whence, r: assume, therefore, Pag यान w2(3-2w), 2(3-2w)2 (3-2w)² (3-2w)2 (3-2w)2 Or w¹+2w35w2-12w+18=0; Assume this (w²+w+2)²= w+2w3+5w2+4w=4, or w= ; which, substituted, gives Then, 12w+18=4w+4, 14 7 16 8 And we shall have r + 129 w2 3-2w · 49 80' and n= -31 80 Consequently, m= " 80 } -31, Or x=80, y=129, and z—— Which are three numbers answering the required con- ditions, and a similar process will give the three num- bers, 9, 73, 328, all positive. S } Cha 175 Ex. 22. Find three squares, a², 62, c², such, that their sum may be a square, or a²+b²+c²=d², That is, assume a²+b²+c²=(c+r)²; from which we a²+62-2 shall obtain c= 2r where a, b, and r, may be taken at pleasure, provided r² be less than a²+b². This being done, let ax, bx, and cx, be the required squares; then, a²x²+b²x²+c²x²-d²x², And we have to find Or, dividing by d², =p; these become d²x²+ax= d²x² + bx d²x² + cx= 1 p2 (2r+m)2 1-2 } and putting x²+mx= xả nư x²+ px=0; Assume x2+mx=(r-x)²= r²-2rx+x², 7.2 Then we have a= > a d2 b m, d2 2r+m This value of x, substituted in the second and third, gives r²+n(2r+m)=v2 12+p(2r+m)=w2; n, and Then we have, r ×{r²+n(2r+m)} × {r²+p(2r+m)} = O; (2r+·m)2 And, since the first two factors of each are squares, we have only to find ! 1 Make +n(2r+m)=(s)²=s—2sr+g*; $2 82 nm K 2(n+s)' C d2 ¿ 176 Which value of r, substituted in the latter equation, gives $2. $2 mn nm ]²+p(· +m)=w², ·2(n+s)· n+s Which, by reduction, becomes -b” I 1 1 m S $ (s² nm)² + 4ps(s+m)(n+s) 4(n+s)2 And x= J a I Or (s2nm)2+4ps(s+m)(s+n)=w2, Assume this [(s2nm)22ps]² = (s² — nm)² - 4ps(s² — nm)+4p²s², Then, this, by reduction, gives (s+m)(s+n)=nm-s²+ps; Whence, again, s= Vlaande Company 4s(s+m)(s+n) Now, if we take a=2, b=6, and c-9, then d-11, 2 b 6 с 9 121 #2 d²¯¯¯¯ 121' p r2 (s2nm)2(n+s) 2r+m 4(s²+ms)(n+s)²) (s² nm)² n: plakultak. (m+n) 2 Jan d2 9-8 121 Whence, ax=2α CO p~~(m+n) 2 A bx-6x- Patagon and 9x p= pleting 1 242 J 4418 62920 13254 62920 19881 62920 Ex. 23. Let x, rx, and r²x be the three numbers in geometrical progression; then, by the question, www.yo I ? Sporting on d² ~ 121° and a K · xrx=x(1+r)=v² rar²xrx(1+r)=w²; progr 2209 62920; " Dividing the second by the first, it is obvious that r 11 177 ㅏ ​must be a square; let, therefore, ry², and it only re- mains to find x(1+ y²)=v2. Which equation we will obtain, if we make x = n z 2 and 1+ y²=nw²; Whence, x=n(z² — w²)+ y²+1, Where all the indeterminates may be taken at pleasure. If x=w, / then, xy²+1, in which, y may be taken at pleasure. If y=2, then x=5, and r= 1; and the three numbers sought are 5, 20, and 80. Ex. 24. Let x and y represent the two numbers; then, it is required to find x+1=m² x-1=n² 20 x+y+1=r2 x-y+1=s². Now, here, it is obvious that the three squares, r², m², and s², are in arithmetical progression, their common dif- ference being y. Let us, therefore, represent these squares as in Ex. 12; viz. s²(2pq-p² - 292)2 m², — ( p²+2q² — 2pq)² · g² = (p² + 2q²)²; Then we have for their common difference y=4p³q — 12p²q²+8pq³, S and all that is required is to find this quantity plus 1 a 4p³q-12p² q²+8pq³ +1=n²; square, or Assume, therefore, in this case; n=1+4pq³, And we shall have, by squaring and cancelling the like parts, 4 p³q-12p² q²-16p²q6 ; Whence, p=4q+3q, in which expression, q may be assumed at pleasure. Q } 1 1 178 Thus the general values of x and y will be determined; viz. by first making p=4q5+39, and then x=(p²+2q²-2pq)²-1, y=(1+4 pq³)² — 1, Where, by taking q=1, we have p=7; whence, x= 1368, and y=840; which numbers answer the condi- tions of the question; for 13681872 840+1-292 1368+840+1=472 1368-840+1=232. 1 But these are not the numbers in the answer in the Introduction; they are, however, the least integral num- bers; and various others may be found by giving diffe- rent values to q. Ex. 25. Let x, y, and z be the three numbers, and s their sum; then, we have to find 202 S x²+s=0, y²+s=0, y² y² - s z²+s=0, O, 22 1 ▸ 1 22-s=0, And x+y+2=s. Now, if we assume for r and s any numbers whatever, and take afterwards a=r252, b=2rs+s², and p² J c = r²+rs+s², we shall have c² — a² + ab +-b2; where a, b, and c, will be known numbers, and will possess the following properties, viz. (c² + b²)² -± 4 abc(a+b), a complete square (c² + a²)²±4abc(a+b), a complete square [c²+(a+b)2]24abc(a+b), a complete square. Then, since a, b, and c are here known numbers, and answer the first conditions of the question, it only re- mains to fulfil the last condition * ↓ x+y+x=s.. In order to effect which, let all the three latter equa- tions be multiplied by x2, and we shall have x²(c²+b²)²-4abc(a+b) x²= x²(c²+a²)²-4abc(a+b)x² x² [c² + ( a + b) ²]²±4abc(a+b)x²=0; 1 179 And it consequently remains to find x(c²+b²)+x(c²+a²)+x[c²+(a+b)2]=4abc(a+b)x²; 3c²+2a²+26²+2ab 4abc(a+b) Whence, x= Or, putting c²+b²=m, c²+a²=n, c²+(a+b)²=p, and 4abc(a+b)=q, we shall have x= m And mx=(m+n+p), q n NX= (m+n+p), px ~(m+n+p), q Which are the three numbers sought. If r=2, and s=1; then, a=3, b=5, and c=7. Also m=c²+b²=74, n=c²+ a²=58, p=c² + (a+b)²=113, and q=4abc(a+b)=3360. Whence, x= Consequently mx= 1 p m + n + p q nx= m+n+p 9 245×74 3360 245 × 58 3360 245 X 113 =245 518 96 406 ; 96 791 96 1 px= 3360 are the numbers sought; and an indefinite number of other answers may be obtained by assuming other values for r and s. Ex. 26. Let x², y², and 2² be the required squares; then, we have to find x²² + y¹+z¹=m² Assume+y1+1= [(x²+y2)-2²]² — .2 x¹+2x²y²+y¹—2z²(x²+y²)+z8 Then, 2x2y2-2z²x²+2z²y², S 180 sure. And consequently z²= 2 Therefore ²+y² must be a square; which it will be if we assume x-p²-q2, and y=2pq; for then x²+ y²=(p²-q²)²+4p²q²=(p²+q²)²'; Hence the required squares will now be x²=(p²-q²)2 y² - 4 p² q² :( p³ — q²)² 4p²q² ; (p²+94)² Where p and q may be any numbers taken at plea- If p=2, and q=1, then x² = ( p² — q²)² = 329 y²=4p2q242-16 (p2-92)2X4p2q2_122_144 (p²+q²)2 52 25 which numbers answer the required conditions; and va- rious others may be found by giving different values to p and q. 1 Ex. 27. Here, if x2, y2, and 22, are taken to repre- sent the three squares, we have to find x²-y², or m² y2 , or n² 1 22 22= x²y 2 23 p Which, by the actual operation, becomes =r; hence z=r; and the sum of the series continued to infinity is T. zo (r-1)2 Now, for the other part, the terms after the nth are, n+1 n+2 22+3 + + rn+1 rn+2 †† +, &c. 1,n - 1 Let z ând s= p2 | 2 3 the sum required. + 3+ + J ܐ Z (r− 1)² 2 4 5 +2+$ 3 4 n+2 n+3 Ju 22 n r ( 1 − 1 ) ~ + 3/3 r - 1)². Whence, by subtracting this from the whole sum be- foré found, we have 1 n —172) { + =-=-= {21+ = 1} ቍ ----- ] P 1 1 1 1 4-6 + 8 + 10 + 1 (r- sum of ʼn terms, as required; where p: 1 1 1 1 1 +=+=+=+ 4 8 1 1 tiat, 12 14 G fit, &c.) 5' face +, &c.) = p3 } N &c. + Panganda (1 —p)r (p² — 1 ) 2 3 1. доть pn r—1 10+, &c. &c. the • 198 7 Whence z= Again, let z Then z- ALC I 4 3 + + + +, &c. 4 2.6 4.8 6.10 8.12 3 1 1 1 1 + t + +, &c. 16 2.6 4.8 6.10 8.12 The infinite sum required. } 1. 1. 1. 1 1 +=+=+=+ 6 3 And + 3 16 And therefore by subtraction. 4 4 4 4 ļ 1 Kamalakaligtas IQ A 31.1 1 2n+2 2 4. 6 1 6 +=+ 1 1 10 12 1 1 1 1 1 1 + 4 2n+2 2n+46 + 8 + + + 12 1 1 &c. 14 2n+4 Whence by subtraction, Ex. 13. Let z Then z to 1. 1 + + + +, &c. to n terms, 2.6 4.8 6.10 8.12 1 5n+3n2 10+ P 1 2n' Consequently by division, 1 1 4 4 4 4 + + + 2n+4 2.6 4.8 6.10 8.12 +, &c. to n terms'; ; 1 8(n+1) the sum of n terms as required. 1 1 1 1 +=+=+ F3 1 12 1 2n' 8(n+2)—16n²+48n+32 te 1 1 1 1 1 +해 ​3-6 And by subtraction, 3 3 3 + 3 + + 12.15 Clas 3 3.6 6.9 9.12 &c. 12 +15+, &c. +, &c. 1 199 1 Therefore, + Or And z Jl. T 1 1 Again, let z= J 1 3.8 3 Then, z 3 4. 2.3 6 6 7+; 2.7 7.12 7.1 13 Then, z 11 Jamal 1 1 1 + 3. + 3n+ 3 = 6 + 9 + 1 1 1 1 + + + 12 3.8 6.12 9.16 1 1 1 1 3 + 1 + 9 + 12 +=+=+ +, &c. . 1 1 2 1 1 1 + + f 3.6 4.9 5.12 1 1 1 1 5+ y + 12 Ex. 14. Let z=-=-= 2 + Whence, by subtraction 1 1 1 1 1 3 3(n+1)=3.2 +63 +9.4+12.5 And, dividing by 4, 1 1 1 + + 6.12 9.16 12.20 +12.20+, &c. to n terms. 1 1 1 5 2 2.7 7.12 +, &c. +, &c. . 15 1 1 + 1 12 15 3n+3 1212(n+1)=12(n+1)=s, the sum of ʼn terms as required. 1 1 1 1 1 ++++, &c. 12 17 n 1 12.20 1 3n 1 1 1 1 + + tat, &c. 12 17 22 7 And, by subtraction, 5 5 5 it 12.17 17.22+, &c. Whence, multiplying by g, we have 6 6 3 + + 12.17 17.22 +, &c. ad. inf. 5 Now, the general term of the series is 6 (5n-3)(5n+2) } 1 12.5+, &c. n terms. +, &c. 1 3n 200° 41- F And therefore, to find the sum of all the terms beyond this, we need only assume { 1 Z= Then z- 1 1 + + 5n+2' 5n+7' 5n+12 5+12 1 1 1 5 öl 1. "1 + 5n+25n+7 5n+12 5n+17 5 + 5n+2(5n+2)(5n+7)' (5n+7)(5n+12) Or, multiplying by, we have 6 6 + (5n+2)(5n+7) " (5n+7)(5n+12) 1 1.2 Mandag + &c. + And, by subtraction, 5 + 6 + &c. = 5 (5n+2) Which latter expression is the sum of all the terms af- ter the nth; consequently 3 3n + Then, z- By subtraction. 6 55(5n+2)-5n+2' Ex. 15. This series may be put under the form 13 9. + 1.2.3.4 3.4.5.6 5.6.7.8 + &c.) Let us therefore assume *** 1 1 1 + + + &c. 3.4 5.6 7.8 1 1 1 1 1 + m) smarte + + +&c. 2 3.4 5.6 7.8 9.10 18 26 10 + 1.2.3.4 3.4.5.6 5.6.7.8 + Or, by dividing by 2, 5 9 + 1.2.3.4 3.4.5.6 13 + 5.6.7.8 + &c. the sum of n terms. 1 + &c. +&c. -&c. =-1/13 12 1 201 } and, consequently, 13 + 5.6.7.8 1 1 1 1 1 STAR SEMIN inf. sum. +, &c. + 6.8 9.10 3.6 12.12 24 Again, if the above assumed series be carried beyond the nth term, it becomes ől 6 5 9. 1.2.3.4 +3.4.5.6 or- Whence, as above, 1 1 (2n+2)(2n+1)' (2n+4)(2n+3) Éx. 16. Assume II Then, z Subtracting, I 3 1 after n terms; consequently, 1 n 24 12(2n+2)(2n+1)-2(3+6n) 4(6+6n) sum of n terms. 1 pad 3 1 1 1 1 1 3-5 +7 9 + 11 J 5 00 + 1 12(2n+2)(2n+1) 8 1 1 + +, &c.) = 7 12 16 + 3.5 5.7 7.9 1 N 1 + 9 11 1 &c. Dividing by 4, 5 9.11 1 13 20 9.11 1 24' +, &c. + +, &c. +, &c. K inf. sum. 2 3 4 + 3.5 5.7 7.9 Again, if the above assumed series be continued beyond the nth term, it will be x=± 24 11.13 1 1 1 ±3+2n +5+2n+7+2n+9+2n ន 1 12 +, &c. inf. sum. ent, +, &c. f 202 * And therefore, as before, ± sum of the terms after the nth: 1 1 Consequently 12+4(3+2n) t sum of n terms requir- ed; the ambiguous sign being + when n is odd, and when n is even. را Then z Subtracting or, 3 4 6 Ex. 17. Assume z= + + + + &c. 1.2 2.3 3.4 4.5 [ 3 ૭ ૩ 3 4 2.3 3 5 6 'my 8 + + + + &c. 2 1.2.3 2.3.4 2.3.4 3.4.5 4.5.6 the infinite sum. Again, if the above assumed series be carried beyond the nth term, it will be n+5 f n+3 n+4 (n+1)(n+2)' (n+2)(n+3) (n+3)(n+4) Whence, as before, the first term n+3 (n+1)(n+2) will be the infinite sum of the terms past the nth. Therefore 2 $ W NI W 2 p 3 210 1 4(3+2n) af n+ln+2 + n+3 2 (n+1)(n+2) I' 喜 ​will be the infinite` 5 6 7 + ++ 3.4 4.5 5.6 1 10 } $ + the sum of n terms. + &c. 1 My در : 203 h 1 Ex. 7. $ Ex. 8. Ex. 9. Ex. 10. Ex. 11. Ex. 12. Į LOGARITHMS. MULTIPLICATION BY LOGARITHMS. Log, of log. of Prod. 117.1347 23.14 1.3643634 5.062 0.7043221 2.0686855 0.6102661 0.9931363 1.6034024 = 2.6974525 = 1.6153892 4.3128417 0.6049541 Log. of log, of Prod. 40.12383 Log. of 498.256 log. of 41.2467 Prod. 20551.41 4.0763 9.8432 Log. of 4.26747 log. of .012345-2.0914911 Prod. .04971016 Log. of 3.12567 log. of .02868 log. of .12379 Prod. .01109705 1 Log. of log. of log. of log. of .0031598 Prod. .0958299 2876.9 .10674 .098762 .098762 -2.6964452 0.4949431. -2.4575791 -1.0926856 -2.0452078 3.4589248 -1.0283272 -2.9945899 -3.4996596 -2.9815015 * 1 { ! } DIVISION BY LOGARITHMS. Log. of 125 log. of 1728 Quot. .0723379 Ex. 8. Log. of 1728.95 log. of 1.10678 Quot. 1562.144 Ex. 7. Ex. 9. Log. of 10.23674 log. of 4.96523 Quot. 2.061685 Ex. 10. Log. of 19956.7 = log. of .048235 Quot. 413719 $ Ex. 11. Log. of .067859 log. of 1234.59 Quot. .0000549648 4 1 204 M RULE OF THREE BY LOGARITHMS. Ex. 5. Comp. log. of 12.678 log. of 14.065 log. of 100.979 Ans. 112.0263 1 2.0969100 3.2375437 2.8593663 3.2377825 0.0440613 3.1937212 1.0101617 0.6959393 0.3142224 4.3000888 -2.6833623 5.6167265 -2.8316075 3.0915228 -5.7400847 J Ex. 6. Comp. log. of 1.9864 log. of .4678 log. of 50.4567 Ans. 11.88262 kad J 8.8969493 1.1481397 2.0042311 2.0493201 9.7019333 -1.6700602 1.7029274 1.0749209 1 1 1 Y 205 Comp. log. of .09658 11.0151127 log. of 24958= log. of .008967 — — 1.3972098 3.9526472 Ans. .02317234 2.3649697 Ex. 8. Comp. log. of .498621 log. of 2.9587= log. of 2.9587= Ex. 7. W 1 } 3d prop. 17.55623 }, 3d prop. .8240212 11 palmstada j Whence, 17.55623, the answer required. Comp. log. of 12.796-8.8929258 log. of 3.24718=0.5115063 log, of 3.24718=0.5115063 -10.3022294 0.4711009 0.4711009 1.2444312 --1.9159384 INVOLUTION BY LOGARITHMS. Ans. .00003432594 8 2 Ex. 5. Log. of 6.05987=0.7824633 Ans. 36.72203 Ex. 6. Log. of .176546——1.2468579 2 1.5649266 3 Ans. .005502674 Ex. 7. Log. .076543-2.8839055 3.7405737 4. 5.5356220 ✓ { 1 Ex. 8. Ex. 9. Ex. 10. } 206 Log. of 2.97643-0.4736957 Ex. 11. } Ans. 233.6031 Log. of 21.0576=1.3234089. Ans. 87187340 7.9404534 Log. of 1.09684 0.0401432 17 Ans. .9918624 # A Ans. 1.909864 0.2810024 EVOLUTION BY LOGARITHMS. Ex. 7. Log. of 365.5674 2)2.5629674 Ans. 19.11981 1.2814837 ↓ } 5 Ex. 8. Log. of 2.987635 3)0.4753276 Ans. 1.440265 0.1584425 Log, of 21- log. of 373- D 2.3684785 Ex. 9. Log. of .967845 4)-1.9858059 A -1.9964515 6 Ex. 10. Log. of .098674 7)-2.9942027 Ans. .7183146 Pala 1 1.8563147 1.3222193 2.5717088 -2.7505105 Multiplying by Dividing by 3)-3.5010210 Ans. .146895-1.1670070 > 2 1 2 ر 207 } Ex. 12. Log. of 112: log. of 1727- } Ex. 1. -2.8119257 Which, being multiplied by 3, and then divided by 5, gives log. 1.2871554, Ans. .1937115. 1 Ex. 3. 1 MISCELLANEOUS EXAMPLES. Log. of 2= 0.3010300 log. of 123= 2.0809051 2)-2.2111249 Ans. .1275153 -1.1055624 Ex. 2. 0-log. 3.14159 3)-1.5028505 .6827842 1.8342835 Ex. 4. ? t Jermantogl G Sapp 17층 ​Hence 0- log. 6 2.0492180 3.2372923 Log. .005633.7505084 .07=100, theref. Log. of Ans. .04279825 Ungg 32 100)—16.2535588 Ans. .6958321 The student will observe, that 84 is borrowed, in this example, to make the 16 up to 100, according to the rule. Hy 1 8. ({})² - (?) = ( - ) + 2 = ( )¹³× - - ; 9.1 52 3 6 X X 16 52 3 6)-1.2218487`· -1.8425355 -1.8703081 -2.7611180 -2.6314261 w ť 1 1 Ex. 5. Log. of log. of log. of .012-2.0791812 -1.9345684 7 log. Ans. .001165713 -3.0665916 Ex. 6. Log. of 1.0457574 J -1.8595867 -2.4771213 0.3939479 -3.7764133 0.8653014 log. of 22 log. 12 = log. .19 = -1.2787536 0.3621199 log. 171 0.3084076 · 208 ¡, HM · Ans. .0009158636 Ex. 7. Log. of & =-1.1549020 -1.8979400 T 21 log. of 11 log. .03 log. .15 = From Subtract p log. 19- p 7 Log. 4 log. 106- 3 Befo S 0.8145825 3.7764133 0.8145825 -4.9618308 -1.2208187 0.6393768 Nat. numb. .7247622-1.8601955 ( -1.7569620 0.5160615 Nat. numb. 1.875096-0.2730235 Log. of 2.5998582: log. of 127 4 0.4149496 1.5017437 1.9166933 In the same manner we find log. (147-283)=0.2230796 Ans. 49.38712 1.6936137 A 1 1 1 OC 12 MISCELLANEOUS QUESTIONS. Ex. 1. Let x be the number of minutes after 8, or the number the minute hand is before it overtakes the hour hand, after the former is at 12, and the latter at 8. Then, will be the number of minutes that the hour hand has advanced in the same time. 11 1 1 40+ 12 p -00 + or 209 x=40, or x= 11 201 viz. the time was 8h. 43m. 38 sec. 11 Ex. 2. Let x represent the digit in the place of tens, and y that in the units; then will 10x+y= the number itself, and 10y+x the number formed by the inverted di- gits. Hence, by the question, ¢°~j°=10c+? 10x+y+36=10y+x S From the latter, we have 9x-9y——36, or x-y= 4, or y=x+4; whence, it appears, that y is greater than x, and our first equation becomes y2-x210x+y; Or, by substituting the above value of y, we have (x+4)²x²-10x+x+4, P 12 } And, by the question, 40 X 12 77 -43- 11 That is, 8x+16=10x+x+4, or 3x=12, or x=4, and y=x+4=8; Hence, 10x+y=10x4+8=48, the number sought. Ex. 3. Let x and y represent the two numbers; then, by the question, 3x-3y=2x+2y) 5x+5y=3xy xy: x+y:: 2:3) }, or x+y: xy :: 3 5 S From the former of these, x=5y, which, substituted for x in the latter, gives 25y+5y=15y2, or 30y=15y²; Whence, dividing by 15y, we have y=2, and, conse- quently, x=5y=10. Ex. 4. Let.x be the number of games won, and y the number lost; then, by the question, x+y=201 2x-3y= 5 1 } 210 I ↓ Multiply the first by 3, and we have 3x+3y=60 Add the latter, 2x-3y= 5 and we have 5x =65, or x=13 games won, And, consequently, 20-x-20-13-7 games lost. Ex. 5. Let x be the number of yards in the four sides; then 3x is the number of feet; and, therefore, by the question, 3x-150 is the number of pallisades. As is also x+70; and, consequently, 3x-150=x+70, or 2x=220, or x=110; Therefore, 3x-150=x+70=180, the number sought. Ex. 6. Let x be the number of hours in which в will 1 fill it; then, will be the quantity thrown in by в in an hour, and 1 volgd. 20 have aptop OC is the quantity a throws in. Also, since the two together will fill it in 12 hours, will be the quantity the two throw in in an hour. 1. 60 3x 50 I Whence, - +20-12' or Cf. X 60x 60x 60x 5x-3x-60; therefore, 2x=60, or x= 30. Ans. Ex. 7. This is not properly an algebraical question; • but the best method of solution is as follows: and Dividing each effect by the time it is produced in, we a b è f a ¿ and therefore ++ M or G 1 12 C C O & for the momentary effect of each agent, g 1 the momentary effect of the three > b C agents; consequently, d÷(++)will be the time in a e 2 which all three will produce the effect d. Ex. 8. Let x be the required number; then, by the question, (x+3)(x+19): : (x+19): (x+51). 1 1 } K 211 Or, since, by the nature of geometrical progression, the product of the means and extremes are equal to each other, we have (x+3)(x+5)=(x+19)², > or x²+54x+153x²+38x+361, or 54x-38x=208, or 16x=208, or x=13. Ans. Ex. 9. Let r be the ratio; then, since 3 is the first term, 3:3r :: 372: 24, are the first four proportionals, and 3r and 3r² the two means sought. Hence, since the product of the extremes and means are equal, we have A 9r³—72, or r³—8, or r=2; and, therefore, 6 and 12 are the means required. In the same way, in the second part, we have 3: 3r :: 3r²: 3r3 :: 3r¹: 96, Whence also we have 3rX 3¹=3 × 96, or 975-288, or r532, or r=2, Therefore, 6, 12, 24, and 48 are the four mean propor- tionals. Ex. 10. Let x, rx, r²x, r³x, rx, rx, be the six pro- portionals. Then, by the question, x+rx+r²x+r³x+r¹x+r³x=315 +12=165 2-6-1 7 1 1 gud. the and x By subtraction, rx+r²x+r³x+r¹x=150 Now, by the rules for the geometrical progression, our equation may be written Stan x=315 and the latter rx-150 r-l By dividing the former of these by the latter, 26-1 315+ 21r. 10 24-1 150 Magdal I ++ Reducing the latter, by dividing both terms by im 1, we have 1 212 ! 1 } 21 p² + 1 10 21r or +7²+1=217 (1²+1) " 10 And by adding r² to both sides 21r god+22+1= ·(r²+1)+r², 10 or (r²+1)² — 21 1" (r²+1)=r² 10 Hence, by completing the square, 21r 441,3 = r²+ 10 400 (x²+1)² Or (r²+1)2 ·(y²+1)+ And, by extracting the root, (r2+1)- 25 ·(2² + 1) + 21r 10 or r2. Let x and -18 # J' 10 1 25 625 25 15 Or r ·1)= 土 ​20 20 20 ±√( 400 But from our first reduced equation we have x + y = a 1 -b + Vi y 4412 400 8412 441,2 400 400 -1 315(r-1) 315 40 = 5; r6 — 1 63 Therefore, 5, 10, 20, 40, 80, and 160, are the propor- tionals sought. Ex. 11. the question, 21 29r nga 20 20 Y be the two numbers; then, by prod 2; From the 2d. x+y=bxy=a, or xy: Squaring the 1st, x²+2xy+y2=a² Subtract 4xy. a, 4a = b 1 } D 1 213 t We have x²-2xy + y²=a² 1 Or *~y=√(a²b—4a b Repeating again x+y=a, the first question, we have, by 1 1 a addition, 2x=a+√(a²b—4α), ), or x= a+ √ (ab—4) 2 z z a2b-4a And by subtraction, 2y=a~√( 1 1 or y=√(ab-4) 2 Whence, 24x16x+256, or x= K Ex. 12. Let x be the number of men employed at first; then, x+16 will be the number in the second in- stance; and since the time in performing the work is re- ciprocally as the number of men employed, we have x: 16x+16: 24 256 8 Kateg 32, the number of men at first, and 32+16=48, the number during the second part of the time. Hence, 32 × 24 × 1=1152s. -£115 4s. Ans. 48×16×13—1152s. Ex. 13. Let x and y be the two numbers; then, by the question, we have x²+y= 62 § y²+x=176} From the first, y=62-x²; and, consequently, y²=622-124x²+x4 This substituted in the second, gives 622124x²+x¹+x=176 or a¹-124x²-176-3844-x or at-124x2x3668. 4a Vid b. 1 Whenco, will be found a 7, and, consequently, y=62 x²=13; but the question cannot, I believe, be in any T 214 manner reduced to a quadratic form, at least while it is considered under the general form 2 x²+ya, and y²+x=b. 1 Ex. 14. Let a denote the number of feet in the cir- cumference of the less wheel, and y the circumference of the greater wheel. 360. Then will be the number of the revolutions of the ་ OC less wheel, and greater. the number of revolutions of the 360 360 Y 360 360 x+3y+3 360(y-x)=6xy From the first, From the second, 360(y-x)=4(x+3)(y+3) The latter of which, by multiplication, gives 360y-360x-4xy+12x+12y+36 360 У 1 Whence by the question or 348y-372x=4xy+36 } And the former, by division, 60xy-60x=xy XC And we have 60x -6 Mult. this by 4, 240y-240x=4xy Subtract 348y-372x-4xy+36 108y+132x=-36 And we have Maka Data 1087-36 9y—3 Whence x= 123 11 Substitute this value of x in the equation 60y-60x=xy, 4. 660y-540y+180-9y2-3y, or 9y²-123y=180, or 540y-180 9y2-3y 11 11 , or 1 1 215 41 2 y² — — — y 3 1681 36 41 y = = = √( ย 6 1 f and, consequently, 9y-3 11. 11 cumference of the greater wheel ference of the less 12. } y=20; whence, +20)===== 84 10.742 41 49 90 6 6 6 Ex. 15. Here, by the question, x + y = a mx+ny = bS Multiply the first by m, and we have mx+my=ma Subtract mx+ny= b 842 10.743 843 135-3 Whence, x= And we have (m—n)y=ma—b, or y= Again, multiply the first by n, and we obtain nx + ny = na Subtract mx+ny= b And we have x(n—m)=na—b, or x(m—n)—b—na. na 744 Kansan yake 843 m N Ex. 16. Here, the several portions of wine drawn off were 10 gals.. remains 74 10.74 b 12; that is, the cir- 15, and the circum- =15 remains 74 remains remains 'ma-b m-n N 84 842 842 744 743 10.743 842 843 843 Therefore, =50.569 gallons remaining, • 10.74 742 84 84 742 10.742 742 ་་ } 1 + 1 | 216 Ex. 17. Let x represent the number of persons, and y the number of pounds each received, then xy is the whole sum divided. Now by the question (x-3)X(y+150)=xy or " (x+6)x(y-120)=xy S xy-3y+150x-450=xy xy+6y-120x-720=xy' 150x-3y=450 -120x+6y=720 S or Multiply the first equation by 2, and we have 300x-6y=900 M "' Add the 2d, -120x+6y=720 Whence 180x=1620, or x=9, the number of persons; 300x-900 And, consequently y=- 300, the sum each 6 received; and 9 × 300-27007. the whole sum divided. Ex. 18. Since the reduced value of the 16 pieces is 81. 8s. and the part taken from them is 16 × 2s. 6d. 21. the ori- 208 =13s. 16 ginal value was 107. 8s. or 208s; consequently the original value of each. B St ľ Then by the question x+y=505, and 11+5 00 Y 41 Ex. 19. Let x be the number of ounces of tin, and y the number of ounces. of copper. We have 17x+17y=505 × 17; Subtracting, gives 4x-20×17; 1 100. The second equation cleared of fractions, is 21x+17y =525×17, and the first multiplied by 17, Therefore x=5×17-85, the ounces of tin; and con- sequently 505-85-420 ounces of copper. 217 Ex. 20. Since the privateer gains 2 miles per hour on her prize, and the latter is 18 miles a-head, 18 9 hours 2 the time before she is overtaken; and consequently 9×8 -72, the distance run. Ex. 21. Let a represent the number of 7s. pieces, and the number of dollars. Y Then by the question 7x+4y-2000 shillings, or 14x+9 y=4000; 4000-14x Consequently y= whole number, 9 { } ĭ Let now Ássume Or y=444-x+ 4-5x 9 4--9p 5 Consequently a= 00 285 9 X 4-5x 9 S =1-2p 1—P=q, and we have p=−5y+1; 5 459-5 5 KATA wh. p; then 9p-4-5x, or 1-P: 5 taken at pleasure. If we take q=1, 2, 3, 4, 5, &c. x=8, 17, 26, 35, 44, &c. But the greatest value of x cannot exceed -9q-1, where q may be * 4000-9 14 therefore 31, the number of different ways. Otherwise. By the rule before given, we find from the equation 14p-9q=1, p=2, and q=3; 2×4000 3×4000 Whence 31, the number of diffe- 9 14 rent ways; the same as before. T 2 { =285, | 218 Ꮳ Ex. 22. Let x and y represent the two numbers; then by the question, " { x + y 2=ad x⁹+y⁹=32=b $ Assume xy=p; then x³+y³ a³-3ap xº + y² = (a³ — 3 ap)³-3p³ (a³—3ap), Or a9-9a¹p+27a5p²-30a³p³+9ap¹=b; pt. Spandekag Or in numbers, by transposing, &c. 40 plakaof — — 3 p³+48p² ~64p+ =0, 80 3 Which equation is resolvable into the factors 28 20 (p² - 4p+4)(p²- x + 1 ) = 0 ; Whence by the solution of these two quadratics we have 14 √136 √136 p=2, p=-2, p= + and p: 3 14 ; 3 3 3 But x+y=2, and xy=p; Whence x-y=±√(4-4p)=2√(1−p): And substituting here the above values of p, we have the following solutions, viz. x=1+/-1, and y-1-1-1 136 x=1+ √ { = 11 =1+√ 3 x = 1 + √ / { y = 1 - √ { = 1} 3 1 -11 MĒ p 11, √136 = 1 + 3 3. and y=1—√{—11+136}=1—§√(6√/34—33), The latter two of which are the only real answers ; the others being imaginary. 1 }, and 3 √136 3 =1+3√(6√/34-33) ན 1 + 219 } 事 ​Ex. 23. Put a=666, and b-406;, then, from the second equation, b23 y= XC and this value of y being substituted in the first equation, gives Co b³-36²x³ +3bx¤ —œ⁹ 003 -b+x³ = a; and multiplying by a³, and transposing, we have x² - (3b+1)x+(36²+b+a) x³-b3; or, in numbers, x9—1219y+937470x³=4063; from whence, by the resolution of a cubic equation, we have x³-343; therefore b-x3 406-343 63 7 17 9, the answer. OC x=7, and y=- Ex. 24. Let x and y be the two numbers; then The geometrical mean = √xy The arithmetical mean =x+y 2xy The harmonical mean x+y Therefore by the question = x + 3y = √xy=13 2xy x+y Villamart √xy- 12; From the first of which equations we have x+y=26+2√xy, And, from the 2nd, x+y= 2xy. √xy-12 Lxy Consequently 26+2 √xy = √xy-12 • Teleportat Į Whence 26√xy+2xy-312-24 √xy=2xy; ** Or 2xy=312, or √xy=156, and xy=24336: Substituting the value of √xy in the first equation, and repeating our last, we have ง 220 } x+y=338 xy=24336, And by squaring the first of these, x²+2xy + y² = 114244 4xy = 97344 By subtraction, x²-2xy+y²= 16900 By extraction, Also, ∞ y=130 x+y=338; Whence by addition 2x=468, or a=234, And by subtraction, 2y=208, or y=104. Ex. 25. Here a³y +y³x 32 = x¹y²+ y²x²=7 $ 4 By squaring the first equation, we have x®y²+2x¹y¹+y®x² = 9 +yox²=7 6 Subt. the 2d, xy2 Saphal 2x¹y=2, or xy=1. Hence, dividing the first by xy, and the second by x2y2, we have x²+ y²=3 x²+y¹=7; 1 From double the last Subt. the square of the first And we shall have 2x+2y=14 +2x²y²+y+= 9 - 5; x¹ — 2x²y²+y¹: Or by extracting 2-y²√5, And by repeating x²+y23; Therefore by addition and subtraction, 3 1 3 1 20² = 2 + √5, y² = 2 - √5; y? Whence, extracting these by the rule for binomial surds, we have x=(5+1), and y(5-1). } Ex. 26. Here the equations are x + x + x = 23 xy+xz+yz=167 to find x, y, and z. xyz=385 { } 釁 ​221 7 From what has been said in the Introduction, relative to the doctrine of equations, it is obvious that these num- bers are the co-efficients of a cubic equation which has its three roots equal to the several values of x, y, and z z; whence we have at once x³-23x²+167x-385-0. The three roots of which equation, by the rules for cu- bics, are found to be 5, 7, and 11, the numbers sought. Ex. 27. Here the given equations are xyz 231=a xyw: 420=b xzw= 660d, yzw=1540=c; Multiplying these into each other, we have x³y³w³z2 abcd, 3 or x y w z = abcd; Whence, dividing this last equation by each of the given equations, we have 3 14 } } W y abcd a abcd OC' b 31 a b c d d Vabcd J -20 :11 = 7 3. C Ex. 28. Here the equations are x+yz=384=a y+xz=237=b z+xy=192=c. } } Ĭ From the first x-a-yz, which substituted in the second and third, gives y+az—y z² = b z+ay—y²z —e; Also from the first of these two equations b- az az b Y 1. z2 22-1. But yo 222 } } 7 Whence, substituting this value of y in the latter, we have a²z-ab z(az-b)2 z+ 1 I 2 or z(z²—1)²+a(az —b)(x²-1) —z (az —b)²=c(2² — 1)², Which by multiplying and involving the several fac- tors, becomes 22-1 (≈² — 1)². 2 z5 — cz¹—2z³+(2c+ab)z²—(b²+a²—1)z=c—ab, or, in numbers, 25—192.24—22³+9124522-203621z— — 90669. An equation of the 5th degree, the integral root of which is z=22; whence az-b y= z² — 1 Ex. 29. Here we have equations x²+xy=108 a y²+yz= 69=b x²+zx=580=c; Assume x=my, and z=ny, and these will become y²(m² + m)=( y²(1 + n)=b y²(n² + nm) = c; b Whence y2- # N Pa 1 ~ a m²+m a(1+n) =b(m²+m) a(n²+mn) =c(m²+m), From the first of these we shall have 8 C, 17, and x=α-yz=10, с 1+n n²+mn b(m²+m)-a ; a Which value of n, substituted in the second equation, { b(m²+m)—a} × {bm² +(b+ a)m-a} gives =c(m²+m), j 1 or a And by reduction, b²m¹+(ab+2b²)m³ + (b² — ab~ac)m²—a(a+2b+c)m+ a²=0: 1 1 ! 223 l Or in numbers, 65331 m². 4761 From which is obtained m=3; 82 m² + m³ 83 1 Qualidad 2 b(m²+m)-a 20 Consequently n= a 3 3, x=my 9, and z=ny 20, as required. Ex. 30. Given x²+x y +y²: પ x²+xy + y²= 5 36 x²+ y²= 5 x¹+x²y²+y¹11 S Here, dividing the latter by the former, we have 11 x²-xy+ y²=5 89208 11664 4761 4761 And By addition By subtraction Also, by adding and subtracting double the latter from the former, we have x²+2xy + y²= Consequently x= x²-2xy + y²== And y 10 14 xy=10 36 28 64 +: 10 10 10 36 28 8 10 10 Whence x+y=√10' x - y = √ TO 10 8 1 64 1 Voot 210 m= y = √( 1 64 1 ✓ 2 10 2 1 1 10 64 2 Or by reduction, = /3/ √ √10+ √5, 1 ; 8 10' a m² + m J / } 224 f 2 1 y=V10 √10~ √5, Which are the values of x and y, as required. Atta Ex. 31. Here the given equation 46x3+13x²- 12x-5, may be put under the form (x23x)2+4(x². 3x) -5, (by Note, page 134, Book,) which being now a quadratic, we find x²-3x-2√9-1, and this is again a quadratic, from which we derive x=√13, the answer. 3d a OC Ex. 32. Let a represent the number of people, and the first year's increase; then at the end of each follow- ing year the numbers will be 1st year, a+ 2d 4th 5th a JE ام ac a a x2 a+2-+ 00 a a a+3 +3 + 002 OC a 00 a a+4+6+4+ 202 a 003 ❤ a+5+10+10+5 + α a 002 004 α a +1002, or OC 201 003 a I +p+, &c. n M 00 a 100th a + m² + n + x100' OC 2099 203 Where m, n, p, &c. are the coefficients of the binomial (1+1)100. Which expression may therefore be written 00 a STWY 006 a a(1+1)100, which by the question is to be equal to 2a; whence 1 1 -100/ 2-1; 賣 ​P 1 225* 1 That is, .00695, or x= =144, nearly. XX .00695 The annual increase must therefore be 11th part of the population. 144 Ex. 33. It is obvious that the least number of weights that can be used to weigh 37b. is two, viz. 1lb. and 37b., and if to these we add a 91b. we shall be able to weigh all the weights, 9±1, 9±2, 9±3, 94, as far as 13lb. Increasing again our weights by 3×976.-2716. we shall be able to weigh 271,272, 27+3, &c. 2713, that is, to 40lb.; and in the same manner, by the addi- tion of three times the last weight, viz. 81, we can weigh 811, 812, 8113, 814, &c. 8140. Therefore 1, 3, 9, 27, 81, &c. are the weights re- quired.* 2 Ex. 34. Let a, b, x, y, be four numbers, such, that a² +b²+x²-y². Assume y=x+1; then a²+b²+x² = x² +202 +2x+1, or a²+62-2x+1; hence x=(a²+b² — 1), where a and b may be taken at pleasure, provided the one be an even, and the other an odd number. If a=2, and b=3, then x=6, and y=7; hence 2, 3, 6, 7, are the least whole numbers that will satisfy the conditions of the question. If a=3, and b=4, then x=12, and y=13, the num- bers in the book. Ex. 35. Let x be the number sought; then by the x-5 x 4 x- 3 x-2 1 X M > 3 2 1.00 question 6 whole numbers. x-5 J 5. Make =p; 6 second, and we have $ > and then x=6p+5; substitute this in the 1.3 > are to be all *The most general method of solving questions of this kind is by moans of the ternary scale of notation.-See Barlow's Theory of Num- bers, chap. 10. U 1 f 226 A 1 1 T 1 1 ↑ 6p +1 5 - ~p +P + 1 =p 5 f =wh, or consequently a=30g-1. Substitute this in the 3d, and we have 30q-4 =wh, or 7q-1+21=wh 4 comes pią matkad Let Whence 9=2r; consequently x=60r-1. This, substituted in the 4th and 5th equations, gives whole numbers; therefore the general value of x= -60r —1, and if r=1; whence x= whence x=59, the least value sought. Ex. 36. Let a be the year required; then if x+9, x x+1, and x+3, be divided respectively by 28, 19, and 15, the remainders will be the cycles of the sun, the golden number, and the Roman indiction. Hence x+9-18 x+1-8· x+3-10 19 28 15 9p+2 19 Consequently Whence p p+1 1 x-7 15 00 .9 x -7 X-7 Or, 28 19 15 must all be whole numbers. x-9 Let -p, then x=28p+9; 28 and substituting this value in the second equation, it be 28p+2 wh; whence 28p+2 9p-+-2 -=p+ wh 19 19 19 =29+ જે A q, or p=5q-1; and and and " 19q-2 9 q; then p 9-2 9 199-2 9 Again, by substitution, 532r+114 =35r+7+ 15 =r, or q=9r+2; } X 2 9 AL 19r+4, and x=532r+121. 2r+9 15 wh. 227 # Or 2r+9 15 Let 5x+1 87 2 s; whence r S 1 Garde Assumé t; then s-2t+1, 2 Je where t might be taken at pleasure; but as the least va- lue is required, let t=0, then s=1, and r=3; and con- sequently x=532r+121-1596+121-1717, the year required. Ex. 37. Given the equation 256x-67y=1, to find the least values of x and y, in whole numbers. By transposition and division, y= =wh. } 256x-1 87 =p; then x= 2p-1 5 =wh=r; Also yz 2 155-9 2 3x Y'z'· 5x+1 87. 87p-1 5 -75-4+ =q; then p= =17p+ : 59+1 2p-1 5 ✓ Assume O x 2 + 1 2 where may be taken at pleasure; if r=1, then q=1, 87p-1 and p=3; whence x-- 52, and y=153, the 5 least numbers that answer the conditions of the equation. Z Ex. 38. Let x be one of the equal sides, y the base, and the perpendicular of the first triangle; and x', y', and z', the corresponding lines in the second triangle. Then 2x+y is the perimeter of the first; and 2x+y the perimeter of the second. the area of the first, s-1 = 2 =wh, 2 r; whence q=2r-1, and the area of the second; Betonarvon §. V. 2 Whence, by the question, we must have 2x+y=2x+y' and yz= y'z' : wh. { 228 ๆ 1 Also 2 Also a²=2*+s², and a²="+s²; y'2 x2 4 4 Y Assume x=r²+s², and g2 — §2, 2 yz 2 Then z= = (x² And r2; Again, assuming ar²+p², and r²-p², we have 2 z"=2rp; and the perimeter of each will be 4r2. yz 2rs(r2-s2), and =2rp(r2-p²); 2 s³ — p³ > S d-s 2/2) + == 32 Whence 2rs(r²—s²)=2rp(r² —p²), s(r² = s²) = p(r² —p²), or =s²+sp+p²; Therefore it remains to find =2rs. میرم s²+sp+p²=; Which latter condition has place, if we assume s=m²-n², and p=n²+2mn, t In which case r—m²+mn+n²; x =29, y =40 | x' =37, y' =24 } 17] ། Whence x= r²+s²= (m²+mn+n²)² + (m²-n²)² y= r²s² = 2(m²+mn+n²)² -2(m² —n²)² x = p²_p² (m²+mn+n²)² + (n² + 2mn)² y' = r² = p² = 2(m²+mn+n²)2—2( n²+2mn)²,, Where m and n may be taken at pleasure. If m-2 and n-1, then x=58, y=80, x'=74, and y' 48, or since all these numbers are divisible by 2, we have as required. Ex. 39. Let x, y, and z represent the base, perpen- dicular, and hypothenuse of the first triangle. *སྐ! 229 1 1 x', y', and z', those of the second, and x", y, and z", those of the third; x²+y 2 then, we have to find x' 2+ y²²=z/2 2 Z x" 2 + y2=22! Also xy=x'y'=x'"y". And, in order to fulfil the three first conditions, Let x=m²-n², and y=2mn ! x'm' -n'², and y'=2m'n' x"=m"2-n'2, and y"=2m"n"; 112 —n"², Then it remains to find J (m² —n²) × 2mn = (m¹² — n'²) × 2m'n' (m² - n'²) × 2m'n' = (m'2-n"²)×2m"n", Which equations may be resolved into the factors (m+n)(mn)=(m trí) (m) n ጎ m n ' (m'"+n'') (m'' — n'')m''n'' =(m'+n')(m' —n')m'n', Where it is only necessary so to equate the factors of each of these equations, that the reduction of them may give the same ratio to two of the quantities. For which purpose, let m+n=2n', and m=m', then, m―n=2(m' —n'), and n=2n' —m' ; But, since the product of the preceding factors are equal, we must have ! P 2n' m' = or 7n'=5m'; Again, in the second equation, we may assume m"+n"=3n' n" —m' 1 m'+n' 4 and 2(m'"'—n")=m'" —n' Then, ¿m" — 3n'—m' 2 } · 1 Where again, because the product of the factors are equal, we must have U 2 230 3n'—m' m' +n' 2 3 the same ratio as before. Assuming, therefore, m'=7, and n'=5, we have m=7, and n=3; m"-8, and n'—'7 ; Whence, x—m²—n²=40; y=2mn=42, x'm'"-n'²=24; y'=2m'n'=70 x''2—m"¹²—n'2—15; y"-2m"n"=112. ..z=58, z'=74, z=113. · $ 1 palak Ex. 40. Given y 1.2655, to find x. Here, a few trials show that x is between 1.3 and 1.4. Where, if x=1.3, then, log. 1.3 =.087649 1.3 And if x=1.4, then, or 5m'—'7n', · log. 1.2655.102262 Error-.014613 } log. 1.4 1.4 ད་ Many J Hence, .01665: 1 :: .014613: .0884 Therefore, 1.3+.08841.3884. Answer. .104307 Ex. 41. This equation is better put under the form yo .6, log. 1.2655.102262 Error+.002115 Octy Or, log. y Xxlog. xxy=log. .6-1.77815. Which, by a few trials, shows us that x is between 4 and 5, and y between 5 and 6; let us, therefore, take one of those quantities, y, as constant, and correct the other, æ. Here, then, assuming y=5.5, let us take x=4.6 and 4.7; then, 1 B L ! 231 +1 1 } : L First, if x=4.6 : log. 5.5 x 4.6 3.405665 log. 4.6 x 5.5=3.645158 log. .6 Error Secondly, if x=4.7, then 1 ► • log. .6 -1.760507 1.778151 .017644 Drag log. 5.5×4.7-3.479704 log. 4.7×5.5=3.696538, H 1 -1.783166 -1.778151 A Madag Error + .005015 Whence .022659 :'.1 :: .017644 : .09; Therefore x=4.69 nearly. By a similar process we may find y=5.51 nearly; and repeating the operations again on x and y, we have x=4.691445, and y=5.510132. Ex. 42. After a few trials we find x-4 nearly, and y a little less than 3. 1 } Let us therefore assume x4.01; then since 4.01 4.01 =262.18, we have y' =22.82. In this exponential equation we may proceed according to the approximating rule given at page 181, from which we determine y to be between 2.8, and 2.9; and by em- ploying these values of y in the second equation, another approximation is found for x, viz. x=4.0166. And employing this again in the same manner, we have y=2.8257; with which two values of x and y, repeating the same operations, we find x=4.016698, and y= 2.825710, as required. Ex. 43. Here, calling 2x the number sought, we have to find 2x+1=0, and x+1=0. Assume x-m²-2m; then x+1=(m-1)2 a square, as N } 1 2 232 J required; and therefore it remains to find 2m²-4m+1 a square. Assume 1-4m+2m²=(rm-1)2 2r-4 Then, 2m²-4m-r2m² - 2rm, or m= r² - 2 Or, since r may be a fraction, let r, q q may be taken at Then, m= pleasure, provided m be integral. 2pq-4q² p² -2q² If we take p 4, and q=3, then m=6, and 2x=48, p=7, and q=5, then m=30, and 2x-1680, the numbers sought; and various others may be found by giving different values to p and q. ; where p and 虐 ​Ex. 44. Here x and y being taken to denote the num- bers sought, we have to find. x² + y² x³+y³=0 - Assume x2+y2-(ry-x)2; then we have x²+y²(ry—x)² x²+ y²—r²y² — 2ryx+x², or y=r²y-2rx, (x² - 1)y 2r or x= Whence, x³+y³ (23—1)³y³ 8p3 which is to be a square, or (2-1)38r3 2r +y³ -y=0=s². 2rs2 And consequently, y= (p² — 1 )³ +8µ³ ³ where r and s may be taken at pleasure. Matang If r=2, then y=91' and x= ; so that, taking s= 4s2 3s2 91 91, we have y=364, and x=273, the numbers required. 1 f 233 ~ Ex. 45. Let x² and y² be the numbers sought, then, the latter two conditions will be fulfilled, and it will only remain to find x²+y² a cube. For which purpose, let x=rz, let r² z² + s ² z² = a cube. Assume r22+s²²= 23 Then, zv³(r²+s²) z=rv³(r²+s²) y=sv³ (r²+s²) Where r, s,' and v may be assumed at pleasure. If r=2, s=1, and v-1, then x=10, and y=5; conse- quently, a²=100, and y2-25, are the numbers sought. And, by giving different values to r, s, and v, an indefi- nite number of other answers may be found. Ex. 46. This is the same question, except a little va- riation in the enunciation, as Ex. 26 of the Diophantine Problems, where we found generally بایرام C 23 (p² — q²) x=(p²;—q²), y=2pq, z=2p ¶ [{ p²+q²) 202 y¹ — x²y² x2 y=sz; then, Hence, if each of these be multiplied by (p²+q²), p¹… q*, 2pq(p²+q²), and 2pq(p²-q²) will be the integral roots in the present question; p and q being taken = any un- equal numbers whatever. If r=2, and s=1; then, 154, 124, and 20¹ are the bi- quadrates required. Ex. 47. Let ax², ay², and -represent the three num- ay¹ 002 bers in geometrical progression; then, by the question,` (y²x²) a Y¹ — x4. -)a=0 ·) a = 1 1 a 1 y² — x² 卓 ​234 Of y² x2 am², and y¹-x¹—an²; n2 Whence, y²+x²=; =n/2. m² Assume, therefore, y=p2-q2, and x=2pq; where p and q may be taken at pleasure. If p=2, and q=1, then y=3, and x=4; and the numbers are 81a 16a, 9a, and padd 16' or 256a, 144a, and 81a, where m² may be any square factor a being m² whatever of y²— x². In the present instance, let m=1, then a=7, and the required numbers are 1792, 1008, and 567. Ex. 48. Let x, y, and z denote the three numbers, and assume x+y=a², x+z=b², and y+z=c²; then, by subtraction,' xz a² c² x-y= b² — c² } all squares. Y-2 = a² - 62 AC= y= Z • We have, therefore, only to find such values of a, b, and c, as will satisfy the latter conditions; for in that case the former must have place. But such values of a, b, and c, have been found in Ex. 27, Diophantine Problems, viz. 6972-485809, 1852 34225, and 1532=23409. Hence, considering these quantities as known, we have, by the common rules, a²+b²-c² 2 a² + c² b² 1 =2483123 2 -a²+b²+c² 2 } 2374961 =214087, - { 235 F \ ▸ Or, multiplying each by 4, in order to avoid fractions, we have x=993250, y-949986, and z=856350. Which numbers answer the conditions of the question; and various others may be had by finding different va- lues for a, b, and c. Ex. 49. Let x and y be the numbers sought; then, we have to find "x + y = a square. x²+y²= a.biquadrate. First, in order to make x²+y2- a square, assume x=p2-q2, and y=2pq, x²+ y² = (p²+q²)², Then shall But when x²+y2 a biquadrate, p2+q2 must be a square; assume, therefore, again, pr2s2, and q=2rs, Then we shall have = x²+y² — ( p²+q²)²=(r²+s²)4, a biquadrate, as required. And it now only remains to find x+y=r¹+4r³s — 6r252—4rs³+s4=0; Hence, in order to reduce this to a more convenient あり ​35 form for solution, substitute r 2. mula, after multiplying by 16, reduces to Zapat J s¹+296s³t+408s²t²+160st³ +16t¹=0; Again, assume the formula = (s²+148st — 412)²=s1+296s³t+21896s2t2-1184st³ 3s 2 +t, and the above for- +164, Then, by cancelling the like terms in both, this reduces to 21896s-1184t=408s+160t; S Whence, 1344 84 21488-1343 ; Assume, therefore, s—84, and t—1343, and we shall have ga +t=1469, and, consequently, Matata x—p¹ — 67¹²§¹²+s4565486027761 y=4r³s—4rs²=1061652293520, for the numbers sought. 4 236 Ex. 50. The solution of this question is intimately connected with that of Ex. 48; for if we here call w, x, and the four numbers, and at the same time make } y, z, w=x+y+z, we shall have to find W ·x= y + z=0 w-y=x+x= w—z=x+y=0; Also x-y=,x−%=0, and y-2=0. It is, therefore, only necessary that x, y, and z, may be such numbers that the sum and difference of every two of them may be a square, which are the conditions of Ex, 48, where we found the three numbers to be / My Trade J M A x= 993250 949986 y=. z= 856350 x=2399057 y=2288168 z=1873432 and w=6560657, 1 十六​十 ​And consequently w=2799586, Which numbers answer the conditions of the question. And if, in our 48th Example, we had taken a= b=2067, and c=2040, we should have found =2165,. which are the numbers given in the answer in the Intro- duction. 1 1 1 1 +-+ + 9 27 81 1 1 27 + 81' 1 1 +27 + 81' Ex. 51. Here the proposed series may be put under the form &c. &c. ✓ &c. &c. 3 1 2X3 1 =2x9 VI W NI W NI W NI W 11 3 1 227 1 81 1 X B • t 1 } 237 4 2 Whence, by the rules and formula for geometrical progression, the whole sum is 3,1 1. 1 1 263 +-+ P tốt, 27 + 81+, &c.) = 33 927 Ma 9 (16 " Whence, the whole series = my 81 729 + 256 4096 + 3 3 1 3 × Ex. 52. This may be separated into two series 327 243 + + +, &c. = 17 x 1 16 3 X 4 64 1024 4 NI W 2 x 3 +, &c.) = 12 9 3 グーグ ​II 16 Ex. 55. This series is the same as } mit Ex. 53. Here, by the rules for arithmetical progres- sion, the nth term =5+(n−1). 9 X 7 16 Therefore, {10+ (n-1)}=(+9) is the sum required. 2 2 Answer. Ex. 54. The 25th term of the progression. 1, 2, 4, 8, 16, &c. 22416777218; therefore, the 25th term of the proposed series is 216777216; that is, the 16777216th power of 2. Now, the log. of 2=0.3010300 Mult. by 16777216 Gives 5050445.3324800 for the logarithm of the 25th term; consequently, the in- dex being 5050445, the number of integers will be 5050446. (22-2)+(43-4)+(62-6)+, &c. viz. § 4(12+22+32 +42+, &c. 1002)- 2(1 +2 +3 +4 +, &c. 100) V } 238 } 1 But the sum of the former − 4 × (n + 1)n(2n + 1 __ 4×101×100×201 =1353400, 6 6 And the sum of the latter =2× =10100; Whence, 1353400-10100=1843300, the sum re- quired. Ex. 56. Here, the general form of the series being (n+1)(n)(2n+1) 6 51 X 50 X 101 6 and n being =50, we shall have na+ Ex. 57. By the differential formula, we have 35, 72, 111, 152, &c. 37, 39, 41, &c. 1st diff. 2, 2, &c. 2d diff. Whence, a 35, d'—37, d''—2, n=25; And, consequently, n(n−1)(n−2) 1.2.3 I 101 × 100 2 n(n − 1)ď + 1.2 42925, the sum required. ·¿'' 35×25+12×25 × 37--25 × 8×23=875+11100+ 4600-16575, the sum sought. ì 1 1 3. * J L 0 239 I ! f X APPLICATION OF ALGEBRA TO GEOMETRY.. MISCELLANEOUS PROBLEMS. PROBLEM I. Let ABCD, be the given semicircle AB, its diameter; a, its centre: and CDFE, the required square. D B Bady M } CG2-GE2+CE2, or x²+x²-d²; Whence 5a2d2, or x=d√}=}d√5. 2. Let ABC be the required right angled triangle; in which take AB=13=h, BC=x, Ac=y, and the given difference BC Ac=7=d; FGE Then, since DF-CE, we have FGGE. Let therefore AB=d, or CG=d; also cE=x, and con-. sequently GE=; then, (Playfair's Geom. Ryan's ed.. Prop. 37, p. 30,) no B Then, (by Ryan's Play. xxxvii. 30,) we shall have x² + y² = h² x-y=d, Where squaring the second equation, and subtracting: it from double the first, we have 3 1 240 x²+2xy+y2=2h²-d² x+y=√(2h²-d²) x-y=d x= √(2h²-d²) + 3 d y= √ √ (2h² — d²) - Id h=13, d=7, we have x=12, and y=5, as required. Consequently, and Whence, and Substituting here Ex. 3. Let AFGc be the given circle, and ABC the re- quired inscribed equilateral triangle. Į Join a and the centre ; also join CE, and produce it A E " 10 D. t K D H I Then, (by Ryan's Play. ii. 87,) the angle D is a right angle, and the triangles ADE and ADC are similar. But ADAC, therefore also DE AE., B F Let, then, AE radius=d; and, consequently, ED= { AE1d; also put AB=x, or AD} x. Then, (by Ry. Pl. xxxvii. 30,)²= } d² — 1d²=1&d²; x 16 Whence, x=d2d3, the side of the inscribed triangle. 1 1 Again, produce CD to F, and AE both ways to G and H; and draw HI, IK, perpendicular to EF and EG. Then, it is obvious, from the propositions above re- ferred to, (and Ryan's Play. iii. 88,) that EF=HE, and нr HI. Let, now, HI (the side of the circumscribed triangle)= y, and we shall have HE²=EF²=HF²; and, since Er=d, HE=d, and the above becomes ノ ​ 1 241 ? Bour } | d2 d²-d²+ y², or y2-3d2, or y=d√3, the side of the circumscribing triangle. Ex. 4. It appears, (from Ryan's Play. x. 93,) that the side of an equilateral and equiangular pentagon inscribed in a circle, is found by dividing the radius of the circle into extreme and mean ratio, the greater part of which is the side of the decagon. • " 1 A D E Hence, calling the radius oв=r, and the 'greater part OD=x, we must have r(r-x)= x², or x²+ræ=r²; B Now, since the 4a, and its area Whence xr+ √5r2, or x={r(−1+√/5), That is, BC or AB, in the above figure, r(−1+√5); Produce, Bo to E, and join Ec; then, (by Ryan's Play. xxxvii. 30.) Ec²—EB²—BC²—r²(5+1√5), or Ec=r√({+}√5); Again, as EB EC :: Ec: ED=r(5+1√5), and ED: CB :: BC: BD=7-√5); Whence, DC=√(EDX DB)=√ {r²(§ +1√5)(} −1√5)}. (10-25), or ac d√(10—2/5), the side of the pentagon required. = 3 4 Ex. 5. Let x= the length of the rectangle, and y the breadth ; then, 2x+2y= perimeter, and xy side of the square a², we have xy=4a² 2x+2y=4a S v 2. area. a, its perimeter 1 A 2 f 242 Whence, x²-2xy + y²=4a² Subtract 4xy =2α² We have x²-2xy + y²=2a², or But x-y=a√/2 x+y=2a; Whence, by addition and subtraction, x=a+}a√/2=a(1+3√2) y=a2a√/2=a(1—1√/2), the length and breadth as required. Ex. 6. Let ABC be the given equilateral triangle, and bisect the two sides AB, AC, by the two perpendiculars D'O, EO: E 0 I D B Į Then shall the point o be the centre of the inscribed circle; and oD its radius. Also, if so, ob, be joined, they will bisect the angles ▲ and â, and o will therefore be the centre of the cir- cumscribed circle, and on its radius, (Ryan's Play. iii. 89,) and (iv. 89.) Again, if on be produced, it follows from these proposi- tions, that it will pass through c, and bisect the angle ACB. Therefore, the triangles ACD and AOD are similar; and since ADAC, therefore Do=0. 1 s, Let now ABS, the given side, or AD=3s; also, do =x, and consequently A0=2x; then, Aо²=AD²+DO², or 4x²=1s²x²; whence, 3x²-s², or x=}s√/}=§√/3—2.8868, and 2x s√20/35.7736, the two radii required. 3 1 * 243 Ex. 7. Let ABCD be the rhombus, and AC, DB, its two diagonals, intersecting each other in E. / 3 D X Also, let the perimeter =4p, that is, each side of the rhombus =p, and the sum of the two diagonals =S+ Then, since the diagonals of parallelograms bisect each other, CE+EBS. { And because the three sides of the triangle DEC, and CEB, are respectively equal to each other, the angles at E are each equal to a right angle. Therefore, calling AC, and DB=y, or CE= EBay, we have 感 ​ B x+y=s 2 x² + y² = p² 5 From 8 times the latter, 2x²+2y28p². Subtract (x+y)2=x²+2xy+y2s2 And we have } E=x, and x²-2xy+y28p2-s2, x-y=√(8p2-s²) or But x+y=s; Whence, by addition and subtraction, x = {} s + { √(8 p²-s2) 1 y = } s — = √(8p² — s²), Or, since s-8, and p-3; these become x=4+8=4+√/2 y—4—1√/8=4—√/2, Which are the two diagonals required. + 13 } 244 1 Ex. 8. Here the three sides of the right angled ´tri- angle are x³, x2, x; and, since the square of the long- est side is equal to the sum of the squares of the other two, (Ryan's Play. xxxvii. 30,) we have 206=x+x²; or x-x-1; X Whence, x²+5=1.618034 And x 1.618034-1.272020; Therefore, } Ex. 9. It is a well known geometrical theorem, that the diagonals of a parallelogram bisect each other; and that the sum of their squares is equal to the sum of the squares of the four sides of the parallelogram. D C 202x xxx 2 =1.029085, the area. ak If, therefore, we represent the given parallelogram by the figure ABCD, and make its side nca, DC=b, DB=d, and Acx, we have from the above theorem x²+d²=2a²+262, or x²=2a²+262 — d², 4 E T Ex. 10. Let ABC be the proposed triangle, AE its per- pendicular, and BE, EC the segments, of which the diffe- rence is given. or x= √(2a²+2b2—d²), Which is the diagonal required.. Let that differenced, the sum of the sides perpendicular AE=P, + cs, and “ 245 1 B Also, put, BE=y+d, and Ec=y-d; then, (Ryan's Play. xxxvii. 30.) 1 √ {(y+} d)² + p²2}=AB, √ { (y — { α)² + p²}=ac; And, by the question, √ { (y+ { d)²+p² }+√ {(y−}d)²+p² } = s; Squaring both sides, and transposing. ? √ { (y + { d)² + p²}× √ {(y-}d)²+p² } —s²—2y²—}d²—2p²; Or, in order to simplify the expression, let s² p² s2-2-2p2-2m; then, √ {(y+ } { d)²+p² } × √ {(y−}d)²+p² }=m—y²; Squaring again both sides, and actually performing the multiplication of the first two factors, we have 2 (y² — ‡ d²)² + 2 p² (y²+}d²)+p¹=m²—2my²+y¹;. And, by involving and collecting the terms, Fod¹ — { d²y²+2p²y²+\p²d²+p¹—m² — 2my², (2m+2p² — } d²) y² — m² — } p²d² —¿¿dª—pª, ‚ m² — { } p² d² — √ ¹ d¹—p¹ 1 1 T6 ·P²); Whence y=√( 2m+2p² - } d² E 1 C In which formula, substituting the values of p, d, and m, we have y=472}; Whence, y+d=720=BE, y-d=225=EC, AB=√(BE²+AE²)=780 1 Consequently, AC= √(E0²+AE²)=375 BC BE+EC945, Which are the three sides required. + 1 t } 246 Ex. 11. Let ABC be the required triangle, and ar, BE, and CD, the three given lines bisecting the three sides, CB, AC, and AB. J A 1 E F Make AF=α, BE=b, CD=c; also cв=x, Ac=y, and ABZ. Now it is a well known property of triangles, that "double the square of a line drawn from any angle of a triangle to the opposite side, together with double the square of half that side, is equal to the sum of the squares of the other two sides ;" that is, * 2a²+3x²=y²+z² 2b²+ y²= x²+z² 262 2 c ² + z z² = x² + y², 2 Or, y²+z2-3x²=2a, + y²+x²+˜¯x²=26², and x²+ y²- z²=2c2 From whence, by taking the former of these equations from twice the sum of the two latter, there comes out 4x²+4x²=2(26²+2c²-a²);, And consequently x= 3√(2b²+2c² —a²). In like manner y=}√(2a²+2c²—b²), and z=√(2a²+2b²—c²) ; Where, by substituting the given values of a, b, and c, viz. a=18, b=24, c=30, we have x=34.176, y=28.844, and z=20, which are the sides required, 247 Ex. 12. Let ABC be the proposed triangle, of which the base AB is given A 50=26. C Д ED B Then, since the area is also given 796; the perpen- 796 dicular =p, is also known. 25 and cp=p, Make, now, AE half base—a, and ED x; then AD=a+x, and BDα-x; Asc=√/{p²+(a+x)² }; and BC=√{p²+(a-x)2}. Whence, calling the given difference 10=d, we have вDa √ { p²+(a+x)² } −d= √ { p²+(a−x)²}, Which equation, squared, gives 1 1 } } duke p²+(a+x)²−2d√ { p²+(a+x)²}+d²=p²+(a−x)” ÷ And this, by reduction, becomes 4ax+d²=2d√ { p²+(a+x)² }. And this, again squared, prodúces 16a²x²+8ad²x+d¹=4d³×( p²+a²+2ax+x²) ; or 16a²x²+d=4d³× (a²+p²)+4d²x²; 4d2x (a²+p²)-d Whence, x=√( -)- 16a²-4d2 Where, by substituting the numeral values for a, p, and d, the answers for the sides will be found. Vamp Ex. 13. Let ABC be the proposed triangle, AR its base 194b; Ix the diameter of the circumscribing circle =200=d, drawn parallel to AB; and De the bisecting line 66( ) .. 248 I S DE, or ax $ Plymouth . a H N C M E Then we shall have HI=GK (IK—AB)=3, And consequently AH=GB=√(IG X GK) √(IGXGK)=√197x3=√591=e. Let now co be produced to meet the circle in E; Then, because CD bisects the angle ACB, it will bisect the arc AEB, and therefore the perpendicular ELM will pass through the centre L; Consequently EM=100+√/591e is also known, as MN=100/591=f. is also Now let DE=x; then, since the two triangles ENC and MDE are similar, we have ME DE: CE: NE, or e : OC :: a+x: d; F FG K a² Whence x²+ax=de, or x +v+de), which thus 2 4 becomes known; and consequently the rectangle CD X 来 ​ 2 a a² +V+de)}=r, is also known. 4 1 a But this latter rectangle cDX DE=ADX DB; therefore ADXDB, and AD+DB, are both known. Assume now, AD=y, and BD=z, then we have = y + z = br y z = r $ Whence is readily found y = b + 3 √(b² - 4r2)=m z = 1 0 - 1 √ (6² - 4r²) ==n; N C 1 249 Substituting Again, calling AC=v, and вC-u, and we have v : U :: m : Ny um ΟΙ U v= um Um n √(1+ y) = $2 Also, x: b I • • N also, uv=a²+mn. 素 ​J ; ņ a²n + mn² m a²m+m² : a² + mn; Whence, u=√(~ VI n N The numerical values of which may be found by sub- stituting those of a, b, and d, in the original equations. Ex. 14. Let ABC be the proposed triangle, AE=a, and DC=b, the two given lines. for v in the second equation, gives ย mu2 裴 ​{ =cos. BAE, and √( and vi¹ + Amer E ), and, consequently, B D Also, let x and y represent the sine and cosine of the angle BAC respectively; then, by trigonometry, we have the sides required. " I 1 1+x. -10 ! 2 )=cos. x b 1+x √(² + " ") = √(1 + ") == √ 2 : ✡C COS. BCD. } )=AC, 1-1 a 1+y 2 and y: a :: √(ty) √(2 + 1) = √(1 + 1 ) == AC =AC; X 1 1 1 1 -? 1 } 1 b Whence, -√ OC เ } Y +x. ах √(1+3=1 by Again, by trigonometry, sin A*--- 1 20 1 250 1+20%) a A VI (5), 2 1—tan² COS A 1+tan² }^ Putting, therefore, tanBAE=t, and substituting 1 12 and y=1+123 2t 1+12' 1+t 2at We have √2-6(1~t²)' b(1+t)(1 − t²)=2at√/2, or 1+2a√2) t = 1. 13+ t²+(- br ED } 2tants 1+tan²Ã' 21 Which is a cubic equation, whence the value of t may be determined; viz. the tangent of the angle BAE: and Hence, also, the angles BDC and BEA become known, and, consequently, the sides AB=35.80737, BC=47. 40728, and Ac-59.41143, as required. Ex. 15. Let ABC be the proposed triangle. C or B ΟΙ ܕ and 1 A denotes any angle; but, in this example, it is put for the angle BAC. See Gregory's Trig. 1 1 251 Then the base AB-8, or AE-base-4-b, CD=4= p, and Ac+BC=12=s; and make also ED-x. Then AD=b+x, and DB-b-x; and consequently √ {(b + x)² + p²}=AC √ {(b−x)² + p² } =BC; Whence, by the question, 2 √ { (b+x)²+p² } +√ { (b−x)²+p² } = s; Squaring both sides, and transposing 2 2 √ { (b + ∞)² + p² } × √ { (b−x)²+p² } =s²-2b²-2x²-2p²; Or, in order to simplify, writing s²-262-2p²-2m √ {(b + x)² + p² } × √ { (b−x)²+p² } = mx². Squaring again, and collecting the terms (b²x²)²+2p² (b² + x²)+p¹— m² — 2mx²+x+, Or b¹— 2b²x²+2p²b²+2p²x²+p¹—m² — 2mx² Therefore 2(2p2-2b2+2m)=m²-2p²b²-b4—p¹, ‚ m² — Qp ² b ² — b 4 —p4 2p²-2b²+2m -) ; Whence x, and consequently AD and DB become known, and hence also Ac=√(AD²+CD²) and CB=√(DB²+CD²), are determined. Or x= √( 65 In the present case, p=4, b=4, m=40, whence x= √5; therefore AC 6+ √5, and BC-6-√5. Ex. 16. Let ABC be the proposed triangle. B 1 * D 45 Let BC-a, AD= b, and BDx; therefore CD- 7/3/ a-x; also the ratio AB AC : 3:2, orm: n; then, AB² = b²+x², and ṣc²=b²+(a−x)²; consequently 62+x²: 6² +(a-x)²; m² n². Whence, we have m2b2+ m² a² - : : 2 : ; 1 1 1 } J 252' F 2m²ax+m²x²=n²b²+n²x², or (m²-n²)x2-2m²ax (n²- m²)b² m²a²; therefore, by solving this quadratic, and substituting the values of a, b, m, and ǹ, the numeral value of x may be determined, and hence those of AB and AC. Ex. 17. Let ABC be the proposed triangle, and make the perpendicular CD=24-p, CE the line bisecting the angle ACB 25-b, and cr, the line bisecting the base, =40=c: } A FEDB. Then, (Ry. Play. xxxvii. 30.) ED=√(CE2-CD2)=7=m, FD=√(FC²—cD²)=32=n; Also And, in order to simplify, let Er=q. Also, let half the base AF-FB-x; then, AE=x+7, CB=x-q, AD=x+n, DB=x-n; Hence, Ac={(x+n)²+p² } 2 BC= √{(x-n)² + p² } And, (from Ryan's Play. iii. 118.) we have I AC: BC:: AE : EB, − q ; 2 or √ {(x+n)²+p²} : √{(x−n)²+p² } :: x+q: x- Whence(x+n)²+p²} × (x−−q)²= {(x−n)²+p² } × (x+q)* Which, by multiplying, cancelling, &c. becomes nx(x²+q²)=qx(x²+n²+p²) qn² + q p² ~ ng² Where x² .1 or Pally 1 n Չ 2x=2 √ (?n² — ng² + qp²), n ~ q the base of the triangle; which, by substituting the proper 250 numeral values of q, n, and p, gives 714; from which i and the given lines the other two sides are readily ob- tained. 253 Ex. 18. Let ABC be the proposed right angled trian- gle, and o the centre of its inscribed circle; and let co -AO 2-d, and Ac=10=h.. A X= E ļ C D Produce co to D, and let it fall upon the perpendicular AD; which put =x. Then, since co and Ao bisect the two angles c and a, so and these two angles together are equal to a right angle, it follows that the two angles oac and oca = half a right angle. But the outward of any triangle, being equal to the two inward opposite S, AOD LOAC+ LOCA. Whence, also AOD half a right angle; and since D is a right angle, DAO is also = half a right angle. Therefore, DOAD=2, and A0=√/2x²=x√/2; and, consequently, co=xy/2+d, and çv=x+x√/2+d=(1+ √2)x+d. Now, AD²+DC²=AC² or x²+{(1+√/2)x+d}²=h², or{1+(1+ √2)²} x²+2d(1+ √/2)x=h² —d², or (4+2√/2)x²+2d(1+ √2)x=h² — d², or x²+2a ¹ + √2 h² - d² 4+21/2 4+2√2 Whence, by the solution of this equation, we have 1+√2 1+ √/2 4+2√/20 {d ± √ { ( 1 + 2 √ √ 2 ) ² d² + 4+2√2 h² - d² 4+21/2 x 2 • A 1 254 Where the proper numeral values of h and d being sub- stituted, those of x, x/2=A0, and of d+xy/2=oc, will become known. Let these be denoted by m and n; that is, A0=m, and oc = n; Then, in the right angled triangle ABC, we have given the hypothenuse Ach, and the lengths of two lines oc and os, drawn from the acute angles to the centre of the inscribed circle, to find the two sides AB, BC. Now, let fall the three perpendiculars, OE, or, oc, and, make AG=y, and Goz; then, it follows, (from Ryan's Play. xxxvii. 30.) that z² — y² = n² m², also, z+y=h. Whence, by division, z-y= h² + n² m² Consequently, z= A H 2h h² — n² + m² y= 2h Where both segments become known, and consequent- ly also OG = √(n² —z²), (h²+n² m²)² E podla or OG=√ {n² 4/2 Whence AG, Gc, and oc being known, we have also ABAGOG=5.87447, and BC=GC+OG 8.08004. Ex. 19. Let ABC be the required circle, AB=α, and DC=b, the two given lines, and Erc, the given distance.· 23 డ n2 B m² h h { 255 I 2 Also, draw the two perpendiculars EH, EG; and join AE, ED. Let AE, or ED=x; then, Eн=x²-1a², and EG² HF²x²-162; but EH2+HF2=EF2-c2; therefore, 002 x² - 1 α² + x² - 16² c², and consequently, x=(a+1 62 +c). Whence, the diameter is given = √(√a²+b² +2c2), the Answer. NOTE. The answer in the Introduction is erroneous. Ca Ex. 20. This question, of which the figure is as fol- lows, does not require the assistance of algebra. · B H D A 70 For BD=√(HB2+HD2)=√(400+14400)=√14800, Whence, ED and BE are cach =√/14800; Consequently, EC=√(ED²-DC2)=√(14800-6400)— √/8400=20/21; and AE=√(BE2-AB2)= √(14800-10000)=√/4800=40/3, as required. Ex. 21. Let ACBF be the given trapezium, Where Ar=6, FB-4, CB=5, and ca= 3. A E D B F Then, draw the diagonal AB, and let fall upon it the two perpendiculars CD and FE, and make CD=p, and FE=p′; also, put the required diameter=a. Then (Ry. Pl. c. 143.) Į ; 256. px=5X3-15, p'x=4×6=24, or x(p+p')=39, In the same manner, calling q and q' two perpendicu- lars falling on the diagonal cr, we have qx=3×6=18, q'x=4×5=20, or x(q+q)=38; t Whence, (p+-p'): (q+q): 39: 38. But the sum of these perpendiculars into double their respective diagonals are equal to the area of the trapezi- um; whence, AB CF: 38: 39. : X CB; Also, (Ryan's Play. D. 144,) ABX CF ACX BF+AF Let, therefore, AB=y, and CF=2; then, y: z:: 38: 39, and yz=42; 42 X 38), and 2=√ Whence, y=√( z= 39 Now, the three sides of each of the triangles ABC and AFB being known, the perpendiculars CD and FE are readily determined. Let, therefore, these be denoted as above by p and p', and we shall have from our first equa- the diameter sought. 39 tion, x P+P" Ex. 22. This question, of which the figure is as follows, is nothing more than having the three sides of a triangle given to find the radius of the circumscribing circle. A 1 B F 42 X 39 38 -). ¡ M Let, therefore, ABC be the given triangle, of which ( 1 7 257 AC 25 α, AB=30=b, CB=20=c, and o the required centre. From which the perpendicular CD is readily found. For, by putting AD=x, and DB=y, we have x²-y² — a² — c², x+y=b; Whence, x- x-y= By addition, x= 26 C (b² + a² — c²) 2 } 462 Again, (Ryan's Play. c. 143,) diam. XCD ACX CB, 25×20 80 80 Whence, diam. = ✓7=30.237116; 25 vy √77 and, consequently, OA, OB, or oc=15.118558, the dis- tance sought. Ex. 23. Let ACB be the proposed semicircle, ECF the given equilateral triangle, whose area =100-a, and whose perpendicular CD, which is the radius of the cir- cle, is required. And hence, CD=√ { a²- a²- c² b b²+a²-c² * (data) A E D F B Let CE, the side of the triangle, x; then, ED=x, and CD=√(x²x²)=x√3. But DCXED-xx√3=100, the area. 400 √ √ 3 201/}. Whence, 2/3=100, or x=√ 25 =22√7. 4 ( ww Consequently, √/3=10/3=radius, and, therefore, 204/ 3, the diameter. " 258 Į Ex. 24. Here, the rectangle of the perpendicular and diameter of the circumscribing circle of any triangle is equal to the rectangle of the two sides from which the perpendicular is drawn. (Ryan's Play. c. 143.) F A GD B Whence, the present question reduces to this,i. e. given the perpendicular p, the radius of the inscribed circle oG, OE, or or=r, and the product of the two sides =2Rp, to find the sides. Let, therefore, the segment AD=z+x, and DB=7—x. Then, (Ryan's Play. xxxvii. 30,) ac=√{(z+x)²+p² BC= √ {(2x)²+p² { AB=2z. 4 2 Also, because ABX CD=(AB+BC+AC) ×OG, we have √ {(='+x)²+p² } × √ {(≈−x)²+p²}=2Rp, and [ √ {(z+x)² + p²}+ √ {(~~∞)²+p²}]×r=2(p-r)z. Whence, squaring the latter equation, and substituting for the double rectangle, we have 2x²+2x²+2p²+4Rp= Or x2: E 2(p—r)² 22 2(p—r)² 2.2 4(p − r)² z 2 p2 کا -z² — z² — p² — 2RP, Apri or, by putting x² - mz²-n. Also, squaring the first equation, and multiplying, (≈ ² — x²)² + 2p² (z²x²)+p¹— 4R²p²; -1=m, and p2+2Rp=n, 3 • * 259 In which, substituting for a² the value found above, we have (22-me²+n)²+2p² (2²+mz2-n)=4R2p2-p¹; That is, 24(1m)²+2n(1-m) z²+2p²(1+m)²+n²? ___ 2p²n=4R² p² -p4; which reduces to 24+ 2n(1—m)+2p²(1+m) 2_4R²p²—p*—n²+2p²n (1 —m)² (1—m)² Where, by re-establishing the value of n, the second side of the equation becomes 0, and there remains 2n(m-1)-2p2(m+1) ; (m-1)2 Or, by substituting for m and n their respective values, we have ! 1 2= Whence, 22- Ex. 25. Let BCA be the proposed triangle, in which BC-2a, and AD-a. 1 r√(2Rp-4Rr-r2) p-2r 2r√(2Rp—4Rr — p²) p-2r · oc=a Then, assuming Bba+x, and DC-a-x, or -(x—α), we have AB- =√(2a²+2ax+x²) Ac=√(2a²-2ax+x²), And, by the question, (2a2+2ax+x2)²+(2a2-2ax+x2)²=24a³. Let 2a2+2ax+x2m, and 2a2-2ax+x²=N, Then, m²+n=24a³ } 2 1 260 < 41 And, by squaring, m³ + n³ + 2m³n²=576aº, Or 4m³n³ (576a6 m³ n³)² Where, substituting the above values of m and n, the equation reduces to this; viz. α=α√, Whence, BD=(1+√)a, and oc⇒(1→√/§)α.. Which, being negative, shows that the perpendicular falls on the base produced. ' "I ŕ Myndata 24 4 ca√ + √ ©) Therefore, AB=a√(²++ 24 4 And ac=d√(√6) Ac 3 31/6) And this, by extracting the roots, gives AB=a(2+√6) Ac=a(2-1√6) Which are the two sides required.. m m²-84 SOLUTION TO EX. 24, PAGE 177. Let x2-1, and y2-1, be the two numbers sought; then each of them, when increased by unity, will be a square we have, therefore, only to make x2+y2-1==v2, and x²-y²+1=0 2 u². From the first we get y2v2 +1, and by adding the first and second, 2x²-v² This last condition will be satisfied by making x=(r2+ s²)n, v-(2rs-r2s2)n, and u=(2rs+r²-s²)n. Then 22-1 will be 4n²rs(r²-s²)+1. Assume its side 2nm-1; then 4n2rs(r2-s²)+14n² m² 4nm +1, and nrs(r² s²)=nm²-m; hence n where m, r, s, may be taken at pleasure. If r=4, and s=3, — 3, then n m² BEAMLINED p Mapen 1 VEZAN 2. 3, (if m be taken 9,) hence xn(r2+2)-75, y=2nm-1=- 55, œ²—1—5624, and y²-13024, the answer in the book. ทา rs(r² 1