IN MEMORIAM FLORIAN CAJORI Digitized by the Internet Archive in 2008 with funding from IVIicrosoft Corporation http://www,archive.org/details/algebradesignedfOOstodrich AN ALGEBRA, faa^ GNED FOK THE USE OF HIGH SCHOOLS, ACADEMIES, AND COLLEGES. BY JOHN F.UtODDAKD, A.M., AUTnOK OF "STODDASD S AKITHMKTICAL SKRIES," E*JO. AND PROF. W. D. HENKLE, OF OREFNMOCNT C O L L E (, E . INDIANA. 4j ^q., are called fractional exponents, and are the signs of evolution. Thus, a^, a^", a*, &c., respectively indicate that the square root, the cube root, the fourth root, &c., of a are to be extracted. (23.) In fractional exponents the numerator denotes a power, and 3. the denominator a root. Thus, x^ denotes the fifth root of the third power of X, or the third power of the fifth root of x. (24.) The symbol V is called the radical, and is the sign of evolu- tion. Thus, V^, V«, Vflt, &c., respectively indicate that the square root, the cube root, the fourth root, &c., of a are to be extracted. The small figure in the angle of the radical is the index of the root. When no index is written, " is understood. Thus, Va is the same as Va. SYMBOLS OF EELATION. (25.) The symbols of relation are :, =, : :, > <, -f- . . ., and -H- : : : . (26.) The symbol : denotes ratio. Thus, a : b denotes the ratio of a to b. (27.) The symbols = and : : are signs of equality. Thus, a=b denotes that a equals b ; and a:b=c:d, ot a:b ::c:d denotes that the ratio of a to 6 equals the ratio of c to d. The symbol : : is not used except to denote the equality of ratios. Thus, we never write a::b hr a=b. (28.) The symbols > and < are signs of inequality. Thus, a>6 denotes that a is greater than b ; and a<6, that a is less than b. (29.) The symbol -i- ... is the sign of an arithmetical series. Thus, -4- a . 6 . c . c? denotes the equality of the difference between a and b, b and c, and c and d, (30.) The symbol -^f : : : is the sign of a geometrical series. Thus, -~a:b:c:d denotes the equality of the ratios of a to 6, 6 to c, and c to d. 2 18 DEFINITIONS. SYMBOLS OF AGGEEGATION. (31») The symbols of aggregation are , | , { ), [ ], and \ j-. (32#) The symbol is called a vinculum, and denotes that the quantities over which it is placed are to be considered as one quantity. Thus, Va + h-\-c denotes that the square root of the sum of a, 6, and c is to be extracted. (33») The symbol | is called a 6ar, and denotes that the quantities, in the column immediately preceding it, are to be considered as one -\-ax quantity. Thus, + h denotes that the sum of a, 6, and c is to be + c multiplied by x. (34.) The parenthesis ( ), brackets [ ], and braces -j j-, denote that the quantities contained within them are to be considered as one quantity. Thus, (6 + c)ar denotes that the sum of b and c is to be multiplied by x; la-\-{b + c)x^g denotes that the sum of a and (6 4- c)x is to be multiplied by y; and •j2 + [a + (6 + c)ic]y|'tt denotes that the sum of z and [a + (6 + cjx^y is to be multiplied by u. SYMBOLS OF CONTINUATION. (35.) The symbols of continuation are .... and , and are equivalent to c&c, and so on, or continued according to tike same law. Thus, a, a'*, a^, a*, a^, a", a', a®, &c., and a^, a^, a^, a^, a^, a„, a„. a &c., may be written a, a", a^ . . . . and a^,a^,a g, . . . . SYMBOLS OF DEDUCTION. (36») The symbols of deduction are .• . and ••• (37.) The symbol .*. signifies therefore, whence, ccmsequentlpy hence, from which we infer, )] + 4y» I ^ V' . o^^ • ^'^ CHAPTER II. \^ ^ ADDITION. (72.) Actdition is finding the simplest expression for the sum of several algebraic quantities. CASE I. (73.) When quantities are entirely dissimilar. RULE. Connect them together by their proper signs. PROBLEM. Find the sum of a, —6, +3c, and —5d, SOLUTION. Connecting these expressions together, we have, a— 6 + 3c~6(?. EXAMPLES. !• Find the sum of Qa and —'76. Ans. 6a— lb. 2. Find the sum of 4a, —36, and -\-5x. Ans. 4a—3b-{-5x. 3* Find the sum of aa;, +bm, — cy, and -{-nw. Ans. ax + bm—cg+nw. CASE II. (7 4.) AVTien the quantities are similar and have the same sign. RULE 1. When the quantities are numerical, add as in arithmetic, and prefix the sign +, or —, as the case mag be. RULE 2. When the quantities are literal, add the coefficients, affix the literal part, and prefix +, or —, as the case mag be. PROBLEM 1. Find the sum of —4, —1, —8, and —3. 24 ADDITION. SOLUTION. Operation. — 4 Adding these numerals together and prefixing the — 1 sign — , we have —22. — 8 — 3 -22 PROBLEM 2. Find the sum of —Zahx, —lahxj —GabXj and —9ahx, Operation, — Sabx — labx — 6abx — dabx SOLUTION. Adding the numerical coefficie sign — , and annexing the comn we have —25abx. mts, prefixing th< ion literal factor — 25abx EXAMPLES, 1. 2. 3. 4. 5. 4a a 4a6 lab — abc — Sabc 5{aJrh) Ha + b) 2|/a+a;' dVa + x^ 6a bob — 4abc S{a + b) 5Va-\-x^ a 2ab — 12abc U{a + b) 1Va + x' 2a ab - 2abe (a + b) Wa+x^ 8a 8a *25a ] dab llab 33ab -Uabc 12{a + b) lOKa + ic' —BQabc 42(a + 6) 6. 7. 8. -4(a' -t^)i Bix^+y)i 4(.r'-y«)* - V(a' -11 (a» -b')i Wx^+y Wx'-^y QVx'-y' Ql^x'-y'' - 8(a' -6^)* l^ix'^yY Vx^-f - 3(a' -b')^ lQ{x' + yY ^{x^-f)^ * When no sign is written + is understood. 9. ADDn 10. 3a;V— 4y' 6a;'y— 3y' 8a:'y— 4y' 11. 3a' + p {a^^hy-\QVrrv'-n^ Ya'+ 3p ^Va'-h-" - AVm'-n^) 9a' + 6i? 14a' + 16i? ^(a'-^Y- 5(m'-ii')2 25 12. 3a'— 7 + 5a6c 4a'— 6 + 7a5c + 4a;'y 5a'— 3 + 8a6c + 5a;''y + 4m7i 6a'— 4: + Qahc + Qx^y+ mn+ cd 7a' — 1 2 4- 3a6c + 4a;'y + Smn + 4cd CASE III. (75.) When the quantities are similar, and all have not the same EULE. Mnd the sum of the similar positive terms, also the sum of the similar negative terms, and then, disregarding the signs, ascertain their difference, and prefix +, or —, according as the sum of the positive or negative terms is greater, PROBLEM. . Find the sum of 6a, —*la, —3a, +4a, +2a. — ? SOLUTION. Operation. + Qa The sum of the positive terms is +10a, and the — Ya sum of the negative terms is —12a. Disregarding — 3a the sio^ns, we have 2a for the difference between 12a + 4a and 10a, to which prefix — , because 12a is greater —2a than 10a and has a minus sign. — 2a , "We may also add the terms successively. Thus, — 2a + 4a is + 2a; which, added to —3a, is —a; which, added to —7a, is — 8a^' which, added to -\-6a, is —2a. 26 ADDITION, EXAMPLES. !• What is the sum of — 4a;y, -^^xy^ +6a;y, and —bxy? Ana, Axy, 2t What is the sum of — 6a6, — taft, +3a6, and -f 4a6.^ -4w5. — 6a&. 3* What is the sum of 3a+xy, —^a—4txy, +1a—5xyj and Qa + xy? Ans. I2a—1xy, 4. What is the sum of 4,xy—ab^ —Bxy + 4abj —Axy—bab^ and + 5xy + 4a6 ? ^ws. 2a;y + 2ah. 5* What is the sum of a + 6 + c+c? + e— /, a + 6 + c + (f— e4-^ a + 6 + c— c? + e+/, a-\-h—c-\-d + e+f, a— 6 + c + e4-/, and — a + J 6« What is the sum of 7a— 5c + 36 and 2a— 3c— lb ? Ans. 9a— 8c— 46. 7t What is the sum of 6a + 46— 3c— Yc^ + S and 3a— 126 + 7c— lOrf— 4? Ans. 8a— 86 + 4c— l7rf + 4. 8. What is the sum of — Y/+3a, 4/— 2a, 3/— 3a, and +2a.^ Ans. 0. 9t What is the sum of 12A— 3c— 7/+3^ and — 3^4-8c— 2/— Q^r -\-bx? Ans. 9A + 6C— 9/— 6^ + 6a;. 10. What is the sum of 16a— 66 + 15c— 9c?, 3a + 186— 5c— Ydf+ 3c, — 7a— 26— 3£? + 5e— 9A, and 11a— 36 + 2c + 8c? + 7A.^ Ans. 23a + 86 + 12c— llc?+8c— 2A. U, What is the sum of 8a + 6, 2a— 6 + c, — 3a + 56 + 2c?, —66— 3c + 3c?, and — 5a + 7c— 2c?.^ Ans.2a—b-\-5c-\-3d. 12. What is the sum of — a + 36— c— 115rXZ+yz-\- 2\/ax + by. 16. What is the sum of ^{a-^b)Vx''-y''-2{a-b)Vx'' + y\ -3(a ^, 6) f^d^'' + {a-b)Vx-'Tf, -{a-Vb){x'-f)UMP'-h){x'' + f)\, %{a + b){x''-y'')\-{a-b){x' + y'')\,lO{a + b)Vx'-y^-5(a-b)(x^ + y')i, and -2(a^b){x^-y^)\ + 4.{a-b)Vx''-\-y'"i Ans, \4:(a-\-b)Vx''-y\ 17. What is the sum of 10f^2 + 5V8-7V5 + 2Va, 6i/2 + V8 + 4V5-3l/a, and-3i/2-9l/8-3V5 + V«+i^a^; ^w*. 12f2 — 3V8— eVS+f'aS 18. What is the sum 'of 6a*6 + 3a-'6'c— Ya6, — 6a*6 + 2a-'6V^- l7a6, and 9a*6— 8a-'6'c— lOaS; ^n». Sa'b—3a-''b\, 19. What is the sum of —3(ax-{■by->^cz)^-^^x'^\■y''^\^a—b, 2 V^^+lyT^+(a:' + y')^-3 {a-b), \/ax+by +CZ- fVTy' + 2 (a-^>),3Vaa;+ 6y +cz + {x'-\-y'')\-{-a-b,6Vax+ by +cz + {x'+y'')i -2{a-b), and {ax + by-\-cz)\-Vx^ + y''-3{a-b)'i Ans, Q{ax-\-by-\-cz)\—i:{a—b), 20. What is the sum of Gasftf— 9cfrd from a + 6. -4w5. a-\-b—c—d, 2. Subtract 6— c—e?+e from a. -4w«. a—b + c+d—e. 3, Subtract — (6+c + «?) from a. Ans. aH-6 + c + or — 2> or +2 taken negatively, or subtractively, — ( + 2)) —23 —3 3 times. Hence, the result is —6, __/_!_ g^ _ _Q _ _Q as is shown by the operation. 3 34 MULTIPLICATION. The expression — {-f 2) is the same as —2, because Operation. —( + 2) means that +2 is to be subtracted. But + 2—2 there is nothing from which to subtract it. Let us, + 2 then, subtract it fi'om nothing, or zero. For zero, we — 2 write +2 — 2, and taking +2 from it we have —2, as is shown by the operation. Therefore, — ( + 2) = — 2. The principle contained in this proposition is generally expressed in the following RULE. Plus ( + ) multiplied hy minus (— ) gives minus (— ). PROPOSITION (84.) 4. When a negative quantity is multiplied hy a negative quantity, the product is positive. DEMONSTRATION. Operatixm. Let us multiply —2 by —3. This means that the —2 is to be or —2 taken negatively, or subtractively, —3 3 times. Hence, the result is +6, (_6) +6~ _|_g as is shown by the operation. The expression —(—2) is the same as +2, because Operation. —(—2) means that —2 is to be subtracted. But + 2—2 there is nothing from which to subtract it. Let us, — 2 then, subtract it from nothing, or zero. For zero, we + 2 write +2 — 2, and taking — 2 from it we have + 2, as is shown by the operation. Therefore, —(—2)= +2. The principle contained in this proposition is generally expressed in the following RULE. Minus { — ) multiplied hy minus ( — ) gives plus ( + ). (85.) The principles contained in these four propositions may be expressed by the following RULE. The multiplication of like signs gives plus ( + ), and the multipli- cation o/" UNLIKE signs, minus (— ). MULTIPLICATION. 85 PROPOSITION (86.) 6. The product of two literal terms may he expressed hy writing them in order^ with or without a sign of multiplication between the consecutive terms, preceded hy + or —, according as the signs are like or unlike. DEMONSTRATION. The truth of this proposition depends upon (19.) Thus, a multiplied by —6 may be expressed by —axb, — a-6, or — «5; (a + h) multi- plied by (c—d) may be expressed by (a-\-b) x (c— c?), {a-\-h)'(c—d), or (a-{-h) (c—d) ', and (a + 8) multiplied by (4—6) may be ex- pressed by (a + 3)x(4— 6), (a + 3)-(4— 6), or (a + 3)(4— 6). PROPOSITION (87.) 6. ^ two terms, when the exponent and sign of each are not considered, have a common part, their product may he expressed hy the common part'uffected hy the sum of their exponents, and preceded hy -j-, or — , according as the signs of the terms are like or unlike. DEMONSTR ATI ON. The two terms +«'' and —a^, when the exponents and signs are not considered, have a common part a ; therefore, we are to prove that the product of -\-a^ and —a^ is — a\ Since, +(Z^= 4-aa and —a' ^=—aaa, we know by the last proposition the product of +aa and —aaa is —aaaaa. But, —aaaaa=:—a^. Therefore, -|-a''x— a'= Remark. — By this proposition, we have -|-2^ x2*=:-j-2^+^=-}-2' =4 ; —a-' X —a-'=z +a-'-'— +a-' ; a-"" x — a«=— a-'^+^zz:— a^ or —a. GENERAL RULE. (88.) In multiplication, coefficients are multiplied, and exponents are added. CASE I. (89.) When both multiplicand and multiplier are monomials. RULE. Multiply according to the principles of the preceding propositions B6 MULTIPLICATION. EXAMPLES. 1. Multiply 4:a'b'cdhjSabc'd\ Ans. l^a'b'c'cf. 2. Multiply 12f ay by ^hx, Ans. 48bxVai/. 3. Multiply 5ix'fz* by exy'z\ Ans. 3SxYz\ 4. Multiply ISa'ftVy by —6abxy^. Ans. — 65a^5V/. 5. Multiply — 20a^6? by 5a'"6V. Ans. — 100a'"+^^>"+^c''. 6. Multiply a"* by a". Ans. a'"+^ 7. Multiply a"* by a"", ^?^s. a"*~". 8. Multiply a"*" by oT. Ans. aT"". 9. Multiply a"*^ by a"". -4%s. a-^''^\ . 10. Multiply 2a~^, Ya"**, auji — ^a" together. ^tis. — ^2a~^ 11. Multiply Z'n~\ n~\ and 4*7' together, ^tis. 12-f . 12. Multiply -7a-'6*c-*by Sa'^^-'c. ^ns. -21a6-^c-*. 13. Multiply ^aF~\ — 3a''~y, and Sa^'ca; together.' Ans. \W^'"i-^ycx. 14. Multiply — 13a~Vby -4a~'5~V. J^5. 52a~'&~''c~\ 15. Multiply a''"^, arb~\ and a"^^6 together. Ans. a'"6^+\ 16. Multiply {a + y)~%n\ {a + y)'^H~'m, and (a + y) together. Ans. mA^(a + y)'"+\ 17. Multiply a^ by ai». -4%s. a. 18. Multiply ai by —a" 2, ^?is. — Va. 19. Multiply ai by a\. Ans. ai. 20 Multiply -xT^hyx^. Ans. -arzk. CASE II. 1 (90.) When the multiplicand is a polynomial, and the multiplier a monomial. RULE. Multiply each term of the multiplicand by the multiplier^ connect- ing them by their proper signs. PROBLEM. , Multiply 6a + 45V - 3d' by 4a'. MULTIPLICATION. 37 SOLUTION. Operation. • Multiplying 6a, +46^c, and —Sd^ by 6a + 4:b'^c —dd^ 4a'', respectively, gives 2 4a^, +1 6a Vc, and 4a'' —12a^d^ which connected by their proper 24a' + 16a=^6'c— 12a''6' signs is 24a' + 16a''6"'c— 12a''V by 3a^bc\ Ans. 6a'6V— 15a'5c' + 2W6V. 3. Multiply 2a'— 3c + 6 by 5c. Ans. 2a'bc—8bc'' + 5bc. 4, Multiply ax'—bx^ + cx—d by — a;^ Ans. —ax^ + bx''—cx^+dx^. 5« Multiply 5mn + 8m^—2n^ by 12abn. Ans. 60abmn^ -\-36abm'n—24:ahn^. 6* Multiply 3ax—5bj/ + 1x7/ by —*labxy. Ans. — 21a'6a:V + 35a6'a;y'— 49a6a?y. 7. Multiply — 15a'6 + 3a6'— 126' by —bab. Ans. ISa^b^—Ua'^b' + eOab*. 8. Multiply a"'af* + b"'y"—(fj/"'—d''af' by afy*", Ans. a"'x'''+^y''-i-b"*x"'y^"—c"af*y"'^—d"x^"'y". 9. Multiply 3x~''—5x'^y~'*+z~' by 2x~*y"'. Ans. Qx'^y"^ — 1 Ox'^~*y'"~'* + 2x~*y'^z~\ lOi Multiply 2a~s^ — 7a;~2y*— llcff by axy~^c^. Ans. 2a\xy~^c%—^axkyc\—Waxy~^c. CASE III. (9 1 •) When both the multiplicand and multiplier are polynomials EULE. Multiply each term of the multiplicand by each term of the muU tiplier, and add the products. PROBLEM. Multiply a' + ah-\-b'' hj a + b. 38 MULTIPLICATION. SOLUTION. Operation. 0*+ ab-\-b^ Multiplying a^-\-ab + V by a gives a' + a'6 « + b + a¥; also, multiplying it by b gives a^J + ab^ a^+ a'^64- aft'* 4-*^ which results added produces a' + 2a''6 + a^b^ ab^ + b^ -\-2ab'' + b\ a» + 2a'^64-2a6' + 6' EXAMPLES. 1 . Multiply a-\-bhja + h, Ans. a" + 2ab + 6.' 2. Multiply a—bhja—b, Ans. a" — 'Hob + 6». 3. Multiply a + bhy a—b. Ans, a'—V. 4. Multiply a"— ab + b^ by a + b. Ans. a' -f b\ 5. Multiplya'' + a6 + 6''by a-6. ^w*. a"— 6^ 6. Multiply a'—a^'b + ab^—b^ by a + 6. Ans. a* — b\ 7. Multiply a' + a'^ + aft' + i^ by a-6. ^ws. a*-b\ 8. Multiply a*-25' by a-6. Ans. a''-2ab'-a*b + 2b\ 9. Multiply »''— 3ar— 7bya?— 2. ^W5. rr' — Sa;'— a; + 14. 10. Multiply a' + a* + a" by a""—!. Ans. a^—a\ 11. Multiply 4a"— 16aa; + 3ar' by 5a''—2a=a;. Ans. 20a'— 88a*a; + 4 taV- 6a V. 12. Multiply a* - 2a'b + 4a'b^ - Sab' + 1 66* by a + 2b. Ans. a' + 326'. 13. Multiply I x^ + Sax—^a''hj2x^—ax—la\ Ans. 5x* + iax^ — L j^^y > cd'' a a" 1 a a ' From this, we see that the reciprocal of a is -, or a~^ ; of a' is —r, or «-' ; of «-' is — r, or a\ Hence, we may obtain the a a reciprocal of any quantity by merely changing the sign of its ex- ponent. PROPOSITION (1 02,) 8. Ani/ qvxintity which has zero for an exponent is equal to unity. DEMONSTRATION. If we prove that a"=:l, we shall prove the proposition; since a may represent any quantity whatever. Weknowthat -=^=1. But, by Prop. 7, (101.),^=«'"'=a''. CL CL €(/ d' Since, then, a° and 1 are each equal to -^^ they are equal to each Cb other, that is, a^^l. GENERAL RULE. (103.) In division, coefficients are divided, and exponents sub- traded. 46 DIVISION. CASE I. (104.) When both dividend and divisor are monomials. RULE. Divide according to the principles of the preceding propositions. EXAMPLES. 1* Divide abc by etc. Ans. b. 2» Divide 6abc by —2a. Ans. —36c. 3« Divide —lOxyz by 5y. Ans, —2xz. 4* Divide ISax^ by Sax. Ans. 6«. 5* Divide — 28a;V by -~4iXy, Ans. *lxy'^. 6« Divide a* by a\ ^ ^?is. oT^. 7. Divide a"* by a~". ^ws. a'''+". 8. Divide ar^ by a". ^»s. «-("+"). 9« Divide, a""* by a"". ^W5. a'"^". ca^* lOt Divide ca** by da~\ Ans. —j-. 11. Divide —Sa'^'b^hj —^M^c\ Ans. — . 12. Divide 6 (a + by by 4 (a+6)-\ Ans. - , ^^., . 13. Divide (a + a:)Xa4-y)-' by (a4-a?)-> + y)-'. ^7j«. (a + a;)"(a+y)*. 14. Divide UOa'b'cd' by 30a'6'(f . ^/w?. Ba^b'cd. 15. Divide 15a"Vy by 3a"*a;'>'*. ^»«. Sa'^aJ^^T^. 16. Divide —^Sa^'b"" by 6a^5^. ^n«. — Sa'^ft"^. 2 1 12 Q^ 17. Divide ea^c?"^ bv 3a~3<^^ -4iw. -p. 18. Divide 12a~^dic~* by 3a8c?~«c^. Ans. —. '' arc 19. Divide (a + a?)"* by 6(a4-a;)^. -^n« 6(a + a;)' k DIVISION. 47 20. Divide {a-\-b)^{x + y)~^ by (a + by^{x-\-yp\ {a + by CASE II. (105.) When the dividend is a polynomial, and the divisor a monomial. RULE. Divide each term of the dividend by the divisor^ and connect the quotients by their proper signs. PROBLEM. Divide 6a'6*-8a''6'c?'4-4a*6'c by 2a'b\ SOLUTION. Dividing Qa^b\ —Sa'b'd^ and Operaticm, +4a*6''c respectively, by 2a'' J*, 2a% ) ''6a''b'Sa''b'd' + 4a*b^c gives Bab% 4bd\ and + 2a'c, which, 2ai)^ —4bd^ -\-2a^c connected by their proper signs, is Sab''—4bd'' + 2a'c. EXAMPLES. 1 . Divide 1 2a'x -\- 4aic' — 1 6a by 4a. Ans, Sax + a?' — 4. 2. Divide 12ay-16ay + 20ay-28ay by -4ay. Ans. —3y^-\-4ay^—5a^y-\-'7a*, 3. Divide 15a'5c— 20acy' + 6ccP by —5abc. 4y» rf" Ans. — 3a + -r 7-. b ah 4. Divide 3;''+'— a?'*+' + af+'— af»+* by x\ Ans. x—x^-^x^—x\ 5. Divide a"+'a;— a"^a;— a"+'a;— a»+*a; by a". Ans. ax—a^x—a^x—a*x. 6. Divide aaf -{■ax''-^' +ax*^'' ^ax"^ by a;". Ans. a-\-ax-\-a^-\-a3^, 7. Divide 6(a;+y)'-8(a; + y)» + 4a^(a?+y)by 2(a:4-y). Ans. Z{x-\-yY-^{x-\-y)-\-2a\ 8. Divide 5(a + 6)»-10(a+6)» + 15(a + 6) by -5{a + b). Ans. -(a + 6)'' + 2(a4-6)-3. 48 DIVISION. 9. Divide asf^^ +hx'^^—Car-^+dx'' by x^^, Ans. ax^+bx''—cx*+dx^^-^. 10. Divide ia'x^—^ax^ + Sab^xhj faV. CASE III. (106,) When both dividend and divisor are polynomials. RULE. 1. Arrange both dividend and divisor according to the ascending or descending powers of the same letter in both. 2. Divide the first term of the dividend by the first term of the divisor ; the result will be the first term of the quotient, by which multiply all the terms in the divisor, arid subtract the product from the dividend. 3. Then to the remainder annex as many of the remaining terms of the dividend as are necessary, and find the next term of the quotient as before, and so on, PROBLEM 1. Divide 6aV + a*— 4a'a: + a;*— 4aa;' by x' + a''—2ax, SOLUTION. Arranging the terms according to the descending powers of a, we have a''—2ax + x')a*~4a'x+6a''x^—4:ax'-{-x\a'—2ax+a^ a* — 2a^x-[- aV — 2a'a; + 5aV— 4aa;' — 2a^a; + 4a V — 2aa;' a^x^—2ax'-\-x* aV— 2aa;'+a;* ANOTHER SOLUTION. Arranging the t^rms according to the descending powers of x, we have DIVISION. x^—2xa+a' — 2x^a + 5x^a^ — 4txa^ x^a*—2xa^+a* x^a*—2xa^+a* PROBLEM 2. Divide 2a»*— 6a''*6'*+6a"6''*-26''' by a"— 5*. SOLUTION. 2ci^"— 2a''"6" 2a"6''*— 26'» _ PRO B LEM 8. Divide a;'— (a4-6+c)a;''+(a64-ac + 6c)a:— a6c by x—e, SOLUTION. a;*-— (a + 6 4- c)x^ + {ab-^ac + bc)x — dbc^ - x—c x^—(a + b)x-i-ab — (a + 6)a;'^ + (a6 +ac-\- bc)x — (a + 6)a;^ + (ac + 6c)a; o5a; — abc abx — abc EXAMPLES. 1« Divide a* 4- 2a6 + 6' by a + 6. Ans. a+b. 2. Divide a'—2ab-\-b^ by a— 6. .4ws. a—b. 3. Divide a?*+4a;V + 6a;y + 4a;y'+y* by a;' + 2a;y + y'. ^W5. aj'-f 2a:y4-y'. 4 50 DIVISION, 4* Divide ic* — ^x^y + ^x'y^ — 4a;y' + y* by x—y. Ans. x^—Sx'y + Sxy^—y*, 5t Divide x^—y* by x—y, Ans. x^ + x^y + x^y^ + xy^ + y*. 6. Divide ic' + y^ by a;+y. ^^. ar*— a;'y + a;y— ary'+y*. 7. Divide a:^— 9a;' + 2 7a;— 27 by a;— 3. Ans. a;'— 6a; + 9. 8. Divide 12a;*— 192 by 3a;— 6. Ans. 4a;' + 8a;'' + 16a; + 32. 9. Divide dx' — 6y' by 2a;' — 2y\ Ans, Sx* + 3a;y + Sy\ lOi Divide a;' + 5a;'y4- 5a;y'+y^ by a;'' + 4a;y+y^ Ans. x+y. 11. Divide a'-Za'¥ -[-^a'h'-h' by a'-3a'6 + 3a6'-6^ Ans. «'4-3a'6+3a6' + 6'c 12. Divide a^-h' by a^ + a%^-db^+h\ Ans. a-b, 13. Divide ia;'+a;' + ia;+f by ia; + l. Ans. a;' + 3, 14. Divide x^-\-y* by a;+y, Ans. x^—x^y-\-xy^—y^ + - ^ x+y 15. Divide a;"'+' + a;"'y+a;y'" + y'"+' by a;"*+y"*. Ans. x+y. 16. Divide a;*" + x^Y^-^-y*" by ar"* -f a;"y~ + y'". 17. Divide a'^6"— 4a"''+^'6'*— 27a'*+'^V* + 42a'»+^'5*» by a^^h^ — '7a"-'b'\ Ans. a'«+3a'^'6"— ea''*-'^'*. 18. Divide a**- a;" by a— a;. a'-^-V— af* ^%s. a '^^ + ar-''x+ a'^ V + • a—x 19. Divide ar* — (a+6+rf)a;' + (a(/ + ^'» 3—^ m— 3 *»^3 3— *» 12. Find the product of fa 2 6 a +fa 2 6 2 and |a 3 6 2 3—m »t— 3 4a"^3 96"*— 3 — |a 262. Ans. — rr— o — 7-;r-o' THEOREM IV. (112.) The difference of two quantities is divisible by the differ- ence of the same roots of the quantities. DEMONSTRATION. Let xn—y» be a general expression for the difference of two quan- m \ t \ m t tities, and {x^)'^—{y~*)'^, or x^—y~> be a general expression for the dijfference of the same roots of the two quantities. m t m t We are to prove that xn—y7 is divisible by x^—y~». m t m t^ Dividing x'^—y'^ by xm—yr,^ we obtain for the m m t_ t 1st remainder x'^~^y~»—y». m 2"* 2' * 2d " X^~ ~Tny~rr — y~s. m 3wi 3t t 8d " x'n 7^y17 — yT. m Ttn, rt t yth " X^ rtTyTB — yT, THEOREMS AND FACTORING. 59 rm m rt t m m J^ t t Since, "5^—"^, and ^=7, this remainder z=zx^~'^y—y^=:afy»^ t L 1 L y.. Because by Prop. 8, (102.) x°=l, we have xy»—y=y t — y7=0. m t m t Hence, x^—y^ is divisible by x^—y^, because, after obtaining r terms in the quotient, the remainder equals zero. The form of the mm m 2"* ' "* S^" 2^ quotient, as we see by division, is x'^~'^-\-x'^ ^y7'3-^xn~ r»y ra m t __2^ t t_ ,, , ,, Xmy « r« -j- y « r» , m 2 Q^n 0/ J mm m 2"* ' »» 3"* 2' 2*'* < 3* _ Hence, —^ j=zX'n~'^ -{- X'n~ '^y'Vi -\- X'^ nTy"^ Xmy» n Xm yrt m t _ 2t t t + Xmyl « +y. « (A.) When — =tt, and -=z^, and r=:w, we have ^=:a;'*~^ + a;'*~V + iP'-y arV-'-hary—' + y—'. (B.) PROBLEM 1. Divide x^—y^ by x—y. SOLUTION . x^—y^ Making, in formula (B.), %= 5, we have =zx^-\-x^y-^x'^y*-{' x — y xy^' + y*. PROBLEM 2. Divide a^—b^ by (a^)^— {bi) 5, or a^^—b'^^. 8OLUTIO N *^ w-^*^ Making, in formula (A.), — =|, -=|, and r=5, we get — =1^ /fr O fit) a. , 7 t a ^ b ^ SL -9 and — =T^5 • Also, putting x=za and y =&, we have — — =a2-i tt -f-ai»6>s+6»5=as +a^ ^b^ ^ +a^b^ ^ +a^ ^b^ +b^ K JO THEOKEMS AND FACTORING. EXAMPLES. 1 • Divide x^—a^hj x— a, Ans, x^ + ax-\- a^, 2. Divide a^—h" by a—h. Ans, a* + a'6 + a'6" + a6H6*. 3. Divide a'— 6" hy a—h, Ans. a" + a'h + a'6' 4- a^h^ + «&* + h\ 4. Divide a'—li' hy a—h. Ans. a' + a'h + a*b^ + a'b' + a%' + ah'' + h\ 5t Divide a^—b^ hj a—h. Ans. a' + a'b + a'b' + a*b' + a'h' + a'h' + ah' + h\ Li 3. i JL 3. 6. Divide x^—y^ by x^—y^, Ans. x^ +xy^ +x^y-\-y^, 2. 2. 7. Divide x^—y^ by x^ — y^. a. A3 £A sfi. 1 8. Divide x^—y" by x^—y^. Ans. x^ +a; 3 y6 4.3:2^3 ^x^y^ +x^y ^ +y ^ , 9i Divide x^—yhj x\ —yh Ans. xz + xyi + ici^ri' + y^. 2. JI _2_ i 10. Divide m^ — ti^ by m^ ^ —n^ . A 1.0. L J_ L 2 3 _4_ i _2_ i. 3. Ans. m'^ +^2 1^8 -|-m2»y4 4-m^2/8+m2»y2 +^212^8 ^^4 11. Divide a;'— y"' by x^—y-^. 2. 3. _3. 3 _3 _£ ^ws. iC* +a;2y * -\-x*y ^ +y *. 12. Divide a~^— a;^ by a~2j_a;3s. _L6 12 _4_ ?_ _8_ 4- -12 J.J8 Ans. a 254.^ 2Sa.35_|_a 25x^5^a ZS^^S^x^^. THEOREM V. (1 1 3») The difference of two quantities is divisible by the sum of the same roots of the quantities, when the index of the root is even, DEMONSTRATION. m t Let x'^—y'^ be a general expression for tbe difference of two quan- tities, and x^n^r'^ \y 3) r^ or xm+yra he a general expression for the sum of the same roots of the two quantities. We are to prove that xn—yi is divisible by xm-\-yr$^ when r is an even number. THEOREMS AND FACTORING. Dividing ic«—y» by xm-^-yra^ we obtain for the m m t t 1st remainder —xn ~ myVa —yl 2d u m 2»» 2« * Xn rn y ra — y » 8d (( |»_3i^ _3*_ *_ —Xn rmy r» — y« 4th u TO 4ff» 4< t Xn my n —y. tn rm rt t rth remainder, x^ '^yVs—yl when r is an even number. This remainder =0 as was shown in the Dem. of Theorem 4th. m t m t Hence, a;"^— y » is divisible by icm + y^, when r is an even number, be- cause after obtaining r terms in the quotient, the remainder equals zero. The form of the quotient, as we see by division, is, a?^~m— ajn""^ t TO 3n» 2< »» 4'» 3< ^ t 2t J_ * yr»-{-a;n m y ra — Xn ntWrj-}-, Xmy a ra — ya ra, — i. /j;„ yj TO TO TO a*** _^ Hence, when r is an even number, -^^ j=zxn~~m—xn~'^yr*-{- Xm-{- yra TO 3»» 2' "» 4"* 3' 3"* J_ 4* 2*» J 3* ajn" fny~rr — iCn~~rny~ra ■\- X m y a~ ra — X my a n, m t 2t t t +a:^yT 7ry'7~7i, (A.) When — =u: -=w ; and r=u, r being an even number, we have n s ^ =a;«-i _a;«-V + x^'-y —x'^-y + — ic'y'^' + xy"*-* x—y -y^^ (B.) PROBLEM 1. Dividea'— 6"by a + ft. SOLUTION. Making, in formula (B.), w=6, we have a-\-o 62 THEOREMS AND FACTORING. PR OBLE2i 2. Divide a^—h" bya«+6«. OLUTION. Making, in formula (A.), -=5, -=5, and r=6, we get — =- and — =-. rs 6 Also, putting x=a, and y=&, we have =a 6— a 6 66 .^a 6 6 —a ^b^+a f^ b ^ —b ^ z=z fO a6+66 a 6 —a 3""66 ^a^hi —a^h^ j^f^^if^- _^- ^ EXAMPLES. 1 . Divide a* — 6* by a + 6. ^ */4ws. a' — a^b 4- a6* — 6' 2. Divide a®— 6^ by a + 6. Ans. a'-a'b + a'b^-a'b' + a'b^-a'b' + ab'-b' 3. Divide a— 6 by ai + 6i. -4?i5. a^—aibi + albi—bi, 4. Dmde a^—b" by a* +6*. Ans. a * —a^b^ -^a^b^—b * 5. 3. j_s. 5. ^ f. 3. a. 5. Divide m^—n^ by m* +w*. ^w*. m *~— m^w* +m*w2_^4 ji i i ajL 6. Divide x^ —y' by a; + y * . Ans, x^ — ic'y * + ary ^ —y ♦ 7. Divide «'— y° by a:^ +y. J A 2 1 Ans, x^—x^y + xy^ + x^y^-\-x^y*—y^, 7. 3. 8. Divide x''—y^ by a;« +3^*. 4JL 2J_ 3_ JL5_ 3. 7. S. 2JL 3 1 JJL 1 a aJL ^ws. x^ — a; * y8 +« 8 y4 _a;2y8 .i.^. a yz —x^y « +a;82/4 _y s , 9. Divide x^—y^ by a;^ +y"TF"^ -4ws. x^—x^y^'^ 4- a?6 y * — y * ^. 10. Divide x-^—y~^ by a;-'+y-'. ^W5. ar"— a;-V~'+a:-V~'— y~'- —3. —3. -_3. —fi. 11. Divide a;*— y~' by a; + y *. Ans. x^—x^y ^ +xy ^—y *. 12. Divide x^—y~^ by a;~^^ + y~*^ T_ 2. 5 _3_ ft_ i 5_ _l_ _10. _a5. Jins. x^—x^y *2_{_./^io^ 2i_-j.sy i4_j_,^i02^ 2^— y '*2. THEOEEMS AND FACTOEING. 6S X THEOREM VI. (114.) The sum of two quantities is divisible by the sum of the swme odd roots of the quantities. DEMONSTRATION. Let «» +y"r be a general expression for the sum of two quantities, and (a;»)^+ \y^p, 0YX~^+y^ be a general expression for the sum of the same roots of the two quantities. We are to prove that xn-\-y» is divisible by a;m+yr., when ris an odd number. Dividing x^-^y'^ by xrn-{-yr»^ we obtain for the m m m t 1st remainder —x'^~ ^y~» + yT. m 2"* 2* * 2d " x'^~~T^y~^+yT, 3d " —Xn my r, +y.. 4th " x^ "^y^ 4- yT. m rm tt t rth remainder —x^ Wy'TT-^ys when r is an odd number. TO t_ This remainder is obviously equal to zero. Hence, a;n+y« is divisible by a;"»+y«, when r is an odd number, because after obtain- ing r terms in the quotient, the remainder is zero. mm m Qm t The form of the quotient, as we see by division, is x'^~'m—x'^~~^yr» m 3m 2' 2jn t 3< m t 2< _< t -j- X^ rrTyrs •\- -\- X m y t ra X^nV a ra -\- y a f», m t_ Xn-{-ya ^_^ ^_2^ i. Hence, when r is an odd number —^ -=xn m—x* r7iyrs-{- x^-\-y^ w* 3CT 2* 2»n « 3t m t 2< * * Xn m y ra ...... ...-j-iC»^y« •"* Xmy a *■* "j" V • '*. (-"••) When — =^*, -=w, and r=u, r being an odd number, — n ' s ' ' " « +y PROBLEM 1. Dividea* + 6'by a + 6. &4c THEOREMS AND FACTORDiTO. SOLUTION, Main Tig, in formula (B.), u=5, we have r-=a*— a"6+a*6'— ab' + b\ PROBLEM 2. Divide a^ + b^ hy a^+b^, SOLUTION. Making, in formula (A.), — =8, -=8, and r=:Sj we get — =-, and — =-. rs 3 Also, puting x=aj and y=^bj we have a3 + 63 EXAMPLES. 1. Divide a' + 6' by a + b, Ans, a^—ab-\-b\ 2. Divide a"4-6" by a' + 6'. Ans. a^-a'b^-\-a'b*-a'b' + b\ 8. Dividea' + 6'by a + 6. Ans. a'-a'b-ha*b'-a'b'-{-a'b'-ab' + b\ L 4. Divide a + bhj a^ +b~' 4. 3.1 2.2. iS. 1. ^Tis. a*— a^ft* +a*6^ +a^6« +65. 5. Divide a« + 6' by o* + 6«. -<4w«. a '^ —a ^ b^ +a ^ b ^ —a^o^ +6 * . 6. Divide a" + 6' by a^ + 6. .4w«. a^ —a^ b + a « 6"— a«6' + 6*. 7. Divide a' + 6* by a + 6*. ^»5. a*— a'6^ ^a^'b^—ab'^ + b^^. 8. Divide a + 6' by a^' + bK A J.2 2. A J.,6 ,8 Ans. a^ —a^b^ -\-a^b^—a^b^ -\-b^, 9. Divide a-'-\-b-' by a-' + b-\ Ans. a-*-a-'b-' + a-^b-'-a-'b-' + b~\ THEOREMS AND FACTORING. 65 10. Divide a^ -i-b^ hj a'^^ +b^\ _8_ _6_ _3_ _4_ _3_ ^_ _9_ J 11. Divide a^ + 6 Hya^^+i ^^^ _8_ _6_ 4_ _4_ 8_ _2_ _JL2. 15. THEOREM VII. (11 5.) The sum of the squares of two quantities, plus twice the product of the quantities, is equal to the square of the sum of the quantities, DEMONSTRATION. Let a and b represent any two quantities, then a^ + b^+2ab, or a' + 2a6 + 6^ represents the sum of their squares plus twice their pro- duct. We are to prove that a' + 6' + 2ab, or a' + 2a6 + 6" = (a + b)\ By Theorem 1, we have (a + by=a'' + 2ab-\-b^\ ,-. a^ + 2ab + b\ or a^ + b'' + 2ab=(a + b)\ PROBLEM 1. Find the factors of oi^+4x+4, SOLUTION. Since x^ + 4:X + 4, or x^-[-4: + 4:X=x^ + 2^ + 2'2x, it represents the sum of the squares of x and 2, plus twice their product. Hence, by the Theorem, we have a;' + 2' + 2-2a;, or x'' + 4x + 4=(x + 2y=:{x + 2) {x + 2). EXAMPLES. 1 , Find the factors of a;" + y' + 2xy. Ans. (a: +y) (^ 4- y). 2« Find the factors ofa^ + 2ay + t/*, Ans. (a-{-y)(a-\-y). 3. Find the factors of a;' + 10a; + 25. Ans. (x + 5){x + 6). 4. Find the factors of ar' + 24a; + 144. Ans. (a; + 12)(a; + 12). 5. Find the factors of m'' + 114m + 3249. Ans. (m + 57)(m + 57). 6* Find the factors x + 2xiyi + y. Ans. {xi + yi) (ar^ + yi). W THEOREMS AJSTD FACTORING. 7. Find the factors of 4x^-\-4:X+l. Ans. {2x + l)(2a; + 1). 8. Find the factors of 169a;'* + 26a: +1. Ans. {lSx + l)(lSx + l). 9. Find the factors of 25ar-'' + 60a;-Vi + 36yf . Ans, (5x-^ + 6yi)(6ar' + 6yi). 10. Find the factors of a;+a+2aiari^. Ans. (xi+ai){xi + ai). 11. Find the factors of xi+a'x^ + 2ax. 3. J 3. A Ans. {x^ +ax^){x^ +ax^). 12. Find the factors of 9xi + 49a;2y~s + 42a;y'"^. • Ans. {3x^-\-1x*y-'^)(dx*+1x*y-^). THEOREM VIII. (116.) The sum of the squares of two quantities, minus twice the product of the quantities^ is equal to the square of the difference of the quantities. DEMONSTRATION. Let a and b represent any two quantities, then a^-\-b^—2db, or a^—2ab + h'^ represents the sum of their squares, minus twice their product. We are to prove that a^ + b^— 2abj or a^ — 2ab -\-b^=(a— by. By Theorem U. we have {a—by=a^-2ab + b^ ; . • . a''^2ab + b% OT a^ + b''-2ab={a-by. PROBLEM. Find the factors of 9a; + 4yi—12a;iyi. SOLUTION. Since 9a; + 4yi— 12a;i3^=(3a;i)' + (2yi)''— 2-(3a;i)(2?/i), it repre- sents the sum of the squares of two quantities, minus twice their pro- duct Hence, by the Theorem, we have 9a; + 4yi— 12a;iyi, or 9a;— 12a;iyi + 4yi=:(3a;i— 2yi)''=(3a;i— 2yi)(3a;i— 2yi). EXAMPLES. 1. Find the factors of a;''—2a;y + 2/^ Ans. {x—y){x—y), 2. Find the factors of m'*— 4 w^?^ + 4?l^ Ans. (m—2n){m—2n), 2x 3. Find the factors of x^-\-^ — -. Ans. (a;— i)(a;— i). THEOREMS AND FACTORING. 67 4. Pmd the factors oia^—2a^x^+x^, Ans. {d^ —x''){a^ —x^y 5. Find the factors ofa'—4aV + 4/. Am, {a^ —2y^)((fi —^y""), 6. Fmd the factors of 9a;' + 9y— 18a;V^. Ans. (3a;*— 32/i)(3a;*— 3yi). 7. Find the factors of :^a;'— 3a;^ + 4. Ans, (|a;2_2)(f a;^— 2). 16 8. Find the factors of 16a;f — 16a;i2r' + 4y-'. Ans. (4a;i— 2y-')(4a;i— 23r'). 9. Find the factors of 3a;f +3y-f— Garfy-i. Ans. (34a;f-3i2r-4) (3Ja;f-3iy-i). 10. Fmd the factors of a;'"" + y'''»— 2a;'"y''. Ans, {f—y%(if^—'f)- 11. Find the factors of 4a;*'»-16a;""y^ + 16y". Ans. (2a;'^'«— 42^)(2a;'""— 4yl). 12. Find the factors of a;^y~T— 2+ a; «y». Ans. {x~2^y 2«— a; 2'»y2«)(a;2n— a; 2«y2#). THEOREM IX. (117,) The difference of the squares of two quantities is equal to the product of the sum and difference of the quantities. DEMONSTRATION. Let a and b represent any two quantities, then a^—b' represents the difference of their squares. We are to prove that a'^—¥={a + b){a—b). By Theorem m., we have (a + 6)(a— 6)=a'— 6"), .*. a'— 6"= {a + b){a—b). PROBLEM. Find the factors of a— 6. SOLUTION. Since a— 6=(ai)'— (H)", it represents the difference of the squares of the quantities ai and bi. Hence, by the Theorem, we have a— 6=(ai + H)(ai— &i). 68 THEOREMS AND FACTORINa. EXAMPLES. 1, Find the factors of a;'— y^ Ans. (x-\-y){x—y) 2, Find the factors of ic*—y'. Ans. (a;* + y')(a;*— y') 3* Find the factors of a?'— y. Ans, (a^+y¥)(a:f--y|-) 4. Fmd the factors of a;»-y^ Ans. (a;t +yf)(a;f-y|) 5. Find the factors of ar-*-2r'. Am. (^ + -)(^'-2rO 6. Find the factors of 9a;''— 4y*. Ans. (3a;+2y')(3a?— 2y') 7. Find the factors of 2a;— 2y. Ans. L(2a;)i + (27y)i](2ia;i-2iyi) 8. Find the factors of 3'a;^— 6y«. Ans. {3M + 52yTV)(3ia;6_5^yTV) 9i Find the factors of 4:X^—y~^. Ans. l^x^ + -A (2x^ r) 10. Find the factors of 16ar-''-25y. Ans. (i+Syi) (--5yA, 11. Find the factors of 4af*— G^r". 2a;2+3y~^j (2a;2— 3y ^), 2m 2m 2* 12. Find the factors of a?"""^- 4a;~Jry T. Xn y» ' Xn y* ', THEOREM X. (11 8.) The expression x^—yT=\x^—y7»)\xn m-{-xn myr,-{- m 3m 2* 2m t 3* m t 2< _£ t\ Xn my r« + -{-X m y 9 n ■\-XTny » ra ■\-y a r$ f ^ whlch, Wi t when — , -, and r respectively equals u, becomes a;"— y''=(a;— y) n s ' (ar-^+a.«-«y+a;«-y + +x^y'^'' + xy'^^+ir'\ DEMONSTRATION. The truth of this Theorem depends on the truth of Theorem IV., and the fact that the divisor multiplied by the quotient equals the dividend. THEOEEMS AND FACTOBINQt, $& PROBLEM 1. Fmd the factors of x^—y^ SOLUTION. In this M=5, and making r=6, we have, by the Theorem, a;*— y* Other factors may be obtained by assigning other positive integral values to r, PROBLEM 2. Find the factors of xl—yh SOLUTION. In this example, — =-, and -=-, and we are at liberty to make r any positive integer. Let r then equal 6 ; whence, — — ^» ^^^ — =-—. Hence, by the Theorem, we have x* —y^ = [x'^^ —y^'^j rs 15 (3 92 34 32 8\ a;5 4-a;2 0yTj^a;ToyT5>|_^2 0ys _|_yTsj, Other factors may be ob- tained by assigning different positive integral values to r. EXAMPLES. 1. Find the factors of a'— a;'. Ans, (a—x){a'' + ax-\-x''), or [a^—x^) [J + x^), 2« Find the factors of a^—x*. Ans. (a—x){a-{-x), or \a^—x^) [a^ + ax^ +aix + x^) , 3. Find the factors of a^—x\ Ans. (a-x){a' -ha'x + a'x' -{-a^x' -{-ax*+x% or {a'-x'){a'+x'). it Find the factors of d'—x\ Ans, {a—x) {a" + a^x + d'x' + a"a;' +a''a;* + ax^ + x\) 5* Find the factors of a'— a;^ Ans.{a'-x')(a' + a'x^+x'\oY {J-x^) {J + a'x^ + Jx' + x^) 6. Find the factors of a'^—x'". Ans. (a-a:)(a''+a'a;+aV + aV+aV + aV+aV + aV + ax'+x'). 70 THEOREMS AND FACTORING. 7. Find the factors of a'"— a;". Ans. (a''-ar')(a« + aV4-aV + aV+a;'), or {a'-x'')(a''+x'), 8t Find the factors of m—w. Ans. {mi—ni){m^ + mhii-\-nf)j or ^mi—ni;){mi+ni), 9» Find the factors oix^—y*. Ans. {x''-y){x'+x'y + xY+y% or {x'-f){x'+y^). 10. Find the factors of x~^—yz. Ans. {ar'-y^){x-'-{-x-''y^+(xr'y^'\-y\ or (ar-''-yf)(^+yf j. 11, Find the factors o{af'—y\ 3 3 12* Find the factors of x^—y *, Ans. I XI s— )ia;i5+ + + + — ). THEOREM XI. (119.) The expression x'^—y'^= \x^ + y m j [x^ ~m—xn ^yV, -f Xn m y rs , ..,.,, X rn y a r» -^X^ny s ra — y a ra ) when r is an even number, which, when — , -, and r respectively equals u, becomes x''—y''={x+y){x'^^—x'*~^y + x'^^y^— — xy-'-^-xy^'-y^K DEMONSTRATION. The truth of this Theorem depends on the truth of Theorem V., and the fact that the divisor multiplied by the quotient equals the dividend. PROBLEM 1. Find the factors of x^—y^. SOLUTION. In this example w=6, and making /=6, we have, by the Theorem, a;"— y°=z(a? + 2/)(a;'— a;*y4-a;y— icy + a?/— /). Making r= 2, we have x^^y^^{x^j^f){x'^f). THEOREMS AND FACTORING. 71 Other factors may be obtained by assuming r equal to other even numbers. PROBLEM 2. Find the factors of a;* * — y^. SOLUTION. In this example, — =-, and -=-, and we are at liberty to make /i Tc So r any positive even number. Let r then equal 6 ; whence, — =-, and — =-. Hence, by the Theorem, we have, rn 8 rs 9 ' -^ ' * Other factors may be obtained by assuming r to be equal to other even numbers. EXAMPLES. 1. Find the factors of a*—b*, Ans. (a + b)(a'^a'h + ab'-b% or (a' + 6»)(a"~6'),&c. 2. Find the factors of a—b. Ans. (ai + bl){ai—aUi + aibi—bl), &c. 3. Find the factors of a'—b^^ Ans. \a + b) (a—b), or [a^ + b^) [a^—a^b^ + ab^—aH -\-ah^—b^), &c. 4. Find the factors of a'—b'. Ans. (a^ + b) {a^-ah + ab'^-^ah' + ah^-b'), &o. 5. Find the factors of ar*—b*. \a /\a' a" a / 6. Fmd the factors of a-'—b-'\ /11\/1 1 1 i 1 1\„ 7. Find the factors of aJ + b\ Ans. {J-b') {J—ah' + ah*-b'), &c 8. Find the factors of 16a^-b'\ Ans. (4a' + 6^)(8a*'-— 4aV + 2a'6«-6"). &c. 72 THEOREMS AND FACTOEINa. 9. Find the factors of «*'"—«"». Ans. (a"* + a;"»)(a''"*— a""a;»'» + a'"a;*"— a;"*), ' + aV4-a&* + &')- 13. a'-b' = {a + b){a'-a'b-{-a'b'-a'b' + ab'-by 14. a«-6« = (a''-6'0K + «'^' + **)- 15. a'-b'={a' + b'){a'-¥), 16. a''-'6«=(a + 6)(a-6)K + a& + 6')(«'-«* + *'). 17. a* + a'^6'' + 6*=(a' + a6 + «>')(«'-«& + ^')- 18. a'-hb':^(a-\-b)(a'-~-ab-^by 19. a^ + a*6 + a'6'^ + a'6' + a6* + 6'*=:(a +6)(a'' + a6 + 6'')(a'-a6 + 6') 20. a^-a*6 + a'6'-a'6' + a^*~2»'=(«-^) (a^' + aft + J") (a'-aft + fi") ={a-b){a' + a'b' + b'). CHAPTER IV. GREATEST COMMON DIVISOB. (126.) A multiple of a quantity, is a quantity that contains it an an exact number of times. Thus, 6 is a multiple of 2 and 3, and ah is a multiple of a and h. (1 J27«) A measure of a quantity, is a quantity that is contained in it an exact number of times. Thus, 2 and 3 are measures of 6, and a and h are. measures of a6. (128.) Kcommmi measure or common divisor of two or more quantities is one that is exactly contained in each of them. (129.) The greatest common measure or greatest common divisor of two or more quantities is the greatest quantity that is exactly con- tained in each of them. THEOREM I. (130.) If a quantity measures another quantity it will measure any multiple of that quantity, DEMONSTRATION. Let a be a quantity, then ta is a multiple of a. We are to prove that if a quantity d measures a it will also measure ta. Let r be the number of times d is contained in a; whence, we have, a—rd and ta^ztrd. Now d measures trd and must, therefore, measure to, which is equal to trd. THEOREM II. (131.) If a. quantity measures two other quantities, it will also measure their sum and their difference. DEMONSTRATION. Let Rd and rd be two quantities that are divisible by d. GKEATEST COMMON DIVISOR. gl We are to prove that Bd + rd and Rd—rd are both divisible by d» Since Bd+rd=d{R-\-r) and Rd—rd=d{R—r\ we see that d is a factor in both these expressions, and is, therefore, a divisor of both. PROBLEM. (132.) To find the greatest common divisor of two or more monomials. RULE. Resolve the monomials into their factors, and the product of the fojctors common to all the mcmomials will he the greatest common divisor. PROBLEM. (133.) Find the greatest common divisor of 3a^b^d\ 6ab^d\ and l^bd', ^ 8 OLUTI O N. Resolving these monomials into factors, we have 3a'b'd*= a'bd''Sbd\ 6ab'd'=2abd ' Bbd% and 12bd^= 4 • 36c?'. From which it is seen that Sbd^ contains all the common factors, and is, therefore, the greatest common divisor. EXAMPLES. 1, Find the greatest common divisor of 12a6'c and 256V. Ans. b*e, 2, Find the greatest common divisor of 3a'6V and 6a*6V'. Ans. 3a'6V. 3* Find the greatest common divisor of 3a^b~^ and 2abcd^, Ans. abh 4. Find the greatest common divisor of 49a^h'^c* and 63a'6'c'. Ans. la'bY. 5* Find the greatest common divisor of ai and ai. 6« Find the greatest common divisor of a-i and ah 7, Find the greatest common divisor of a-i and -i. Remark. — The last three examples are left unanswered to call out thought. 82 GEEATEST COMMON DIVISOR. PROBLEM. (134.) To find the greatest common divisor of two polynomials. RULE. 1. Find the greatest monomial factor that is contained in both poly- nomials, and reserve it. 2. Meject the remaining monomial factors from each polynomial. 3. Arrange the terms of the resulting polynomials according to the powers of some letter in both, and consider that polynomial of which the leading letter in the first term has the least exponent as the divisor, and the other polynomial as the dividend. 4. Multiply the dividend by the lea^t monomial that will render the first term of the dividend exactly divisible by the first term of the divisor, 6. Divide the dividend by the divisor^ and continue the division until the highest exponent of the leading lefter in the remainder is less than the highest exponent of the leading letter in the divisor. [^If the coefficient of the first term in any remainder is not divisible by the co- • efficient of the first term in the divisor, to avoid fractions multiply the remainder by such a number a^ will render its first coefficient exactly divisible by the coefficient of the first term of the divisor^ 6. Reject from the remainder its greatest monomial factor, and then consider the result a new divisor, and the former divisor a new divi- dend, proceed cw before, and continue the process until the remainder is zero. 7. Multiply the last divisor by the reserved monomial, and the pro- duct will be the greatest common divisor of the given polynomials. DEMONSTRATION. Let A and B represent two polynomials of which we seek the greatest common divisor. Let C and D represent two other polynomials, neither of which can be divided by a monomial. Suppose A=a^bC audi B^zo^cD ; a, b, and c being monomials. 1. It is evident that the greatest monomial factor common to a^bO and a^cD is a^, which must be reserved, because it evidently is a factor of the greatest common divisor of A and B, or, which is the same, of «'6(7and aVD. GREATEST COMMON DIVISOR. 83 2. We have left ahC and cD. We now seek the greatest common divisor of these polynomials. Since a is not a factor common to both these polynomials, it can not be* a factor of their common divisor, and therefore may be rejected. For the same reason h and c may be re- jected. Hence the greatest common divisor of ahC and cD is the same as the greatest common divisor of C and D. 3. Suppose that the terms of the polynomials and D are arranged according to the powers of the same letter in both, and that the ex- ponent of the leading letter in the first term of C is less than the ex- ponent of the same letter in the first term of D. Therefore, consider (7 as a divisor and i> as a dividend. 4 Suppose the first term of C contains the monomial 3m and that the first term of D does not. Then, in order that the first term of the quotient should not be fractional, the polynomial D should be multi- pHed by 3?7i. The greatest common divisor of C and D is the same* as the greatest common divisor of C and 3mi>, since the introduced monomial 3m can form no part of the greatest common divisor of C and 3mZ), because by hypothesis 3m can not be a factor of C. Operation. 5. Divide Bml) by C, and suppose the first ferm C)3mD{q of the quotient to be q, and the first remainder qC E. Again, suppose that the first coefficient of E C contains the factor 2, and that the first co- 2 eflScient of E does not. Then multiply E by 2 C)2E{q' and divide the result 2E by C and let q' repre- q'O sent the first term of the quotient, and i^ the bp) F remainder. Also suppose that the exponent of G)2nC{q" the leading letter in the first term of i^is less q"G than the exponent of the same letter in the first term of C. 6. Suppose that the greatest monomial factor of ^is bp. Reject this factor, or, in other words, divide F by 5p, and let G represent the result, which consider as a new divisor and C as a new dividend. Suppose, then, the first term of G contains the monomial 2n and the first term of C does not. Then, in order that the first term of the quotient should not be jfractional, the polynomial O should be multi- phed by 2n. Let q" be the exact quotient of 2nC by G. 7. 6^^ is the greatest common di/isor of C and i>, and multiplying 84 GREATEST COMMON DIVISOR, it by the reserved monomial «", we have a^Q for the greatest com- mon divisor of A and B, We prove that G is the greatest common divisor of C and D as follows : — Let G' be the greatest common divisor of C and B. We have al- ready shown that the greatest common divisor of C and ^nD is the same as the greatest common divisor of C and D. Since G' is a measure of (7 and D, it must (130) be a measure oi qC and 3mi>, and it must also ( 1 3 1 ) be a measure of JS, the dif- ference between qC and 3mi). Since G' is a measure of C and ^, it must also(130) be a measure of gf'Cand 2E^ and it must also (131) be a measure of i^, the difference between q' C and 2E. Because, by hypothesis, C and D contain no monomial factors, it follows that G', their greatest common divisor, is neither a monomial nor divisible by a monomial. But G' measures F, and consequently must measure G, which re- presents F after its monomial factors are rejected. Hence, the great- est common measure of C and D can not be greater than G, and, therefore, if (r is a common measure of and i>, it must be the Since G measures 2nO^ and is not a monomial, it must also mea- sure (7, and consequently must measure 5'' (7. But G also measures F^ therefore it must measure 2E, which is the sum of F and q'C. Because, G measures 2-fi', and is not a monomial, it must also measure F : G must also measure qC^ and therefore must measure 3wiZ>, which is the sum of E and qC. Since G measures 3mZ>, and is not a monomial, it must also measure D, Hence, G is a common measure of C and i>, and must, therefore, as shown above, be the greatest. Now let C—QG2indiD= Q'G, and we have A=a^bQG and JS=a^cQ'G ; whence, we see that a^G must be the common measure of A and B, because Q and Q' may be rejected, since they can have no common measure. Therefore the truth of the rule is established. PROBLEM. (135») Find the greatest common divisor of the polynomials, 6a'b~6a'bi/—2by' -{-2aby''{A) and lia'b + Sbf-ldabt/ (B). €^REATEST COMMON DIVISOR. 85 SOLUTION. I* We see by inspection that b is the greatest monomial that is common to both polynomials, which reserve. 2. By rejecting the monomial 2 from (^), and 3 from (J5), we have 3a^—Sa^j/—y^ + ay^ and 4a'' + 2/'* — hay. 3* Arranging these results according to the powers of the letter a, we have 3a^ — Za^y + ay"^ — y^ and 4a'' — hay + y'. Hence, the latter must be the divisor. Operation, 3a'- 3aV+ oy'~ f 4 4a'— 6ay+y»)12a'— 12aV+ 4ay*— 4/(8a 12a''— 15a V+ ^ay"" Za^y^ ay'- 4/ 4 12aV+ 4ay^-16yX8y 12aV— 15ay''+ 3/ 19y')19ay^l92^ a— y)4a'— 6ay+y*(4a— y 4a'— 4ay — ay + y' 4( Since the first term of the divisor is not contained in the divi- dend, we multiply the dividend by 4, and then dividing, we have for the first remainder Sa'zz + ay''— 4?/^, which must be multiplied by 4 to avoid a fractional result in the quotient. Continuing the division, we have for the next remainder 19ay' — 19y', in which the exponent of the leading letter a is less than the exponent of the same letter in the first term of the divisor. 5t Rejecting from this remainder IQy'*, the greatest monomial con- tained in it, we have a— y, which constitutes the new divisor, which divided into 4a'— 5ay + 2/^, leaves zero for a remainder. 7t Hence, the last divisor a—y multiplied by the reserved mono- mial 6, gives h(a—y) for the greatest common divisor of the given polynomials (A) and (B). ,86 GREATEST COMMON DIVISOB. EXAMPLES. 1. Find the greatest common divisor of ic*— 8a;' + 2 la;'— 20a; + 4 and 23;"— 12a;' + 2 la;— 10. Ans. x—2, 2. Find the greatest common divisor of Sa'*— 2a— 1 and 4a" — 2a'^— 3a + l. Ans. a—1. « 3« Find the greafest common divisor of a^—b^ and a^ + 2a'6 H-2a6' + 6'. Ans. a'' + ab + b\ 4. Find the greatest common divisor of a*-\-a^b—ab^—b* and a^+d'b' + b\ Ans. a'' + ab + b\ 5t Find the greatest common divisor of a'' — 2aa; + a;'' and a^—a^x — ax^+x^. Ans. a' — 2aa; + a;'. 6« Find the greatest common divisor of a;' + 9a;^ + 27a^— 98 and a;' + 12a;— 28. Ans. x—2. 7. Find the greatest common measure of Va'— 23a6 + 66' and 5a* — 18a''b + llab^—6b\ Ans. a—Sb. 8. Find the greatest common divisor of 6a® + 15a*6— 4a''c' — lOa'bc'' and 9a'6— 2'7a'6c— 6a6c'' + 186c'. Ans. 3a»— 2c'. 9. Find the greatest common divisor of SGa'^'- 18a'6'— 27a*6' + 9a'6' and27a'6'— 18a'6'-9a'6'. Ans. 9a%\a—l). 10. Find the greatest common divisor of {c—d)a^ + {2bc—2bd)a -{-(b'c-b'd) and {bc-bd + c'-cd)a + (b'd + bc'-b'c-bcd). Ans. c—d. PROBLEM. (136.) To find the greatest common divisor of two or more polynomials. RULE. Resolve all the polynomials into their simplest factors^ and take the product of those which are common, for Hie greatest common divisor. DEMONSTRATION. The truth of this rule is a consequence of the definition of the greatest common divisor. PROBLEM. (137.) Find the greatest common divisor of a;'— 6a; + 2aa;— 2a6 and x^ 4- ax^ + bx^ — 2a^x + bax — 26a'. . GEEATEST COMMON DIVISOR. 87 SOLUTION. Arranging x'^—hx + 2ax—2ah^ we have x"^ -\-2ax—hx~2ah. Factoring, we obtain {x-\-2a)x—h(x + 2a) = (a; + 2a){x—h), Arranging, x^ -\-ax'^ + hx'^—2a^x-\- hax—2ha^, we have x^-{-ax^ — 2aV + hx"" + bax—2ba^. Since, ax^=2ax^ — ax^ and bax=:2hax — bax^ we have x^-\-ax^ — 2a'x + bx'^ + bax—2ba^=x^ + 2ax^—ax'^ — 2a^x + bx^ + 2bax—bax—2ba^. Factoring, we obtain {x + 2a) x^—ax{x-\-2a) + bx (x + 2a) — ab (x + 2a) = (x + 2a) (x"^ — ax + bx— ab) . But, x'^—ax + bx—ab=(x—a)x-\-b{x—a) = (x—a)(x-\-b). Hence, we have a;'H-aa;'' + 6a;''— 2a''a; + 6aa;— 25a''=(a; + 2a)(a;— a) (x + b). We also have x^—bx + 2ax—2ab=(x-\-2a){x—b). Whence we see that a; + 2a is the only common factor. EXAMPLES. 1. Find the greatest common divisor of a:'— a", x*—a*, and x^^a^, Ans. x—a. 2. Find the greatest common divisor of a*— a;*, a'— a^r— aa;^ + aj', and 2a^b—2abx^. Ans. a^—x^. 3. Find the greatest common divisor of 5a^ + 10a*b + 5a''b^j a'S^- 2a'b'' + 2a6' + b\ and a^—b\ Ans, a + b. 4. Find the greatest common divisor of x^ + ax^^a^x—a* and a;*4-aV + a*. Ans. x'' + ax + a\ 5. Find the greatest common divisor of ar*— 1, x^+x^^ a;*— 1, and a:* + 2a;' + l. Ans. x^ + 1. 6. Find the greatest common divisor of x^—b^x and x^-j-2bx-\-b^. Ans. x + b. 7. Find the greatest common divisor of Sa'—3a'6 + a6'— 6' and 4a'b—5ab^ + b^. Ans. a—b. 8. Find the greatest common divisor of a' — 5a6 + 46', a^—d'b -\-3ab'^ — Sb^,aiida*—b*. Ans. a—b. 9. Find the greatest common divisor of 12a;*— 24a:^y + 12a;y and 8ajy— 24a;y4-24a'y*— 82/'. Ans. 4:(x^—2x7/ + f). 10. Find the greatest common divisor of 2a;'' + 3aa; + a", 2aa;'*— a'a; —a", and 6a;'^4-3aar. Ans. 2x+a. CHAPTER V. LEAST COMMON MULTIPLE. (138.) The LEAST COMMON MULTIPLE of two OT more quantities is the least quantity that can be measured by all of them. PROBLEM. (139.) To find the least common multiple of two or more quantities. RULE. Resolve the given quantities into their simplest factors, and find the continued product of all the different factors, taking each factor the greatest number of times that it occurs as a factor in any of the given quantities, and the result will he the least common multiple sought. DEMONSTRATION. The least common multiple must contain as many factors as are contained in any of the given quantities. PEOBLEM. (140.) Find the least conmaon multiple of a'— a;', a'— 2aaJ+ar*, and a'^^x^. SOLUTION. Factoring these quantities, we have {a-\-x)(a^x), {a—x){a—x), and (a — x){a + x){a^ -\- x^). Hence, the least common multiple is (a + a;)(a— a:)(a— ar)(a' + a;') = a' — ax* — a*x + «*. EXAMPLES. 1. Find the least common multiple of 2a'ar and lOaV. Ans. lOa'a;'. 2i Find the least common multiple of Sax', 4:a^h^x, and ^a^h^xy. Ans. 1Aa^h^x*y. LEAST COMMON MULTIPLE. 89 3i Find the least common multiple of x^—y"* and «'— y'. Ans. (a;+y)(a;'— y"). 4* Find the least common multiple oia^—h^ and a^ + &^ Am. (a-6)(a«+6»). 5* Find the least common multiple of a;'— 8a; + '7 and x^ + ^x—Q, Ans. ic^— 5*70: + 56. 6« Find the least common multiple of a;'— aj'y—a^y' + y", x^—x^y + xy^—y^ and a;*— y*. ^W5. x^—xy*—x*y+y^. 7. Find the least common multiple of (a + 6)', a^—b^^ {a— by, and 8. Find the least common multiple of a +6, a— 6, d' + ct6 + 6', and a'^ab + b'. Ans. a'—b\ 9i Find the least common multiple of a^+^a'b + ^ab^ + b^, a' + 2a6 + 6'', and a'— 6'. Ans. a* + 2a'&— 2a6'— 6*. 10. Find the least common multiple of a;*— Sa-'+Qa;"— '7a; + 2, and x^—Qx' + Sx—Z. Ans. a;*— 2a;*— 6a:' + 20a;''— 19a;+6. CHAPTER VI. ALGEBRAIC FRACTIONS. (141.) An Algebraic Fraction is an algebraic expression de- noting the di\dsion of one quantity by another, and is written in the same manner as a common fraction in arithmetic. An algebraic fraction may also be considered as a certain part of unity. THEOREM I. (14:3«) The value of a fraction is not changed when both of its terms are multipHed by the same quantity. DEMONSTRATION. A Let -^ represent any algebraic fraction. We are to prove that -=r-= — =r. ^ B mB By (101), we have — g= — ^ — ~'~B~~~W^ because m°=l. THEOREM II. (143.) The value of a fraction is not changed when both of its terms are divided by the same quantity. DEMONSTRATION. Let -5- represent any algebraic fraction. We are to prove that — -=— -^ — . B B-\-m Since, A-\-m= — and ^-T-m= — ^^ we have, by (101) — r7»= mr^m^A m^A A ^ „ ^ ALGEBRAIC FRACTIONS. 91 PROBLEM. (144.) To reduce a fraction to its lowest terms. RULE. Resolve both terms into their simplest factors, and cancel those that are common. Or, divide both terms by their greatest common divisor, DEMONSTRATION. The accuracy of this rule depends upon (1 43). PROBLEM. (145.) Reduce —. ^ 5 x to its lowest terms. ^ ^ x^+ax^—a^x—a* SOLUTION. Factoring, we have a;* + aV 4- a* _{x^ + a^ + ax){x^-{-a^—ax)__ x^-\-ax^—a^x—a^ ~ x^—a^ + ax^—a^x ~~ (x^-\- J 2a;' + 3a;^ + a;^ v i .. a ^^ + 1 13* Reduce — = 5 — -— to its lowest terms. Ans, — -— . x^—x^—2x x + 2 %M -n n ac-\-bd+ad-\-bc ^ .^ , , . II. Reduce -^ . ^, - — r-r? to its lowest terms. a/^+26a;4-2aa;-|-6/^ 15. Reduce — -. — 5 — -7 to its lowest terms. a^—b^—c^—2bc, . a-hb+e Ans. = — . a — 6 — c ^o T. J a'-3a&4-ac + 26'-26c ^ .^ , 16. Reduce 5 — ,„ . ^, 5 to its lowest terms. a — +26c— c a— 26 Ans, r — . a + 6— c — ^ , (a + 6)(a + 6+c)(a + 6— c) , ., , "• ^^^^ 2a'6- + 2aV + 26V ^a^3Fi:?-^" ^^ ^^^^^' *^"^- Ans, (c + a— 6)(6— o + c)* ALGEBRAIC FBACTTOKS. 93 18. Reduce — — 77-^ — ( =- to its lowest terms. Ans, x^-^(b + c)x + bc *A 1. J 2x'-\-x'-Sx + 5 ^ .. 19. Reduce t-^ — — — to its lowest terms. ^x^ — l2x-\-5 Ans. — — . PROBLEM. (146.) To reduce a fraction to an entire or mixed quantity. RULE. Divide the numerator by the denominator, and when there is a re- mainder place it over the denominator for the fractional part, and connect it with the entire part of the quotient by the sign of addition, or change all the signs of the numerator of the fractional part, and connect it with the entire part of the quotient by the sign of sub- traction, DEMONSTRATION. The accuracy of this rule is an obvious consequence of the accuracy of the process of division. » PROBLEM. (1 47 .) Reduce ^y — y +a;y— y— a;+ ^ ^ ^^^ quantity. SOLtTTION. Dividing the numerator by «'— 1, we get y' and the fractional , xy'-y- x-{-\_ {x-\)y-l{x-l) _{x-\)(y-^l)_y--\ connected with y' by the sign of addition gives y^ + ~~Tj ^^'^ X -x" X. with the sign of subtraction gives y' -, or y^—z — -, X -\- A 1 -x'X EXAMPLES. 1, Reduce — ^ — to a mixed quantity. Ans, a+j. 2* Reduce to a mixed quantity. Ans. aH , a ^ J a 94' ALGEBRAIC FRACTIONS. 3* Reduce to a mixed quantity. Ans. 5a + l-\ — — . , ^ , 4a + 2ax + b . , . j . ^ ^ 4* Reduce to a mixed quantity. Ans. 44-2a;H — . ., -.^ , x'y—xy—x+1 . , . ^ 1 o* Reduce —^ to a mixed quantity. Ans, x , xy—y ^ y 6. Reduce •— to a mixed quantity. Ans. 1— a — . \^x-^*J a;-4-2 7. Reduce — - — to a mixed quantity. Ans, Zx-V\-\ — — -. 8. Reduce to a mixed quantity. ox Ans. ^x -. 6x T9. Reduce to a mixed quantity. oCI/ Ans. a— 3c— 2H , 3a 10. Reduce to an entire quantity. Ans. 5{x^-{-xy-{-y'*), X y PROBLEM. (1 48.) To reduce a mixed quantity to a fractional form. RULE. Multiply the entire part by the denominator of the fractional paA and add the numerator to the product when the sign of the fraction is plus, and subtract it from it when the sign of the fraction is minus ; this result placed over the denominator will give the required form, DEMONSTRATION. This rule indicates a process whicli is just the reverse of that in the last rule, and must, therefore, be accurate, because the problem to be solved is just the reverse. PROBLEM. (149.) Reduce 2a to a fractional form. ALGEBRAIC FRACTIONS. 95 SOLUTION. Multiplying 2a by c gives 2ac, and from this must be subtracted Bx—bj because the sign of the fraction is minus, whence we have 2a—{3x-b) 2a-Sx + b . ^, ... ^^ -= for the required form. c c Scholium. — It should be remembered that the 3a; in this example is plus, and that the minus sign before the fraction belongs to the whole fraction, or what is the same thing, belongs to the whole numerator as indicated in the solution by the parentheses. EXAMPLES. d^ /^a 2a^ !• Reduce 1 H — = r to a fractional form. Ans. -= r. a +a; a'+aj' x xix -4- 2/1 2. Reduce x-\ to a fractional form. Arts, — ^. y y n T. T . 2a;— 5 - .. , ^ . 6(2a; + l) 3, Reduce 4a; — to a fractional form. Atis, ^ — '-. 3 3 4i Reduce 3a;— 9 ; — to a fractional form. Ans. a;+3 a; + 3 5. Reduce 1 H 7 to a fractional form. 26c 26c 6. Reduce 1— ^ — tk- to a fractional form. Ans, lab a^-Vb'' a" + 6" 7. Reduce 1 z to a fractional form. 25c Ans (« + 5-c)(g-6 + c) 2bc a? + a;' 2a;* 8, Reduce a—x to a fractional form. Ans, a + a; a+x 9. Reduce 2a— a; H-^^ — to a fractional form. Ans, — . X X x^ 10, Reduce a^—ax-^x^ to a fractional form. a+x a' Ans. a-\-x 96 ALGEBRAIC FRACTIONS. PROBLEM. (150.) To reduce fractions to equivalent fractions having a common denominator. RULE 1. Multiply both terms of each fraction hy the prodiict of all the de- nominators except its own* RULE 2. Find a commmi multiple (generally the least) of all the denominr ators, and divide it by each denominator, and then multiply both terms of each fraction by these results resptectively, DEMONSTRATION. The accuracy of these rules depends upon the fact that the value of a fraction is not changed when both of its terms are multiplied by the same quantity. PROBLEM. ( 1 5 1 •) Reduce , -3, and ^ to equivalent fractions having a common denominator. SOLUTION. Applying Rule 1, we have (1+^)(1_^^)(1_^3) (1^^.)(1_^)(1_^S ) _^( l+^3)(l_^)(i_^.) (1 -x){\ -x'^){\ -x^y (1 -x'){i -x){i -xj ^'"''{i -x')(i -x){i -xy Applying Rule 2, using the least common multiple of the denom- inators, which is, (1 +x)(l-{-x^), or (1 -\- x){l + x){l -{-x + x^), we have (l+x)(l+x){l+x + x') {i-x'){i+x-x') {l+x')(l+ x) (^l^x){l+x)(l+x-\-xj {i-x'){i+x + xj ^""^ (l-rc^)(l4-ajy'''* by putting the denominators in the same form, we obtain (^i^xY{l-{-x + x') {l+x'){l+x + x') (l+a;)(l+a;^) (l+a:)(l-a:^) ' {l+x)(l-x') ' ^'"'^ (l+a:)(l-a:«y EX amples. a c 1, Reduce ~ and -5 to equivalent fractions having a common de- . , . ad . be nommator. Ans. j-^ and -=-=. oa bd ALGEBRAIC FRACTIONS. 97 2« Reduce and — ^— to a common denominator. x+y x—y x'—xy xy-\ry^ Am, -- — \ and -V"^- 3« Reduce r, -, and -^ — tt, to a common denominator. it Reduce , , and — to a common denominator. a+x' 3 ' 2x 18a;' 2a'x—2x' , 3a + 3a; j\fi8 ^——^—^——~ and. — — ^^— -^— 6aar + 6a:" 6aa; + 6a;" 6aa;-f6a;'* ^ -, , a 2c' , 3a— a;' , . ^ da Reduce t, -r» and x-\ to a common denommator. b a x adx 2bc'x Sabd bdx^ bdx ' hdx ' X X -I- \ X \ 6* Reduce -, — : — , and to a common denominator. 3 4 \-\-x 4a;' + 4a; 3a;' + 6ar4-3 ^ 12a;— 12 Ans, , , and . 12a;+12* 12a; + 12 ' 12a; + 12 7. Reduce and ^ ^ to a common denominator. x 3ao 6abx-\-9ab , 5a;' -{-2a! Ans. — —-Z and — — -; — . 3a6ar 3aoa; 8. Reduce r, — , and to a common denominator. a^—x^ 4a— 4a; a -fa; 4a 3a6-f 36a; 20aa;— 20a;' •^'''' 4a'-4a;'' 4^^=ix^' ^""^ ~4a'-4a;' ' -^ T> , ^ ^'—5 j^ 4a— 15^ 9« Reduce — , a , and 7 H — to a common denommator. 5y Sx 2 6x^ SOaxy—lOx'y -]- 50y . QOaxy—lbxy Ans, -— — , — , and . 30a;y 30a;y ' 30a;y 1 1 6y 10* Reduce -= z, -5 =, and -i rto a common denomnator. a: + 2ax-\- a;' a' — a;' a* — x* a^-[-ax'^—a'^x~x^ a^ + ax^-\-a*x + x^ . 5ay-{-5xy Ans, -^ T—-. -., -T ;—- 1 ., and a*— <3u;*-fa*a;— a;'' a'— aa;*4-a*a;— a;"' a^ -aa;*-f a*a;— «'* 7 98 ALGEBRAIC FRACTlONaf. PROBLEM. (152.) To reduce an entire quantity to a fractional form having a given denominator, RULE. Multiply the entire quantity hy the given denominator^ and place it over the denominator, PROBLEM* (153.) Reduce 1 + a; to a fraction with l—x for its denominator, SOLUTION, Multiplying l+ar by l—x, and placing the product over l—x, we 1— a;" have for the required form. 1— a? ^ EXAMPLES, !• Reduce 2 to a fraction, having 4 for its denominator. Ans, {• 2» Reduce a to a fraction, having 5 for its denominator. . ab Ans. -7-r 6 3« Reduce 2xy* to a fraction, having a;' ior its denominator. Ans. — f-, X* 4, Reduce a^+ab + 6" to a fraction, having a—b for its denominator, Ans. =-, a—o 5» Reduce a*+a^x^ + x* to a fraction, having a^—x' for its deno- minator. Ans. a'-x' 6* Reduce a-f 5 + c to a fraction, having a + 6— c for its denomina- {a + by-c' tor. Ans. ' . a + o — c PROBLEM. (154.) To convert a fraction into an equivalent one having a given denominator. ADDITION OP FKACTIONS. 99 RULE. Divide the given denominator by the denominator of the frojcticmy and multiply both terms of the fraction by the quotient, PROBLEM. (155.) Convert - into a fraction, having ab for its denominator. SOLUTION. Dividing ab by 6, and multiplying both terms of ^ by the quotient a, gives — = for the required form. EXAMPLES. 1* Convert | into a fraction, having 6 for its denominator. Am. ^. 2. Convert f into a fraction, having 32 for its denominator. Ans. f|. 3i Convert into a fraction, having x^—y^ for its denominator. Ans. ^'. 4. Convert —5 ; — rr into a fraction having a* 4- 6" for its denom- a^—ab-\-h^ ® inator. Ans, -^ — ;V« a' + ft a^-\-x* 5. Convert -7 r-r 7 into a fraction having a'— a;" for its de- nominator. ^tw. a»-aj« ADDITION OF FRACTIONS. PROBLEM. (156.) To find the sum of two or more algebraic fractions. RULE. Reduce the fractions to a, common denominator^ and add together the numerators^ and write the sum over the common denominator. 100 ADDITION OF FBACTIONS. PROBLEM. (157.) Add — T-, — , and ■= together. 2o o 7 SOLUTION. Performing the operations indicated by the rule, we have Sa' 2a b_105a^' 28ab 106'' _ 105a'+28a6 + 10y '2b"^T'^1~~10b ~lOb"^'lOb~ 10b EXAMPLES. - ,^^a 2a , 56 , ,, . 20a'' + 156» ^- ^^^ V Sb^ ^^^ 4^ ^^'^^''' ^^'- -l2^6~ • a + b a—b , ^ 2(a'' + 6') 2. Add r and -— y" together. -4«s. -^^^ — =j^. 3. Add T and ^ together. Ans, — ^^ . 4* Add and together. Ans. ~z — -. 5. Add -, -, and j together. -4w«. a:+— . 6. Add 4a, 5a+— -, and a— ^ together. ^w«. 10a——. 7. Add 3c+— -, , and ^together. 5 a—x a ° Ans. 3C+24 Sa'a;— Sor' + Sa:* 5a (a— a;) x—2 2a?— 3 8. Add Ix-i — -— and dx — together. o ^x A -.« 5a:'— 16a;+9 16a; A .,,a4-ar , a—x^ . . 4ax 9« Add and together. Ans. -= -. a—x a-\-x " a'— a;' -^ * ,, 2x ^ 5x ^ . . llx lOt Add 5a: — =- and — — 4a: together. Ans. a:+— — . 7 y oo 11. Add and together. Ans. 2-\ — =- . a—x a ° a^—ax SUBTRACTION OF FRACTIONS. 101 11 2 12. Add and together. Ans, ,, 13. Add ^ and -——^ together. Ans, -^ f, 1 — X 1 -7-2? 1 — X 14. Add together y, — , -, and -. Ans. —^ . 15. Add -^„ -^, and -i- together. Ans. 2^+^y-y-+y a , y — Qamy^ 16. Add together ^^^^^^^ and ^^^y._^y ^''^- {zmf-xy • -« .,, , 2x 1x , 2a; + l . « 1.49a; 17. Add together — , — , and — - — . Ans, 2a?+-+— -. a' 6 , ah 18. Add together ^^^, ^^, and -^^^p^. (a +6)' --..,, , 3a 2a+ir , 5 19. Add together -. — ^, -7 — —^7 r-r, and ; — . ^ (a— 2a;)''' (a 4- a?) (a --2a;)' a+a? 2(10g-lla;)a; ' (a + a;) (a— 2a;)'* 3 3 1^ 1-a; 20. Add together ttt— rj, ^tt— r, ^77— r, and 4(1 -a;)"' 8(1 -a;)' 8(H-a;)' 4(l+a;») . l+aj+a;" Ans, l_a;~a;*+a;» SUBTRACTION OF FRACTIONS. PROBLEM. (158.) To find the difference of two algebraic fractions. RULE. Reduce the fractions to a common denominator^ and subtract the numerator of the fraction to he subtracted from the numerator of the other fraction, and place the result over the common denominator. 102 SUBTBACTION OF FRACTIONS. P&OBLBM. (159*) Subtract from -— — . SOLUTION. Performing the operation indicated in the rule, we have l--2a;4-a;' l+2a; + g' _— 4a; _ 4a; 1-a;" l-a;» T^'" \^^' EXAMPLES^ !• Erom t subtract ^. .4»«. ^ _, a-\-h -^ ^ a— 6 . 4a6 Z* Prom ^ subtract r« ^w«. -^ — rr, a— a + 6 a —0 « „ 1+3;"^^^ l-a;" . 4a;' 3« From :; z subtract , — -5. Am, -.. 1— a;' 1+a^ 1—x* 4« From subtract . Ans, -z — r. a—x a-\-x or— or *' -TL. « . ^ IX X ^— « ^ « ca;+&a;— aft Ot From 3a; 4-- subtract x . Ans, 2a; H j . Q c be - „ r . ^—2 ,x X ^ 2a;— 3 . 11a;— 19 6* From 5a; H — subtract 4a; — . Am. x-\ , 3 6 16 7« From —. — -^+a subtract -7 r. Am, a- a(a+x) a(a—xy ' a'— a;** Q t:^. o v* ^3a + 12a; . 3a;— 3a 8* From 3a; subtract . Ans. , 5 5« 16a; + 23 9* From 2a;-J 3^ subtract 8a; ^—, Am, 42 lUt rrom ax-{ — subtract x — — . Am, ax , 11, From -^ subtract -^. Ans, x+—, 7 6 35 MULTIPLICATION OF FRACTIONS. 103 1 + 2?/ 6v 1 12, From 9y subtract — — ^. Ans, Sy+~ — . 8 8 13« From subtract , Ans. x—3 X x^—3x 1,1 .26 14* From r- subtract =•, Ans. tz =. ^ ^-r . Id* From - -\ subtract - H — . on 9. y anqy-\-mbqy^pbny-'Xhnq bnqy 16* From — ; — I — ; — subtract . Ans 0. ab be ae . 17. From — !-^ subtract — -— + -^ ^. .^tw. 1. y x-\ry xy—y'' 18. IYoml^^+^ subtract ,^+^.. ^. ^ 3(l-y) 1-y 3(l-y) 1-y 19. From subtract ^, . ^. +777-rTT\- -^^- 4—, • a;— 1 2(a; + l) 2(ir''4-l) «*— 1 1 1 rc-1 20. From — +-3+-m subtract - + 7-7— t\5- uln^. x\x' + lf 1^ . ■ » > . »■ MULTIPLICATION OF FRACTIONS. PBOBLEM. (160.) To find the product of two or more algebraic fractions. RULE. Multiply the numerators together for a new numerator, and the de- nominators together for a new denominator. 104 MULTIPLICATION OF FRACTIONS. DEMONSTRATION. A C Let -^ and -yr represent any two fractions. •«r ,. ^ A C AC We are here to prove that -^ ^'f)—~WjT' A C A C Since —==AB-' and -^= CD-\ -^ x -^ must equal AB-' x (7i>-*. But by Prop. 1 in Multiplication, we have AR-^ x CI)-'=AB-^ AB~^ CD~^ CD~^= which, freed from negative exponents by Prop, AC (101), is ^. Q.E.D. PROBLEM. (16 1.) find the product of -r — and -5 — r- — . SOLUTION. To save labor, let us resolve the terms in both of the fractions into their simplest factors, and proceed as directed by the rule, merely in- dicating the multiplication, and then canceling the fectors ic+l, jc+l, and a; + 4, common to both terms, we have (ar+l)(a; + 2)(a;+l)(a!4-4)_a: + 2 («+ l)(a;+ l)(a; + 3)(a; + 4)~'a:+3" EX AMP LE S. 1« Find the product of—- and — - Ans, — , ^ 6 2a 5a 2. Find the product of — , , and — ~, Ans, 15ax, ^ a c 2b 3* Find the product of -^, — , and — ; — . Ans 3 ' 7 ' a-rx' ' 21(a-f a?)* - _ , , , /. ar-i- 1 ^x—1 , Sarfx'— 1) i, Fmd the product of 3a;, -— — , and 7. Ans. -—^7 -/. ^ ' 2a ' a-\-b 2a{a + b) 5* Find the product of 2aH and 3a . ^ a ax Ans. Qd'-hSbx r. X Q^ MULTIPLICATION OF FRACTIONS. 106 6t Find the product of -,-, -„, and a-\- ^ a + h ax + x^ < ax a—x a'(a—h\ Ans. — ^ /. X It Find the product of -z — i-r and r. Ans. -, — -7^,. ^ a^—b^ a+b {a + o) « ^. , , , ^a;'— 9:r + 20 , a;'— 13a? + 42 8. Find the product of 5 — and 5 — . X — \)X X — ox a;'— 11a; + 28 Ans. X' -X ^. , , , ^a—a; , Va;' , 1(a-hx)x^ 9* Find the product of — - — and . Ans. - ^ ^ ' . ^ 3a a^x Za X X (2a + x\x 10. Find the product of a + - and a—- Ans. a^—- — r— -- . 00 15 ^- ^. , , , ^ 3a;' , 15a;— 30 ', , .x 11. Find the product of — - and — Ans. 4a;+-. oX'~~ x\J JdX i -«. ^. , , , „2a—2x , 3aa; " . 2x 12. Find the product of -—- = — and —. Ans, —r, ' 3ao 5a— 5a; 66 13. Find the product of 7-, > and «-. a + o a + x ax — x . a(a—li) Ans. -i ^. _ X a* "— a;* a A- it a — — ti 14. Find the product of -5 5, — „ =, and . Ans. a + x, ^ a^—7/^a^-^x* a—x ax ax ax 15. Find the product of a-! and x . Ans. -= z. * a—x a+x a—x^ x^—h^ 3;"+ 6' a;*— 6* 16. Find the product of — 7-- and -7 . Ans. r— ^ — r, OC u •\- C OC\0 + Cj *w T^. J XT. 1 X i.5a 13c Qh . , ,3a 17. Find the product of -= --; r-+ '« and — ;. ^ 6 2a bog 5d 3a' 39ac 18aA 21a ^* 'bd ~~10c?' ~25^^'^~6~* tQ T?; A,i. A . f^' ab b' , 3a'' 2a6 , b' 18. Find the product of —i—- h— ^ and — 1 — ;. * X* 2xy y a;' 5a;y y 3a* 19a^5 21a'6' 9a6' ^ a;* T0^"*"^5^~T0^«"*"y 106 DIVISION OF FRACTIONS. 19. Find the product of -^^ ^ and ^r_^y Ans. 2(a + «) + }. , ia"— 166' , 5a 20. Find the product of ^_^^ and 20a' + 80a6 + 806»- Ans, a + 2b' DIVISION OF FRACTIONS. PROBLEM. (162.) To divide one fraction by another. RULE. Invert the terms of the divisor, and proceed as in Multiplication, DEMONSTRATION. A Let -^ and -^ represent any two fractions. ^ ^ A A I) AD We are to prove that "^-^77 ="^ ^ ~r~~Wn' A G A C Since ^^AB-^ and -^=CD-\ -g-^-^ must equal AB-^-r- CD~^, or , which, freed from negative exponents, is -577- e. ^. i>. PROBLEM. (163.) Divide ^-^ by—. SOLUTION. By the rule, we have 2a;' x-\-a_2x''{x-\-a)_ 2x'{x+a) _ 2x a'+a;' x ~x(a^-{-x^)~x{a-\-x)(a^—ax-^x^)~a^—ax-\'a^' EXAMPLE s. < -rv. ., a;-}-a , a; + 6 . 5x^-{-Qax-{-a* !■ I^vide 2^-24 by j-^^. Jns. -^^^_^ 2. Divide ^^7 ~E'' ^^*' ^^* 8 9 DIVISION OF FEACTIONS. 107 3* Divide — - — by -— . Am, -^ \ 4» Divide by . Ans. 2x, 5 '' 6x Cb^'~'X^ CL X 5» Divide by — — r. Ans. a^ + 2a*x + 2ax'-\-x\ a-\-x '' a* + 2ax-\-x^ ^ ^. .-, a c , e a . (ad-\-bc)fh « ^. ., x-\-a , x + b , 6x^ + Qax-\-a* 7. Divide r by — — - Ans, ^ — 75 . x—b -^ 5x + a x^—¥ _ _. ., 2ax-^x\ X . 2a + x 8. Divide —5 r by Ans, -3- -— ,. a b o, b 9. Divide —-7-+ r by —r -:, Ans, 1. a + b a^b '' a--b a+b 10. Divide , o/. ■ >.a by ^. Ans, x+—. x^ — 2bx-\-b^ "^ x—b x 11. Divide — ; — byaH . Ans, \, , / . «A Tx- -J 3a— 3a; , 5a— 5x . 3 12. Divide -r- by — -r-- Ans. -. a + 6 •' a-\-b 5 18. Divide 12 by i^+^-a. ^w. -j-i^. •^ a? a +ax-\-x ^M Tx. ., a*-2aV+a;* a'— rr' . a'-a:" 14. Divide T—r — i — by -7-7—5. ^ Ans. . a^x+ax '' a* + x^ ax 15. Divide H — ^ by 1 -. Ans, n. n-\-l '' n + \ «A -i^. .1 2ax—l . x—a 16. Divide a-\ 7 — by ax-^1 a\bx + 2a;') + a{x + b) — l Ans. 77 -r . b{x—a) 17. Divide — ^ + by —. Ans. — . a—xa + x ^ a—x a+x 2ax a'—h' ^ a^ + ah + b^ . , , , 3 18. Divide , ^ . „ by 7^ » ^^. «' + *'. 108 MISCELLA]?TEOUS PROPOSITIONS. «A T^. ., «'c a*c la^c 3a*c a\^ 2aV ,3 , a 3a* 19. Drnde^^+jr— j^-- jr+-^— J— «V by -^, + -^, +c*. Arts, Y 1 — "^• MISCELLANEOUS PROPOSITIONS. PROPOSITION (164.) 1. If the same quantity be added to both terms of a fraction the resulting fraction will be greater or less than the given fraction, according as the numerator of the given fraction is less or greater than the denominator. DEMONSTBATION To be supplied by the student. PROPOSITION (165.) 2. If the same quantity be subtracted from both terms of a fraction, the resulting fraction will be less or greater than the given fraction, according as the numerator of the given fraction is less or greater than the denominator. DEMONSTBATION To be supplied by the student. PROPOSITION (166.) 3. If two fractions added together equal unity, their dif- ference is equal to the difference of their squares. DEMONSTRATION To be supplied by the student. PROPOSITION (167#) 4. The sum or difference of two quantities divided by their product, is equal to the sum or difference of their reciprocals. DEMONSTRATION To be supplied by the student. MISCELLANEOUS PROPOSITIONS. 109 PROPOSITION X (168.) 5. If the difference of two quantities is equal to-, their sum multiplied by x equals the difference of their squares multiplied byy. DEMONSTRATION To be supplied by the student. PROPOSITION (169*) 6. Zero divided by a finite quantity equals zero. DEMONSTRATION. Let b represent any finite quantity. We are to prove that r=0» It is evident that the value of the expression - will become less by diminishing a when b remains constant. Therefore, when a is as- sumed to be less than any assignable quantity, the value of the ex- pression - must also be less than any assignable quantity, or in other words, when a=0, we have t=0. Q.B,D. PROPOSITION (170.) Y. A finite quantity divided by zero equals infinity. DEMONSTRATION. Let a represent any finite quantity. We are to prove that ^=QO . It is evident that the value of the expression - will become greater by diminishing 6, when a remains constant. Therefore, when b is assumed to be less than any assignable quantity, the value of the ex- pression T must be greater than any assignable quantity, or in other words, when 6=0, we have -= oo, Q. JEJ. D, 110 VANISHING FRACTIONS. PBOPOSITION (171,) 8. Zero divided by zero, considered without reference to the expression from which it is derived, may represent any value whatever. DEMONSTRATION. Since, a x 0=0, by making the last zero the dividend, and the other the divisor, we have 0=" Since a may be of any value whatever, the truth of the proposition is estabhshed. VANISHING FRACTIONS. (172») A vanishing fraction is one which becomes equal to -r when certain suppositions are made. PROBLEM. (173.) To find the value of a vanishing fraction. SOLUTION. Let — be a fraction whose value is sought when it becomes equal to -. By (112), we have ^z^x'^' + x^^y + x^Y ■\'X'y'^ -\-xy^^ ^y'^K x—y If now we make y=a;, we have ^ ^ — =a;^^ + a;"*-' + x"^^ + .r™-' + a;*^' + a;*^'. x—x Since there must be m terms in the quotient, each of which, equals af*~\ the whole quotient must be ma;"*-\ Hence, we have ^=-=ma;"^\ when y=^x. x—y In this example it may be seen that we first obtained an expression for the value of the given fraction, before making the supposition VANISHING FRACTIONS. Ill which reduced it to a vanishing fraction, and it was this process that enabled us to find the value of the given fraction when it be- came -. Hence, to find the value of a vanishing fraction the following RULE. Mnd an expression for the, value of the given fraction^ and then make the supposition necessary to reduce the given fraction to-. EXAMPLES. 2 1.8 !• Find the value of — when 6= a. Ans, 2a. a—o x^—a^ 2* Find the value of when xz=.a, Ans, 3a*. x—a x*—a* 3* Find the value of when x=:a, Ans, 4a'. x—a x*—a* it Find the value of -; — - when x=a. Ans, 2a" x^—a^ /p3 Q^ Of, 5* Find the value of —^ 5 when x—a, Ans, — . x^—ar 2 /p Q^X^ 1 6t Find the value of when x-=^a, Ans, -. x—a 2 x^—d^ 7f Find the value of , r^ when x^^a. Ans, oo . {x-ay 8t Find the value of —^ — 4" when xz=:a, Ans, 0, a;^— a' 9. Find the value of -^^ ~ when x=a. Ans. (2a)f , (x—a)^ X — a;' 10. Find the value of , when x=l, Ans, 4. 1—x x^ cc"* 11. Find the value of when x=a, Ans. ma'^K x—a I /p» 12. Find the value of when x=zl. Ans. n. 1—x I 112 VANISHING FRACTIONS, 13. Find the value of — — -— - — -— — — when x=z(l Ans, -, 3 14. Find the value of — -— when x=:^a, Ans. 0, 15* Find the value of —. — — -r — - — — - when xz=z±a, a^ — 2a'x—4ax^ + 8x^ ^ Ans, 00 . 16« Find the value of :r—r- when x=a. Ans, 3a, a — a-tx't 17. Find the value of 7^ ^ when x=l. Ans, 1—x^ 18» Find the value of 7-5 — ~ — ^ when x=a, {x^—a^)i Ans, 2^iari, 19. Find the value of {^-'^)^+n^'-^)^ ^^^^ ^^^ (x'-l)i Ans. l±^i, 20. Find the value of ) , ^ . \ \ — -f=- when x=zl. (x^—2x* + x^ + x' — 2x+l)i Ans. -r- 2* (Tx -\- cic 2cicx n 21. Find the value of -^-r — -7 — r when x—c. Ans. -, bx^—2bcx + br b X^ — ttx'^ — Cl^X -\- c^ 22. Find the value of when x=a. Ans. 0. 23. Find the value of —. — -—z — r -r when x=a. Ans. go . a* — 2a^x + 2ax^ — a;* y 24. Find the value of ^5 — when a;=0. Ans. — . ic' 2a CHAPTER VII. INVOLUTION. (174*) Invohjtion is raising a given quantity to a given power, PROBLEM. (175*) To raise a monomial to the nth. power. RULE. Multiply all the exponents contained in the monomial hy the index of the power, and prefix + to the result, except when the monomial is negative and the index of the power is odd, in which case prefix —, DEMONSTRATION. Let Bah^c^ be a given monomial whose nth power is sought. We know by the definition of a power that the wth power of Zah^c^ equals 3a6V' taken n times as a factor, or, in other words, equals the product of 3\ a\ h"^, and c', each taken n times as a factor. Therefore, {z'a'h\y=z''a\hy{cy. Since h^ taken n times as a factor is the same as h taken 2n times as a factor, we have (6'*)** =6'". In the same way we get (c^)''=:c''*, whence (3aJ'c^)'*=: S^a^i^ "^3"^ which result agrees with the rule. When a positive monomial is raised to any power, it is evident that the result must be positive, since, any number of positive factors multiplied to- gether will produce a positive product ; also, when a negative mono- mial is raised to an even power the result must be positive, since, an even number of negative factors multiplied together will produce a positive product. But, when a negative monomial is raised to an odd power, the result must be negative, since, an odd number of negative factors multiplied together will produce a negative product. PROBLEM (176.) 1. Involve 2ai6ic'" to the 4th power. 8 114 INVOLUTION. SOLUTION. By the rule, we have 2*a2btc*"'=iea'bc*\ Whenever one of the factors of the monomial is numerical, its power may be found arith- metically. Thus, 2* =16. PROBLEM 2. Kaise Va"* to the ^th power. SOLUTION. 'PntVar=BsiTidar=A. Since,by (22) and (24),wehave VV*= (a"*)'*, we may write B={A)'' or i2=^". By applying the rule, we obtain H^ for the pth power of jR\ and An for the ^th power of -4" ; therefore, i2^=:^«, or i2^=(^^)-«=V^^. Since B=Va"'= VA, we have i?^r=(V^^ ; but I^=V'Ap, therefore, (V^^=V3^ which expression, because A=a"' and AP'=q!^^^ becomes (Va"*)^-'= HencBy to raise a radical to any power ^ we have only to raise the quantity under the radical to that power. This principle gives (Vay-V'^, {Vo^y-VaF, (Va)'=^/a^ (Vai)' Since V-4=-4", we have {^/A)^=An, If, now, we suppose 2? =w, we obtain {yAY=A'^=A^=Ay which expression, because -4=a'", becomes (V«'")'*=a"*. Hence^ to raise a radical to h power equal to the index of the root, we have only to remove the radical. This principle gives {yay=a, {Vc^y^ia^ (4/^)'=— a, (V~l)' =-1, {y^y=a\ (y/I2^)''=-2% ' + 224a;y« + 128y'. 6. Cube a; — . Ans. x' i—^3 — )• X x^ \ xl 7. Cube a;'^'*-yi Ans. a;""*-3a;"»'*y^ + 3a;^'*y-yi 8. Cube |a-|6. Ans, |a»-^^^6^-x««6 + |a6«, ^ ^ , a-b ^ a'-Sa'b+Sab'-b^ 9* Cube -r-. Ans, —^ — ^ ^i , in ra — ^s* a— 26 a'— 6a'*6 + 12a6''— 86' 10. Cube x^ + f, Ans. x' + 3a;y + 3a;V' + y\ 11* Raise a;' + a' to the 4th power. Ans, a;" + 4a;V + 6a;V + 4a;V + a*. BINOMIAL THEOEEM. 119 12* Cube2a-36. Am, 8a'— 36a'6 + 64a6'— 276'. 13* Raise ax-^hy to the 6th power. Ans. a'x'—5a'bx*j/ + 10a'b'xy—10a''b'xY + 5ah*xy*—by. 14. Cube 2r— 6m. Ans. 8r'— '72wr=' + 216wV— 216m'. 15. CubeV2+V5. Ans. Itf2 + llf6. 16. CubeV^+l^"^. Ans. -Yj/^-Sf"^. 17. Raise aVa—bVb to the 4th power. Ans. a'—4ayabVb+6a'b'—4aVab*Vb + b\ 18. Raise ar-f — y~* to the 4th power. 19. Raise —2V^^—Sybto the 4th power. ^ws. 16a''-96aV'-af^6-216a6+216i/^6V^6 + 8l5'. 20. Cube (-l)i-(-l)i. Ans. 4/-^-2. PROBLEM. (180.) To square a polynomial. B U L E. Square each term^ and annex twice its product irHo each of the fol- lowing terms. PROBLEM. (181.) Square 2a'— 35 + 4c—5(?. SOLUTION. Squaring 2a', we have 4a*, and annexing twice the product of 2a* into each of the following terms —36, +4c, and —5dj we obtain 4a* — 12a''6 + 16a'c— 20a''(?. Also squaring —36, we have 96', and an- nexing twice the product of —36 into each of the following terms, 4- 4c and —5c?, we obtain 96'^- 246c4-306c?. Also squaring +4c, we have 16c', and annexing twice the product of +4c into the follow- intcy{d + e) + lb{a + b-\-cY{d-^ey + 20{a + b^cy{d^-ey + \b(a-\-b + c)\d + eY + Q{a + b + c){d + ey + {d + ey, In the same way, we find (a4-64-c)''=:(a + 6)" + 6(a + 6)'^c + 16(a + 6)V+20(a+6)V + 16(a + 6) V + Q{a + by + c\ (a+6-fc)^=(a+6)^ + 5(a + 6)V + 10(a + 6)V + 10(a + 6)V4- 6{a^by-\-c\ (a + 6H-c)* = (a+6)* + 4(a+&)''c + 6(a4-6)V + 4(o + 6)c» + c*. (a + 64.c)«=(a4-6)'' + 3(a + ft)»c + 3(a + 6)c» + c». \a-^b + cy-~ {n^by -\-2(a + b)c-\'c\ m BINOMIAL THEOEEM. 123 Inserting these values in tlie above expression, we obtain (a+ft4.c4-rf + e)«=(a-f6)'' + 6(a + 5)*c+15(a + 6)V+20(a+6)V + 16(a + 6)V+ 6(a + &K + c«+6[(a + 6)'+ 5(a + 6)V + 10(a^-6)V^- 10(a+6)V 4- 5(a + %* 4- c']{d + e) + 15 [(a + b)* + 4(a + 6)'c + 6(a+ 6)V + 4(a + by + c*] {d+ef + 20 [ (a + 6y + 3(a + 6)'c + 3(a + bfc + c^] (c^ + e)^ 4- U[(a + bY + 2(a + 6)c + c»] (c? + e)* + 6(a + 6 + c)(c^ + e)^ + ( I a;°— 63.-^ + 12a;"— 8 a;' 3a;') 9a;* a;'— 6a;'4-21a;*— 44a;' + 63a;'— 54a; + 27. Remark. — ^The student should observe that it is not necessary to bring down any terms of the remainder except the first, as was done in the above opera- tion. SOLUTION. 1. We aiTanged the given polynomial according to the decreasing powers of x. 2. We extracted the cube root of a;' and placed the result, «', as the first term of the root. 3. We subtracted the cube of x^ which is x* from the given poly- nomial. 4. We divided —Qx^^ the first term of the remainder, by 3 a;*, or three time the square of a;'*, and placed the result, —2a;, as the second term of the root. 5. We cubed a;''— 2a;, which is the sum of the first two terms of the root, and subtracted the result, a;"— 6a;^ + 12a;*— 8a;', from the given polynomial. 6. We divided 9a;*, the first term of the second remainder, by 3a;*, or three times the square of a;", and placed the result, 3, as the third term of the root. 7. We cubed a;" — 2a; -f 3 which is the sum of the first three terms of the root and subtracted the result, a;^— 6a;^ + 21a;* — 44a;' + 63a;' —64a; + 2 7, from the given polynomial, which left no remainder. EXAMPLES. 1 . Extract the cube root of a' + 3a^6 + ^ah^ + 6'. Ans. a + h, 2. Extract the cube root of 8a'a;'— 84a'*fta;* + 294a6V— 3436'a;". Ans. 2ax—1bx', 3. Extract the cube root of 8a;' — 36aa;'4-102aV — lYlaV 4-204aV— 144a'a;+64a«. Ans. 2x''-3ax^4a\ 4. Extract the cube root of a;"— 9a;' + 39a;*— 99a;' + 156a;''— 144a; + 64. Ans. a;^ — 3a; 4-4. EVOLUTION. 137 5. Extract the cube root of aj' + 6a;*--40ir'H-96ar— 64. Ans. a;'+2a;— 4. 6. Extract the cube root of a;"— 6a;' + 15a;*— 20a;' + 15a;''— 6a; + 1. Ans, a;''— 2a; + l. 7. Extract the cube root of a^+Sa'b + 3a'c + 3a6' + 3acH 6a6c +b' + Sb''c + Sbc'' + c\ Ans. a+b + c, 8. Extract the cube root of 27a;'— 54a;'' + 63a;*— 44a;' + 21a;'— 6a? + 1. Ans. 3a;''— 2a;+l. 9. Extract the cube root of {a + 6)' + 3(a + bye + S{a+b)c^ + c'. Ans. a + b+c. 10* Extract the cube root of 1 — 6a; + 1 2a;* — 8a;'. Ans. 1 — 2a?. PR OBLEM. (195*) To find the mth root of a polynomial. BULE. 1. Arrange the polynomial according to the powers of a certain letter. 2. Extract the mth root of the first term of the polynomial, and place the result as the first term of the root. 3. Raise the first term of the root to the wth power, and subtract the result from the given polynomial. 4. Divide the first term of the remainder by m times the (m— l)st power of the first term of the root, and place the quotient as the second term of the root. 5. liaise the sum of the first two terms of the root to the mih power, and subtract the result from the given polynomial. 6. Divide the first term of the second remainder by m times the (m—l)st power of the first term of the root, and place the quotient as the third term of the root. Continue thus until all the terms of the root are obtained. DEMONSTRATION To be supplied by the student. PROBLEM. (196.) Extract the 4th root of a*— 4a''a; + a;*-|-6aV— 4aa;". 188 EVOLUTION. Operation. 4a'') —4ta'x a4_4a^c + 6aV— 4aa;'+a:* SOLUTION. 1. We arranged the polynomial according to the decreasing powers of a. 2. We extracted the 4th root of a* and placed the result, a, as the first term of the root. 3. We raised a—x to the 4th power, and obtained a*— 4a'a; -h6aV— 4aa;'H-a;*, which, subtracted from the given polynomial, left no remainder. EXAMPLES. 1 • Find the second root of x^ + 2xy + y' + Qxz 4- Qyz + 9z'. Ans. db(ar + y + 30). 2. Find the third root of a»— 6aV-hl2a«''— 8a:'. Ans. a—2x. 3. Find the fourth root of 16a* — 96a'a; + 216aV— 216aa;' + 81a;\ Ans. ±(2a— 3a:). !• Find the sixth root of x* — 6a:' + 1 bx" — 20a:' + 1 5a:' — 6a;+ 1 . Ans. rfc(a:— 1). 5. Find the eighth root of a:' + 8a:' +28a:" + 56a:' + '70a;*4-56a;' -f-28a:'' + 8a: + l. Aris. ±(a;+l). CHAPTER IX. RADICALS. (197.) A radical expression is one which contains one or more radical signs, or fractional exponents ; as, Va^ ai, Va", : If A does not equal 0, let it equal Cdzt, then C±t+VB'= C-^VD, or i^t+VBzzzVD, which shows that the square root of D is equal to a quantity partly rational and partly a quadratic surd, which by the last Theorem is impossible, imless ±Ms equal to nothing; whence VB=VD and A=a Q. E. D. THEOREM. (205.) If the square root of a+Vb equals x-\-y, then the square root ofa—Vb equals x—y\x and y being supposed to be one or both quadratic surds, DEMONSTRATION. By hypothesis a+Vbzzzx^ -{-^xy -{•y'^=x'^ -\-y'^ + 2xy. It is evident that x^+y"^ must be rational, since both x^ and y" are rational, whether x ot y are supposed to be rational or quadratic surds. Hence, by the last Theorem, we must have a=x^+y'' andi^6==2ay, whence a—Vb=a^-{-y''—2xy=x''-2xy+y'^, or i/a— 1^6 =a?— y. REDUCTION OF SURDS. PROBLEM. (206,) To reduce a rational quantity to the form of a proposed surd. RULE. Involve the given rational quantity to a power denoted by the index of the proposed surd, and then represent the corresponding root of the result by means of a radical sign or fractional exponent. EEBUCTION OF SUBDS. 141 PROBLEM. (207.) Reduce a to the form of a cubic surd. SOLUTION. The cube root of a' or VaF={a^)^ is evidently the required form. EXAMPLES. !• Reduce 3 to the form of a quadratic surd. Ans. V9 or 9^. 2. Reduce 3a;' to the form of a cubic surd. Ans, V^lx', 3* Reduce ; to the form of a cubic surd. a+b+x ^ (a + b+it)'' I. Reduce a+a; to the form of a quadratic surd. Ans \ya' + 2ax+x\ 5. V2 Reduce — - to the form of a quadratic surd. 6 Ans. /A. 6. Reduce -^ to the form of a cubic sui'd. V4r AnsYf. 7. Reduce a* 6* to the form of the fifth root. Ans. (ab')i. 8. Reduce Va to the form of a quadratic surd. Ans, JuVa^ 9i Reduce —a to the form of a quadratic surd. Ans, Va^=.V{-d)\ 10» Reduce —a to the form of the fourth root. Ans, V(— a)*. PROBLEM. (208») Reduce a^h to the form of a quadratic surd. SOLUTION. Since a=f/a^, we have (]f^h-=S/ayh=S/€Fh, EXAMPLES. 1, Reduce 2V3 to the form of a quadratic surd. Ans. VVl, 2« Reduce 3\/2 to the form of a cubic surd. Ans. 1/54, 3* Reduce a\fh to the form of the mth root. Ans, "^JcS^h, 142 REDUCTION OF StJEDS. 4, Reduce (a—b)Va^ + b'* + 2ab to the form of a quadratic surd. Ans. Va' — 2a'b^-^b\ PBOBLEM. (209.) To reduce two or more radicals having different indices to equivalent ones having the same index. RULE. Represent the given radicals by the aid of fractional exponentSj and reduce these fractional exponents to equivalent ones having a common denominator ; then raise ea^h quantity respectively to the powers de- noted by the numerators of these fractions, and the common denomina- tor will be the index of the root of each, PRO BLEM. (210.) To reduce 2\/3 and 34^2 to surds expressing the same root SOLUTION. We have 2l/3=V24 = (24)3 and 3f 2=Vl8=:(18)i. But i=f and|=^; whence, (24)i=(24)«=(24'')»=576^=V676,and (18)^ =(18)°=(18')e=(5832)«=:V6832. Therefore, 2V3=V576 and 3f^2=V5832. EXAMPLES. 1« Reduce V2^and V4 to surds expressing the same root. Ans. V8 and VI 6. 2* Reduce Vx^ and Vy to surds expressing the same root. Ans, V"^' and "v//. 3. Reduce Vax and X/ba^ to surds having a common index. Ans, VaV and V6V. 4i Reduce ^3 and \/2 to surds having a common index. Ans, V27 and V4. 5^ Reduce 6^ and 5* to surds having a common index. Ans. Ve^and V5^ REDUCTION OF SURDS. 143 6* Reduce 2*, 3^, and f 5 to surds having a common index. Ans. ^512, V6561, and V15625. 7» Reduce aVa—x and bVo,^—x^ to surds having a common index. Ans. l/a'-3a'a;H-3aV— aV and Va*b' — 2a'b'x^ + b'x\ b 8, Reduce aVx—y and T7== to surds having a common index. ^»«. VaV— 2a*a:y+ay and V^X^Typ'. 9t Reduce Va^—x^ and Va*+ic* to the form of the eighth root. Ans. X/ia^'—xy and l/a' + 2aV+?. lOi Reduce {a + x)\ and (a— a;)f to surds having a common index. Ans, (o' + 3a'a; + 3aa;' + a;')i and (a*— 4a'a; + 6aV— 4aa;'+a?*)^. PROBLEM. (21 !•) To reduce surds to their simplest form. BULB. Separate the quantity under the radical into two factors^ one of which must be the greatest perfect power corresponding to the root indicated that is contained in the given quantity. Extract the root of this factor^ and place the product of it by the coefficient of the radical part, as the coefficient of the other factor affected by the given radical sign, PROBLEM (212.) 1. Reduce 4\/5(a' +a*6) to it simplest form. SOLUTION. Since, V5(a'4-a*6)=V5(l +a6)a'=Va' V5(l +a6)=al/6(l +a6), we have, 4V6(a'' + a*6)=4aV5(l +a6). PROBLEM 2. Reduce ^Vf to its simplest form. SOLUTION. Since, Vf =VM=V2V • 15=ViV Vl5=iVl5, we ^*^® W^- 144- REDUCTION 01* StJRDS. EXAMPLES 1, Reduce Vl6a^x to its simplest form. Ans. 4:aVx, 2. Reduce \/a¥x to its simplest form, Ans, hl/cLX. 3« Reduce V81 to its simplest form. Ans. 3V3. 4i Reduce 1^288 to its simplest form. Ans, 12V2. 5* Reduce V^ to its simplest form. Ans, \VQ, 6* Reduce V2^a^x^ to its simplest form. Ans, 3ax^V3ax, 7. Reduce V5ax*—Sb'^x^ to its simplest form. Ans. xV5ax—3b^. 8. Reduce Vo^'"'^^ to its simplest form. Ans, aVa"&. 9. Reduce V(a-\-bxyxi/ to its simplest form. Ans, (a + bxy^xy. 10« Reduce Via + xyb^ to its simplest fonn. Ans, (a + x)b^. 11. Reduce y- 7^—^ to its simplest form. 12. Reduce n/135 to its simplest form. Ans. 3\/5. 13t Reduce 51^64 to its simplest form. Ans, 15VQ. 14. Reduce 3v/108 to its simplest form. Ans, 9\/4. 15. Reduce \/ax^-\-bx^ to its simplest form. Ans. x\/a + bx\ 16. Reduce ^V^ to its simplest form. Ans, t^^I^21. 17. Reduce 6\/| to its simplest form. Ans, f VIS. 18. Reduce -ji/ -r- to its simplest fonn. Ans. -r^^d. 19. Reduce y j-, r to its simplest form. / CLOl/ X ————— 20. Reduce 4/ to its simplest form. Ans. Va(a—x). ^ a—x ^ a—x ^ ' ADDITION OF RADICALS. 145 ADDITION OF RADICALS. • - » > ^ PROBLEM. ' (21 3.) To add radicals. RULE. Reduce the radicals to their si?nplest''/6rm, and proceed as in addition. PROBLEM (214.) 1. Add together V500 and VTOS- SOLUTION. V500=V125 . 4=V125V4=:5V4 and V108~V27 , 4=V2l V4.=SV4. Therefore, V600 + \/T08 == 5 V4 + 3 V4 = 81/4. PROBLEM 2. Find the sum of SVf and 2V^. SOLUTION. and 2V^=2V^~=Wji^ro=2^rU^lO=2 . ^yf^lQ:^ 14/13 Therefore, 3 V| 4- 2yjr=l^l^ + if 10 =|f/10. EXAMPLES. 1, Find the sum of yTs and 4/f. Ans. 5V2, 2. Find the sum of |/T5 and y48. ^ws. 9|/3. 3« Find the sum of |/aV and 4/c'ar. Ans. {ai-c)Vjs. 4, Find the sum of |/T50 and -^64. ^ws. 2 4/6. 5. Find the sum of Va'"^ and V&a;"". ^tw?. (a + a:')V6. . $, Find the suna of j/iax' and 3xV9a. Ans^ llx^a, 10 146 ADDITION OF RADICALS. 7. Find the sum of 3a;V2aV, SaV^d'x^ and 2ax\/2a^x\ Ans. 13a^V2aV. ^ ^. , , . /a*x /a^x^ , /a'c'^x 8. Find the sum of y — , |/ --^, and y -r^- la^ ax ac\ /x ». Rud the sum of |/-,_^_-^ and |/_--^-_^. An. .g-±5).^. 10. Find the sum of V32 and 2 V40. A'ns, 2V2 + 4V6. 11. Find the sum of ^^24, \/bl, and — 1^6. ^tw. 4|/6. 12. Find the sum of 2VS, -7fT8, 54^72, and -^^50. ^«s. 8^2. 13. Find the sum of 3V32 and 21/54. ^«5. 6(^4"+ V2). 14. Find the sum of V24, 24/^2, and aV^\ Ans. 2{V^^-QVi) + axVQ, 15. Find the sum of 8^1, VOO, — Y'^^is'and Vf u4««. 44^3. 16. Find the sum of V81, —2^24,' ^28 and 2V63; Ans, Sf'Y—VS. 26 ^^ 26 17. Fmd the sum of y —-=-^ and —y — . Ans. (3a- 1) |/— . 18. Find the sum of Sb'VaFc, +- i^oV and— c*|/^. 19. Find the sum of V54a'-»-*'6', -Vl6a^-'b% V2a*^% and V2c'a'". Ans. {sa'b-— +a'^+^+c\V2^. 20. Find the sum of V2'"flt'"^+'6'""+', V3'"a"'''-^+''6'"+', and -Va'6V». ^W5. (2a''6''-}-3a''-"6-c')Va^^"^. SUBTRACTION OF RADICALS. 147 PROBLEM. (215.) Find the sum of V^8 and 1^32. SOLUTION. If we square V8 + 1/32 and then take the square root, we would have a quantity equal to the sum of V8 and 1^32. The square of V8 i-V32 is 8 + 2.V'8.V^32 + 32, or 40 + 2 . |/8.32, or 40 + 2 1^256, or 40 + 2 . 16, or 40 + 32 = '72. The square root of 72, or V'72=:6f'2, which is the sum of V8 and f 32. The application of this mode gives exercise in the involution of radicals. EXAMPLES. 1, Find the sum of Vl2 and V2l. Ans. Sf^S. 2. Find the sirni of V16 and V54. Ans. 5V2. SUBTRACTION OF RADICALS. PROBLEM. (216.) To subtract radicals. RULE. Change the sign of the radical to he subtracted and proceed as in addition of radicals. PROBLEM. (2 1 T .) From V^ subtract V}. SOLUTION. and Vj = ¥~^\=^ V^\ . 6=} V6 = j\ V6 Therefore, V^— Vi= j\ V6 — j\ V6= j\ V6. EXAMPLES. 1, From Vl08ax^ subtract V48ax\ . Ans. 2xVSa. 14£( MULTIPLICATION OF RADICALS. 2. From 9a Vb^ take bxVc^b, Am, 4axVh. 3. From Va"6 take Vb^- Am, (a— a;*) V6. 4. Subtract aVbc^ from Vl6a*6V. -4w5. od^ft. 5. Subtract 3^45 from 61^20. ^n«. V5. 6. From Vi92 subtract 1/24. ^w«. 2V3. 7. From SV'§ subtract 2V^. Am, f VlO. 8. From ||/| subtract |V^. Am. }^Ve, 9, From y ——- subtract y ttt- Am. l3ax—-\ y -^ ax 2b' tA 17 /a'b + 2ab^ + b' ^, .A^— 10. From a/ — 7 ^ ^ „ subtract i/ —^- f a''—2ab + b^ ^ a' + a'b-2ab^+b^ 2ab-\-b^ , 4cabVh Ans, a'-b'' 11, From V66 subtract — Vl89. Am, 5 VI. 12. From SVa'b subtract —SViea*b, Am. (12a» + 3a)f'6. ^ *» ♦ >' MULTIPLICATION OF RADICALS. PROBLEM. (21 8.) To multiply by radicals. RULE. Reduce the radicals to equivalent ones expressing the same root^ and multiply the coefficients together for the coefficient of the product, and the parts under the radicals for the radical part. PROBLEM. (2 1 9.) Multiply Va by V&. SOLUTION. V«*='"Va'* and V6=:'V6"', whence Var-V6='^ V«* ••"%/&-= "Va'*" MULTIPLICATION OF EADICAL8. EXAMPLES. 149 1. Multiplyy2 by^/S. 2. Multiply i/2 by V2. 3. Multiply |/2 by V4. 4. Multiply 5 |/6 by 2 4/3. 5. Multiply 4/8 by Vl6. 6. Multiply 2V| by 3V|. 7. Multiply 4 4/3 by 3l/i. 8. Multiply 2 |/27 by 4/3. 9. Multiply 5a^ by 3aa. 10. Multiply together Va, V* and Vc. 11. Multiply together V4, ^Ve and ^1/5. 12. Multiply together 4, 2V8 and VY2. 13. Multiply j^xhj \/ax{a''—x^), 14. Multiply together y 4/aa;, -^ V<^y and ^Vcz . ^7W. xfVa'bVxyz\ 15. Multiply a64- V«^V^+«&+ |/a V6*+ W by 4/a -4n«. |/10. -47W. V32. u4n5. V128. Ans, 30 4/2. -4w«. 4V32. An9, 2V15. ^n«. 12V432. ^W5. 18. Am, 16^/a^ Ans, |Vi20. -4»«. 8 j/6. Ans, a'x jJq^—q^, c'd. — -Vb. 18. Multiply 4/2+ 4/3 by 2 4/2- 4/3. 19. Multiply 2 4/6-3 4/5 by 4 4/3- 4/T0 ^W5. 1+ 4/6. Ans, 39 4/2 — 16 4/I5. 20. Multiply 5 + Vl + 2 V5 by 4/6 + 4/5. ^ns. 5 4/6 + 5 4/5 + 2Vl25+2Vi80-f 2V54 + V2000. 150 DIVISION OF RADICALS. DIVISIOlSr OF RADICALS. PROBLEM. (220.) To divide one radical by another. RULE. Reduce the radicals to equivalent ones expressing the same root, avd divide the coefficient of the dividend hy the coefficient of the divisor for the coefficient of the quotient, and the radical part of the dividend hy the radical part of the divisor for the radical part of the quotient, PR OB LE M. (221.) Divide J 4/5 by |V2. SOLUTION. i |/5=iVi25, and |V2=|Vi. Dividing | by f, we obtain f, and Vi25 by Vi, we get V^^. Therefore, J y5^|V2=|VH^ =tV-^6T^=|V2000. examples. 1, Divide ^V5 by %V2. Ans, im. 2, Divide 4V32 by VI6. Ans. 4Vl, or 2V2. 3, Divide 6^12 by 1/24. Ans, 6V3. 4. Divide 4Vax by Wxy, Ans. - — Va^xy, 5, Divide ^ax^ by j^bx, Ans. Wab, 6. Divide \/ax—x'' by Va^—x\ Ans. VxVO'—x %/(i->tx) 7. Divide a/^-^^ by 6|/?^. ^^. V^fll^. ^ c ^ y d ar hc{d—x) 8. Divide a^x —Vhx + af^y — V% by Vx +Vy. Ans. a —Vb, INVOLUTION OF KADICALS. 151 9. Divide a + b—c + 2Vab by fa -[-Vh —Vc. Ans. 4/a +V6 -f Vc. 10. Divide 34/15-1^20+1/10-7 by 2Vb. Ans. ?f 3- 1+1^2-^^^246. 1 1 , Divide ^8 + 1/12.+ V2 by W2. Ans, 1 + ^ VTs + ^ VB. 2 12. Divide ^/a"— ar' by a— a;. -4?is. 4/- +2- a — X 13. Divide ^ab^—b^c by 4/a— c. ^»s. 5. 14. Divide i|/I by >i/2 + 3fi. -^r^s. yV- 15. Divide 4/72 +i/32— 4 by |/8. ^**s. 5— 1/2. 16. Divide 4^12 by 2^3. Ans. 2 VY- n. Divide |/^ by |/|. ^m j/^. 18. Divide V64 by 2. ^?^s. \/2. 19. Divide a by \/a. Ans. ^a. 20. Divide %/ah^'c' by |/ ^-^^i- -^»*»- V i — . INVOLUTION" OF RADICALS. PROBLEM. (222.) To raise a radical expression to a given power. RULE. Proceed according to the method given in Involution^ observing the rules just given in reference to radicals. PROBLEM. (223.) 1. Square SVB. 162 INVOLUTION OF RADICALS. • ' SOLUTION. It is evident that the square of any quantity is equal to the product of the square of its feclors. If, then, we multiply the square of VS by 9, we must have the desired result. We know from the nature of fractional exponents, that the square of the cube root of a quantity is eqtial to tlie cube root of its square ; or, in algebraic language, (Vcf)- = \/a', because {aiY=ah Hence, the square of V3=V9 and consequehUy (3V3)' = 9l/9. / PROBLEM . 2, Raise |/3— 1^2 to the 4th power. - , SOLUTION. By the Binomial Theorem, we have {v3-V2y={v~3y-4{V3y{v~2) +fi{vsy{V2y^4:(Vs)(V2y + {V2y. ^ Sin^lifying this by the rules for radicals, we obtain {VS—V2y=9—12y6-\-SQ — SV6 + 4t=49 — 20V6, EXAMPLES. 'i. Raise }V6 to the 4th power. Ans, ^. .2. Raise —v/a=' to the 4th power. Ans. a^X/a^. 3. Raise 1*71/21 to the 3d power. Ans. 1031731/21. 4, Cube j^x + dVy. Ans. xVx + 2lyVx + QxVy + 2 1yVy> 5t Cube —A/Va—Vhc. Ans. \/bc—Va. 6. Square Vaa;*. Ans. j^ax*. 7. Cubei^i Ans. j/ («+^)(^^ > VaVa+x «' ah 8. Raise — ^ V{c + xy to th6 4th power. X Ans. — g-(^+^)*» X 9. Cube aj/ft"— a;' + 3a;V««'. '• Ans. a'6«— aV + 3a'a:Va?. 10. Square |/6+V3. Ans. S^WU. EVOLUTION OF RADICALS. 158 EVOLUTION OF RADICALS. PROBLEM. (224.) Extract the mth root of a given monomial radical. RULE. Extract the mth root of its coefficient, atid multiply the result hy the mtli root of the radical, which is obtained hy multiplying the index of the radical hy the index of the root to he extracted, leaving the quantity under the radical sign unchanged ; or hy extracting the mth root of the quantity under the radical sign, leaving the radical sign unchanged. PROBLEM. (225.) To extract the mth root of 6V«^ SOLUTION. m. / rr \ 1 " /~7 Following the rule, we get i/6v/a'=6'»'"Va'— S«y ««. When r=m, we have juhVa'^=.h'^ Va, EXAMPLES. 1, Extract the square root of |/a. Ans, \/a. 2. Extract the cube root of |/a\ Ans. Va, 3* Extract tlie square root of ym'n, Ans. i/rnVn. 4* Extract the square root of |/a*6'. An^. af/\ 5* Extract the square root of %\Va. Ans,^^\/a. 6. Extract the square root of 91/3. Ans. 3V3. 7. Extract the 4th root of jf Vo^ Ans. \Va. 8. Extract the cube root of (5a'' — 3a;'*) 2. Ans. ^/Ea'^—Sx''. 9. Extract the cub6 root of i«' Vb. Ans. ^aVh. 10. Extract the 4th root of iea" Vx} Ans. Wa^^ 164 EVOLUTION OF EADICALS. 11. Extract the 5th root of 32V^ Ans, 2VS. 12. Extract the nth. root of VaV. Ans. am x mn, 13. Extract the cube root of -y -. Ans. ^Sa. 3 S 14. Extract the cube root of i 1/2. Ans. iV2. 15. Extract the cube root of {a-i-x)Va+x, Ans. \/a + x. 16. Extract the cube root of "V8^'. Ans, ^V^. 17. Extract the cube root of 27 V27a». Ans. 3 V3a'. 18. Extract the 4th root of 81a" V4?7 Ans. 3a'' \nx. 19. Extract the 18th root of Va'°6". ^tw?. a '\/h\ 20. Extract the 5th root of 4/320;'". ^rw. a;v/2. PKOBLEM. (226.) To extract the square root of a binomial surd, one of whose terms is rational and the other a quadratic surd. SOLUTION . Let \/x-^Vy represent the square root oi a-^Vh. Then, x + 2Vxy + y, or a; 4-2/ + ^Vxy^^a + ^h ; whence, by Theorem, (204), we have x-\-y^=.a. We know, by Theorem (205), that if |/a; + f/y = i/a +4/6, that /^x —Vy= ju a — 1/6. It is evident, then, that the product of j^x^Vy by i^x—Vy^ which is x—y^ must equal the product of A/a+Vb by A/a—Vb, which is ^a^—b. This gives us a;— y=V'a'^ — 5. If we extract the square root of half the sum of x-^y and x—y, the result must equal the square root of half the sum of a and >^V— 6, that is ^x=y . Also, if we extract the square root of half the difference of x-\-y and x—y^ the result pust equal the square root of half the difference ot m o? ,li !?i ■-':?r wl < j . 2 . - ■ ■ . • ■ ' '-' ^'--y 'hnc)ltE^M\ • ■ • ' :.^:- ;„/.\ (432.) To nnd'sucK a multiplier, orsucHinuIilpBere ^ will ifilake any binomiarslird rational. . - o-n i I'o -^v'.o All binonaial siifds, not imaginary, may be represebteS ty^'^aitvi. EVOLUTIOJSr OF EADICALS. 161 Let us then seek a general expression for a multiplier which will ren- der "y/aztiy/h a rational quantity. We know by Theorem 5, (1 1 3), r r dm — On . -, . that =V> ^^ exact quotient when r is an even number. am-\- bn 1 \. L L § is a multiplier which will make a^-^bn equal to a»«— 6». But we r r wish am—b^ to be a rational quantity, which it will be when r is a multiple of both m and n. In this case, however, we must take for r a multiple of m and n, that is also an even number. r r By Theorem 6, (114), we know that — -=Qi an exact quotient when r is an odd number. If r is taken, an odd number, such that it is a multiple of both m and w, §, is the multiple 1. JL necessary to render am-{-bm rational. r r We know by Theorem 4, (112), that -^ ^=§2 ^ ®^^* dm — On quotient, r being any positive integer. If then r be taken, any multi- i_ i_ pie of both m and n Q^ is the multiplier necessary to render a«— 6i». In each of these cases it is evident that r may be assumed to be any of the indefinite number of values which can be found to fidfill the required conditions. The least of these values, however, is the one generally used. PROBLEM (233.) 1. Find a multiplier that will render 6+1^7 rational. SOLUTION. A simple inspection shows that the multiplier is 6— VI, for (6+V1){6—V1)=36 — 1=29, a rational quantity. r r r r r Afifain, we have = = , rmust be assumed to am-if-b'^ 61 + 7^ 6 + '7* be some even number which is a multiple of 1 and 2. Making r=2, we have -=6— T^ = Q as before. 6 + V^ 11 162 EVOLUTION OF RADICALS. PROBLEM 2. Find a multiplier that will render 2V3— 3V2 rational. SOLUTION. r r r r alk—lTn 122—543 Here 2^/3 — 31/2 =1^1 2 — V54 ; hence, we have am— 611 122—543 122—543 12^— 54^^ = =z . Applying the formula given in the de- 122—543 122-543 12^-54' 5 mostration of Theorem 4, (112), we get — j ^=1224- 122—543 122543 + 122543 -f 122543 + 122543 4- 54^ = V12* + 144V54 + V'T2'V5? + 648+Vi2V54' + V54^ which simplified is 28»^'3 + 432V2 + 2161/3^4 + 648 + 324V3V2 + 486V4. EXAMPLES. 1, Find a multiplier that will render Vh-\V^ rational. Ans. |/5— f^3. 2. Find a multiplier that will render 5+1^3 rational. Ans. 5— V3. 3« Find a multiplier that will render V2—Vx rational. Ans. ^2+Vx. it Find a multiplier that will render aVh + hVa rational. Ans. aVh—hVa. 5« Find a multiplier that will render 1— V2a rational. A71S, l+V2a + V4a^ 6. Find a multiplier that will render V3— ^1/2 rational. Ans. V9 + iV6 + iV4. 7. Find a multiplier that will render V5 + \/3 rational. Ans. (V5-V3)(t/5+t/3). 8i Find a multiplier that will render VoN- V^rational. Ans. V'cf-V'a'^+V^'-V^- EVOLUTION OF RADICALS. 163 PROBLEM. (23 4t) To reduce a fraction whose denominator is a surd quan- tity to another that shall have a rational denominator. SOLUTION. A simple fraction which has a monominal surd for its denominator, may be represented by — =, in which a may represent any quantity whatever. Now, if we multiply both terms of —^ by V^"~S we have Vx =zr-=: , which is a fraction having a rational denominator, if X is rational. If ic is a binomial surd, it must be rendered rational as in the last problem, PROBLEM 2 (235,) 1. Reduce — r to a fraction having a rational denominator. SOLUT ION. — 2y'3 Multiply both terms by |/3, and we have . 3 PROBLEM 2. Reduce — - to a fraction having a rational denominator. 24/2 + ^/3- 8OL UTION. 1^8+^48 (8 — f48)i/8+V48 21/2 + 1/3 f8+V'48 8+V/48 64-48 4(2-1/3)21/2 +|^3 ^ (2-|/3)|^2+l/3 _ ,^-— ;^ 16 ~ 2 ~^ XAMPLES It Reduce -^—^ to a fraction having a rational denominator. Ans. |V125. 164 .^n^EVOLUTION OF RADICALS. 3 2, Reduce -^r =. to a fraction having a rational denominator. Ans, 4/5 + ^2. 3* .Reduce — — -- to a fraction having a rational denominator, B-V2 . 2 + 3^2 Ans. — . 7 Q 4. Reduce -^ to a fraction having a rational denominator. V5-V2 _ Ans. V26 + \/io + V4. 5, Reduce ^^+^ -- Va—x ^ ^ fraction having a rational do- j^a+x +Va—x . a—Va'-^x^ nominator. Ans. • 6« Reduce y ^ +x+ l +y x x ^^ ^ fraction having a rational x^-ir^x'^—x^-'lx—X denommator. Ans. — • 2 7, Reduce — = = = to a fraction having a rational denom- 4/6 + 1^3-^2 _ _ _ . 2V3-3t/2 + V30 mator. An%. — •• 8# Reduce -^ to a fraction having a rational denominator. V3-V2 Ans. V9 + V6+V4. 8 9, Reduce — = = to a fraction having a rational denomina- 4/3 + ^2 + 1 tor. AnB. 4 + 2y2-2f^6. 10. Reduce — =: to a fraction having a rational denominator. V6+V3 _ _ _ Ans. V126-V'76+V45-V2Y. PROBLEM. (236.) Transform 2—^3" to a general surd. SOLUTION. Squaring 2 — f3, we have 7 —41/3; if now we indicate the ex- IMAGINARY QUANTITIES. 165 traction of the square root of this quantity, we shall have 2 — VS expressed in the form of a general surd, that is 2— VB=y1—4:VS, EXAMPLES. 1. Transform |/a— 2 Va; to a universal surd. Ans, y a + 4x—4:Vax, I 2. Transform SVi+V1^ to a general surd. Ans, SVS. 3* Transform |/2 7 + 1/48 to a universal surd. Ans. ^VS, 4. Transform V320— \/40 to a general surd. Ans. 21/6. 5« Transform |/2 + VS to a general surd. Ans. 1/5 + 2 V6. 6« Transform v/2 + \/4 to a general surd. ^7is.|/6(l+V2+V4). 7. Reduce 4/2— 21^6 to a general surd. Ans. y2e—SV8. J;, j8. Reduce 4—^7 to a general surd. Ans. A/ 23— SVI. 9* Reduce j^2+V3 to a general surd. Ans. |/ 5+2i/6. 10. Reduce 2V3— 3v/9 to a general surd. Ans. |/l62V9-108V3-219. ^ '« ♦ •' IMAGINARY aUANTITIES ADDITION OF IMAGINARY QUANTITIES. PROBLEM. (237.) To add y-a' and ^-b^ together. SOLUTION. Since, i/-a'=Va\-l==aV-l, and ^-b'=Vb\-l=hV-ly we have |/^'' -f V'lIb'—aV—l. -^iV^l =z{a-\-b)V—i. 166 SUBTRACTION OF IMAGINABY QUANTITIES. EXAMPLES. , 1. Find the sum of |/— 4 and 4/— 9. Ans. ol^— 1. 2. Find the sum of 2 + f/-l and S — V—M. Ans. 5 — IV— I. 3. Find the sum of ^—^ and 4/— 18. Ans. 5^^. 4. Find the sum of ^f — 27 and 2V—\2. Ans. ISf— 3. 5. Find the sum of 4/~6 and 4/^9. ^?is. (3 + f 6) V^. 6. Find the sum of 4/ -5 and 4/~7. ^»wf. (1^5+ f' 7) l/"^. 7. Find the sum of V— 4 and V— 9. ^n«. (t/2+ y3)V^. 8. Find the sum of a -\-V—h and a-^V—c. Ans. 2a-\-(Vb-\-Vcy'^. 9. Find the sum of V— a and 2V— «. Ans. SV—a. 10. Find the sum of V-1 and V-16. Ans. 3V— 1. ^ '« ♦ »« » SUBTRACTION OF IMAGINARY QUANTITIES. PROBLEM. (238.) From 4/— a" subtract 4/"^'. SOLUTION. Since, 4/— a'^^a 4/— 1, and 4/— 6'=6 4/— 1, we have aV-^h —6 4/^=:(a— 6) 4/— 1. EXAMPLES. 1. From 4/— 9 subtract 4/— 4. ^ns. i^— 1. 2. From 3— V— 64 subtract 2 +V^. ^»*«. 1 — 94/1. 3. From 4/— 18 subtract 4/— 8. ^w«. 4/— 2. 4. From 4|/~27 subtract 24/— 12. ^rw. 8V^. 5. From a + V^ subtract a + 4^^ ^ws. (f^6 — V^c) V— 1. 6. From V-16 subtract V^. -4ws. V-^i. MULTIPLICATION OF IMAGINARY QUANTITIES. 16? MULTIPLICATION OF IMAGINARY QUANTITIES. THEOREM. (239.) An imaginary expression of the form ^—A can always be reduced to the form r^ —\. DEMONSTRATION. Let r denote the square root of ^-A. It is evident that r can always be obtained either exactly or approximately when ^ is a posi- tive rational quantity. Then, ^^A=V + A. -l=VAV—l=rV—l, PROBLEM. (240.) To ascertain the rule of signs in the multiplication of imaginary quantities. SOL UTION. Let |/— a be multiplied by y— a. Multiplying, as in the case of surds, not imaginary, we have |/— a xV—a=V—a x —a=Va^=—a. We put the square root —a, because the d^ was obtained in this case by multiplying —a by —a, or, what is the same thing, by squaring —a. It may also be observed, that |/ — a x V — a =:{V— ay. We have already shown, in Evolution, that (V^— a)^=— a, which result agrees with that just given. Let us now ascertain the product oi \/—a by V—h. Since, \/—a^=VaV—\^ and \/—h=^VhV—'\.^ we have V'^ xV'^=VaV'^ xVlV'^=V^V^-iVaVb== — lVab=-4^ab, Whence, we see that the same rule applies in multiplying quadratic imaginary surds as in other surds, provided, however, that the result must be affected by a minus sign. The principles already developed will enable the student to multiply any imaginary quantities in which the index of the root is even. In all these operations great care must be taken that all the steps be rigid. 168 MtJLTIPLICATIOK OF IMAQINABY QUANTITIES, Note. — The importance of carefully scrutinizing all the operations in which imaginary quantities are concerned can not be better set forth than by showing the positions assumed by different distinguished mathematicians : " The first idea that occurs on the present subject is, that the square of ^—3, for example, or the product of v/— 3 by >/— 3, is —3; and, in general, that by multiplying ^—a hy y/—a, or, by taking the square of y/—a^ we obtain —a. * * * * * * " Moreover, as y/a multiphed by ^yh makes ^aJ, we shall have y/Q for the value of ^Z— 2 multiplied by v/— 3 ; and'-v/4, or 2, for the value of the product ofy^— 1 by ^—4. Thus, we see that two imaginary numbers, multiplied together, produce a real, or possible one. " But, on the contrary, a possible number, multiplied by an impossible number, gives an imaginary product; thus, ^—3 by v' + S, gives y/ — 15." — Enter's Al, p. 43. But Emerson makes the product of imaginaries to be imaginary ; and for this reason, that "otherwise a real product would be raised from impossible factors, which is absurd. Thus, y/ —aY.y/ —hz=^ — a&, and ^~a X —y/ — 6 = —y^—ah, &c. Also, y/ — a x y/ — a = — a, and y/—ax — y^— a=+a, &c." — Emerson's At, p. 67. From a dissertation "On the Arithmetic of Impossible Quantities," by Mr. Playfair, in the PhU. Trans, for 1778, p. 318, we learn from some operations there performed, that he makes the product oiy/—\ hj y/—l, or the square ofy'— 1, to be —1, and, in another place, he makes the pro- duct of y/^1 by y/U^ to be y/-l+Z\ The authors just quoted not only differ from each other, but each one seems to be inconsistent with himself. Thus, Euber sa-js, y/—aXy/—a= — a, but y/ — axy/—b=y/ab, and Emerson says, y/—axy/—a= — a; but y/—axy/—b=y/—ab. Now, the formula for the product of x/ — a by y/—b ought to be true whatever values may be assigned to a and b. Let, then, a=b. Whence, Euler's formula for the product of y/—a by y/—h^ gives +y/a»=+a, and Emerson's formula gives +^—a^=ay/—l' But they both say that v'— ax v'—a= — «• Mr. Playfair makes y/^Xy/l—^=y/—l+z^, which we conceive is not a correct result when z is more than 1. Let z=y/2 then 2'=2 ; whence, the above expression becomes ^— 1 X v/1— 2=x/— 1+2, or V— lX\/— 1^=\/1~— Ij which result does not agree with his other position unless he takes x/l =—l, which we know would be proper ; that is, when z is more than 1, we have yZ-^i X yi^2^= — v^— 1+Z3. MULTIPLICATION OF IMAGINARY QUANTITIES. 169 PROBLEM. (241.) Multiply XTa by j^\, SOLUTIONv Since |/^=(— 1)2=(— 1)1 =V(— 1)', we have, by the ordinary rule, V(— l)"xVa=Vl x Va=Va. But Va=±>i/±Va; that is, the 4th root of a is +4/+V'«, +|/-Va, — 4/+Va, or — |/ — l^a. Now, to which of these forms must Va, in the present case, be made equal ? We have ViP^y X Va=|A'Plf X |/^^=|/i^pr)Va=|/M^ Since v'(— 1)'=— 1. Therefore, |/^ x Va=|/—f^a. • EXAMPLES.* 1, Multiply 21^32 by 3V^-^. ^JW. —61^6. 2. Multiply 4V^— 3 by 94/--I2. ^ -4w5. — 216. 3. Multiply —2^"^ by — 3V^. ^W5. — 6V^6. 4. Multiply —21/^ by 3 f^^. -4>w. +6V6. 5. Multiply l+f/Hiby 1+*^^. ^tw. 2^^. 6. Multiply 1+i^^ by l-V^^. Am, 2. 7. Multiply a+i/-6' by a+f/-6'. ^n«. a»-6' + 2a6V'-l. 8. Multiply 5+ 2|/"^ by 2-V^. ^ws. IG-*/"^. 9. Multiply V^ by V^. Ans. 2t/^. 10. Multiply 2V— 4 by 3V— 16. Ans. —12. * A glance at these examples shows that the results are the same as would be obtained by the following RULE. To midtiply one quadratic imaginary surd by another, multiply the quantities under Ihe radical signs, according to the rule for signs given in multiplication, hut the coefficients of the radicals, axxording to the reverse of that rule ; thai is, marking the product of like signs minus, and unlike plus. This rule will not hold when but one of the quantities multiplied is an imagin- ary quadratic surd. 170 DIVISION OF IMAGINARY QUANTITIES. DIVISION OF IMAGINARY QUANTITIES, PROBLEM f242.) 1. Divide |/^ by |/^. SOLUTION. _ , |/^^ i^axV'^ /a . , . . We nave =-^= :===i/ -: since the unaennary quantities^ i/-b nxV-1 ^ ^ s / 1 cancel. PROBLEM 2. Divide -V^ by —V'^. SOLUTION, =^=-^Vi. -V-h -VhxV-\ ^ EXAMPLE S- 1, Divide QV^ by 2f'==4. Ans. ^VZ, 2. Divide -V~-i by -QV^. Ans, ^jVl, 3. Divide 1 +V~^ by 1— V'^. Avis. V^, 4, Divide 4 ^-V'^ by 2—V^. Ans, 1 +V^. 5* Divide 1 by |/— 1, -4?is. — V'— 1, 6. Divide a by 4/af^^. Ans, ^^aV—l. 7, Divide 2V8— i^^O by —S/'^. Ans, 4/6 + 4*^— 1. 8, Divide b—V^ by l+i^^. ^/w. 1— 2f/~2r 9. Divide l4-|/15-(7f'3 + 2V'5)f^^ by l-^-V^, 10. Divide a by hV—l. Ans, --?K-1 ^7W. 2—V—9. \ IKVOLtTTION OF IMAGINARY QUANTITIES. 171 INVOLUTION OF IMAGINARY QUANTITIES. PB O BLEM. (243.) Cube a-^^\ SOLUTION. Calling ^—b^—c, we have (a— c)''=a''— 3a'c + 3ac'— c'. If now, we ascertain the value of c' and c^ we can put these values for c' and c" in the above expression. The square of c, or of |/— 6'= W^iB evidently -b% and c'=c\ or -b' y-^=^bW—l. Hence, {a—V^y=a'—Sa^bV^—Sab^ + b^ V^=a^— 3a6' + (b'-Ba'b)i^'^. PROBLEM. (244.) To find formulas for the powers of y— 1. SOLUTION. Let the index of the powers be 4w, 4w + i, 4n+2, and 4w+8, which comprise all positive integer numbers. Let a=V—l, and we have (|/irr)«»+>=a*M^=a*» . a'=a' =-1. Thus^ to obtain any power of \/—\^ it is only necessary to divide the exponent of the power by 4, and the power of j^—\ indicated by the remainder will be the result required. EXAMPLES. It Raise |/—1 to the 66th power. , Ans. —1. 2i Raise 4/— 1 to the 103d power. Ans, —V—l. 3. Raise \/—\ to the 400th power. Ans. +1. 4. Raise a^— 1 to the 4wth power. Ans. a*\ 5. Raise aV^— 1 to the 4w+ 1st power. Ans. «*"+' /— 1. 6. Raise oi/^ to the 4n + 2d power. Ans. — a*"+'. 172 EVOLUTION OF IMAGINARY QUANTITIES. 7t Raise aV — 1 to the 4n+ 3rd power. Ans. — a''^' /^. 8. Square azhV^. Atis. a'— 6±2aV'^. 9. Cubeadbf/^. Ans. a'— 3a5±[(3a'— 6)V5]f/^. 10* Raise a+V— 1 to the 6th power. ♦ •' »■ EVOLUTION OF IMAGINARY QUANTITIES. PEOBLEM. (245.) To extract the square root of a binomial surd, one of whose terms is rational and the other an imaginary quadratic surd. SOLUTION. Let Vx-{-V—y represent the square root of a-\-V—b. Then x-\-2,V—xy — y, OT x—y + W—xj/^a+V—b; whence, by Theorem. (204), we have x-'y=a. We know by Theorem (205), that if ^x+V^=ya+V^b,thaiVx---V—y'= l/a—V'^, It is evi- dent that the product of |/ic+V— y bjVx—V^y which is x-\-y, must equal the product of 4/a-\-V—b by i/a—V—b, which is Va^-\-b, This gives x-^y^Va' + b. If we extract the square root of half the sum of x—y and a;+y the result must equal the square root of half the sum of a and f^a' + ft; that is, Vx=T . Also, if we extract the square root of half the difference of x—y and x-^-y^ the result must equal the square root of half the difference of a and Va^+h; that i8,V— y=r . Whence, we have the for- mula Letting j^a—V—b=:i^x—V—y, we would have EVOLUTION OF IMAGHNARY QUANTITIES. 178 PBOBLEM. (246.) Extract the square root of l + QV"^, SOLUTION. Since, 1 + 6V—2 = 1 -{- V—12, we have by putting 1 for a, and Y2 for b in the general formula, f 2 ^ We could also solve this by inspection. Thus, since 1 + 6V—2 =9 + 2 • SV^—2, we see that twice the product of the square roots of 9 and —2 is equal to the second term, and therefore, 3 H- i^— 2 is the square root. EXAMPLES. 1, Extract the square root of 31 +421^^. Ans. 1 + SV^. 2. Extract the square root of —3+1^—16. Ans, 1+21^"^. 3i Extract the square root of 4*^^—2. Ans, 2+V^. 4i Extract the square root of 2 + 4 V — 42. Ans, >^14 + 2 V^, 5* Extract the square root of 4/— 1. Ans, i4/2+-J-|^— 2. 6* Extract the square root of —2—2 V—15, Ans, \/%—V—5, 7t Extract the square root of 2cdV^l. Ans. (1 +^111)4/^ 8» Extract the square root of 8/^. Ans, 2 + 2t^— 1. 9. Extract the square root of —V—1. Ans, i.f/2— ^V— 2. tt'c etc ^4cd — — 10* Extract the square root of — —cd -{ ——V—\. A71S, -Vc+V—cd, PROBLEM. (247 •) To extract the cube root of a+V^, BOMBELLl'S RULE. First find Va" + 6, then, hy trials, search out a number c, awd a square root \/d, such that the sum of their squares c^-\-dbe ■=.\yd^ ^ ^j and also, c^ — 8cd be=a; then shall e+V—d be the cube root vf a-\-V—b sought. 174 MODULUS. PROBLEM. (248.) Extract the cube root of 2 + 111^^. SOLUTION. Since, 2 + 11 K— 1=2+ 1^—121, we have a=2 and 6=121; whence, s/a'' + 6=\/125=6 ; then taking c=2, and c?=l, we obtain c* + c?=5=\/a'^ + &, and c^— 3cc^=8 — 6=2=a, as it ought; there- fore, 2+1/— 1 is the cube root of 2 + llf^— 1. This may also be solved by Tartalea's rule. Thus 2 can be separated into two parts 8 and —6, one of which is a perfect cube, and the other divisible by 3. Therefore, r^=8, or r=2 and 3s= — 6, or s= — 2; whence, r+|/-=2 + y -— , or 2+1^—- 1, the same result as before. EXAMPLES. 1. Extract the cube root of 2 + 2 f^^, Ans. — 1 + f^. 2. Extract the cube root of 2 — V— 121. Ans. 2 — f — 1 . 3. Extract the cube root of SldzSOV^. Ans. -3db2|/'^3. 4. Extract the cube root of — 10 + 9^"^. Ans. 2 + V^. 5. Extract the cube root of —5 — 1^—2. An^. 1 — |^— 2. 6. Extract the cube root of — 4 + lOi/^^. Ans. 2 + V^. 7. Extract the cube root of 9 + 25V^^. Ans. 3 + V^. ^ »• ♦ «. »i MODULUS. (249.) The modulvs of a+/?^^ is +>fV+^«, or the square root of the sum of the squares of a and ^. Thus, + i/16 + 9= + 5 is the modulus of 4 + 3 V—1. (250.) Two imaginary expressions are conjugate, when they differ only in the sign of 4/— 1 ; as, a + ^V—l, and a—^—l. These conjugate expressions have the same modulus. MODULUS. 175 THEOREM. (251,) In order that a-\-^V—l be equal to zero, it is necessary and sufficient that its modulus +Va^ + f^^ be equal to zero. DEMONSTRATION. If the modulus \/a^ + i^' is not equal to zero, a and ^ will not be €qual to zero at the same time, and, consequently, a + ^V—l will not be equal to zero. But, ifi/a' + i?" is equal to zero, a and ^ must each equal zero; whence, a + ^V—l would also be equal to zero. THEOREM. (252.) The modulus of the product of two imaginary factors is equal to the product of their moduli. DEMONSTRATION. The product of a + ^V^l and a'-\-^'V-^l is {aa'-§§')^ ^(i^' ^^a')V—\, and the modulus of their product is, therefore, equal to ^{aa^ - §^'y + {a§' + §afY = Va'a'' + ^'§^' + a'^^' + ^'w' = Va''' + §'\ Q. K D. THEOREM. (2 53.) The modulus of the quotient resulting from the division of one imaginary quantity by another is equal to the quotient of their muduli. DEMONSTRATION. Let a" -{-^''V—l represent the quotient of a-\-§V—l by a' + |9Y— 1, since it can be easily proved that the quotient must be of this form. Hence, we have, a + 19|/^= (a' + (^ Y-T) (a" + ($' Y~-I1). By the last Theorem, we have |/^M^'= Va'^-[-tifWa^''+§"\ We may consider ^/a"'* + ^""^ as the quotient arising from dividing VcT^ by |/a'^ + . 176 EXAMPLES IN RADICALS. THEOREM. (254.) In (wder that the product of imaginary factors be equal to zero, it is necessary and sufficient that one of the factors be equal to zero. DEMONSTRATION. It can be easily shown that the product of two or more imaginary fectors must be of the form a-\-?V—l. In order that this product may be equal to zero, we have seen that its modulus must be equal to zero ; but this modulus is the product of the moduli of the several factors, and these moduli are real quanti- ties, and consequently their product can not be zero, unless at least one of these moduli be also equal to zero ; but when a modulus is equal to zero, its corresponding imaginary factor must also be equal to zero. Hence, the Theorem is true. MISCELLANEOUS EXAMPLES IN RADICALS. 1. Extract the cube root of 20 + 14^2. Ans. 2+f2. 2. Simplify L ^^^Q-^ . Am, -{5 + VlO). 22-74/10 3t Extract the square root of —1 +4V'^. Ans, 2+1^—5. 4. Simplify 3 via" + 2l/2a. Ans, 6V2a. 5, Simplify VSa'b +16a*—Vh* + 2ab\ Am. (2a — 6) l/2a + b. 6. Simphfya./^3 + 276^- Am, -Va + 2K 7. Simplify 4/4a'6'—20a'6' + 25a6*. Am, {2d'—b)bVa, 8. Simplify V81-2V24+f2"8'+ 2^/63. Am, 8f7-V3. 9. Sunplify 4/12+2m + 34/76 + 9V48. Am, b^Vl. 10. Simplify . Ans, ——Vx, -- ^. ,.^ ct—b Vac' . Vac 11, Simphfy r • — . Am, ——j-. EXAMPLES IN RADICALS. 177 a-\-b /a—b J . A+^ 12. Simplify _±_/__. ^-Y^- 13. Simplify V2 X Vi X V3. ^«s. "x/'^i. 14. Simplify 1/4 X "v/3 X V6. ^rw. 'V3981312. 15. Find the sum of Vi and VS- ^^^- F^^. 16. Find the difference of V| and VV-- ^^^- f j V75. 17. Find a factor that will make a"' rational. ^%5. a\ 18. Simplify -T= r:. Am. dVlsVd. V5 4-^3 19. Extract the square root of b€-^2bVbc—b^ Am. db(6+V6c-6'). 20. Extract the square root of wp+ 2^"— 2mVwi?+wi". Am. ±(t/w23 + m'— m). 21. Simplify ^16 + 30|/-l+|/l6-30V/~l. Am. ±6l/^. 22. Reduce i/l6 + 30V— 1 +i/l6 — 30i^— 1 to its simplest form. ^TIS. ±10. 23. Simplify |/ 3 1 + 1 21/'^ + 1/— 1 -h 4 V-^. 24. Reduce i/s 1 + 12V'— 54-|/— 1+4V— Stoits simplest form. ^W5. rb4. 25. SimpUfy ^bc-{-2bVbc-b' -{.^bc-2bVbc-b' Am. ±:2Vhc—b\ 26. Reduce A/bc + 2bVhc—b' ■\-A/bc—2bVbc-b'' to its simplest form. Ans. ±2&. 28. Divide 2i^3 x \/l by iV2 x V3. -4w«. 4 V288. 12 178 EXAMPLES IN RADICALS. 29. Divide ^f/j by |/2 + 3 V^. Am. J^. 30.- Multiply |/2 X VS by Vi x Vj- ^^^' '^^8. 31. Multiply V| X Vi by "v/6~ ^n*. *V^. 32. Dividei/Vi X 2 V3 by i/4V2 xf 3- ^7i5. ^Vf. 33. Divide 1 by Va-\-Vb. Va'-V^b + Vab'-Vf' Ans. ; . a — 31. Multiply 41/f +5^1 by i/i + 2Vl. Ans. ^+^-^^2. 35. Divide Va + V6 by Va-V6. a-{-b + 2Vab + 2Varb + 2Va^ Ans. a—b 36. Cube Ans. 1. 2 37. Simplify^^?^^-^^. Ans. 4-V3, 38. Simplify ) 2(2FV 3 f ^ ^^^^ ^rlr-Vi. ( 2V2(3)^f 3«* 39. Simplify ./ ) (i^jtl^ ( '. ^,,5. i/i(\VQ'hV2i). ^ ( 2i/2-(i)2 ^ ^ — 3 1^2 4-2 — 40. Find the sum of 6^/2 — 1 and =:. Ans. 3 + 8t/2. _5^-3t^2 'V'f4+Vl2 V6—V4c 41* Find the difference between — = =- and -^^ 1_. VI- V6 Vs + V2 Ans. SV2 + 4V2i + 4VS. 42. Divide 2 V'S x 4Vi by 4Vi x 4V4. Ans. 1. 43. Extract the square root of 1 — VlS. Ans. i|/26-^f2. 44. Extract the square root of H-fVi — ^V^- Ans. iV^3-iV6. CHAPTER X. EaUATIONS. (255.) An equation is an algebraic statement denoting the equality, in value, of two algebraic expressions; as, ax-\-h=c, 6a;'* + 2a;='7, and aa?— 6=0. (256.) The absolute term of an equation, is that term which is completely known, or is considered as known, and which is equal to the sum of all the unknown terms. Thus, in the equations 6a: =9, 4a;'^ + 3a;— 7=0, and ax=b—c', 9, 7, and (6— c) are the absolute terms, (257.) ThQ first, or left hand member of an equation, is the part of the equation which precedes the sign of equality. (258.) The second or right hand member of an equation is the part of the equation which follows the sign of equality. THEOREM. (259.) Any term of an equation may be transposed from one member to the other by changing its sign. DEMONSTRATION. Let aa; + 6=c be an equation. This equation may assume the fol- lowing form ax-\-b=^c-\-b~b. It is evident that we should still have a correct equation if we should omit +6 in both members. Thus ax=zc—b. Again, let ax—b=:d. Because ax—b=d+b—b, we have ax'=d-^b. These results show that if we transpose any term of an equation from one member to the other, its sign must be changed in order not to destroy the equality. The truth of this Theorem may be proved as follows : Letting aa; + 6 =c and subtracting -\-b from both members, we have ax-\-b=zc +b^b ax ' =c— 6, 180 SIMPLE EQUATIONS. or by adding— 6 to both members, we have ax-^b=c -b=-b ax z=.c—h. Letting aa;— 6= (^ and subtracting —6 from both members, we have ax—b=d -b=-b ax =zd + bj or by adding + ^ to both members, we have ax—b=d -\-b=+b PROBLEM. (260.) Transpose all the known terms of a?*4-6a;"-~4ar— 4+rf —6=0 to the second member. S OLUTION. Transposing —4, +c?, —bhj changing their signs, or by adding + 4— c? + 6 to both members, or by subtracting — 4+ ai ^ 8a; + 25 17— 6a; ^, . 9a;+40 ,,. 14. Solves ;p- __=2J^+a; 1—. (1) SOLUTION. 20-8.-26-«?=?^=8J + 4.-?^ (2)=(1)X4 68— 24a; ,^, . 9a;+40 ,^. ,^, 5 =131^4- 7a; ^^ — (3)=(2) transposed. -68 + 24a;=118 + 63a;--ii^^tl5£ (4)^.g^x9 2 81a;+360 =39a;+186 (5j=(4) transposed. 81a;+360=78a; + 372 3a;=12 «=4. PROBLEM 15. Given — - — : — - — : : 7 : 4 to jfind the value of x, 2 4 SOLUTION. In a proportion, we have the first term divided by the second, equal to the third term divided by the fourth. By performing these divisions, we have 10a;+8 _7 18-a;"~4 40a;+32=126--7aj 47a;=94 «=2 SmPLE EQUATIONS. 189 PROBLEM 4bX 4-3 16. Given — : 1 : : 2.r + 19 : 3a;— 19 to find the value of a?. ^ 6a;— 43 SOLUTION. Putting this proportion in the form of an equation, we have 4a; H- 3 2a; + 19 6^=43=37=19- ^^'^"^^ ^^ ^^*^^^» we have 12a;''4-9a;— 76a;-67=12a;''+114a;—86a;— 817 — 95a;= — 760 a;=8 PROBLEM 17. Given (a+a;)(6+a;)— a(6+c)=— +a;'(l) to find the value of a?. S OLUTION. ah'\-(a+b)x+a^'-cLb^ac=—+x^ (2)=(1) expanded. (a+h)x=—+ac (3) =(2) transposed. / , rx _«'c + a5tf ,._..) with tenns of 2d ^ ^ ~ b ^ ''"^ M member added. ac (a + b)x=z~(a + b) (5) = (4) factored. EXAMPLES. 1, Given 8a;+7=52 — 7a; to find the value of a;. An^, x=S. 2* Given 18a;— 13= 6a; -f 36 to find the value of a;. Ans. a; =4. 3* Given 19a;+13=:59— 4a; to find the value of a;. Ans. a;=2. X 4* Given 3a; + 4— -=46— 2a; to find the value of a;. 3 Ans, a;=9. 5, Given x^ + 15a;=35a;— Sar' to find the value of a;. Ans. x=5. 6« Given 4a; + 36=5a;-f-34 to find the value of a;. Ans. a; =2. 7. Given 3a;''— 10a; = 8a; + 3;" to find the value of x. Ans. x=9. 9» Given Sax—4cab=2ax—6ac to find the value of x. Ans. a; =46— 6c. 190 SIMPLE EQUATIONS. 9. Given ax^ + abx=cdx to find the value of x. . cd—ab Am, x=: , a , X X 10* Given rr— 7 =--f- to find the value of a:. Ans. a?=15. 6 3 XXX 1 1 » Given -4--=-+7to find the value of x. Ans, a;= 1 2. Z O He /g K 284 X 12t Given \-6x= to find the value of x, 4 5 Ans. x=:9 J J ^ jg 2; 13. Given x-] — = — - — to find the value of a:. Ans, x=5, o 2 II* Given 3a? H — =5H to find the value of x, 5 2 Ans. a;=7. «.r r,' ^, 8a;— 11 6a?— 5 9l—1x^ n . ^t. i i? 15, Given 21 -\ — — = — 1 — to find the value of x. 16 8 2 Ans. a? =9. g/p 4 2g ^/p 16t Given 2= — \-x to find the value of a?. Ans. x—4, 17. Given ~_- + 10=-— - + 11 to find the value of a?, o 4 3 2 Ans. arrr 12. /g I 1 2a; 3 18i Given — - — 1-3= — - — to find the value of a?. Ans. x=9, 5 3 '7a;4-2 5x 6 19t Given — h 5a?= 28 H z — to find the value of a?. Ans, a;=4. Sx4-4: 22 — X 20. Given — -^ + 2a; =— - — + 16 to find the value of a?. 6 5 Ans. x=1. 9t n- 7a;— 8 16a; + 8 ^ 31— a; ^ c j ^v i ^ 21. Given -— - — I — — =3a; — to find the value of a;. 11 13 2 Ans. x=9. 22. Given 4a; — =.16 to find the value of a;. Ans. x=S. SIMPLE EQUATIONS. 191 ^^ ^. 21— 9x bx + 2 61 2a; + 5 29 + 4a: ^ « , ^, -+• 23. Given « + — Q~~='\2 3 \2 value of a;. Ans. x=:5. „. ^. 31 + 4a; 3a; + 47 3a:— 19 ., 16 — 10a? 6a;+20 24. Given -^ ^=41^ + -^^ ^ to find the value of a;. Ans, xz=il1. 25 1 Given 5 =- to find the value of a;. Ans. x— — - — . XX 4 26. Given ° — W n /=i4^^^^ ^ of a;. ^^. a;=2|. «« ^. 4a;-l7 3|-22a; 6 j , a;' i xj , ^v , 27. Given — ^-—- — =a; — -{1— rr}- to find the value 9 83 X \ 54 ) of a;. Ans, a;=3. 28. Given \ \ ?a?4-4 [ --^1?=^ -j ?-l [ to find the value of a;. 2*3 • 3 u.\X ) Am, a;=3. 29. Given 3*25a;— 5*007— a;=0-2—0-34a; to find the value of ar. * Ans, a;= 2-0104247. 7*53aj 2aJ x 30. Given -—- 100=— — 3-86— - to find the value of x, 18 5 6 Ans, a;=4-519-675. -~ 31. Given — n + / . rxs + / . rxa =3ca;+— to the value of a;. . ah Ans, x 32. Given r-~4 to ^^ t^e value of a;. a+bx d + ex Ans, x- af—cd ce—hf' t« Given Va' + c =4/^77 r to find the value of a;, y d(x-\-g) d{x-\-g) Ans, a;= —g. dX/a^+c 34. Given Va+a;='Va:' + 6aa:+&' to find the value of a;. Ans, x=—- — . 3a 192 SIMPLE EQUATIONS- bx (Sbc-}-ad)x Sab _(8bc—ad)x ^ 5a (2b—a) ^* ^^^^ 2b^^'~~2ab(a-{-b) d^^" 2ab{a-b) ~~a^-b^ to find the value of a;. Ans. x=-—^ =-^, 36. Given -= :; —z — :; — to find the value of a?. a?— 1 x-\-l x^—1 Ans, «=— IJ. X. Ans. x=8. 9a; +20 4a;— 12 x Zl* Given — -- — =-- +- to find the value of x. 36 6a;— 4 4 38* Given =- to find the value of a;. 21 4a;— 11 3 Ans, a;=8. 39. Given — - — + — — = — — - to find the value of x. 9 Da;+3 3 Ans, a; =4. H^40. Given — 14 = ^^~^ — 15 to find the value of x, * a; + 2 2a;— 2 Ans. x=2. /p 2 2Vx 41* Given — =-=— - to find the value of a;. Ans. x=6» |/a; 3 42. Given a;-H/2aa;+a;''=a to find the value of a;. Ans. x—-. 4 43. Given 2Va*+a;'=4(a— ^a;) to find the value of a;. J 3a Ans. x=z—. 44* Given a-\-x=A/a^-\-xVb^+x'' to find the value of x. b'^4a' Ans. x=- 4a 45. Given y'3a;— 1=2 to find the value of a;. Ans. a;=f. 46« Given |/a;4-a;'=a; + i to find the value of a;. Ans. x=:^. 47. Given \/3a; + 13— 4=0 to find the value of x. Ans. a?==l7. 48. Given + a;=g + 2a; to find the value of x. Va—x Ans. a;=l— a. 49. Given y4-\-Vx*---x^= x—2 to find the value of a;. Ans. a;=2|. SIMPLE EQUATIONS. ' 198 50. Given (2+a;)2+a:2=4(2+a;)"2 to find the value of a;. Ans. 51. Given 3/2^6 + 3 = 15 to find the value of x, Ans. x=5. Ans. x=^. 52. Given |/a; + 3=f^21 +a; to find the value of rr. Ans, a; =4. a 53. Given x-\-Va—x— to find the value of a?. Va—x Ans. a;=a— 1. 54« Given -^^ -/=— — to find the value of a;. 3a; 3a; + 2 11a; Ans. a;=yYT' CL C 55* Given a;+T«-|-r-a;=w» to find the value of a;. O ^ ' bm Ans. X 56f Given \/a-^x-{-Va~-x=Vax to find the value of a;. Ans. x=- a" + 4* 57* Given i/— -— + |/— f_ —a/ ^ to find the value of a;. " a4-a; ~ a—x ^ a^—x^ Ans. 58* Given \fx-\-az=.c—Vx-\-h to find the value of a;. Ans. a;=l 1 V 2c / 59* Given i^a+Vx^zVax to find the va%ie of x. Ans. a;=: — =^- {Va-\f 60* Given |/a;— 16=8— f^a; to find the value of a;. Ans. a;=25. 6)t Given |/a;-f 40=10— Va; to find the value of a;. Ans. a; =9. 62. Given 4/a;— 24=V^a;— 2 to find the value of a;. Ans. a; =49. 63. Given 4/5 xVx-\-2=V5x-\-2 to find the value of a;. Ans. x=-^. 6I« Given ax-{-aV2ax + x^=:ab to find the value of a;. A ^' ^n«. a;= 2(a + 6y 18 194 SIMPLE EQUATIONS, Vx + 2a_Vx-\-4:a Vx + b Vx-hSb 65* Given -= :=-— to find the value of x. Ans. X: (s) 66« Given \/4:a+x=z2Vb+x^Vx to find the value of x. (b-ay Ans, X: 2a— b re rtj' \/ X 1 67. Given — zr-= — to find the value of x. Ans. x: J^X X 1— a 68. Given = — = to find the value of ar. Vax+b SVax + Bb Ans. x=z — , a >, Given — —— -=9 to find the value of ic. Ans. x=^, V4:X+l — 2Vx 70. Given ^« + ^+^^L-l::^& to find the value of x. Va + x—Va—x 2ab Ans. x=.^^. 71. Given ^ '^ ^ i/'^^—l=2 to find the value of a;. Ans. x=l^, x—1^ x-\-l 72. Given V^6^-2 ^ 4V^6^-9 ^^ ^^ ^^^ ^^^^ ^f ^^ ^^^ ^^^^ V6.r + 2 4V^6a;+6 73. Given a + 6Va?+rf=#to find the value of a;, Ans. x=l-^\ —d, Ofi 74. Given 4/3^+ i/rr — Q = to find the value of ar. ♦/a;— 9 Ans. a:=25. 75. Given a/\+W^'+\2 — \ +a; to find the value of a;. Ans. a;=2, 76. Given i/ar + V'^— |/a;— V5=-( =1 to find the value of a?. Ans. xJ^. SIMPLE EQUATIONS. 195 77« Given : — = — : : 14 : 6 to find the value of x, 5 7 Ans, a; =4. 78. Given 5a; +t^ =9H — ;; ;:- to find the value of a;. 4a; + 3 2x + 3 Ans. a;=3. 79* Given \/9ar— 4 + 6 — 8 to find the value of a?. Ans, a: =4. 2Y 4x 15+ 2x 80» Given — - — : 2x : : 5 : 4 to find the value of x. Ans. x=S* 4x4- 14 81. Given 16a; + 5 : ; : : 36a: + 10 : 1 to find the value of a?. vX-x" 31 Ans. x=5» Vx + Vh a , ^ , ,, V . . . ./« + ^\'* Vx-Vh 82. Given -r= ==t to find the value of x. Ans. x=h\ = ) . \/r.—\/h \a—b/ 83. Given — = = — = to find the value of a;. Ans. x=4. Vx+2 fa;+40 84t Given a VbX'-c=d Vex+fx^g to find the value of x. _ a^c—d^g 85« Given -=- • to find the value of a:. 3 6^--3a: 3 x—2 A 1* An^. x=^. 36 86. Given1/a:+Va;— 9=-z= to find the value of a;. 4/a:— 9 Ans. a; =25. J,. _,. 3a; 81a;»-9 „ 3 2a:'-l 57- 3a; ^ ^ . 87. Given— —7 ^, r-=3a;— - • — to find 2 (3a;— l)(a;4-3) 2 a; + 3 2 the value of a;. ^w«. a;=10. 88. Given Ai|/a;' + 39a; + 374-i/a;»4-20a; + 51 i =|/?i^? to 19 I ) '^ a;+ 17 find the value of a;. Ans. a; =78. 89. Given a;» + 2a;=(a; + a)» to find the value of x. Ans. X: 2(1— a)* 90. Given a;'' + 2a;'4-aJ=(a;* + 3a;)(a;— l)-f 16 to find the value of a;. Ans. a;=4. 196 SIMPLE EQlTATlOIfS. *91. Given — - — : x—5 : : - : - to find the value of x, Am^ x=:5, 4 3 4 92. Given ^ ^\ — ^— 3a= t 2a; + — t — to find the a — a + o a' + Sa'b+4a''b''—6ab*-\-2b^ value of a;. Ans, x= —tf::-^ — t T^n • 26[2a' + (a— 6)6] 93. Given t-H-i — I- -7-+^=^ to find the value of x, bx ax fx hx adfh + bcfh + bdeh + bdfg Ans. X- bdfhk ^. _. I0a;+17 12a;+2 5a?-4^ ^ , . , . 91 1 Given — — — --= — - — to find the value of a;. 18 \oX — Id y Ans. a;=4« ^^ ^. 2a;+8i 13a;-2 x 1x x+\Q ^ ^ . . . 95. Given— ^ ___- + _=_____ to find the value of a?. Ans, a?=4. -- _. 7a: +6 2ar + 4f a; 11a; a;-3^ . , . . . 96. Given -28 23^ + 4=Tf """42- *^ ^^^ *^' ^"^^^ ^^''- Ans. x=4. ^ ^. 6-5x 7-2a;' l4-3a; 2a;-V- 1 , « . ,. 97. Given -^---^^^—^^-^^ ^+-- to find the value of a;. Ans* a;=4. 98. Given -7 dc=bx—ac to find the value of a;. Ans. .=ffcfJ-Z£). a"— 6' + oc ^ ^. 18a;- 19 llar+ 21 9a;+ 15 ^ „ ^ . 99. Given — -— \--- — r-TT-= — tt — to h^d the value of x, 28 6a;4-14 14 Ans. a; =7. 2a;— 3 3a; — 1 100. Given =- • to find the value of a;. 2 a;— 1 2 3a;— 2 2 Ans. X——-, 3 * There is a peculiarity in tEis example. Any quantities whatever, whether in the ratio of | to f or not, when substitued for § and |, wiU give the answer. Can the student explain the reason ? QUESTIONS INVOLVING SIMPLE EQUATIOJ^S, ETC. 197 QUESTIONS INVOLVING SIMPLE EQUATIONS CONTAINING ONE UNKNOWN QUANTITY. QUESTION (266.) 1. What number is that, the double of which exceeds its half by 6? SOLUTION. Let x= the number. Then, by the conditions of the question, we must have the equation X 4tx—x=l2, 8ar= 12, x=4, the number required. Another Solution, Let 2a? = the number. Then, by the conditions of the question, we must have the equation 4x—x=6, 3a; =6, x=2, 2a; =4, the number required. QUESTION 2. A person employed 4 workmen ; to the first of whom he gave 2 dollars more than to the second ; to the second, 3 dollars more than to the third ; and to the third, 4 dollars more than to the fourth. Their wages amounted to 32 doUais. How much did each receive ? SOLUTION. Let x= the sum received by the. fourth, then a; + 4= " " " third, x + 1= " " " second, and a; + 9= " " " first. By the conditions of the question, the sum of these must equal 32 dollars. Therefore, 198 QUESTIONS INVOLVING SIMPLE EQUATIONS, ETC. 4a;+20=32, 4rc=12, x= 3j the sum received by the fourth, x + 4=: 1, " " " third, a; + '7 = 10, " " "- second, and a;+9 = 12, " « " first. QUESTION 3. What two numbers are to each other as 2 to 8 ; to each of which if 4 be added, the sums will be as 6 to 7 ? SOLUTION. Let 2x and Sx be the numbers. Then, by the conditions of the question, 2X'\-4:3x + 4:::5:1. Whence, 14aj + 28 = 15a: + 20, ^ 8^x, 2a: =16, the first number. 3a;=24, the second number. QUESTION 4. A person being asked the hour, answered that it was between five and six ; and the hour and minute hands were together. What was the time ? SOLUTION. Let x= the time past 5. Then, since the minute-hand goes 12 times as fast as the hour- hand, it follows, that 6 4- a; is 12 times x, .' . \2x=b-\-x \\x=b x=j\ of an hour = 27 minutes 16 j\ seconds, the time past 5 o'clock. QUESTION 5. What number is that to which if 1, 5, and 13, be severally added, the first sum shall be to the second as the second to the third? QUESTIONS INVOLVING SIMPLE EQUATIONS, ETC. 199 SOLUTION. Let x= the number required. Then, by the conditions of the question, x + \ :a; + 5 :: x-\-b : a;+13. Whose solution gives a;=3. QUESTION 6. A shepherd, in time of war, was plundered by a party of soldiers, who took \ of his flock, and ^^ of a sheep ; another party took from him i of what he had left, and -5 of a sheep ; then a third party took ^ of what now remained, and | of a sheep. After which he had but 25 sheep left. How many had he at first ? SOLUTION. Let x= the number he had at first. X 1 Then -+-= the number the first party took away. 4 4 Which, being subtracted from x, gives 3a; 1 — — -= the number remaining. 4 4 The second party took away \ of these, + J of a sheep, which left 3\T~4/ 3'^^ 2 2' Of these the third party took one-half + i of a sheep, which left 2\2 2/ 1 x ^ 2'"^4-4- X 3 ••• 4-4= '' a--3r=100 x—\0^. QUESTION Y. A man and his wife usually drank a vessel of beer in 12 days ; but when the man was gone, it lasted the woman 30 days. In how many days would the man alone drink it ? SOLUTION. Let X = the number of days it would take the man alone to drink it. Then, the man, in 1 day, would drink - of it. X The woman, " " " 3V " The man and woman together " y»^ " 200 QUESTIONS INVOLVING SIMPLE EQUATIONS, ETC. ... i-=i+i 12 x^S( 5a:=60 4-2a: Sx=z60 ic— 20, the time it would take the man alone to drink it. QUE STION 8. A person engaged to reap a field of 35 acres, consisting partly of wheat and partly of rye. For every acre of rye he received 5 shillings ; and what he received for an acre of wheat, augmented by one shilling, is, to what he received for an acre of rye, as 7 to 3. For his whole labor he received 260 shiUings. What was the number of acres of each sort ? SOLUTION. Let x= the number of acres of wheat ; Then 35— x= the number of acres of rye ; Then 1 75 — 50:= the price of reaping the rye. By the question, 3:7 : : 5 : 1 + , the price of reaping an acre of wheat* But 3:7::5:11|. Therefore, the price of reaping 1 acre of wheat is lOf^^^^- shillings, 32a; And " " " ic acres " " shillings. 32a; ^ ^ .-. -— + 175— 5a;=260, 3 32a; + 525 — 15a:=780, I7a;=r255, ar=15, the No. of acres of wheat. 35— ar=20, " ". rye. QUESTION 9. The hold of a ship contained 442 gallons of water. This was emptied out by two buckets : the greater of which, holding twice as much as the other, was emptied twice in 3 minutes ; but the less^ three times in 2 minutes ; and the whole time of emptying was 12 minutes. How much did each hold ? SOLUTION. Let X = the number of gallons the less held. Then 2a; = the number of gallons the greater held. QUESTIONS. 201 4a?= the gallons thrown out by the greater in 3 minutes iex= " " " " 12 '* 18a:= " " " less " 12 " ,-. 16a;+18a;=34a;=442, a;^=13, the No. of gallons the first bucket held. 2a:=26, " " second " " QUESTIONS. 1. What number is that, from the treble of which if 18 be sub- tracted, the remainder is 6 ? Ans. 8. 2. What number is that, the double of which exceeds f of its half by 40? Ans. 25. 3. In fencing the side of a field, whose length was 450 yards, two workmen were employed ; one of whom fenced 9 yards, and the other 6, per day. How many days did they work ? Ans. 30. 4. A farmer sold 13 bushels of barley, at a certain price; and afterward 17 bushels, at the same rate; and at the second time received 36 dimes more than at the first. What was the price of a bushel ? Ans. 90 cents. 5. A draper sold two pieces of cloth, by one of which he lost $6 more than by the other : and his whole loss was $5 less than treble the less loss. What were the losses sustained by each piece ? Ans. $11, and $17. 6. A company settling their reckoning at a tavern, pay $8 each ; but observe, that if there had been 4 more, they should only have paid 17 each ? How many were there ? Ans. 28. 7. 'I'wo workmen received the same sum for their labor ; but if one had received $15 more, and the other $9 less, then one would have had just three times as much as the other. What did they each re- ceive ? - Ans. $21 each. 8. What number is that, the treble of which is as much above 40, as its half is below 51 ? A7iS. 26. 9. A person has a certain number of ijorses at a livery stable, and 3 times as many at grass. He keeps 15 in constant employment; and his whole number is 7 times the number in the stable. What was the whole number ? Ans. 35. 10. Two men at the distance of 150 miles set out to meet each other ; one goes 3 miles while the other goes 7. How much of the distance does each travel? Ans. One 45, and the other 105 miles. 202 QUESTIONS. 11. A person put out a certain sum at interest for 6i years, at 6 per cent, simple interest ; and found that if had put out the same sum for 12 years and 9 months, at 4 per cent., he would have received $185 more. What was the sum put at interest. Ans. $1000. 12. From two casks of equal size are drawn quantities, which are in the proportion of 6 to 7 ; and it appears that if 16 gallons less had been drawn from that which is now the emptier, only half as much would have been drawn from it as from the other. How many gal- lons were drawn from each ? Ans. 24 gallons from one, and 28 gallons from the other. 13. Out of a certain sum, a man paid his creditors |96 ; half o» the remainder he lent his friend ; he then spent i of what now re- mained ; and after all these deductions had j\ of his money left. How much had he at first ? Ans. $128. 14. Six hundred persons voted upon a disputed question, which* was lost by a certain number. The same number of persons having voted again upon the same question, it was from some change in cir- cumstances carried by twice as many as it was before lost by ; and the new majority was to the former one as 8:7. How many changed tht;ir minds ? Ans. 150. 15. A sportsman, keeping an account of the number of birds he killed, found that each succeeding season he wanted 50, in order that the number killed might bear the proportion of 3 : 2 to the number killed in the preceding year. In the fourth year he found that he had killed 170 fewer than three times the number killed in the first year. How many did he kill the first year ? Ans. 180. 16. Several detachments of artillery divided a certain number of cannon balls. The first took 72, and ^ of the remainder ; the next 144, and ^ of the remainder ; the third 216, and ^ of the remainder; the fourth 288, and i of those that were left ; and so on ; when it was found that the balls had been equally divided. What was the number of balls and detachments ? Ans. 4608 balls, and 8 detachments. 17. Two persons, A and B, start at the same time for a race which lasted 6 minutes. Now after galloping 4 minutes at the same uniform pace at which each started, the distance between them is j^-^ the part of the whole length of the course. They continue to run for I minute more, at the same speed as at first ; and then B, who is last, QUESTIONS. 203 quickens the speed of his horse 20 yards a minute, and comes in exactly two yards before A, whose horse had run at the same uni- form pace throughout. What was the length of the course ? Ans. 3 miles. 1 8. A packet sailing from Dover with a fair wind, arrives at Calais in 2 horn's; and on its return the wind being contrary, it proceeds 6 miles an hour slower than it went. Now when it is half way over, the wind changing, it sails 2 miles an hour faster, and reaches Dover sooner than it would have done had the wind not chano^ed, in the proportion of 6 : Y. What were the rates of sailing and the dis- tance between Dover and Calais ? Ans. On its return it sails 5 and 7 miles an hour, and the distance is 22 miles. 19. A farmer has a stack of hay, from which he sells a quantity I which is to the quantity remaining in the proportion of 4 to 5. He then uses 15 loads, and finds that he has a quantity left which is to the quantity sold as 1 to 2. How many loads did the stack contain at first? Ans. 45. 20. In a naval engagement, the number of ships taken was 1 more, and the number burnt 2 fewer than the number sunk. Fifteen . escaped, and the fleet consisted of 8 times the number sunk. Of how many ships did the fleet consist ? Ans. 32. 21. At the review of an army, the troops were drawn up in a solid mass, 40 deep, when there were just J- as many men in front as there were spectators. Had the depth, however, been increased by 5, and the spectators been drawn up in the mass with the array, the number of men in front would have been 100 fewer than before. Of what number did the army consist ? Ans. 180000. 22. A and B playing at billiards, A bet 5 dollars to 4 on every game, and found that after a certain amount of games, he had won 10 dollars. Had B won one game more, the number won by him would have been to the number won by ^1 as 3 to 4. How many did each win? Ans. A won 20, and ^14. 23. A besieged garrison had such a quantity of bread as would, if distributed to each at 10 ounces a day, last 6 weeks, but having lost 1200 men in a sally, the governor was enabled to increase the allow- ance to 12 ounces per day for 8 weeks. What was the number of men at first in the garrison ? Ans 3200. 204 QUESTIONS. 24. During a panic, there was a run on two bankers A and B. B stopped payment at the end of 3 days, in consequence of which the alarm increased, and the daily demand for cash on A being trebled, A failed at the end of 2 more days. But if A and B had joined, their capitals, they might both have stood the I'un, r.s it was at first, for 7 days, at the end of which time B would have been indebted to A $4000. What was the daily demand for cash at ^'s bank at first? Ans. $2000. 25. There are two numbers in the proportion of ^ to |, which being increased respectively by 6 and 5, are in the proportion of | to ^. What are the numbers ? Ans. 30 and 40. 26. A gentleman meeting 4 poor persons, distributed 60 cents among them : to the second he gave twice, to the third thrice, and to the fourth four times as much as to the first. How much did he give to each ? » Ans. 6, 12, 18, and 24 cents respectively. 27. A farmer has two flocks of sheep, each containing the same number. From one of these he sells 39, and from the other 93 ; and finds just twice as many remaining in one as in the other. How many did each flock originally contain ? Ans. 147. 28. Four places are situated in the order of the four letters A, By C, D, The distance from ^ to i> is 34 miles ; the distance from A to B ; distance from (7 to i> : : 2 : 3, and } of the distance from A to B added to half the distance from C to i> is 3 times the distance from B to C. What are the respective distances ? Ans. AB=12, BC=4, and CD=18. 29. In a mixture of wine and cider, half of the whole + 25 gallons was wine, and i of the whole —5 gallons was cider. How many gallons were there of each ? Ans. 85 gallons of wine, and 35 of cider. 30. A footman who contracted for £8 a year, and a livery suit, was turned away at the end of 7 months, and received only £2 3s. 4rf. and his livery. What was its value ? An^. £6. 31. A cistern into which water was let by two pipes, A and B, will be filled by them both running together in 12 hours, and by the pipe A, alone, in 20 hours. In what time will it be filled by the pipe B alone? Ans. 30 hours. 32*. A person has two sorts of wine, one worth 20 pence a quart, QUESTIONS. 206 and the other 1 2 pence ; from which he would mix a quart to be worth 14 pence. How much of each must he take ? Ans. He must take ^ of the iSrst, and | of the second. 33. A hare, 50 of her leaps before a greyhound, takes 4 leaps to the greyhound's 3 ; but 2 of the greyhound's leaps are as much as 3 of the hare's. How many leaps must the greyhound take to catch the hare? Ans, 300. 34. Two pieces of cloth of equal goodness, but of different lengths, were bought, the one for |5, and the other for $6^. Now, if the lengths of both pieces were increased by 10, the numbers resulting would be in the proportion of 5 to 6. How long was each piece, and what was the cost a yard ? Ans. One piece was 20 yards long, and the other 26. Cost, 25 cts. a yard. 35. Two persons, A and B, have both the same annual income. A lays by J of his : but B, by spending $80 per annum more than Af at the end of 4 years finds himself $220 in debt. How much did each receive and spend annually ? Ans. The annual income of each is $125, and ^'s annual expendi- ture is $100, and ^'s $180. 36. An egg-merchant meeting with three customers, sells to the first of them half his stock and 1 egg more ; to the second he dis- poses of half the remainder and 2 eggs more ; and to the third half of ^hat he then had left and 3 eggs more ; and afterward discovers that he has parted with his whole stock. What number had he at first? Ans. 34. 37. A person disposes of turkeys at as many dimes each as the number he has, and returning 1 dime finds that if he had had one more to sell on the same condition, and had returned 2 dimes, he would have received 20 dimes more from his bargain. What num- ber did he dispose of? Ans. 10. 38. A gentleman bequeaths his property as follows : To his eldest child he leaves $1800, and } of the rest of his property ; to the second, twice that sum and ^ of what then remained ; to the third, three times the same sum and ^ of the remainder, and so on ; and by this ar- rangement his property is divided equally among his children. How many children were there, and what was the fortune of each ? Ans. 5, and $9000, the fortune of eachi 206 QUESTIONS. • 89. ^ and B possess certain sums of money, such that if they gain %a and %h respectively, A will be m times as rich as B ; but if they gain %c and %d respectively, A becomes possessed of n times as much as B. How much money has each ? ^ m {nd — c) — n{i7ih — a) Arts. ■ A^s money at first. m — n Cnd—c) — {mb—a) ^ ' =Bs money at first. fourth dav, less than the nrst. f. 67, 62; 58, and 53 milesX-^^^^ flf ^9.1000 ia fn ha rlivirlAflV. m — W 40. The crew of a ship consisted of her complement of sailors and a number of soldiers. Now there were 22 sailors to every 3 guns and 10 more. Also, the whole number of persons was 5 times the number of soldiers and guns together. But after an engagement, in which the slain were i of the survivore, there wanted 5 to be 13 men to every 2 guns. What was the number of guns, soldiers, and sailors ? Ans. 90 guns, 6*70 sailors, and* 5 5 soldiers. 41. An express set out to. travel 140 miles in 4 days, but in conse- ^^quence of the badness of the roads, he found that he must go 5 miles the second day, 9 the third, and 14 the fourth day, less than the first. How many miles did he travel each day Ans. 42. The estate of a bankrupt, valued at 121000, is to be divided^ among four creditors proportion ably to what is due them. The debts due to A and B are as 2:3 ; ^'s and C's claims are in the ratio of 4:5; and C"s, and i>'s in the ratio of 6:7. What sum must each receive ? Ans. A $3200, B $4800, (7 $6000, and D $7000. 43. There are two towns, A and B, which are 131 miles distant from each other. A coach sets out from A at six o'clock in the mom* ing, and travels at the rate of 4 miles an hour without intermission, in the direct road toward B. At 2 o'clock in the afternoon of the same day, a coach sets out from B to go to A, and goes at the rate of 5 miles an hour constantly. Where will they meet ? Ans. 76 miles from A, and 55 from B. 44. A waterman finds by experience that he can with the advan- tage of the common tide row down a river from A to B^ which is 18 miles, in 1 hour and a half, and that to return from B io A against an equal tide, though he rows back along the shore, where the stream is only | as strong as in the middle, takes him just 2 hours and a qu'-.rter. At what rate does the tide run per hour in the middle, \y]iQTfi it is the strongest ? Ans. 2i miles per hour. QUESTIONS. 207 45. The ingredients of a loaf of bread are rice, flour, and water, and tlie weight of the whole is 15 lbs. The weight of the rice augmented by 5 lbs. is | of the weight of the flour, and the weight of the water is I of the weight of the flour and the rice together. What is the weight of each ? Ans. Rice 2 lbs., flour 10^ lbs., water 2\ lbs. 46. Suppose two hands of a watch, (a) and (&), were together on Sunday noon, and the motion of each was such that (a) moved round the horary cfrcle in one hour, and {b) in l^-^ hour. When will they be together again for the first time ? Ans. 6 1 hours. 47. The rent of an estate this year is greater by 8 per cent, than it was last year. This year's rent is $1890. What was the rent of last year? Ans.%\l50, 48. A merchant increases his capital yearly by 20 per cent., and takes from it every year llOOO for the support of himself and family. After he had carried on his business, in this manner, for three years, he finds, after deducting the usual sum, $1000, that his capital has increased $200 more than | of the original sum. What was the original capital ? ^ws. $30000. 49. A person has 4 wine casks of different sizes. When he fills the 2d empty cask from the first full one, there remains in the first only I of the wine ; when he fills the 3d empty cask from the 2d full one, there is left in the 2d only | of the wine ; but when he attempts to fill the 4th empty cask from the 3d full one, then only -^-^ of the 4th is filled, and if he wished to fill the 3d and 4th empty casks from the first full one, then these would not only be filled, but^ he would have 1 5 gallons remaining. How many gallons does each of these four casks contain ? Ans. The 1st, 140 ; the 2d, 60 ; the 3d, 45 ; and the 4th, 80 gallons. 50. A father leaves a number of children, and a certain sum of money, which they are to divide among them as follows : The first is to receive $100, and a 10th part of the remainder; after this, the second receives $200, and a 10th part of the residue ; again, the third receives $300, and a 10th part of the remainder ; and so on. At last it is found that all the children have received the same sum. What was the fortune left, and how many children were there ? Ans. $8100, and 9 children. 61. What number is that which, if it be increased by 7, the square 208 QtTESTlONS. root of the sum shall be equal to the square root of the number itselt and 1 more ? Ans. 9. 62. A person wishes to give 3 cents a-piece to some beggars, but finds he has not money enough by 8 cents ; but if he gives them 2 cents a-piece, he will have 3 cents remaining. What is the number of beggars? Ans. 11. 53. What ai-e the two parts of 60, such that their product is equal to three times the square of the less ? Ans, J5 and 45. 54. In the composition of a quantity of gunpowder, the nitre was 10 lbs. more than | of the whole, the sulphur 4^ lbs. less than } oi the whole, and the charcoal 2 lbs. less than | of the nitre. What was the amount of gunpowder ? Ans. 69 lbs. 55. A person engaged to work a days on these conditions : For each day he worked he was to receive b cents, for each day he was idle he was to forfeit c cents. At the end of a days he received d cents. How many days was he idle ? a ^ /I'-i?. days. 56. What must the fortune and number of children be, when, in general, the first receives a dollars, together with the nth part of the remainder; and each succeeding child a dollars more, together with the nth part of the remainder, and it is found, at last, that they have all received the same sum ? Ans. The fortune, =(71— 1)V, and children, =n — l. 5*7. A person wishes to make the following payments at 4 different periods ; one sum a in /, a sum 6 in m, a sum c in n, and a sum d in p months. K he wishes to pay his whole debt, a-hb + c-^d, at once, at what period must he do it ? , , Ans. , ^-\JL months. a + b + c-^d 68. A master mason has engaged a number of masons for the erection of a building. He finds, after entering into a calculation, that if he gave each man m shillings a day, he would daily expend a shillings less than was assigned for that purpose by the estimate, and that he would lose b shillings, if he should give each n shillings. How many men did he hire, and what was the daily wages of each ? a-\-b Ans. ■< The number of masons was n—m The daily wages of each was ~ shillings. SmULTANEOtJS EQUATIONS, ETC. 209 59. What number must be added to each of the two given num- bers, a and 6, that the sums may be as ??i : w ? . Tfib — nd Ans. . n—m 60. Three masons are employed in building a wall. The first builds 8 cubic feet in 5 days ; the 2d, 9 cubic feet in 4 days ; and the 3d, 10 cubic feet in 6 days. How much time will these masons need, when they work together, to build 756 cubic feet of the wall ? Ans. ISTJ/y days. i^ . ■ ♦ > . ^ SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE, CONTAINING TWO UNKNOWN QUANTITIES. (267.) Simultaneous Equations are such as must exist at the same time, the values c£ the unknown quantities in each equation being restricted by the other. PROBLEM. (268.) To discuss the nature of simultaneous equations. DISCUSSION. Let ic + y=:6. This equation can be satisfied by the following positive integral values for x and y. x~\ and y=6, x=2 " 2/=4, a;=3 " y=3, x=4: " y=2, or, x=b " y—1. Also, if a;=0, y must = 6, and if a;=6, y must = 0. We might assign negative values for a;, and the corresponding values of y would be obtained by subtracting the value of x from 6. Thus, a;=— 2, y=6-(-2) = 8. We can also assign, for the value of x, any proper or improper fraction, and the corresponding value of y must be this value sub- 14 210 ELIMINATION. tracted from 6. There are, then, an infinite number of values which may be assigned to x and y which will satisfy the given equation. Suppose, now, we have ^x + 2y—\4:. In this equation, also, x and y may have an infinite number of values. But, if we 'wish x-\-y=Q and 3a: + 2y=14 to exist at the same time, or, in other words, that each of these equations must be restricted by the other, x and y can only have those values which will at the same time satisfy both equa- tions. "We have, then, this problem, to find what values of x and y will satisfy a; -fy = 6, provided 3a; + 22/ = 1 4. If a; =2 and y=:4, the first equation will be satisfied, and these values will also be found to satisfy the second. It will hereafter be found, that of all values of x and y which will satisfy the first equation, a* =2 and y=4 are the only ones that will, at the same time, satisfy the second equation. rc + y=6 is called an indeterminate equation when it has no other equation to limit it. In like manner x + y-\-z:=^ is an indeterminate equation, when it has no other equation to limit it. In general, that equations may he determinate there must be as many equations as there are unknown quantities. It must be carefully observed that the equations must all be inde- pendent, that is, that no equation be the result of an addition, sub- traction, multiplication, or division performed upon one of the others. Thus a; + 3/ = 6, and a; + y + 3 := 9 are not independent equations. The same may be said of x-{-y=Q, and a^ + y— 3=3 ; a; + y=6 and 2x + 2y=zl2; x + y=6 and ^x + lyz=z2. ELIMINATION. (269.) Elimination is the process of deducing fi-om two or more simultaneous equations, a single equation containing one unknown quantity. (270.) Simultaneous equations are of the first degree when the equation deduced from them is of the first degree, of is a simple v^quation. ELIMINATION. 211 ELIMINATION BY SUBSTITUTION. (27 1 •) Mimination by substitution is finding an expression for the value of an unknown quantity in one equation, and substituting this value for the same unknown quantity in another equation. PROBLEM. (272.) To eliminate by substitution from two simultaneous equations. RULE. J^ind an expression for the value of an unknown quantity in one equation, and substitute this expression for the same unknown quantity in the other equation, and there will result a single equation contain- ing but OTie unknown quantity. PROBLEM (273.) 1. Given \ ""^ +^2/=^ (1) ) ^ ^^ ^^^ ^^^^^^ ^^ ^ ^^ ^ ^ f mx-\-ny=d (2) ) SOLUTION. y= — 7— (3)= value of y in (1). nc—nax , , , . nyz=: — (4) = (3)xw. nc—nax , / x ( value of ny in (4) sub- "'^+— r-^"^ <^)=istitutedin(2). hmx+nc —nax=bd. hmx — nax =:bd — nc. (bm—na)x=bd—nc. bd—nc bm—na (8). abd—acn .. .. ax—— ~ (9) = (8) X a. om — na ' ^ ' abd—nac i _ /-iA\-_i value of ax in (9) bm—na ^ ^~~{ substituted in (1). , abd—nac by=c . bm—na c abd — nac b b'^m—abn cm— ad bm—na 212 ELIMINATION. PROBLEM ( 4x4- 9y=51 (1) ) 2. Given j g^_ ^g _ q /o( f *<> ^J^d the values of x and y. SOLUTION, 4a;=61— 9y (3) =(1) transposed. . 8a;=102-18y (4)=(3)x2. 102-18y-13y==9 (5)= \ ^^lue of So; in (4) substituted (in (2). — 3 ly = — 93. (6) = (5) transposed and added. y=3. (7) = (6)-^-31. Since, y=3, we have instead of 4a;=51— 9y the equation 4ic= 5 1 — 27 = 24 (8) = value of 9y substituted in (8). x=ze. PROBLEM 3. Given j ,_ 3_o Lx f to find the values of x and y. SOLUTION. The second of these equations is not of the first degree, but the equation found by eliminating will be of the first degree. The second equation is in fact the product of x—y=l by 3, for it may be put in the form {x-\-i/){x—y)=3, and we know that x-\-7/=B from the first equation. We have then to find the values of x and y in the equations x + 7/=3 (3). x-i/=l (4). x=l+y (5) = (4) transposed. (6)= value of a? in (5) substituted in (8). 2ar— y PROBLEM 4-14=18(1) 4. Given ^ gvlar ^2H:^ 4. 16=19 (2) ► to find the values of x and y. ELIMINATION. 21t SOLUTION. 2a?— y=8 (3)=(1) X 2 and terms transposed. 2y+a;=9 (4) — (2) X 3 and terms transposed. x=9—2i/ (6)— (4) transposed. 2a?=18— 4y (6)^(5) X 2. 18-4y-y=8 (7)= value of 2x in (6) substituted in (3). -6y=:-10. y=2. a;=9— 2y. a:=:9-4. a;=6. EXAMPLES. !• Given 2, Given 3« Given !• Given 5. Given •I « « . « f to find the values of x and y. ^7J5. fl;=3, y=2. iQ^ Qt/z^ 9 ) *, . ^ f to find the values of x and y. ^ns. a;=3, y=l. ■j ^ y~ r.^ f to find the values of a; and y. \ 15a; + 2y=66, i ^ Ans. ic=4, y=3. », ^ , « f to find the values of x and y« Ans, a;=:l, y=l. 1 4y . *lx—2y ^ to find the values of x and y. ^rw. a;=4, y=ll. 6« Given ^ o 4 to find the values of x and y. ^W5. a;=4, y=ll. 1 I7a;+lly=:40 ) 7. Given i o _ q i *^ ^^^ *^® values of x and y. ^W5. a:=-Tr:r, y = 179 69' 137 69* /fir )c *>&^ -^1^ 214 ELIMINATIOK BY COMPARlSOiT. w ^ ,.,'!• to find the values of x and V. 9. Given ] o _oo' f *^ ^^^ ^® values of a; and y. ( *^ — 23/ — 22, J ir ^4715. a:=4, y=3. f 6a;-4- 4?/r=26 ) 10. Given \ ^ / ' >• to find the values of x and y. ( 6a;-}-4y=28, ) ^ -47W. a:=2, y=4. ELIMINATION BY COMPARISON. (274.) Elimination by Comparison is finding an expression for the value of an unknown quantity in one equation, and also, an expression for the value of the same unknown quantity in another equation, and putting the expressions equal to each other. PROBLEM. (275.) To eliminate by comparison from two simultaneous equations. RULE. Mnd an expression for the value of one of the unknown quantities in the first equation, and put it equal to an expression for the value of the same unknown quantity found from the second equation.) PROBLEM (276.) 1. Given I ll'^l^Zn (2)' [ *"" ^""^ *^^ values of a; andy. SOLUTION. ax=zm—hy (3)=(1) transposed. x=^ (4)=(3)^a. cx=n—dy (S)= (2) transposed. _n'—dy c --* m — by n — dy a ~ c cm — hey := an — ady ady—hcy^=^an—cm {ad—bc)y=an — cm an— cm ^ ad— be The value of x may be obtained by equating the expressions for the (6)=(5)-c. (7)= value of x in (4) and in (6) equated. ELIMINATION BY COMPAEISON. 215 value of y, or it may be obtained from either of the given equations by putting in it instead of y its value as found above. Proceeding by either of these methods, we would find ^=— i — r-» PROBLEM (277.) 2. Given i ^^^^l^^^^^^^ f}\ to find the values of X and y. 8 OLUTION. 100-3y ^=-11- 4+7y _l Q0— 3y 4 ~~ 11 44 + 77^=400-- 12y 89y=366 4+7y y=4. Since, a;= — — ^ and y=4, 1, 4+28 „ we nave «= — ^ — = 8. PROBLEM 8. Given \ ^'':^l'f^^,l ^J^ \ to find the values of a; and y. ( 64a;— 3yVy=12 (2), ) SOLUTION. 3y^y=69— 7a; (3) = (1) transposed. S^y=*lx— 59 (4) = (3) with signs changed. — 3yV2/=12— 64a: (5) = (2) transposed. 7i;— 59 = 12 — 64a; (6)= value of —3yLy in (4) and in (5) equated a;=l. Since 3yVy=59 — ^a; and a;=l, we have 3^^=69-7=52 52y=l7.52 y=11 Hemark. — This solution shows that all that is necessary in eliminating by comparison, is to find an expression for the value of one of the unknown quan- tities when affected by a certain coefScient, and then find from the other equa- tion an expression for the same, and then equate them. 216 ELIMINATION BY COMPARISON. !• Given \ , r to find the values of x and y. ( x—y=b, ) a-^b a— 6 Ans. «=-2-» y^-g-- *2« Given i ' ' {• to find the values of x and y. ( a; + 6y=191, ) Ans. a;=16, y=:36. 3. Given •! ~ !■ to find the values of x and y. ( x-^y=zc ) be ac Ans. x= — — r, y= r. ( Yv^=2iC— 3t/ ) 4. Given i ,^ ^^ ^^!. , !■ to find the values of a; and y ( 19a;=60y + 621i^, ) Ans. a;=88f, y=lT|. ^ ^. j lBx-\-Ty—S41=1ly-{-4SlXj ) to find the values of x ^•^^^^^ j 2x + ^y=l, \ andy. ^n5. a:=: — 12, y=60. ig^ gx^j^Yv 44 ) « ^ = f to find the values of x and y. 2a:=y + 4, J ^w». a;=4|, y=8^. 7. Given i « ~«>, f to find the values of a; and y. i y + Sx=2lf ) Ans. a;=8, y=3. {4ic4-9v=5l ) ,^ f to find the values of x and y. 8a;— 13y=9, ) ^ Ans. a;=6, y=3. 7 + a; 2a;-- y to find the values of x and y. 9. Given 5y— V 4a;— 3 + — ^r— =18-5a;. 10. Given 2 3a;— 1 5 3y-5 6 + 3y— 4=15 + 2a;— 8='7| ^n5. a;=:3, y=2. ► to find the values of x and y. ^ws. a;=7, y=z5. Multiply the 2d equation by 3, and equate the values of 3aj. ELIMINATION BY ADDITION AND SUBTRACTION. 217 ELIMINATION BY ADDITION AND SUBTRACTION. (278.) Elimination hy addition and subtraction is multiplying or dividing two equations so as to make the coefficients of one of the unknown quantities the same in both equations, and then subtracting or adding these equations according as the signs of these terras are like or unlike. PROBLEM* (279.) To eliminate by addition and subtraction from two simul- taneous equations. RULE. Multiply or divide^ if necessary, in such a manner as to cause one of the unknown quantities to have the same coefficient in both equa- tions ; and then add these equations if the signs of these terms are unlike, or subtract one from the other if the signs are alike, PROBLEM (280.) 1. Given ]'^'^*^^'^^jHtofindthevaluesofa;andy. V cx-y-dy — n (2K ) SOLUTION. a€x-\-bcy=cm (3) = (l)xc, acx-]-ady=an (4) = (2)xa, {bc—ad)y=cm—an (5) = (3) — (4), cm— an ., , . ,, ^=^^3^ (6)=(5)-(6«-a<0. The value of x may be obtained in the same way by multiplying the first equation by d, and the second by b, and subtracting, or by substituting in either of the given equations the value of y as already found. By adopting either of these modes, we would get x= — - — 7--. Remark. — ^We should always eliminate those terms which require the least preparation. In the example just given there is no preference, as the operation can not be shortened, because a and c, and b and d are prime to each other. 218 ELIMINATION BY ADDITION AND SUBTEACTION. PROBLEM 2. Given 1 .,_?+i. , +!^ (1), 3 2a;+l (2), to find the values of ot and y. SOLUTIO 48y— l'7a;=156 2y + 30a:=160 48y+'720a;=3840 '737a:=:3685 (3)=(l)x20, &c, (4)=(2)x6,&c., (5) = (4)x24, (6) = (5)-(3), ('7)=:(6)-'737, 2?/ = 1 60 — 30a; (8) = (4) transposed. .2y=160— 150=10, y=5. PBOB LEM 8. Given \ ^ c. ^^ rJ\ \ ^ ^^^ ^^ values of a; and y, ( 6x-^y=^ (2), 3 SOLUTION. 16a;-10y=20 (3) = (2) x 3, 31a;=62 a;=2 10y=42 — 16a; .10y=42-32=10, (4) = (l) + (3), (5)=(4)-31, (6)=(1) transposed, EXAMPLES 1, Given - 2t Given 6 + 4-^' ?+^=5^ 4^6 ^ to find the values of x and y. ^ws. a;=12, y=16. 9a; + :70, >- to find the values of x and y, Ans. a;=6, y=10. S. Given 1(" + ") (^ + ^)^("+'^(^-'> + "'•[ to findthe value, ( 2a?+10=3y+l' ) of a; and y. ^^i«. ar=3, y=5. ELIMINATION BY ADDITION AND SUBTEACTION. 219 f a _ b 4« Given < b +y'^Sa-\-x }- to find the the values of x and y. [aa?+26y=c Ans. x= 26'— 6a' + c Sa'—b^+c 5* Given 3a Bi±4x 9y + 33 6a;— 4y lly— 19 y—3 —-^=x —, ^2 4 y= 86 to find the values of X andy. Ans, x=6, y=6. f 3a; + 4y + 3 2a; + '7-y ^ , y-8 ' 6. Given * 10 15 6 9y + 5a;— 8 x-\-y 1x-{-G 12 11 ' to find the values of X and y. Ans, x=1f y=9. 7, Given - '6a; + 13 By— 3a;— 5_ ^a;— 3y+l " 2 6 ""^"^ 3 ' x + l 3y— 8 3 4 values of x and y. 4a;::4:21, 8. Given ^ 21— Qy 3 „ 21-4y 18a;+13 „, to find the Ans. x=z5t y^4i. to find the values of X and y. ^W5. a;=7, y=5. 9« Given r,« . , 128a;''-18y'* + 2l7 ^ 16a; + 6y— 1 = ^-^ , ^ 8a;— 3y + 2 ' =5 54 10a; + 10y— 35 2a; + 2y + 3 ' 3a; + 2y— 1' 4a;+3y + to find the values of X and y. 10. Given 2a; + 4=3y-+ Ans. a;=6, y=6. 24 + 5^ y_16a;''4-12a;y— 8a; + 5y + 28 2a;4-l"~ 4a;-2 ' 8a;''— 1 By'' + 108 to 4a;4-6y + 3 ' find the values of x and y. Ans, a;=3, y=2. 220 MISCELLANEOUS EXAMPLES IN ELIMINATION. MISCELLANEOUS EXAMPLES IN ELIMINATION PROBLEM X (281.) 1. Given -f7y=99(l) ^ + 1x=5l(2) > to find the values of x and y. SOLUTION. (3)=(l)x7. (4)^(2) X 7. (5)=[(3) + (4)]-f-60. (6)=(3)-(5). (7). (8) = value of y in (1) substituted in (6). a;-f.49y=693 49x-\-y=S5l x+y=:21 48y=672 y= 14 a; + 14= 21 x=1 The above artifice can always be adopted when the coefficients of X and y in the first equation are the same as those of y and x in the second. PROBLEM 2. Given ^ ■ to find the ralttes of tc and y. SOLUl ION. 1 1_4 X y~21 (8) =(1)-U7. 17 17__68 X y~2l (4)=(3)xl7. 13 IS y~ "^ (6)=(2)-(4) y=1 («)• 1__4 1 x'^2\'^y (7)=(3) trans. 14 11 •*• x~2l'^1~3' x=S, EX A MI LBS. 1, Given h the values of x and y. Ans. x=e. «=:4 MISCELLANEOUS EXAMPLES IN ELIMINATION. 221 it Given •} ^ , ^' r to find the values of x and y. ( 3a; + iy=29, ) Ans, a;=9, y=6, 3. Given \ 2^+3^=14, ) ^^ ^^ ^^ ^^^^^ ^^ ^ ^^^ ^ri«. aj=24, y=6. !• Given ■! ® , ~.„.' r to find the values of x and y. ( 8a;H-iy=131, ) Ans, a:=16, y=24. ( 4ic4-v=34 ) 5« Given •< / , « f to find the values of x and y. (a; + 4y=16, ) ^7w. a; =8, y=2. 6« Given i w T r to find the values of x and y. C — a; + '7y=33j ^W5. x=z2, y=^6. {a?4-y=8 ) o o !. „ f to find the values of x and y. a;'— y'=16, ) Ans, x=5, y=8. 8. Given \ • , \ }• to find the values of x and y. ( x^—y^—h, ) a' + J _a}—b 2a *^""~2a' ^IW. x•=—:r—^y^=' 9« Given 10. Given 2^ 3 + 6y=23, Vy 4 h to find the values of x and y. JItw. a;=— 3, y=6. ^-12=1 + 8, __4-__8_-^ + 27, to find the values of a: and y. ^7i«. a;=:60, y=40. 11, Given i 2 + 32/ 8, ) ^ ^^^ ^j^^ ^^^^^ ^^ ^ ^^^ Ans, x=6, y=16. 4a; 5y 9 , ] I-— = 1 , x^ y^ y 12. Given '( ^ ^ h « r t<^ fi^d t^® values of a? and y. - + -=- + -, a; y a? 2'J Ans, x=z4, y=2. I 222 MISCELLANEOUS EXAMPLES IN ELIMINATION. 13* Given - 2y— a; ^^ 69— 2a; x—-^ =20 — , 23— a; 2 ' to find the values of x J^^^ ^^3-3y f andy. ^"^a;-18 3 Ans, a;=21, y=20. f 3a; + 2y 5a;— |y+ 1 _ y— 2a; _ 4a;— y ^ 14. Given <5 3 ""^"^10 Y'V ( y + 2a; : y— 2a; : : 12a; + 6y— 3 : Qy— 12a;— 1, ) to find the values of x and y. 16+60a; 16a;y— lOt 15* Given ^ 8a; 3y-l 5 + 2y ' « « « 2'7a;»--12y*4-38 ^W5. a;=l, y=4. to find the value of a; and y. ^7W. a?=2, y=3. 16i Given •< + 6y + 1 6a;'+130— 24y' ' 2a;— 4y + 3 ' 3a;- 151 — 16a; _ 9a;y— 110 4y-l "^ 3y-4 ' 4a;*-y(a; + 3y) to find the valuei of X and y . (3y + ll= "^-y^^ + ^^^ 4-31-4., 17. Given ^ ^ a;-y + 4 ' ( (a;+'7)(y-2) + 3 = 2a;y-(y-l)(a; + l), values of x and y. 18. Given Ans, a;=9, y=2. to find th< An9, xz=4c, y=3. ( ^y—Yy — X=V20 — X, I Vy^ : |/20— a; : : 3 : 2, to find the values of a; and y, Ans, a;=16, y=26. •-. ^. (168H-19a;4-fVy=12fa;4-1084, ) ^ . , ^, , 19. Given ] ,, J J,9^^3i;^_ jy, [ to find the valu. of a; and y. Ans. a;=— 2*71, y=136J. 20. Given 3a; + 5y ab^c {Sa—2b)ab '' a'-b' ' to fin( a'x z + (a + 6+c)6y=&'a; + (a+26)a6, Or •J- the values of x and y. -4»«. a;= — r^ y ab _ ab a—h' ^"a + b' MISCELLANEOUS EXAMPLES IN ELIMINATION. 223 ( hcx=^cy—2b, 21, Given I ,, a(c^—b^) 26\ , [ to find the values of a; and y. 6c a a + 26 Am. ^— , y= , be c 22» Given -j a b — I — =m. X y c d X y >■ to find the values of x and y. be— ad be— ad Ans. x= — z, y: nb—md' me— ma 23, Given W9_^3^j^^^^ 3^+4 ^ 4 4a;— 6 * 2 ' I to find the values of 8y + 7 6a;-3y _ 4y-9 j x and y. 10 "^2y-8 "* "^ 6 'J ^»s. a; =7, y=9. 24. Given ^ ^ f ^ ^ ^ U-2y— ^-^— =Y(^+y)-3(a;-yj, to find the values of a? and y. Ans, x=-^\ y=A ^ . 11 25* Given - 12y ^-I— — £- + i0a; + 13, 4 + 11 X 11 — =y+ 3a; 2a; 3a;— 5_4a;y + ^5^ IT'' y + 1~ 6y+2V ' the values of a; and y. to find ^5. arzrT, y=2. to find the values of X and y. 7a; ^ 3y + 6 3a;— 2 4 ^ 5 10 _g yg 26. Given -j 5 ~" 8 "16' l2^3^^^*2 s^''^ ^ ''i Ans. x=4:, y=3. 4a;— 2y + 3 18— a; + 5y_a; y 1 3 ^ - 27. Given ^ to find the values of x and y. .____ 7yV, X V S V X 1 2.-y+15:y-2.+15::^ + -:|--+-, J[w«. a;=18, y=24. 224 SIMULTANEOUS EQUATIONS, ETC. ^ V 11a? 28. Giv«n ^ —-J^-— -^=-— ^- » 7y 24 6 42 56y 12a;— 15y + -V-: lOy— 8a;4- V • = 93 — 9a; : Qx—i^^ to find the values of a; and y. Ans. a; =9, y='7. 29. Given < 3a; -5y 2a;-8y-9 _y 3 12 "12"^='"*"*' ^4.| + li:4a;-|-24::3i:3i to find the values of X and y. Ans, a;=7, y=4. f. Given ^ 3y— 2+a; 15a; + -J^ o 1+- 11 ■ 33 ' 3a; + 2y y— 5 lla;4-152 3y + l 6 12 2 ' to find the values of X and y. a;=8. i# .. ♦ .. »* SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE CONTAINING THREE OR MORE UNKNOWN QUANTITIES. PROBLEM. (282.) To eliminate from three or more simultaneous equations. SOLUTION. Let (A)^ (B), and (C), represent three simple equations, each of which contains three unknown quantities, as a;, y, and s. Suppose it most convenient first to eliminate z. Then, according to one of the preceding methods, eliminate z from (A) and (B), and there will result an equation which will contain only x and y as unknown quan- tities, which equation designate by (D). Next, eliminate z from [A) and ((7), or from (B) and ((7), as may be most convenient, and there will result another equation, which also will contain only x and y as unknown quantities. Call this equation (IH). We have now nothing SIMULTANEOUS EQUATIONS, ETC. 226 more to do with (A), (-B), and (C7), but must operate on the result- ing equations, (D) and (U). Now, let us eliminate 7/ from (D) and {jS)j and there will result an equation (F) which will contain only x. Q, K F. PROBLEM 2a? + 3y+ 52—23 (1),) 3a;+2y+ 1z=28 (2), C to find the values of 5x + 1y + Uz=52 (3),) SOLUTI ON. (4)=(l)x7. (5)=(2)X5. (283.) 1. Given ic, y, and z. 14.r + 21y + 352: 15a; + 10y + 352: 161 140 or-lly 22a; + 33y + 552: 25a; + 35y+650: -21 253 260 (6) = (5)-(4). (7)=(l)xll. (8) = (3)x5. 3a; + 2y 3a;— 33y 1 63 (9)=(8)-(7). (10) = (6)x3. 35y == y = 3a; = .-. 3a; = YO 2 1-21/ 7-4=2 (11). (12) = (13) = 1 (14)= = (9)-(10). = (11)^35. = (9) transposed. = value of y substituted in (13). X = J + 6+52= 1 23 (16) = (16) = = (U)-3. = values of a; and y substituted in (1) 52 = 15 * 2 = 3. • PROBLEM C ax-{-hy -{-cz =c?, ) 2. Given < a'x + h'y -\-c'z =c?', V to find the values of a;,y,and2. ia"x^h"y + c"z=d'\) SOLUTION Eliminating as directed in Problem (282), we shall obtain, after arranging the terms in the separate results, dh'c" + d'h"c + d"bc'-d'hc"-'dh"c'-d"h'c x= z= ab'c" + a'b"c+ a"hc'-a'hc"-ah"c' '--a"h'c ad'c" -\-a'd"c + a"dc' -a'dc" -a d"c' -a"d'c 'ab'c" +a'b"c -{-a" be' -a' be" —ab"c'—a"Vc ab'd" + a!b"d + a"bd'-a'bd"-ab"d'-a"b'd ab'e" + a'b"e +a"be' - a'bc" - ab"c' - a"b'c 15 226 SIMULTANEOUS EQUATIONS, ETC. If tlie student can only recollect these general values, he need only make the proper substitutions without being compelled to go through the process of elimination. The denominators are all alike, and do not contain any of the ab- solute terms. This denominator can be formed as follows : Write the coefficients of the unknown quantities as in the problem, repeating the first two rows ; thus. a b c a' V c' a" b" c" a b c a! b' c' Omitting , at the right hand con \ hand comer below, we have, 1st. a 2d. a' b' 3d. a" b" c" b c c' ,, at the The product of the terms in each of these lines will give the posi- tive terms of the common denominator. Again, omitting , at the left hand comer above, and ,, , at the right hand comer below, we have b' e 6th. a" b" c 5th. a b 4th. a' The product of the terms in each of these lines will give the negative terms of the common denominator. We can write the respective numerators from the common denom- inator by changing a into d^ a' into d\ and a" into d" for the nu- merator in the value of a; ; 6 into j passing upward to the right, the terms a'hc'\ ah"c\ a"h'e. In the same manner, by putting d for o, d' for a\ and d" for a" we should have from d h c d' h' c! d" b" c" d b c d' V c' by passing downward to the right, db'c'\ d'b"c, and d"bc' for the positive terms in the numerator of the value of a?, and by passing up- ward to the right, d'bc'\ db"c\ and d%'c for the negative terms in the same numerator. In the same way from a d c a' d' c' a" d" c" . a d c a' d' c' we can get the numerators in the value of y. Also, from a b d a' b' d' a" b" d" a b d a' V d! we can get the numerator in the value oiz PROBLEM C 3a;-|-2y4-52=59, ^ 8. Given < 4a;+ y + 3s=41, V to find the values of a;, y, and z, (Sar + Vy-f 2^ = 75, ) SOLUTION. . From 3 2 6 ^: 4 1 3 8 '/ 2 3 2 5 4 13 228 SIMULTANEOUS EQUATIONS, ETC. we get 3x1 x2 + 4x7x5-f8x2x3— 4x2x2 X 5 for the common denominator of a;, y, and z. From 69 2 5 41 1 3 75 1 2 69 2 5 41 1 3 we get for the numerator in the value of a;, 69x1x2+41x7x5 + 75x2x3—41x2x2 3x7x3—8x1 59x7x3—75x1x6. These expressions simplified, give «= 2003 — 1778 225 194 — 119 -^=^- From 3 69 6 4 41 3 8 76 2 3 69 5 4 41 3 We get for the numerator in the value of y, 3x41x2+4x75x5+8x59x3—4x59x2—3x76x3—8x41x5=3162-2787. ^, ^ 3162-2787 375 ^ Inereiore, y. Also, from 194-119 75-^ 3 2 69 4 1 41 8 7 76 3 2 69 4 1 41 We get for the numerator in the value of z 3x1x75+4x7x59 + 8x2x41-4x2x75-3x7x41-8x1x59=2633-1933. 2533 — 1933 600 ^ Inererore, g= _^ . — rT^-=-;rTr=8. 194 — 119 75 PROBLEM C 3a;— 6^+0=- 32, \ 4. Given ^ 2a:+8y = 16, > to find the value of a;, y and z, i 5x-\-1y—2z=z 6, ) SOLUTION. 3 -6 1" — 32 -6 1' 3 —32 1' 3 -6 —32 2 8 16 8 2 16 2 8 16 5 7 -2 -, 5 7-2 -,5 5- -2 - , and 5 7 6 3 -6 1 —32 —6 1 3 —32 1 3 -6-32 2 8 0- 16 8 0. 2 16 0- 2 8 16 ^ y= SIMULTAISIEOUS EQUATIONS, ETC. 229 We get by proceeding as before (-32x8x-2+16x7xl+5x-6x0)-(16x-6x-2+-32x7x0+5x8xl) ~(3x8x-2+2x7xl + 5x-6x0)-(2x-6x-2 + 3x7x + 5x8x1) _624 — 232_ 392 _ — _34_64~— 98~~ * (3 X 16 X -2+2 X 5 X 1+5 x -32 x 0)-(2 x -32 x -2+3 x 5 x 0+5 x 16 k 1 ) _ — 86 — 208 —294 ^ D -98 (3x8x5+2x7x— 32+5x— 6x16)— (2x— 6x5+3x7xl6+5x8x— 32) — 808 + 1004 196 6. Given D ~— 98 PROBLEM U+y+2=3i, (1)) < a;+y— 2=25, (2) >• to find the values of a;, y and z, ( x—y—z= 9, (3) ) 6. Given SOLUTIO N. 2a;=40 (4)=(l) + (3)' a;=20 (6). 2^= 6 (6)=(l)-.(2). .= 3 (7). 2y=16 (8)=(2)~(3). y= 8. PROBLEM 2a;— 3y + 20=13, (1) 42;--2ar=30, (2) 4y + 22!=14, (3) 5y + 3i;=32, (4)J SOLUTION. 1y-2x= . 1 (5) = (3)-(l) 12v — 6a?= 90 (6) = (2)x3. 20y + 12t;rr:128 (7) = (4)x4. to find the values of v, «, y, and z. 20y +6a:=- 38 21y — 6ar= 3 (8) = (7)--(6). (9) = (5)X3. 41y= 41 (10) = (8) + (9). y= 1 (11). 2a;=:7y— 1 (12) =(5) transposed. 2x= 7-1=6 (13). a;= 3 (14). [forward 230 SIMULTANEOUS EQUATIONS, ETC. (2) transposed. 36 (15): (16). (l7) = (16)-4. (18) =(3) transposed. 4:V= 6 + 30: v= 9 22=14—4?/ 22=14-4=10. 2 = 5. Note. — The student should accustom himself to apply the general formula which has been given, so that it may be called up at any time. It will often save much labor, as will be proved to be the case in some of the following examples. There are many curious proper- ties which belong to the general formula for the values of the un- known quantities as deduced from three simultaneous equations. But we leave them to be discovered by the student. EXAMPLES. rx+y +z =29,^ 1. Given 2. Given 3* Given to find the values of ar, y, and z, C2 + 2a; + 3y=12, ) Am. x=l, y=2, 2=4. ra;+a(y+2)=m, ) 10* Given < y + b(x-\-z)=nj > to find the values of ar, y, and 2. ( 2 4-c(a;+y)=^, ) m + nca +pab — na — mcb —pa l + 2a6c — ba — cb — ca n -{-pah +mbc — pb— nac—mh l-\-2ab—ba—cb—ca p-\-mbc-\-nca — 7nc—pba—nc l-\-2abc—ba—cb—ca Am. \ C a;— y+ s=30, ) 11. Given < 8ar— 4y + 2s=50, > to find the values of a?, y, and 2. C 27a;— 9y4-32=64, J 2 Am. a?=-, ^=7, 2=36^-. 3 232 SIMULTANEOUS EQUATIONS, ETC. ' 4x-\-Sr/-\-z 2y-\-2z — x-\-\ x—z—5 — =5+- 12. Given ^ 10 15 5 9a;+6y— 225 2x-\- y—Sz_1y-\-z -\-S 12 4 ~ 11 +i, to 5y + 30 2a;+3y-2 3a; + 2y + '7 — + J2— y— IH , 4 12 find the values of ar, y, and 0. 6 ^7w. x=9, y=*J, g=3. 13* Given ^ >- to find the values of x^ y, and z, Ans, a;=120, y=60, 0=12. 14, Given ^ 12y— ll2 11a;— 10y=-^^ , > to find the values of ar, y, and 2. x-\-z — 2y z—y—1 3^ ~ 2 ' . 3ar = y+^ + V, J ^7W?. «=10, y=ll, 0=12. f 3a;— y+ 0=15, ') 15* Given < 6a; + 3y— 22=16, > to find the values of a;, y, and z, ( 7a; + 4y— 52=11, ) Ans. a;=4, y=2, 2=6. r 2a; + 4y— 32=22, ) 16« Given < 4a;— 2y + 52=18, > to find the values of a;, y, and 2. ( 6a;4-'7y— 2=63, ) Ans. a;=3, y=T, 2=4. r 3a; + 2y— 2=20, ^ 17. Given ^ 2a; + 3y + 62=70, S to find the values of a;, y, and z. I x— y + 62=41, ) Ans. x=5, y=6, 2=7. :=128, ) 18. Given ^ 3a;+ 3y4-72= 60, > to find the values of a;, y, and 2. 68, \ Ans. a;=8, y=5, 2=3. f 6a; + 3y— 42=22, ) 19# Given < 4a;— y-f 62=20, > to find the values of a;, y, and z. (5ar + 2y— 62=11, ) Ans. a;=3, y=4, 2=2. ( 7a;+12y + 42: riven < 3a;+ 3y4-72: (ex+ y + 52: SIMULTANEOUS EQUATIONS, ETC. 233 20. Given - Sy=u-\- x+z, 4z=u-{- x + j/j x=u+ 14 , >■ to find the values of u, x, y, and z. Ans. w=:26, a;=40, y=30, 0=24. to find the values of x, y, 2, «, and t C a;— y— z=—a, j 21. Given } -2x+ y— 22=- a, > to find the values of a?, y, and 0. ( — 3a;— 3y+ 2=— a, ) Ans, ar= Jyttj y=TT^» z=^ja, 2x+ ^-22=40,"^ 4y— a; + 30=35, 22. Given ^ 3w+ <=13, y+ u+ ^=15, 3a;— y-\-St— w=49,^ Ans, x=20, y=10, 0=5, w=4, f=l. fM+v + a;+y=10,'| w + i; + fl;+s=ll, I 23. Given -{ u+v+y+z=l2j }. to find the values of w,v,a;,y, and & u-\-x-^y+z=lS, v+x-\-y+z=l^, Ans. u=l, v=2, a;=3, y=4, z=6. 24. Given a;+y 3 + 225= 21, . ^±i -30.= -65, ^ 2 3aJ4-2/— s _ 38, <■ to find the values of a:, y, and 2. Ans. a;=24, y=9, 2=5. ^+^y + i25=32, 25. Given < ia;+4-y + 42=15, > to find the values of a;, y, and z. V a:+iy + i25=32, < Ja; + i2/ + i^=15, ( ia; + iy + i2=12, ^ws. x=12, y=20, z=SO. C 3a:-9y+8.= 41, ) ^ ^^^ ^^^^ ^^^^^^ ^^ 26. Given -5a: + 4y + 2.= -20, f lla;-7y-62= 37, ) Ans. x=2, y=— 3, 2=1. ( 1x + 5y + 2z=^ '79, ) 27. Given < 8a;+ 7^ + 92= 122, > to find the values of ar, y, and 2. ( a; + 4y + 62= 56, ) ^ws. a;=4, y=9, 2=3. 234 28. Given Ans, x= SIMULTANEOUS EQUATIONS, ETC. x-\-y+z=a ^ my=nx > to find the values of a?, y, and z, pz=qx ) anp amq 29. Given — 1 y=- mp-\-np+mq mp-\-np+mq X y a ,2: mp-^np-\-mq - + -=- )■ to find the values of a?, y, and z. X z h 111 y z c\ Ans. X-. 2abc y 2abc z=- 2abc 30. Given ac—ab-\-bc' ab—ac + bc' ac + ab—bc x+ly-{- lz=pA m>x-\- y+mz=q^\ to find the values of x, y, and z, nx+ny+ z=r, ) Ans, I l-l l-l q m 1—m 1— m -r 1 m 1—1 1 + 7 +■ m ■I 1—m . 9 , 1 + 1— / 1—m 1—n I m n l—n I 1- 9 l^i - + 1— » r 1—i 1 + 1-^ 1— ^ 1—1 Note. — This example is the same as Ex. 10, but the answer is in another form, indicating that it has been solved in a different manner. The student may observe that the expressions in the brackets are identical. Also, in the value of X, the letters which are not included in the brackets are only those which occur in the first equation ; in the value of y, only those that occur in the second equation ; and in the value of z, only those that occur in the third equation. Because these values are symmetrical, being derived from symmetrical equa- tions, we can, after getting the value of x, deduce the value of y from it, and, having the value of y, we can, in hke manner, deduce the value of z. If in the ralue of a;, we change each letter to that which is in advance of it in the circles, QUESTIONS INVOLVING SIMULTANEOUS EQUATIONS. 235 oa we shall get the value of y. In like manner, changing the letters in the value of y, we should get the value of z. The quantities in the brackets would stiU be the same, though thus permuted, only the terms would not occur in the same order. QUESTIONS INVOLVING SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE. QUESTION (284.) 1. A man and his wife could drink a barrel of beer in 16 days. After drinking together 6 days, the woman alone drank the remainder in 30 days. In what time would either, alone, drink a barrel? SOLUTION. Let a; = the number of days in which the man could drink it, And y = « " « woman " " a; y 15 In 6 days both drank j%=^, leaving j. It took the woman 30 days to drink this f ; therefore, in 1 day she would drink ^ of Hence, i= ^ y= 60 X 50 16 a;~"l60 236 QUESTIONS USrVOLVING SIMULTANEOUS EQUATIONS. QUESTION 2. A number consisting of 2 digits when divided by 4, gives a cer- tain quotient and a remainder of 3 ; when divided by 9, gives another quotient and a remainder of 8. Now, the value of the digit on the left hand is equal to the quotient which was obtained when the num- ber was divided by 9 ; and the other digit is equal to y^ of the quotient obtained when the number was divided by 4. What is the number ? SOLUTION. Let a; = the digit in the tens' place, And y = « " units' " Then, in consequence of our system of notation, \Ox-\-y must rep* resent the number. Since lQx-\-y divided by 4 leaves a remainder of 3, if we subtract 3 from lOx+y the result is exactly divisible by 4, and by the question the quotient is 1*1 y. Hence, we have the equation, -^ =17y. In like manner, we get 10a; + y— 8 9 =" 10a;-67y =3 10a;+10y =80 11y =11. y =1 x+1 =8 X =1 Therefore, 71 is the number, because it is equal tp lOar+y, QUESTION 3. What two numbers are there in the ratio of 6 to 7, to which, if two other required numbers in the ratio of 3 to 5 be added, the sums shall be in the ratio of 9 to 13, and the difference of these sums shall be 16 ? SOLUTION. Let 5x and 7a; be the numbers in the ratio of 5 to 7, And 3y and 5y " " ♦♦ 3 to 6. QUESTIONS INVOLVING SIMULTANEOUS EQUATIONS. 237 Then, 6ar + 3y : Ya; 4- 5y : : 9 : 13 Or, 65a; + 39y=63a; + 45y 2a;=6y But (Yaj+Sy)— (6a; + 3y), or 2a; + 2y=16 a;+ y=8 Whence, because a;=3y, 3y+ y=8 4y=8 y=2. But, since x=zZy and y=:2, we have a;=6. Hence, bx and 7a;, or the first two numbers are 30 and 42 ; and 3y and 6y, or the other two numbers are 6 and 10. QUESTIONS. 1. What two numbers are there, the greater of which is to the less as their sum is to 42, and their difierence is to 6 ? Ans. 32 and 24. 2. A person expends half a crown, or 30 pence, in apples and pears, buying his apples at 4, and his pears at 5 for a penny ; and afterward accommodates his neighbor with half his apples, and one- third of his pears for 13 pence. How many did he buy of each ? Ans, 72 apples, and 60 pears. 3. A farmer sells to one person 9 horses and 7 cows for $600 ; and to another, at the same prices, 6 horses and 13 cows for the same sum. What was the price of each ? Ans, The price of a cow was $24, and of a horse $48. 4. A farmer hires a farm for $245 per annum ; the arable land being valued at $2 an Mcre, and the pasture at $1.40 ; now the num- ber of acres of arable is to half the excess of the arable above the pasture as 28 : 9. How many acres were there of each ? Ans. 98 acres of arable, and 35 of pasture. 6. There is a number consisting of two digits, the second of which is greater than the first ; and if the number be divided by the sum of its digits, the quotient is 4 ; but if the digits be inverted, and that number divided by' a number greater by 2 than the difference of the digits, the quotient becomes 14. What is the number 1 Ans. 48. 6. What fraction is that, whose numerator being doubled, and de- nominator increased by 7, the value becomes | ; but the denominator being doubled, and the numerator increased by 2, the value be- comes J ? Ans. f . 238 QUESTIONS INVOLVING SIMULTANEOUS EQUATIONS. Y. Two persons, A and B can perform a piece of work in 16 days. They work together 4 days, when A being called off, B is left to finish it, which he does in 36 days more. In what time could each do it separately ? Ans. ^ in 24 days, and -B in 48 days, 8. There is a cistern, into which water is admitted by three pipes, two of which are exactly of the same dimensions. When they are all open, -^ of the cistern is filled in 4 hours ; and if one of the equal pipes be stopped, ^ of the cistern is filled in 10 hours and 40 minutes. In how many hours would each pipe fill the. cistern ? Ans. Each of the equal ones in 32 hours, and the other in 24. 9. Some hours after a courier had been sent from A to B, which are 14*7 miles distant, a second was sent, who wished to overtake him just as he entered B ; in order to do this he must perform the jour- ney in 28 hours less than the first did. Now the time in which the first travels IV miles added to the time in which the second travels 56 miles is 13 hours and 40 minutes. How many miles does each go per hour ? Ans. The 1st 3, and the 2d Y miles an hour. 10. ^ and B playing at backgammon; A bet 3 dimes to 2 dimes on every game, and after a certain number of games found that he had lost 17 ^mes. Now ha4 A won 3 more from B, the number he would then have won would have been to the number B would have won as 6 to 4. How many games did they play ? Ans. 9. 11. ^ and B engaged to reap a field of corn in 12 days. The times in which they could severally reap an acre are as 2 : 3. After ^ some time, finding themselves unable to finish it in the stipulated time, they called in C to help them; whose rate of working was such, that if he had worked with them from the beginning, it would have been finished in 9 days. Also, the times in which he could severally have reaped the field with A alone, and with B alone, are in the ratio of 7 to 8. When was C called in ? Ans. After 6 days. 12. A vintner has 2 casks of wine, from the greater of which he draws 15 gallons, and from the less 11 ; and finds the quantities re- maining to be in the ratio of 8 to 3. After they become half empty, he puts 10 gallons of water into each, and finds that the quantities of liquid now in them are as 9 to 5. How many gallons will each hold ? Ans. The larger 79, and the smaller 35 gallons. 13. At an election for two members of parliament, three men offer themselves as candidates, and all the electors give single votes. The QUESTIONS INVOLVING SIMULTANEOUS EQUATIONS. 239 number of voters for the two successful ones are in the ratio of 9 to 8 ; and if the first had had seven more, his majority over the second would have been to the majority of the second over the third as 12 : Y. Now if the first and third had formed a coalition, and had one more voter, they would each have succeeded by a majority of Y. How many voted for each ? Ans. 369, 328, and 300, respectively. 14. A wine merchant has two kinds of wine. If he mixes a gal- lons of the first with b gallons of the second, the mixture is worth c dollars per gallon ; but if he mixes / gallons of the first with g gal- lons of the second, the mixture is worth h dollars per gallon. What is the price of each kind of wine per gallon ? (a+b)cg-(f-^g)bh The price of the first kind is Ans i ""^"^^ . u « secondkind (^±^l^-±^ dollars per gallon. 15. A banker has two kinds of money ; it takes a pieces of the first to make a crown, and b pieces of the second to make the same amount. Some one gave him a crown for c pieces. How many pieces of each kind did the person receive ? Ans. — — r^ pieces of the 1st kind, and -^^ — ~- of the 2d kind. a—b ^ a—b 16. What two fractions added make |, and the sum of whose nu- merators is equal to the sum of their denominator ? Ans. I and |i. 17. A purse holds 19 crowns and 6 guineas. Now 4 crowns and 5 guineas fill ^f of it. How many will it hold of each ? Ans. 21 crowns, or 63 guineas; 18. |500 was to be lent out at simple interest in two separate sums, the smaller, at 2 per cent, more than the other. The interest of the greater sum was afterward increased, and that of the smaller diminished by 1 per cent. By this, the interest of tfie whole was augmented by one-fourth of the former value. But if the interest of the greater sum had been so increased, without any diminution of the less, the interest of the whole would have been increased one-third. What were the sums and the rate per cent, of each ? Ans. $100 and $400, and 4 and 2 per cent, respectively. 19. Some smugglers discovered a cave, which would exactly hold the cargo of their boat ; viz. 13 bales of cotton, and 33 casks of rum. 240 QtJESTiONS INVOLVING SIMULTANEOUS EQUATIONS. 1 "Whilst they were unloading, a custom-house cutter coming in sight, they sailed away with 9 casks and 5 bales, leaving the cave two- thirds full. How many bales, or casks would it hold ? An8. 24 bales, or 72 casks. 20. A merchant finds that if he mixes sherry and brandy in quan- tities which are in the ratio of 2 to 1, he can sell the mixture, at 78 dimes a dozen ; but if the ratio be as 7 to 2, he must sell it at 79 dimes a dozen. What is the price of each per dozen ? Ans. Sherry 81, and brandy 72 dimes per dozen. 21. Round two wheels, whose circumferences are as 5 to 3, two ropes are wrapped, whose difierence exceeds the difference of the cir- cumferences by 280 yards. Now the longer rope applied to the larger wheel wraps round it a certain number of times, greater by 12, than the shorter round the smaller wheel ; and if the larger wheel turns round three times as quick as the other, the ropes will be dis- chai'ged at the same time. What are the lengths of the ropes and the circumferences of the wheels ? Ans. The ropes, 360 and 72 yards ; and circumferences of the wheels 20 and 12 yards. 22. If A and B together can perform a piece of work in 8 days, A and C together in 9 days, and B and C together in 10 days, in how many days can each alone perform the same work ? Ans. A in 14ff days, B in I7f^ days, and (7 in 23/7 days. 23. Three brothers bought a vineyard for |100. The youngest says, that he could pay for it alone, if the second would give him 1 the money he had ; the second says, that if the eldest would give him only the J of his money, he could pay for the vineyard ; lastly, the eldest asks only ^ part of the money of the youngest to pay for the vineyard himself. How much money had each ? Ans. The oldest had |84, the 2d |72, and the youngest |64. 24. Three persons. A, B, and C, play together. In the first game A loses to each of the other two, as much money as each of them had when they commenced. In the next game, B loses to each of the other two, as much money as they each had at the commencement of the 2d game. In the third game, C loses to each of the other two as much as they each had at the commencement of the 3d game. On leaving off, they find that each has an equal sum, namely, $24. With how much money did each commence ? Ans. A $39, B $21, and (7$12» Ans. QUESTIONS INVOLVING SIMULTANEOUS EQUATIONS. 241 26. Three laborers, A, By and (7, are employed to do a certain piece of work. A and B can do the work in a days ; A and (7 in 6 days ; and B and (7 in c days. How long would it take each to do the work alone, and how long when they all work together ? 2abc , A requires -z r days, ^ oc+ac — ao . 2abc B requires -. days, ^ bc + ab—ac -^ ' . 2abc C requires — ;- days, ^ ab + ac—bc ^ ' A. jB, and (7 require -; r- days. ^ ' ' ^ ab-\-ac^bc -^ 26. A cistern containing 210 buckets, may be filled by 2 pipes. By an experiment, in which the first was open 4, and the second 5 hours, 90 buckets of water were obtained. By another experiment, when the first was open 7, and the other 3^ hours, 126 buckets were obtained. How many buckets does each pipe discharge in an hour? And in what time will the cistern be filled, when the water flows from both pipes at once ? Ans. The first pipe discharges 15, and the second, 6 buckets in an hour, and it will require 10 hours for them to fill the cistern. 27. A person has two horses, and two saddles. The better saddle cost $50, the other $2. If he places the better saddle upon the first horse, and the worse upon the second, then the latter is worth $8 less than the other ; but if he puts the worse saddle upon the first, and the better upon the second horse, the latter is worth 3f times as much as the former. What is the value of each horse ? Ans, The 1st |30, and the 2d $70. 28. A company at an inn, expended a certain sum of money ; for the payment of which *they agree to contribute equally. Had there been 5 persons more, and had each spent 12^ cents more, then the bill would have been $10.17^ ; but had there been 3 persons less, and had each expended 5 cents less, the bill would have been $8.25. How many were there in the company, and what did each spend ? Ans. There were 11 persons, and each spent 80 cents. 29. A work is to be printed so that each page may contain a cer- tain number of lines, and each line a certain number of letters. If each page should contain 3 lines more, and each line 4 letters more, then there would be 224 letters more on each page ; but if there 16 242 INTERPRETATION OF NEGATIVE RESULTS, should be 2 lines less on a page, and 3 letters less in each line, then each page would contain 145 letters less. How many lines are there on each page, and how naany letters in each Une ? Ans. 29 lines in a page, and 32 letters in a line, 30. A coach set out from Indianapohs to Cincinnati with a certain number of passengers ; 4 more being on the outside than within. Seven outside passengers could travel at 50 cents less expense than 4 inside. The fare of the whole amounted to $45. But at the end of half the journey, it took up 3 more outside and 1 more inside passen- gers ; in consequence of which, the whole fare was increased in the ratio of 17 to 15. What was the number of passengers, and the fare of each ? Ans. There were 5 inside, and 9 outside passengers. The fare of each inside passenger was $4.50, and of each outside passenger 12.50. .^ »« » t« ^ ■ INTERPRETATION OF NEGATIVE feESULlS. THEOREM. (285*) When the value of an unknown quantity in an equation is found to be negative, this result indicates that an absurdity is in- yolved in the enunciation of the problem. DEMONSTRATION. Find a number which, added to + 4, makes 2, Let X = the required number. * Thenar-|-4=2 a;=-2 This result indicates that the problem was incorrectly enunciated ; the words added to being used for subtracted from. The problem is, however, worded correctly, if we consider the word added in its ex- tended algebraic sense. Let us word this problem differently : Find a number to which if 4 be added the sum will be 2. The result x=—2 indicates that the problem is absurd in an arithmetical sense, but definite in an algebraic sense. INTERPRETATION OF NEGATIVE RESULTS. 248 Let us take another problem : A father whose age is 42 years has a son whose age is 12. In how many years will the aye of the son be I that of the father^ This question gives the equation ,^ 42+aJ a;4-12=— — -. 4 The solution of which gives «=— 2. This result shows that the son's age will never be, in the future^ J the father's age ; but that 2 years ago the son's age was \ that of the father. That part of the question which we have put in italics should have been, How many years since the age of the son was I that of the father^ Let us take still another example. A man worked 1 days, and had his son with him 3 days ; and received for wages $2.20. He afterward worked 5 days, and had his son with him 1 day, and received for wages $1.80. What were the father's daily wages, and what was the effect of the son's presence ? Letting x = the father's daily eflfect, And y = " son's " We get the equations, 72:4-3^=220, ) , and 5X+ y=180, S ^ ^' provided the daily effect of each is productive of wages. But, if the daily effect of the father adds to the amount of wages, and the daily effect of the son diminishes the amount of wages, the equations must be and 1x-Sy=220,] . 5x— y=180. P ''' If the reverse were true, we should have -1x+Sy=220,) and —5x+ y=l80.)^ ^' Which of these three couplets is the one which belongs to this problem ? This question may be answered after we have found the values of x and y in^ach. The (1) gives a;=40 and y=— 20; the (2), a;=40 and y=20; and the (3), ir=— 40, y= — 20. Now, the (3) results can not be true, because they make the daily effect of each dicffinish the amount of wages, while the equations from which these values were derived considered this to be true only of the father's daily effect 244 INTERPRETATION OP NEGATIVE RESULTS. The (1) results, in like manner, indicate that the daily effect of the son is to diminish the amount of wages, while the equations from which they were derived considered both to be productive of wages. The results derived from the (2) couplet are the only ones that fulfill the arithmetical conditions. The form of the (2) couplet, also, shows how the question should be stated ; or, in other words, that the son's presence diminished, each day, the father's earnings by an amount equal to 20 cents. We may suppose the father had to allow his employer 20 cents a day for his son's board ; or that he was hindered one-half a day in his work every day the son was present ; or that the father, out of his own wages, requested the employer to allow the son 20 cents for every day the son worked. Guided by such ideas, we are enabled generally to give an arith- metical explanation of negative results. QUESTIONS. 1. A man, at the time of his marriage, was 60 years old, and his wife 40. When was he twice as old as she ? Ans. 30 years before marriage. 2. What fraction is that which becomes | when 1 is added to its numerator, and becomes ^ when 1 is added to its denominator ? Ans. 5^. 3. A man, when he was married, was 30 years old, and his wife 15. How many years must elapse before his age will be three times his wife's age. Ans. He was three times as old as she Y^ years before their marriage. 4. What fraction is that which, if 2 be added to its numerator, its value is zero ; but, if 6 be added to its denominator, its value is infinite ? Ans. — -. — 6 6. What fraction is that which, if 3 be added to its numerator, its value is nothing; but, if 4 be subtracted froni its denominator, its value is 1 ? Ans. — -. % 1 6. A laborer working for a gentleman during 12 days, having with him, the first 7 days, his wife and son, who occasion an expense to GENERAL DISCUSSION, ETC. 245 i him, received $4.60 ; he afterward worked 8 days, during 5 of which his wife and son were with him, and received $3.00. What were the wages of the laborer per day, and also the expense, per day, of his wife and son ? Ans, His daily wages 50 cents, and expense of wife and son 20 cents. 7. Two men, A and £, commenced trade at the same time ; A had 3 times as much money as B ; A gained $400, and ^ $150 ; now, A has twice as much money as £, How much had each at first ? Ans, A was in debt $300, and £ $100. 8. A man worked 10 days, his wife 4 days, and his son 3 days, and their wages amounted to $11.50 ; at another time, he worked 9 days, his wife 8 days, and his son 6 days, and their wages amounted to $12.00 ; a third time, he worked 7 days, his wife 6 days, and his son 4 days, and their wages amounted to $9.00. What were the daily wages of each ? Ans. Husband's daily wages, $1.00 ; wife's, ; and son's, 50 cents. 9. The sum of two numbers is 120, and their difference is 160 ; what are the numbers ? Arts, 140 and —20. 10. What number is that whose | part exceeds its i part by 12 ? Ans. —144. i^ «> ♦ >. »■ GENERAL DISCUSSION OF CERTAIN RESULTS OBTAINED IN THE SOLUTION OF SIMPLE EQUATIONS. PROBLEM. (286.) To interpret the result 0=0. SOLUTION Let us endeavor to solve the problem : Given 0=0 > to find the values of x and y. (3)=(l)xa (4)^(2)-(3) 246 GENERAL DISCUSSION, ETC. Hence, We ate unable to obtain the values of x and y. By looking at the second equation, we see that it is not independent of the first, it being the first multiplied by a. We have, then, only ^H — =c to find the values of x and y. We have already shown that when there are two unknown quan- tities and but one equation, the values of x and y are indeterminate, or, in other words, that there are an infinite number of corresponding values of x and y, which will satisfy the equation. Again, let 2a? + a= a; 4- 1« + ^ + !«. This equation, also, reduces to 0=0. This ought to be the case, because 2x + a=2x + a, or 2a;=2a;, or a:=a;, is an indeterminate equa- tion, since x may be any quantity whatever. Hence, the result 0=0 is a sign of indetermination, and when obtained from two simultaneous equations, indicates that there is but one con- dition. When the values of x and y are indeterminate, it is customary to indicate it by ^=x and y=r« PRO BLEM. (287.) Find a number such, that if 9 be added to 8 times the number, and this sum be divided by 6, the quotient will be equal to 13 augmented by 12 times the number, and this sum divided by 9. SOLUTION. Let X = the required number. Then we have, by the conditions given, 8a;+9 13 + 12a: 6 9 (1) '?2a; + 81 =78 -}-72a; (2)=(l)x64. '72a;— 72a:=78 —81 81 -78 =l2x-'I2x 3 =0 This result is an absurdity, and indicates that the conditions of the question are incompatible or contradictory. The conditions, we have seen, give 72a: + 81 = 72a; + 78. This may assume the form 72a; + 78 + 3 = 72a; + 78. This equation can be satisfied only on the condition 3=0; but 3 does not equal 0, and therefore the equation is impossible. GENEEAL DISCUSSION, ETC. ^7 PBOBLEM. (288.) To interpret x=^ . SOL UTION. We may consider that oo==- when a=iO, and, also, 00 =-7 when 1 b=0. Then, ^=— , becomes iP=5Q-, when both a and h are boA - b equal to 0. a 1 I I h h But {»=-=— f--=-x-=-. 1 a b a I a h Now, if we make a and b each equal to 0, we have x=7> We see by this that we can get ^=S and «=- from the same 1 equation, a;=-, by making both a and b equal ; Hence, ^ is equivalent to -, or, in other words, is sometimes a symbol of indetermination, for we have already shown that - is sometimes a sign of indetermination. In vanishing fractions the value of - is determinate. PROBLEM. (289.) To interpret x=0 . 00 . SOLUTION. Letting ar=ax -7, and making both a and b equal to 0, we get a;=Ox-, or a;=:0 . 00. But x=a X T is the same as a:=Tj or, when a and b are both equal to 0, x=-. 248 GENERAL DISCUSSION, ETC. We see, then, that . oo is equivalent to -. PROBLEM. (290.) To interpret a;= oo — go. SOLUTION. Let (c= -. a If, in this equation, both a and h are 0, we have 1 1 x=-—~, or x= 00 — 00 . But a:= 7 is the same as a h — a x=z . ab Which equation becomes, when a and h are each equal to 0, _ 0-0 or, ^=3. We see, then, that 00 — 00 is equivalent to -. PROBLEM. (291.) Two couriers traveled on the same road; when one was at A the other was at B, When were they together ? SOLUTION. Let ' ^ represent the road and the given places. This problem is stated in very general language, and, therefore, embodies many conditions. It will, however be found not to be so general as the algebraic formula to which it gives rise. To obtain this formula let us take one of the many cases which may occur, viz., that the faster traveler was at A when the other was at Bj and that they were together at D. A B D Let a = the distance from Ato B ; m •=. the number of miles the faster traveler went per hour ; and n = the number of miles the other went per hour. GENEBAL DISCUSSION, ETC. 249 Let x=^AB^ and y—BD. We have then — the number of hours that it took the faster cour- m ier to ffo the distance x, and - = the number of hours it took the ^ n other to go the distance y. Since these terms must be equal, we have X y m ~n But x—y^a The solution of these equations give am x=. and y= m—'i an m—'i X XI Whence, — , or ■-: m n m—n These results are applicable to a great number of separate supposi- tions, or we may consider them as the solution of the general problem. In order to adapt these results to all the supposable cases, we must first establish certain conventional signs. When m is a positive quantity, we shall consider that the traveler who went m miles per hour, went eastward, or to the right ; but when w is a negative quantity, we shall consider that he went westward, or to the left. We shall, also, attach the same ideas to n. When x is a positive quantity, we shall consider that D is the eastward, or to the right of A ; but when it is a negative quantity, we shall consider that D is westward, or to the left of A. Also, we shall consider D as eastward or westward from B, according as y is positive or negative. X 11 When — T or - is positive, we shall consider the travelers were X 11 together after they were at A and J?, but when — or - is negative, that they were together before they were at A and B. But how shall we establish the meaning of 4- a and — a, since, according to the above ideas, the distance a would have a different sign according as ^e refer it to ^ or i? ? We shall let the point A rule ; and shall call AB^ or a, positive 250 when B is to the right of A ; and AB, or a, negative when 5 is to the left of A, Thus -k i ' ■ B A "We shall consider AB in the first equal to +a, but in the second line equal to —a. "With conventional ideas, we can make a particular problem out of each of the following suppositions, and the corresponding values de- duced from the above formulas will be the correct results for each particular supposition. 1st When m=+2, w=-fl, and a=+3, we get a;=+^, and y = 4- 3, — = + 3, and -= + 3. 2d. "When m=:— 2, w= — l,and a=+3, we get a;= + 6, y=+3, X V —=—3, and -= — 3. m n 3d. When m=4-2, 7i= — 1, and a=+8, we get x=: + 2, y= — 1, — =+1, and -= + 1. m n 4th. "When m=—2, ?i= + l, and a=-|-3, wegeta;= -}-2,y= — 1, — =-l,and ^=-1. m n 5th. "When m=+2, 7i= + l,and a=— 3,we get ar=— 6, y=— 3, X V —=—3, and -=—3. m n 6th. "When m=— 2,w=— l,and a=— 3, we get aj=— 6, y=--3, X V — =+3, and -=+3. m n Yth. "When wi= 4-2, 71= — 1, and a=:— 3,we get a;=— 2, y= + l, -^=-1, and^=-l. m n 8th. When m= — 2, w= + l,and a= — 3, we get a;= — 2, y= + l, X V — = + 1, and :^=: + l. m n In the first four suppositions B is east of A^ and in the last four, B is west of A. ■ ' ' ' A D' B D In the first two suppositions the couriers were together at Z>, in the GENERAL DISCUSSION, ETC. 251 Ist 3 hours after they were at A and B^ and in the 2d 3 "before. In the next two suppositions they were together at D\ in the 3d 1 hour after they were at A and B^ and in the 4th 1 hour before, ' ' ' D" B D'" A In the next two suppositions, they were together at D\ in the 5th 3 hours before they were at A and -6, and in the 6th 3 hours after. In the next two suppositions they were together at D"\ in the Vth 1 hour before they were at A and B^ and in the 8 th, 1 hour after. 9th. When m=-|-c, w=+c, and a=4-c?, we get ic=:-fQO, y= -f- oo, — = -}-QO , and -= + Qo. m n 10th. When m=—c^ n=—c, and a=z+d^ we get x=+ oo, y=+oo , — =— oo, and -=—00 . m n 11th. When m=: +c, n= — c, and a= +c?, we get x= +-, y= — -, a; c? . y ' d —— +— , and -= +— . m 2c n 2c 12th. When m= — c, w= + c, and a= + a, we get a:= + -, y = — -, 2 2 X __^ d y _ d m~ 2c* n" 2c' 13th. WTien m=+c, w=+c, and a=—dy we get a;= — cx), y=— oo, — =— QO , and -=— oo. 14th. When m=:—c, n:=z-^c, and a=—dj we get «=— oo, X 1/ y=— oo, — =+oo , and -= 4- oo. 15th. When m=: + c, w= — c, and a= — tf, we get x= — -, y = + -, i2 2 a? _ (? y_ c? w" 2c' w 2c* rf d 16th. When m=— c, w= +c, and a= —a, we get x= — -, y= +-, Z 2 X d . y d — =+— , and-=+— . m 2c n 2c ^ D"" B In the 9th suppo^+ion the couriers will be together at a point jn- 252 GENERAL DISCUSSION, ETC. finitely distant to the east of A and B in an infinite number of hours from the time they were at A and B. In the 10th supposition they were together at the same place an infinite number of hours before they were at A and B, The reverse of these results is found in sup- positions 13 and 14. To say that two couriers were together an in- finite number of hours ago, is the same as saying they never were together, and to say they will be together in an infinite number of hours, is the same as saying they never will be together. It should be remembered that in the 13th and 14th that A is east oiB, It will be seen that to obtain these infinite results, we have consid- ered at one time as +0, and at another —0. We started out with the idea that the position A^ or that of the faster traveler, should rule the sign of a. But wfien both travel equally fast, there seems to be a difficulty. To remove this, we consider m disregarding its sign- as greater than n by an infinitely small quantity. Hence, 7?i— w=+0 and — m + w= — 0. In suppositions 11, 12, 15, and 16, the couriers were together at i)"", a point midway between A and j5, A being the east point in 11 and 12, and the west point in 14 and 15. In these four suppositions, the number of hours was the same, — hours. + — indicating that they were together — hours after they were at A and B^ and — — that they were together — hours before they were at A and B. lYth. When m=+c, w=+0, anda=+Q^,we geta;=+(f, y=-f 0, X d ^ y — =+-, and^=+-. m c n 18th. When m=— c,7i=— 0, and a=. +(^,wegeta;=4- to find the values of ar, y, and 0. (a;+y + 22=-l, ) A 256 SIMPLE mEQUATIONS. Note. — ^The results in the last two examples were obtained by tbe checker-board process, and show that - is not always a symbol of ir determination, for in these examples the equations ar^ incompatible. • ♦>'»■ SIMPLE INEQUATIONS. (29 2«) An iiiequation is that which denotes that one algebraic ex- pression is greater or less than another. Thus a>6, and c6 and c>c? are inequations in the same sense ; so also, are a<6 and c<(?. (294.) Two inequations subsist in a contrary sense when the sign of inequality has not the same position in both. Thus o>6 and c<(/ are inequations in a contrary sense ; so also, are a<6 and c>c?. THEOREM. (295.) The addition or subtraction of an equation to or from an inequation results in an inequation in the same sense. THEOREM. (296.) The addition of two inequations which subsist in the same sense, results in an equation in the same sense. THEOREM. (297.) The subtraction of an inequation from another which exists in the same sense, does not necessaiily result in an inequation in the same sense. DEMONSTRATION. Subtracting 4>3 from 8>5, we get 4> 2 which is true. But if we subtract 8>5 from 4>3, we get — 4>— 2, which is not true according to the conventional idea sometimes attached to negative quantities, that they are less than nothing. SIMPLE INEQUATIONS. 257 But this inequation is still true if we limit the reasoning to negative quantities, for —4 is evidently a greater negative quantity than — 2. A debt of 4 dollars is greater than a debt of 2 dollars. On the other hand, we say that a man who is in debt 2 dollars, and has nothing, is worth more than the man who is in debt 4 dollars and has nothing. Hence considered in the light of wealth —2 is greater than —4, or in other words, —2 comes nearer being a positive quantity than —4 does. Again, let us subtract 8>2 from 11 > 9, and we have 3 < 7, an inequation which subsists in a contrary sense. Let us put these inequations in a different form : 9 + 2>9 and 2 + 6>2 Subtracting we get — 4>0, dropping equal quantities from both sides. But we have just seen that this inequation must subsist in a contrary sense, that is — 4<0. This shows —4 must be considered less than nothing. We can also prove this as follows : Since 3<7 it follows that 3-'7<'7-'7 -4<0 THEOREM. (298.) The multiplication or division of an inequation by a posi- tive quantity, results in an inequation in the same sense. THEOREM. (299») The multiplication or division of an inequation by a nega- tive quantity, results in an inequation in a contrary sense. DEMONSTRATION. Let a>&. Now if we multiply this by — c, we get —ac<^—bc, because the greater a negative quantity is numerically the less it is considered. Thus, 4<5 gives, by multiplying by —6, — 24> — 30. Since, multiplying by —1 is equivalent to changing signs, we derive the COROLLARY. WTien the signs in both members of an inequation are changed^ the sign of inequality must be reversed. IT 258 SIMPLE INEQUATIONS. Thus clianging the signs in —a + b—c'^d—m, we get a — b + c<^m—d. THEOEEM. (300.) When both members of an inequation are positive, they may be raised to the same power, and the result will be an inequation existing in the same sense. ^ THEOREM. (301*) When both members of an inequation are positive, the same root of each may be extracted and the result will be an inequa- tion existing in the same sense. PBOBLE (302*) Find the limit to the value of x in the inequation V.-f >|+5 (1). SOLUTION 21a;— 23>2a;+15 19a;>38 iP> 2 (2)=(l)x3 (3) =(2) transposed, (4)=(3)-19. EXAMPLES. !• Given x + ^x-\-^x^ll to find the limit of x. Ans. ic>6. 2. Given S^c + Yar— 30>10 to find the limit of x. Ans, ar>4. I. Given > to find the limit of x, Ans. a;>2c?. ed ^ Sx 4« Given ^a; + 3a;— 5>16 to find the limit of x. Ans. a;>6. 5« Given Ya;— 1>34 to find the limit of x. Ans. a:>5. 6* Given i ^ . ^ . « ^' r to find an intej2:er value of x. ( 2a; + 4>16 — 2a;, ) * Ans. ar=4. •• to find the limit of x. Ans. a;>a and a;<6. 7. Given ax , , ^ «* ^ bx \ b' _ — aa;+a62 when a and h are bolJb positive or both a SIMPLE INEQUATIONS. 259 8. Prove that a^ + 6' is equal to, or greater than 2a6, according as a and h are equal or unequal. 9. Prove that a' + 1 is equal to, or greater than a' + a, according as a is equal to 1, or is a positive quantity, that differs from 1. 10» Prove that a^ + l«' + « when a is a negative proper fraction. 12. Pr negative. 13( Prove that j--\ — <2 when a and h are not both positive or both negative. 14, Prove that a— 6>{|^a— 1^6).' when a>6, and both are con- sidered positive. 15t Prove that -r — r;,> — when a and h are unequaL a^ + lr a-\-h ^ cc u \ \ 16« Prove that — + — > — I — when x and y are unequal. y X X y 17. Given j ,_ , ^a (" *^ fi^^ *^® l^^i* of a;y. ^?zs. xy^ac-^hd. 18. Prove that a' + 6' + c'>a6 + ac4-6c when a, 6, and c are not all equal. 19. Prove that 1^3+3 V2 > VQ + VV, ft _J_ /J _1_ fi Q Q. Q /T 20. Prove that , >^ and <-, -.being the least, and ^ the greatest of the fractions, -, -, and 7:. 21. Prove that xyz'y{x + y—z){x + z—y)[y-\-z—x) when x, y, and z are not all equal. 22. A shepherd being asked the number of his sheep, replied, that double their number diminished by 7 is greater than 29, and triple their number diminished by 6 is less than double their number in- creased by 16. What was the number of his sheep ? Ans, 19 or 20. 260 SIMPLE INEQUATIONS. 23 « A market woman has a number of oranges, such, that tnple the number increased by 2, exceeds double the number increased by 61 ; and 5 times the number diminished by 70, is less than 4 times the number diminished by 9. How many oranges has she ? Ans. ,60. 24, The sum of two whole numbers is 25 ; if the greater be divided by the less, the quotient will be greater than 3 ; and if the less be divided by the greater, the quotient will be greater than |. What are the numbers ? Ans. 20 and 5. 25. What whole number is that which, if doubled and diminished by 6, is greater than 24 ; but, if tripled and diminished by 6, is less than double the number increased by 10 ? Ans. There is no such whole number. 26. The sum of two whole numbers is 32, and, if the greater be divided by the less, the quotient will be less than 5, but greater than 2. What are the numbers ? Ans. 24 and 8. 27. Twice a certain number, increased by 7, is not greater than 19 ; and thrice the same number, diminished by 5, is not less than 13. What is the only number, whether whole or fractional, that will satisfy these conditions ? Ans. 6. 28 • Four, added to five times a whole number, is greater than 19, added to twice the number ; and 4, subtracted from five times the number is less than 4 added to 4 times the number. What is the number ? Ans. 6 or Y. Ans, x=z4e. bers. Ans» a?=5. ^o ( 2x 4:3fl;+13, ) Ans, X ^•^^'' USS'JtfoK^^lHx;} to find, in whole nuza. CHAPTER XI aUADRATIC EaUATIONS.* (303.) Quadratic Equations are divided into Pure Quadratics and Affected Quadratics. PURE QUADRATICS. (3 04.) A pure quadratic equation is one in which the unknown quantity appears in but one term, and is affected by the exponent 2, or by a fractional exponent, which, when reduced to its lowest terms, 2 _4_ has 2 for its numerator ; as, a;' =9, a; ^^ =4, and a;' <> =16. (305 •) Every pure quadratic equation can be reduced to the form a;^=a', in which a^ may represent any quantity, whether real or imaginary, positive or negative. Thus, a;f =4 is the same as (arJ^)'=4, which becomes (a;)' =4, or a;'' =4, by putting x for xk, I PROBLEM. (306.) To solve a pure quadratic equation. SOLUTION. Since every pure quadratic equation can be reduced to a;'=a', we have only to find the solution of this equation. Taking the square root of both members, we have x=±ia. We do not place the double sign before a;, because we seek only the plus value of a;. . ANOTHER SOLUTION. By transposition, x^^a^ becomes a;''— a'=0, which, being factored, gives (a:— a)(a: + a) = 0. It is evident that this equation will be satisfied by putting either factor equal to zero, * Quadratic equations are also called equations of the second de&bee. 262 QUADRATIC EQUATIONS. . • . x—a=0 or x-\-a=0 which being solved, give x=a and x=^a. Scholium. — These solutions show.^at a pure quadratic equation may be satisfied by substituting for the unknown quantity two values which are numerically equal but of opposite signs. FBOBLEM. (307.) Given a:'— 17=130— 2a;» to find the values of x. SOLUTION. «'— 17 = 130— 22:*, 3a;'=147, x^= 49, a;=±7. EXAMPLES. 1, Given 4a:'' + 5=a;' + 8 to find the values of a?. Ans. a;=dbl. 2* Given 3x^-{-3=x^-\-6 to fiiid the values of a;. Ans, x=-±^V6. 3. Given x^ + ab=5x^ to find the values of x. Ans. x=z±:^Vab, 4« Given ax^-\-d=bx^+c to find the values of x. Ans, x^=.-^a/ — ; /J.9 2 a;' 3 5f Given -r + o="^ + o ^ ^^ *^® values of x. 4 3 8 2 Ans, x—±L\V\h, Ans, aj=db3. 7. Given 4a;'— 8a;"=l to find the values of x, Ans, a;=±-J. 'ar. Ans. ar=±|V77. 2ar a;' 4-3 6* Given — =— -— to find the values of x, Ans, a? =±3. 3 2a; a;' 4- 2 8. Given 3a;'— 4=—— r- to find the vallies of a;. ox PROBLEM 2a' (308.) 1. Givena;4-V'a'+a;'=,7z= (1) tofindthevaluesofa;. > QUADRATIC EQUATIOl^S. 268 SOLUTION. «V^H^+aM^'=2a' (2) = (1) xVa' + x\ x^a'^x^=a''—x' (3) = (2) transposed. aV + a:*=a'-2aV + fl;* (4) =(3)'. 9x^=Sa'' _ (7). _ Saj^iaf/S^ (8)=f'(7). PROBLE M 2. Given \/^,-\-b^-j/~-h^=h (1) to find the values of «. SOLUTION. i/4+6^=i/-^-6' + ft (2)=(1) transposed. f^ X ^ X ^+y=^-6' + 264/^-6'+6' (3)=(2)'. or X ' X' *«= 2hi/~—b^ , (4)=(8) traiffiposed. \=\/^V (6)=(4)^2i. ^=«;_6. (6)=(5)'. (7) =(6) transposed, ice. 4 ~ »' S («)=(^)^£ 4a^ (9). 256' ^ ' «=±?^6 (10) = »^(9). EXAMPLES. 1. Given ^^ ♦'^'-^' ^^ to find x. Am, x=ztV2ab^b\ X X b ' 2. Given ^ =6 to find a?. ^w*. x=±-r--V2b-\-\. Ya' + x'-^x 26 + 1 264 QtTADRATIC EQUATION'S. ^ 3« Given 4/^+3—4/^—3—1/-^ to find jr. Ans, z=±9V2. r 4 r 4 r 3 4, Given V^x + 2 — V^x—2 = Vx+S— Vx—3 to find x. Ans. a;=±5. 5t Given 1 =ax to find x, x+V2—x^ x—S/2—x'' 1 Ans. a;=±-V^a(a + l.) 6« Given =— r to find a?. -4/i5. a:=±4. 7. Given {x-\-dfz=i to find x. {x-^aY ^^^^ a;=±(2a* + 2a6 + 6')i 8. Given =0 to find x, Ans. x=:±:- . »• Given -=:=:r= =-, to find x. ya'-x'-^Vb' + x' d Ans. x-±y 2{d' + c^) ' 10. Given i^(^ + ^+_VcL-x ^/x ^ ^^ ^^ ^^^^ ^^ ±2VS6=^. SIMULTANEOUS EQUATIONS. PROBLEM x + yix—i/iiaib, \ to find the values of x and y. SOLUTION. From the proportion, we get aa;— ay = 6a; + 5y which becomes, by (309.) 1. Given j;t!--^- = ^-} c substituting - for y as obtained fi'om xy=c'y ac^ . be* .. ax =bx H (1). X X ^ ^ {a—b)x^=c\a + b) (2)=(1) x ar, &c QUADBATIC EQUATIONS. 265 PROBLEM 2. Given < o , ^ )J^ r to find the values of x and v. ^'^ GLUT I ON. This problem may be solved without reducing the equations to the general form of pure quadratics. x^^2xy^f=. 36 (3)=(l)-(2). x-y^^^ (4)=V(3). ^=±9 (5) = (l)-(4). y=±3 (6)=(2)-(4). EXAMPLES. C 3? -I- i/*'2/**3*l i 1. Given i ^ '; * * ' ' [• to find the values of x and y. ^?2«. a;==h6, and y=dr3. 2* Given ■{ , . / f * * * ' r to find the values of x and y, Akis. a:=:±9, and y=±6. 3* Given •] , aZo ' [ to find the values of a; and y, ( X — y — y, ) Ans, a;=db5, and y=zfc4. . *Q ,\«* r to find the values of x and y. ajy+y''=126 ) ^ Ans, ic=dbl5, and y==t6. C doi? "}- hxv ^n c' ) 5« Given •{ ^ ' >■ to find the values of x and y. { x^y : a; : : m : w, ) ^ -47W. nc 7ha + nh—mV c{n — m) mb' o(w-m) / nc n y na-{-nb— 6. Given ] ^ + 2/ • a;;. • » j. ^^ g^^j the values of a; and y. ( a;y=6, ) ^ Ans. a;i=zfc:3, and y=db2. f a;'' + a:y=12, ) !• Given •{ rt i f to find the values of x and v, ^W5. a;=±2, and y=:±4. ix^ -{-y Vx^— 9 ) V^4-xVxz=l^ \ *^ ^^^ *^® ^^"^^ ^^ ^ ^^ y- ^rw. a;= ± 1^ and y= ±4. 266 QUESTIONS PRODUCING PURE QUADRATICS. ^ ^. (a;' + a; V^= 208,) 9, Given < , , -4- _^ ' }• to find the values of a; and y. ^?w. a;=:±8, and y=db27. 3 __Q . C to find the values of a; and y. QUESTIONS PRODUCING PURE QUADRATICS. QUESTION 1. There are two numbers in the proportion of 4 to 6, the diflferenoe of whose squares is 81. What are the numbers ? SOLUTION. Let 4a? =one of the numbers, then 5x =the other number, .-. 25a;'^— lear'^^Sl, 9a;' =81, X =±3, 4a; =±12, one of the numbers, bx =±16, the other number. QUESTION 2. What two numbers are those whose sum is to the greater as 10 to 7 ; and whose sum multiplied by the less produces 270 ? SOLUTION. Let 10a; = their sum, then 7a; =the greater number, and 3a; =the less •* .-. 30a;'' = 270, a;" =9, X =±3, 7a;=±21, the greater number, 3a;=± 9, the less " QUESTION S . 1. What two numbers are those whose difference is to the greater as 2 to 9, and the difference of whose squares is 128 ? Ans. ±18, and ±14. QUESTIONS PRODUCING PURE QUADRATICS. 267 2. What three numbers are those which are in the proportion of ^, f, and f, and the sum of whose squares is 724 ? ^ws. ±12, ±16,and d=18. 3. A merchant bought apiece of cloth for |324 ; and the number of dollars he paid for a yard was to the number of yards, as 4 to 9. How many yards did he buy, and what was the price per yard ? Ans. 27 yards, at $12 a yard. 4. A detachment from an army was marching in regular column with 6 men more in depth than in front. The front was afterwards increased by 845 men, and by this movement, the detachment was drawn up in 5 lines. How many men were in the detachment ? Ans. 4550. 5. Two partners, A and B, divided their gain, $60, of which B took $20. ^'s money had been in trade 4 months, and if 50 be divided by the number of dollars A had in, the quotient will give the number of months that ^'s money, which was $100, had been in trade. How much money had A in trade, and how long had ^'s been in trade. Ans. A had $50, and B^s had been in trade 1 month. 6. A and B invested some money in speculation. A disposes of his bargain for $11, and gains as much per cent, as B invested; jB's gain was $36, and the gain upon ^'s investment was 4 times as much per cent, as upon B\ How much did each invest ? Ans. A invested $5, and ^ $120. 7. A dog started in pursuit of a hare which was 7 rods ahead of him, and after running 20 rods, he observed that the hare stmck off at right angles to her former course. He then changed his course so that he might overtake her without another tack. How far did the dog run, provided he ran 10 rods a minute and the hare 8, in the same time ? Ans. 25 rods. 8. What two numbers are those whose difference multipliphed by the greater produces 40, and by the less 15 ? Ans. ±8 and ±3. 9. What two numbers are those whose difference multiplied by the less produces 42, and by their sum 113 ? /o J . Ans. ±13 and ±6. 10. A man bought a field whose length was to its breadth as 8 to 6. The number of dollars that he paid for 1 acre was equal to the number of rods in the length of the field ; and 13 times the number 268 ATTEOTED QUADRATICS. of rods round the field equaled the number of dollars that it cost. What was the length and breadth of the field ? Am. Length, 104 rods ; breadth, 65 rods. 11. A stack of hay, whose length is to its breadth as 5 to 4, and whose height is to its breadth as 7 to 8, is worth as many cents per cubic foot as it is feet in breadth ; and its whole value is 224 times as many cents as there are square feet upon the bottom. What are the dimensions of the stack ? Ans, Length, 20 feet ; breadth, 16 feet ; and height, 14 feet. 12. One number is m' times as much as another, and their product is n^. What are the numbers ? . , , . w Ans, r±imn and ±— . m 13. What two numbers are those which are in the ratio of 8 to 6, and whose product is 360 ? Am. ±24 and ±16. 14. What two numbers are those whose sum is to their diflferenoe as 8 to 1, and the difference of whose squares is 128 ? Am, ±18 and ±14. ^ »> ♦ «« ^ AFFECTED QUADRATICS. (310.) An Affected Quadratic Equation is one which con- tains the unknown quantity in but two terms : its exponent in one being double that in the other, and the least exponent being either one, or a proper fraction whose numerator is one ; as, *5a;' + 7a;=21; (a— %'' + (c4-c?)a;=e+/; a; + 3a;2z=8; 4a;3 +6a;3 = 13; Remark. — Various plans may be adopted for ascertaining the values of the unknown quantity in an affected quadratic equation. In order that the most expeditious mode of solution may be adopted, special regard must be had to the peculiarities of the problem An aptness in observ- ing the elements of a problem which indicate its best solution can only be ob- tained by practice. The student, by a careftil study of the illustrative solutions which follow, will be able, it is hoped, to solve all the problems that are appended to them. - 1 - 1 -I * The student should bear in mind that y/x=x- ; X/x^^x"^ ; ^05=0;*, _2 31 2 &c., and ^ar'=a;^ ; \fx^=^x^=x'^\ \/o^—x^^&q. AFFECTED QUADRATICS. 289 PROBLEM 1. Given a;'— 2a!=— 1 (1) to find the values of «. SOLUTION', jr*— 2a?+l=:0 (2)=(1) transposed. Taking the plus value of zero, «— 1= -f 0, or a:=l, " " minus " " a;— 1 = — 0, or a;=l. This solution shows that x has two values which, in this case^ are identical. This fact may be better illustrated by the following SOLUTION. a;'-2a?4-l=0 (2). (a;~l)(a;-l)=0 (3) =(2) factored. This equation may be satisfied by placing either of the binomia- factors equal to zero. Whence, we obtain the two simple equations, ar-l=0 and a?~l=0 "Whose solutions give «= 1, and x= 1, PROBLEM 2. Given 3^—2ax+a'=:b^ (1) to ^^ ^^ values of a?. SOLUTION. x-a=±:b (2)= 1^(1). x=adtb. Whence, we find that x has two values ; viz., a-\-b and a— 6. The same result may also be obtained by the following SOLUTION. Sincea;''—2aa;+a''=(ar— a)'', Equation (1) transposed, becomes (a:— a)"— 6'=0 (2) {x-a + b){x-a-b)T=0 h)=(2) factored. This equation may be satisfied by putting either of the above fac- tors equal to zero. Whence, we obtain the two simple equations, x—a—b=:0 and «— a + 6=0 Whose solutions give x=a^b and ar=a--6, the same as before. 270 AFFECTED QUADRATICS. EXAMPLES. ]• Given a;'-.4a?+4=0 to find x, Ans, a;=2, or 2. 2* Given 3?' + 2a: + 1=0 to find x. Ans, a;= — 1, or —1. 3* Given a;'' + 4a;4-4=0 to find x. Ans. a;= — 2, or —2. 4. Given a;* + 6a; + 9=0 to find x, Ans. a;= — 3, or — 3. 5« Given a;'— 6a? 4- 9=0 to find a?. Ans. a:=3, or 3. 6« Given a;'— 8a; + 16=0 to find x. Ans. a?=4, or 4. 7. Given a;'+8a; + 16=0 to find x. Ans. a;=— 4, or —4. 8. Given ar* + 10a; + 25=0 to find x. Ans. a;=— 5, or —5. 9» Given a;' — 10a;+25=0 to find x. Ans. x=5, or 6. 10. Given a;'— 12a; + 36=0 to find x. Ans. x=e, or 6. 11. Given a;'+12a;+36=0 to find x. Ans. x=—6, or —6. 12. Given a;' + 14a; + 49=0 to find x. Ans. x=—1, or —1. 13. Given a;"— 14a; +49=0 to find x. Ans. x—1, or 1 14. Given a;' + 16a; + 64= 81 to find x. Ans. a;=l, or —IT. 15. Given a;"— 18a; + 81 = 100 to find x. Ans. a;=19, or —1. 16. Given a;' + 20a; + 100 =64 to find x. Ans. a;=— 2, or —18. 17. Given a;'=22a;— 121 to find x. Ans. a;=ll, or 11. 18. Given a;'— 169=24a;— 144 to find x. Ans. a;=25, or —1. 19. Given 43;"— 4a; + 1=4 to find x. Ans. a;=l|^, or —\. 20. Given 4a;'— 8a;+4=9 to find x. Ans. a;=2i, or — i. 21. Given 253;"— 20a; + 4=16 to find x. Ans. a;=li, or — |. 22. Given 2a;''— 2a; + 1=3;'' +9 to find x. Ans. a; =4, or —2. 23. Given 4a;''— 4a; + 4=33;" +25 to find x. Ans. x=1, or —3. 24. Given 6a;'— 6a;+9=6a;'' + l to find x. Ans. a;=4, or 2. 25. Given 9a;'— 8a; + 16 = 83?' + 25 to find x. Ans. a;=9, or — 1. 26. Given 5a;' — 1 2a; + 9 =a;' + 36 to find x. Ans. a;=4|, or — 1^. 27. Given 100a;'— 36a; + 4= 19a;' + 49 to find the value of a;. Ans. a;=l, or — f. 28. Given a;'— 4a; + 1 = — 2a; + 9 to find x. Ans. a;=4, or —2. 29. Given 4a;' + 9a; + 16=4 — Ya; to find x. Ans. a;= — l,or —3. AFFECTED QUADRATICS. 271 30* Given Sic'— 6aJ + 4=4a?'— 2a; + 16 to find the value of x. Ans. a;=6, or -2. 31. Given 9a;'-f 12aj— 25=6a;''4-4a;— 4 to find the value oi x, Ans. a;=l^, or — 3|. 32. Given 26ar' + 20a;— 81 = 9a;' + 4a:— 4 to find the value of x, Ans, x=li, or — 2f . 33. Givena;"— 2aa;-f-a':^6 tofind a;. Ans, x=ia^Vh. 31* Given iaj' + ^a;- 9=— | to find x. Ans, a;=7^, or — lO^. 4^ x'-Yx + l 36. Given J^a;"+a;4-16=16a:' to find x, Ans, a;=lJy, or — 1|. 37. Given — + - + -=12 to find x, Ans, a;=-|±6V^3. ^V 2arB f^ f^ f* 38. Given — +=^=0 to find x. Ans, x='^—, ot<-, r 9 f ^9 ^9 - Vcd 39. Given cx^—2cxVd=dx''—cd to find x. Ans, x i. Given — — ;; — 7-; =1 to find x, Ann, x=5, or —3. Vmn 40. Given ma;* + WW =2ma?Vw + fia;' to find a;. Ans, x: Vm-j^ Vn 41. Given ahx^—2x{a-\-b)Vahz=z{a—hy to find the value of a;. a + 6±i/2aN^^ Ans, a;= =r= . Vah 42. Given aa;'' + 6''+c'=a' + 26c + 2(6— c)a;Va to find the value of x, Ans, x= 1^ — . 43. Given a;'— a;=— :i to find the value of a;. Ans. a;=i, or |. 44. Given a;"— 10a;=— 25 to find x. Ans, a;=5, or 5. (3 1 1 •) In order to solve the following problems, the student should be familiar with the modes of eliminating radicals, and should also be able at sight to detect the squares of binomials. PROBLEM . 1. Given i/^^^ + 2l/-^=b^4/-^ (1) to find the values of a;, ~ X ^ x-\-a r x+a ^ ' 272 AFFECTED QUADRATICS. SOLUTION. '^Wh^' (2)=(i)x4/^4:« The student might not be able to see that the left hand member of this equation is a perfect square, and not observing this, the solution would be much more difficult for him. Let us endeavor to render this fact more apparent. . 1 +-+2|/-=6» (3)=(2) divided. l + 2-^-+-=6' (4)=(3) arranged. (1=.*)'-^ (i=F*y PROBLEM g/p \ VSx 1 2. Given -= =1H — to find the values of a?. VSx+ 1 2 Ans, a; =3, or J-, SOLUTION Since, Bx-l = (V3x + l)(VSx^l), y^x 1 whence, V3x — 1 = H — ^ 2 2f3^=4+V8^— 1, 3ar=9, x=S, , But how is the value i obtained? By putting ^^33; + 1, which is used in reducing the fraction in the first member of the equation, equal to zero. AFFECTED QUADRATICS. 273 ThllS, ^3^+1=0 But if we attempt to verify the given equations by substituting ^ for the value of x, we obtain --=!+-. Is this a true equation ? It seems not ; for by dropping - from both members, we obtain 0=1, an absurdity. Are we then to conclude that a; =i is not true ? In substituting i for ic, we assumed that V^3a;=4-1, whereas, we leam from the equation that V'3V= — 1. If we substitute, keeping in mind the feet that VSx must be taken equal to —1, we shall obtain 5=0, or 0=0, from which we leam that a;=i is a true value of a?. From this solution we may infer the following PRINCIPLE. Ani/ expression containing the unknown quanity that mil divide both the numerator and the denominator of any fractimi in the equa- tion^ being placed equal to zero^ will give as many values of the un- known quantity of the given equation as the unknown quantity in the equation thus formed has values. Since, a factor of the kind referred to in this principle may by mul- tiplication become common to both members of the equation, we ob- tain the following PRINCIPLE. Any expression containing the unknown quantity that will divide both members of an equation^ being placed equal to zero, will form an equation of which every value of the unknown quantity will also be a value of the unknovm quantity in the given equation. 18 274 AFFECTED QUADRATICS. EXAMPLES. 1, Given _f~ =1 -i ^^^^ to find the values of x, V5X + S 2 Ans. x=5, or }. 2« Given ^^ =c-\ — to find the values of x. Vax + b ^ Arts. a;=-( h H v I , or — 4-4-:c 3, Given V64+a;'— 8a;=-===- to find the values of x. Ans. x=3, or —4. ft _L. /ij 4. Given \/a'' +it'—2axz=-^=:z. to find the values of x, \/a-\-x a(a— 1) ulw^. a;=— ^— — ^, or —a* 5. Given 5Ltfifl^±?=:6 to find the values of x. a-\-x a Ans, a;=-a±^jirj-y26— 6*" PROBLEM (312.) 1. Given !^''+^-'J^Jl]u>&xiihevBiueso{xmiy. S OLUTI ON. ^^+2xy\-f=s' (3)=(ir. 4xy=4.a' (4)=(2)x4. x-^2xy + f=s-'-^ (5)=(3)-(4). x-y=^Vs '-^a' (6)= 4/5. 2x=s±Vs^-4a' 0)={e) + {l). 2y=8^Vs'^ -4.a' (8)=(l)-(6). a;=i(sitV«^-4a^). y=i(s:p|/s«-4a«). PROBLEM 2. Given | Jl^.'l^^ [ to find the values of x and y. AFFECTED QUADRATICS. 275 SOLUTION. a.«+ary4-y'=7 (3)= (2)--(l). a;''-2rry + y»=l (4)= (l)^ ^xy=Q (5)= (3)-(4). xy:=2 (6)= (6)--3. s?-¥2xy+f=^_ (1)= (3) +(6). a; + y=±3 (8)=V(Y). 2a;=4, or— 2 (9)= (8) + (l). 2y=2,or-4 (10)= (8)-(l). a;=2,or-l (ll)= (9)-f-2. y=l,or-2 (12)=(10)-j-2. Remabk. — Different artifices may be employed in eliminating one of the nn- known quantities in the following equations. The student should, however, be careful not to find the value of one of the unknown quantities in one equation, and substitute it in the other, for an equation would then arise which he is not yet prepared to solve. EXAMPLES. !• Given \ ^^ !• to find the values of «, and y. ( xy=2 J Ans, a;=2, or 1 ; y=l, or 2. 2t Given \ ^~ ^ [• to find the values of a;, and y. { xy=15 ) Ans. x=5, or 3 ; y=3, or 5. 3. Given \ ^~ ^ !• to find the values of a: and y. ( xy=266 ) Ans. a;=19, or 14 ; y=:14, or 19. I. Given \ , «~^« r to find the values of x and y. i x^—y'=26 ) Ans. x=S^ or — 1 ; y=l, or --3. ,__ ,_ j- to find the values of x and y. ^Sb + b-a' Ans. 2 ^y 3a '36 + 6— a' 6* Given •< , , ^ f to find the values of x and y. Ans. x=2, or 1 ; y=l, or 2. 276 AFFECTED QUADRATICS. yfix a_ a f to find the values of a? and y. B , |._-i q f to find the vabies of x and y. -47W. x=2l, or 8 ; y=8, or 21, - ^_ ^ >- to find the values of a: and y, Ans. x=2j or —1 ; y=l, or —2. 1 X IS — ft 1 10* Given -j „ . „ , r to find the values of x and v. Ans. i^=4«±i^2i3^ 11. Given "{ 3 , 3 / . va f to find the values of aj and v. ^W5. a?=2, or 2 ; y=2, or 2. 12. Given "S ^ , , ~ r to find the values of a; and y, ( a;2^y2 =2 ) Ans, ar=4, or 1 ; y=l, or 4. 13t Given -J ^^ ~~^^ ~ V to find the values of x and y. ( a;2— ^2=26 ) ^ri5. a;=9, or 1 ; y=l, or 9. lit Given -j ~ ^\ to find the values of a; and y. {x -\-y —b\ ^ Ans. x=4, or 1 ; y=l, or 4. 15. Given i ■ Z c f to find the values of x and y. ( a; +y — 5 ) ^ws. ar=4, or 1 ; y=l, or 4. I a;^ 4-^^=4 f 16« Given ■) 3 3 , 1 (" to find the values of x and y, ix^+y^=(x-^-{-y^Y) Ans. ar=4, or 4 ; y=4, or 4. 17« Given J ^" +2^® — L to find the values of .t and y. ( x8yT=2 ) Ans. ar=256, or 1 ; y=l^ or 266. AFFECTED QUADKATICS. 277 18. Given i ^""^y' "^ I to find the values of a; and y. V X8 — ye 1 3 3 \-y} = l I \+yi = 5 ) Ans. a;=:6561, or 1 ; y=l, or 6561. 19t Given ^ "*'" "~"^° ~ ^ J- to find the values of x and y, Ans. ar=256, or 1 ; y=l, or 256. (313«) Every affected quadratic equation can be reduced to the following form : it'd:zAx=:±zB, In which A and B may represent any quantities whatever, whether real or imaginary, whole or fi*actional. PROBLEM. To solve x'±:Ax=dbB, SOLUTION. First, let us examine the case in which ^ is an even whole number. Putting A=:2aj and B=b, and using the^^ws signs, since the princi- ple upon which the solution depends is the same as when A and B are both minus ^ or either one plus and the other minus, we have a;» + 2aa;=6. (1) We see by the principles of binomial squares that the first mem- ber of (1) would be a perfect square if it were increased by c^. Let us then add c^ to the first member, and, to preserve the equality, we must also add it to the second member. We then have a;'' + 2aa; + a''=o' + 6 (2) Extracting the square root, x-\-az=zrkL i/a^ + b x=—adtzVa'-\-h This problem may also be solved in the following manner : By transposition (2) becomes {x + ay — (a' + 6) = (3) Considering a^ + h as the square of Va^ + h, the left hand member of (3) is the difference of two squares, and consequently may be fac- tored thus : This equation may be satisfied by putting either factor equal to zero, whence result the two simple equations : a; + a— VVT6=0 and x-\-a + Va^ 4- ^=0 278 AFFECTED QUADRATICS. The first of these simple equations gives And the second, x=—a— Va^ + h, the same as before. EXAMPLES. 1, Given a;'' + 8aj=48 to find x, Ans. .i;=4, or —12. 2t Given ic' + 4a:=140 to find x, Ans, a:=10, or —14. 3* Given «'— 6a;=— 8 to find x» Ans, a;=4, or 4-2. it Given a;' + 8a;=33 to find x. Ans. x=S, or —11. 5* Given a;'— 10a;= — 21 to find x, Ans. a;=7, or 3. 6* Given aj' + 8a;=65 to find x. Ans, x=5, or —13. 7. Given a;'*— 2^a;=rg' tofind a;. Ans. x=p±:Vp' + q. 8. Given a;'' + 12a;=108 to find x. Ans. x=6, or —18. 9. Given a;'—14a;= 51 to find a;. Ans. a;=l7, or —3. 10. Given a;*— 8a; =48 to find x. Ans, a;=12, or —4. 11. Given a;' + 10a;=— 24 to find x, Ans, x=—4, or —6. 12. Given a;* + 16a?=— 55 to find x, Ans, x=—5, or— 11. il JL (314.) Equations of the form a;'*±2aa;'' = =fc6, n being integral, are affected quadratics, and should be solved as the preceding ex- amples. In such equations, if we consider, primarily, that a;" is the unknown quantity, as we should do, the above equation becomes of the general form, which we have already discussed; for, putting y=a;'', y'=a;", and substituting these values, we get the equation, ?/^±2ay=±6 which is the general form referred to. The student should observe that, in some of the following exam- ples, a;** = ±Va;, when n is an even number, should be taken with the minus sign, in order to verify the equation. PROBLEM 1. Given a; + 8ari-=48 (1) to find the values of x. AFFECTED QUADRATICS. 270 SOLUTION. a.+ 8an^+16=64 (2)=(1) with 16 added to both members, ar^ + 4=db8 (3) = l^(2). ici=4, or— 12 (4) =(3) transposed. a;=16,orl44 (5)=(4)'. The equations (4) and (5) show, that in attempting to verify the original equation with 16, that ark must be taken equal to 4 ; but, when attempting to verify it with 144, ark must be taken equal to — 12, and not +12. Let us solve another PROBLEM 2. Given a?— 6a?J=— 8 (1) to find the values of ar. SOLUTION. a?— 6ar^ 4-9 = 1 (2) == (1 ) with 9 added to both members. xi-S = dbl (3) = |/(2). iri=4, or 2 (4) = (3) transposed. a;=16, or4 (5)=(4)'. In this example, in verifying the values of a; ; for a;=16, ari^ must be taken equal to 4 ; and for a;=4, ari- must be taken equal to 2, and this is just what the student would be likely to do. We learn from these solutions that, in verifying equations, the values of a?*, (n being even, as in the above problems,) must be care- fully observed. EXAMPLES. 1. Givena; + 2ari=8 to find ari. Am, xi=2, or —4. 2. Given a; + 6ari= 16 to find xl. Am. a?i=2, or —8. 3. Given a; + 4arj^=12 to find art. Am. ari=2, or —6. 4. Given a;— 8ari=48 to find ark. Ans. ari= 12, or —4. 5. Given a;4-10a;i=3 to find xi. Am. ark= — 5dt2V1. 6* Given arl4- 12art=:13 to find ari. Am. ark=l, or —13. 7. Given a;i+14a:i=15 to find xi. Am. a:i=l, or —15. 8, Given arf +16a;5 =17 to find xi. Am. a^=l, or —17. 280 AFFECTED QUADRATICS. 9. Given a;5V4-18a;T7o=19 to find the values of arrio, Ans. arT^o=l, or —-19. 10* Given a; 4-201^^=21 to find Vx.^ Ans. Vx=\, or —21. 11, Given V^4-22 V^=23 to find \/~x, Ans. V^=l, or —23. 12. Given Vi + 24 \/x=2b to find X/x. Ans. V^=l, or —25. (3 15.) The solution of these examples may be somewhat ■ abridged by omitting the formality of completing the square. PROBLEM . Given a;*— 6a; +19= 13 (1) to find the values of ar. SOLUTION. a;*— 6a;= — 6 (2) =(1) transposed. ar»-6a; + 9=3 (3)=(2) with 9 added to both members. a:-3 = ±f3 (4) = 4/(3). «= 3 ± V^3 (5) = (4) transposed. Equation (5) may be written immediately from (2) ; since x is found to be equal to half the coefficient of x taken with a contrary sign^ PLUS or minus the square root of the known term after it has been increased by the square of half the coefficient of x. To prove this fact to be general, let us assume the four general equations : a;' + 2aa;=6 ar'— 2aa;=6 a;'' + 2aa;=— 6 a;'— 2aa;= — 6 A solution of these four equations gives the following values of a?, respectively : x=—a±:Va^b x= arbV'oM^ a;=— adb^/a'— 6 x= aiV'a'*— 6 From an examination of these four equations and the values of the unknown quantity in each, we derive the following RULE. WTien an affected qtiadratic equation is reduced to the form AFFECTED QUADRATICS. 281 a;'db2aa;=±6, the value of the unknown quantity may he found hy putting it equal to half the coefficient of its first power taken with a contrary sign, plus or minus the square root of the unknoivn term after it has been added to the square of half the coefficient of the first power of the unknown quantity. Remark. — The student should apply this rule in the solution of the following examples after they are reduced to the proper form. E X A M P L E Sv 1. Given a;^—6a; + 19=11 to find x, Ans. x=2, or 4. 2. Given x^ + 6bx=c^ to find x. An^. x='-Sb±i^9b^+c\ X a 3. Given — h -=- to find the values of x. Am, x=zldtVl —a'. a X a O/u o 2jj— — 2 4. Given ~+_=ic + - to find x, Ans, x-2:ii2V2. 2 dx 3 5. Given a;"-!- 12a;— 16=92 to find x. Ans, x=e, or —18. 6. Given a;" + 6a; + 4=59 to find x, Ans, x=:5, or —11. 7. Given a;''— 8a; +10= 19 to find x, Ans. a;=9, or —1. 8. Given a;'— 12a;+30=3 to find x, Ans, a;=9, or 3. 9. Given a;'' + 6a;=2'7 to find x. Ans, x=S, or —9. 10* Given 8a;' + 3 2a; =3 60 to find x. Ans, x=5, or —9. 11. Given a;''— 8a;=14 to find x, Ans, x=4±^3d, 12. Given 2a;' + 8a;— 20= TO to find x, Ans, a;=:6, or — ». 13. Given 4ar»— 8a;+6=326 to find x, Ans, a;=10, or —8. 14. Given a;+6a;i=27 to find x, Ans, aj=9, or 81. 15. Given Vx—4: Vx=9 to find x, Ans, a;=49'7 ±1361/13. 16. Given a;— 8^5=14 to find x, Ans. a;=46rt8i/30. " (316*) Every affected quadratic equation may be reduced to the form ca;'±2aa;=db6, in which c, a, and b are whole numbers or surds. The simplest case of this general form, which is when c=l, has al- ready been treated of. PROBLEM 1. Given ca;' + 2aa;=6 (1) to find the values of a;. 282 AFFECTED QUADEATIOS. SOLUTION. cV+2aca:=5c (2)=(1) X c c"a:'+2aca:+a''=a' + 6c (3)=:(2)witha''addedtobothmember8. cx+a=^V^i^Tbc (4)=|/(3). ca>=—a±.V^Thc (5) =(4) transposed. «= (6)=(5)-7-c. PROBLEM 2. Given a;'— 3aJ=40 (1) to find tlie values of a?. SOLUTION. As the coeflScient of x is odd, this equation is not of the requsite form, but in all such cases it may be made so by multiplying by 2. 2a;»-6a;=80 (2)=(l)x2. ' 4a;'-12a;=:160 (3)==(2)x2. 4»* — 1 2a; + 9 = 1 6 9 (4) = (3) with square completed. 2a;~3=±13 (5)=V(4). 2a;=16, or —10 (6)=(5) transposed, x=8, or —5 ('7) = (6)-f-2. EXAMPLES. 1, Given 3a;' + 2a; =85 to find x. Ans, x=5^ or — 5|. 2, Given 3a;'+4a;=340 to find x. Ans. a;=10, or —11^. 3, Given 5a;'' + 6a;=63 to find x. Ans. x=S, or — 4i. 4, Given 3a;'— 14a;= — 15 to find x. Ans, x=S, or If, 5, Given 4a;''— 6a;=108 to find x. Ans. x=6, or — 4|. 6* Given 3a;'— 2a;=65 to find x. Ans. x=5j or — 4i. 7, Given 15a;''—622a;= — 6384 to find a;. ^W5. a;=22|, or 18|. m 8. Given — -\ 19= 15^ to find x. Ans. a;=:9, or — llf. 3 5 ^ , , 118±^13724 9. Given 118a;— 2ia;'=20 to find a;. Ans. x= . 10. Given ^a;'- 20a;=32 to find a;, Ans. a;=4, or — 14. 11. Given 5a;'' + 4a;=273 to find x, Ans. x=1, or —7a. 12. Given afta;"— 2aj(a + 6)Va6=(a— &)' to find x. . ^ a-\-h±iV2a' + 2ly' Ans. r:= , i^ab AFFECTED QUADKATICS. 283 (317.) The student, after solving the examples in the last article, is presumed to be ftdly acquainted with the principles of their solution, and is, therefore, prepared to omit some of the intermediate equations. The general form, may be divided into the four following equations : cx^ + 2ax=by ca^— 2aa:=6, cx^ + 2ax=—bf cx^—2ax=: — b, whose solutions give, respectively, the following values for x : —a-±^a^^hc c ff — a±Va:' + hc c ^a±:Va'-bc c X- a±Va'--bc c A comparison of these values with the equations from which they are derived, gives the following E U L E. WTien an affected quadratic equation is reduced to one of the four forms indicated by the general equation, ca;''±2aa:==t6, the values of the unknown quantity may be found by putting it equal to half the coefficient of x, taken with a contrary sign, plus or minus the square root of the joroduct of the known term by the coefficient of x^, after this product has been increased by the square of half the coefficient of X, and then dividing the whole by the coefficient of x^. Ebmark. — In the following examples, when reduced to the proper fonn, the student should write immediately the values of x, being guided by the rule just given. PROBLEM. Given 3ic'— 30=4a; + 2 to find the values of a?. SOLUTION 2±10 a:=- =4, or -2|. 284 AFFECTED QUADRATICS. EXAMPLES. 1. Given z^—^lx— — 3| to find the values of «. Ans, x=^\^ or i. 2. Given 23?" — 10aJ+7 = — 6 to find x. Ans, ar=3, or 2. 3. Given 3a:" + 4a;— 7=88 to find x, Ans. x=5, or — 6i. 4. Given 11a;'— 100a;=~201 to find ar. Ans, x=3, or 6yV. x^ 5. Given — + 20a;=i3a;''— 80 to find a?. -47W. a;=10 or —24. a; 6 6. Given -+-=6^ to find a?. Ans, a;=25, or 1. o a; 7. Given21a;"— 1616a;=--20'748tofinda;. .4?wf. a:=60|,or 16f. o ^. 18a;» 18078aJ ^^^^ ^ ^ ^ 8. Given -— •+ — — — =—4728 to find x, o 00 -47W. a;=— 251^, or — 52, 9. Given 4a:'— 9a;=5a;'— 265J— Sa; to find the values of x, Ans, 151, or —16^. 10. Given {U^--9cd^)a^ + {4a''c' + ^abd')x—-(ac'+bdy to find ac' + bd^ the values of x, Ans. x=z- 11* Given 6a;'' + 2a;=14 to find x, Ans, x= 2adzSdVc -ld=i^"85 6 12. Given32a'""c*^' +4a'^''c'*-*(ac«— 2)a:=a'c"+"a;» to find tlrff o^m — * values of x, Ans, a:=4a'^^ or -—. c" (3 1 8.) Let us now examine the general equation ca;'±aa;=db6, in which a is an odd number. It is evident that the rule given in the last article is also applicable here, but in order to avoid multiplying the equation by 2, or in case this should not be done, to avoid firao- tions, we seek another mode of solution. PROBLEM. Given cx^+ax^zb to find the values of a?. SOLUTION. If we multiply this equation by 4c, we shall have 4c V + 4aca: = 45c, 4cV+4gca; + ct'=«' + 45c, 2ca; + a=dbVa' + 46c, 2ca:= —a±: Va'' + 46c, -a±|/a' + 46c '= 2i — • AFFECTED QUADEATICS. 285 This mode of solution is found in the Bija Ganita, a Hindoo Treat- ise on Algebra, which has been translated by Mr. Colebrooke. PROBLEM. Given 2a;"— 5fl?=ll'7 (1) to find the values of x, SOLUTION. 16a;'— 40a;=936 (2)=(1) X 8. 16a;»—40a; + 25=961. 4a;— 6 = db31. 4a;=6±31=36,or— 26. a;=9, or —6^. EXAMPLES. 1, Given a;*— 84=:Ja; to find x. Ans, a;=6, or — 6|. 2. Given a;'4-3a;=72 to find x, Ans, x= , 3* Given 5x^-{-x=4 to find x. Ans, a;=|, or —1. 4* Given 2a;"— a;=21 to find a;. Ans, a;=3|, or —3. 5* Given ax^—oxz^c to find x, Ans, x= . 2a p-±: Yp' 4- 4(7 6. Given ar'-f^a;= g' to find «. Ans, x=—±- — -£- ±, 7. Given 3a;'4-6a;=42 to find x, Ans, a;=3, or — 4|. 8. Given a;" 4- 6a; + 4= 22— a; to find x, Ans, a;=2, or —9. 9. Given a;"— 5fa;=18 to find x, Ans, a;=8,or —2\, 10. Given a;"— 3a;=10 to find x, Ans. x=5, or —2. X 11. Given 6a;"— -=78 to find a;. Ans. a;=4, or — 3^^. 2 12. Given 4a;3 +a;« =39 to find x. Ans. a;=729, ot(-^\ , (319*) This mode of solution may be abridged by omitting the intermediate equations. The relation of the unknown quantity to the known terms of the general equation ca;'±aa;=db6, may be observed by solving the following equations : cx^-\-ax=bj cx'*—ax=b, cx^-\-ax=: — b, cx^—ax=i^b. 286 AFFECTED QUADRATICS. The values of a; in these four equations are —a±:VaF+Abc x= , 2c ' a±:Va'-\-4bc ''= ¥c ' —a±:Va'—4:hc azhVa'—4:hc aj= . 2c By a comparison of these values with the equations from which they are derived, we obtain the following RULE. When an affected quadratic equation is reduced to one of the forms indicated by the general equation ca:'±aaj=±6, the value of the unknown quantity may he found by putting it equal to the coefficient of ar, taken with a contrary sign plus or minus the square root of the square of the coefficient of x, after it has been added to four times the product of the coefficient of x^ by the known term, and dividing the whole by twice the coefficient of x^, FROB LEM. " Given Sx^—^x= —34 to find the values of x, SOLUTION. «= . 16 EXAMPLES. 1, Given ea;"— a;=92 to find x, Ans. fl;=4, or — 3|. 2, Given 8x^—1x=lQ5 to find x. Ans. aj=5, or — 4|. 3, Given 3rc'— 3a;4-6=6^ to find x. Ans. x=^, or |. 4, Given Ufa;— 3^0;'=— 411 to find x. Ans. x=—2\, or 5^. 5, Given 9ia:'— 90ia;= — 195 to find x. Ans. x=z6^, or 3^. d h 6« Given adx—acx* ^zbcx—bd to find x. Ans. a;=-, or — . c a m r^' / ,S Q ««^ . /. , A C±lVc^ + 4:a^ 7. Given ia-\-b)cir—cx-\ ^ to find a?. Atis. x^=.—~. — — . AFFECTED QUADEATIOS. 287 8. Given 1/9a;4-4=3a:to find a;. Am, a;=lJ,or— i. 9* Given ax^—hx + c^ca;" 4- 2c to find the value of x. 6=b|/6^ + 4c(a-c) Ans. x= -, ^^ \ 2(a— c) 10. Given ^^ + -^~(a-6)(2c+ac?)^=(a4-6)-|'-(a»-6>'tofind the values of a?. Ans. x=—- — — -, or d{a-^hy d{a-h)' 11. Given aoa?" H =- = to find the values of x, c c c . '2a— h 3a+26 Ans. x= . or , ac he 3a;» 21a;— 2* 7782 T"^ 12 of a?. * -4ws. aj=— 46, or 24J. (320*) It is frequently advisable to consider several tenns as one in the solution of afieoted quadratics involving radicals. PROBLEM. 12. Given 80a; + -^ + r:^ = 18591— 3a;» to find the values Given Va;-f 12+\/a?+12=:6 to find the values of a?. SOLUTION If we put y=X/x -h 12 , y' will equal j^x + 12. y= — -—=2, or -8, and y^— Vx + 12=4, or 9, aj + 12 = 16, or81, a;=4, or 69. In verifying the value a;=69, we must take, as the solution indicates, y=V«Tl2 = — 3. EXAMPLES. 1, Given ^/aj + lO— Va; + 10=2 to find x. Ans. a;=6, or —9. 2. Given4/a; + 2H-Va; + 21 = 12tofinda;. ^tw. ar=60,or 235. 3. Given 4/20:4-6 4- V2a;+6 =6 to find x. Ans. x=5, or 3*7^. 4. Given a; 4- 1/^+6= 2 4- 3 VW^ to find a;. Ans. a?=10, or —2. 5. Given a;4-5=V'a; 4-54-6 to find x. Ans. a;=4, or — 1. 288 AFFECTED QtJADRATICS. 6. Given a;+16~7iV+T6=10— 4 V^+16 to find x, An^, a;=9, or —12. 7. Given ^x + a-\-b Vx+a=:2b'' to find of x. Ans, x—h^—ay or 166*— a. (321.) It is sometimes advisable to complete the square without reducing the equation to any of the forms given above4 PROBLEM. Am? Sx Given — +—=8 to find the values of x, ol lo SOLUTION. Adding 1 to both members of the equation, and we have 4a;" 8a; ^ ^ 81 +18 + ^=^- 2,x y=2,or-4. 2a;=18, or —36. a;=9, or —18. EXAMPLES. 1. Given _— _4.5=0 to find x. Ans, a;=9(l±2i^— 1). a;" 4a; 1 2. Given — _— + _=o to find a;. Ans. x=S, or 1. « ^. dx* 16a; 3. Given — + — + 4=0 to find x, Ans, a;= — IJ, or — 5f x^ 223/ 4. Given — — _=_ 32 to find a;. Ans, a;=152, or 76. 5. Given ^ - 4aa; + 36''=0 to find x, Ans, a;=— , or -. ^ a a 4a;* 6* Given — — 4a;=:7 to find x, Ans. a;=10i, or —1^. ^ _. aV Sbix 126* ^ , , 6bi 2bi 7, Given -^ ^ + —^=0 to find a;. Ans, a;=-^, or — . a a a* ^ n* AFFECTED QUADRATICS. 289 8. Given —-+—=61 to find x. Am. x=1^ or — 11|. PROBLEM. (322.) Given x^+x=b to find tlie value of x, b being the pro- duct of two consecutive whole numbers. SOL UTION. By the conditions, we have by putting a equal to the least of the consecutive numbers . x^ i-x=b=a{a-\-l) x' + x=a^-\-a. A bare inspection of the last equation shows that one value of a; is a, but to get both, we complete the square x'' + x + l=a^ + a-\-l, . aj + i=±(a4-i) x=a, or — (a + 1). From which we observe that in an equation of the form x^-]-x=b=a{a + l)j X has a positive and a negative value, the positive value being equal to the least of the consecutive numbers, and the negative one equal to the other. If the equation were a;"— a;=6=a(a+l), the values of x would be found to be the same with opposite signs, namely, a;=— a, ora + 1. Scholium. — Since, a^-\-a + \=a{a-\-\)-\-\^vTQ conclude that the product of any two positive consecutive numbers increased by J is a perfect square, therefore, 6i=(2i)». 12i=(3i)'. 20J=(H)'. &c. &c. PROBLEM. Given a;' -fa?=20 to find x. SOLUTION. ar»-fa; + i=20J. a;=4, or -—6. We might write immediately the value of x thus a;=-i±:4^=4,or-6, or decide its values by the principle mentioned above. 19 290 AFFECTED QUADBATICS. BX AMPLE 9. 1« Given a;'' + «=2 to find the value of x. Ans, a;=l, or —2, 3» Given x^^x=::6 to find the value of x. Ans. ic=— 2, or 3, 3* Given x"^ -\-x-=\2 to find the value of ar^ Ans, a;=3, or —4. 4* Given x'^—x^:z20 to find the value of x. Avis. a;=— 4, or 5. H, Given ar' + a;=:30 to find the value of ar, Ans. x=5, or —6. 6* Given a;'— ar=42 to find the value of ar. Ans. x=:'—6, or 7, 7. Given x''-\-x=56 to find the value of ar. Ans. a; =7, or —8. 8* Given a;'— a;=V2 to find the value of ar. Ans. x=—8, or 9. 9* Given a;' + a?=90 to find the value of a?. Ans. x=9, or — 10, 10« Given a;"— a;=110 tofindthe valueof a;. Ans. x= — 10,ot 11. 11, Given ar* + a;=132 to find the value of x. Ans. ar=:ll, or —12. 12. Given a;''— a;=306 to find the value of a:. Ans. a;= — 1*7, or 18. (3!23.) The student is not always limited to the modes of solu- tion which have already been given, P R O B L.E M . Given a3C^-\-hx=c to find x. SOLUTION. Any two terms can be made a square by adding to them the square of the quotient arising from dividing one of the terms by twice the square root of the other. Hence, ax^ 4- hx will become a square, if ,, . , . hx h ^. ^ . b' we add to it the square of =:, or which is -— . 2xVa 2V'a *« Adding — to both sides of the given equation, we have , ^ b' b' 4:ac + b^ 4a 4a 4a r^x_ . , /- * ,y4ac-hb' Extracting square root, we have xva-i ==± = — . 2Va 2 Va 2aX'hb=±:V4ac-\-b\ 2aa;= — b±:V4ac + b\ _ '-'b±V4ac + b^ ~~ 2a ' AFFECTED QUADRATICS. 291 Remark. — ax^ + hx will become a square when —r- is added to it. a^x^ a^x^ That is,6a;+aa;^ + — ^, or —-^-{-ax^-^hxh a perfect square. So also, will aar'+fta? become a square when 2xVahx is added to it. 2xVahx is obtained by considering ax^^ and hx as the first and last term of the square, and finding the middle term which is equal to twice the pro- duct of the square roots of the other two. Hence, two terms being given, we can find three terms, any one of which being added to the given terms will produce a square. The first one is the only available one in the solution of quadratics. EXAMPLES. !• Given 3a;'4-4a;=7 to find x. Ans, a;=l, or — 2i. 2« Given 6ar* + a?=22 to find x. Ans, x=2j or — 2J. (324«) Problems of a special character may sometimes be solved by modes different firom any that have been given. A few are given a& a matter of curiosity. PROBLEM. Given 3a;" + 4a;= — 1 (1) to find the value of x, SOLUTION. 3a;' 4- 4x + 1 = (2) = (1 ) transposed. K the coeflScient of x^ were 4 instead of 3, the first member of this equation would be a perfect square. In order to render it so, let us add x^ to both sides, and we have 4a;' + 4a;+l=a;'. 2a;+l = ±a?. 2a;if:a;=: — 1. a;or3a;=— 1. ar=-l, or-f EXAMPLES. 1, Given 3a;' + 8a;= —4 to find x, Ans, a;=--2, or — f. 2* Given 16a;'4-8a;— — 1 to find x. Ans. a;=— i, or —J. 3« Given 1 53;" + 16a: =—4 to find a;. Ans. a;=— |, or—f. 4* Given 12a;'H-8ar= — 1 to find x, Ans. x=—^, or —}. 5« Given 12a;' 4- 16a; =—4 to find x. Ans. a;= — 1, or —^, 6* Given 8a;'— 12a; =—4 to find a;. Ans. a;=l, or |. 292 AFFECTED QUADRATICS. (325.) The student should observe carefully the solution here given of the equation 3a:''-+-10a;=— 8, not because the plan given is the best, but that he may be able to solve in a similar manner the examples which follow, which are intended to give him a foretaste of I a principle which is employed in the solution of cubic equations. PROBLEM. Given Saj' + lOa;:::— 8 to find the value of a?. SOLUTION. 3a;' + 10a;+8=0 (1). Adding 1 to both members of (1), makes the last term of the first member a square. Thus, 3a;'' + 10aJ4-9=l (2). Adding now «* to both members of (2), makes the first term of the first member a square. Thus, 4ar' + 10a; + 9=a;'' + l. By examining the first member, we see that it would be a perfect square if the middle term were 12a; instead of 10a:. But we can make it 12ar by adding 2a; to both members, which being done, not only renders the first member a perfect square, but also the second one. Thus, 4a;» + 12a;+9=a;»+2a;+l, 2a;+3 = ±(a; + l), a;=-2,-H. EXAMPLES. 1* Given 3a;' + 8a;= — 5 to find x. Am, a;= — 1, or — If. 2. Given 8a;' + 10a;=— 3 to find x. Am, a;=— ^, or — f. 3* Given 6a;'' + 8a;= — 3 to find x. Ans. a;=— |, or —1. 4, Given 3a;''-M8a;= — 15 to find x, Ans. x= — l, or —5. 5. Given 12a;''— 3 2a; =—5 to find a;. Am, x=2ly or }. 0* Given '7a;''— 18a; = — 8 to find x. Am, a; =2, or 4. 7. Given 15a;'— 16a;='7 to find x, Ans, x=—i, or If. 8i Given 15a;'— 32a;=7 to find x, * Am. a;=2i, or —J. 9. Given 9a:' + 22a;=16 to find x. Am. a;= — 3, or f. 10. Given 16a;'4-8a;=8 to find x. Am. a;= — 1, or 1. 11. Given 24ar'— 2a;=15 to find x. Am. a;=— f, or f. 12. Given 20a;' + 10a;=24 to find x. Am. a;= — 1 i, or f . AFFECTED QUADRATICS. MISCELLANEOUS EXAMPLES. 1, Given -r — l=a?+ll to find x, Ans. a;=12, or —6. 6 6 2 2, Given — -—-\ — =3 to find x. Am, x=:2, or — ^. x+1 X 3« Given 6aJH =44 to find x, Ans, a:='7, or f. . ^. 16 100— 9a; „ , « , . . « , 4* Given —5 — =3 to find x, Ans, «=4, or 2^^^. X 4:X d — X X b 5« Given 1 =- to find a?. «; a-^x c ^^' ^=4^^5(2^/^'-4 - -. i^4a; + 2 4-Va; ^ . , ' . «^ 6* Given == =r- to find x. Ans, a;=4, or -y. 4 +V^ar f'a; V^a'aj + ft a— 1/a; 7« Given r-= =- to find x, a+Vx Vx /-6±V'4a' + 4a''+6V Ans. x^=\ —. -T I \ 2(aH-l) / 8, Given (ya? + 6)(t'a: + 12)=:12 to find x. Ans. a;=4, or —21. 2 x—8 6 2 3^ ^ ^. 8— a; 2a;— 11 a;— 2^ . , . « , 9, Given — ZI^~~T~ ^' ^* ^~ ' ^^ ^* 10. Given r-=22| to find x, Ans, a;=49, or ^^ _ 21 11 . Given f/2a; + 1 + 2i^a;= - to find x, V2x + l Ans. a; =4, or —26. 12. Given V^+4^ip^=6V« to find x. Ans, a;=2, or —3. 13. GiveniV— 21^^— a;=0 to find x. Ans. x=4, or 1. II, Given Vx^—a^=x—b to find x, Ans. a;=:^dr-i^V48a'6-126*. 3 1/^__ 2 1 15. Given ^^ = — to find x. Ans. a; =3 49, or 25. x—5 20 294: AFFECTED QUADRATICS. 16, Given x+V5x-\-lQ=^S to find x. Ans. a;=18, or 3. 17. Given a;4Yl0a;+6=9 to find x. Am. a;=25, or 8. _______ ___ '7/» _1_ K/*; 18. Given 2Vx-'a + W2x= to find x. Vx-a. Ans, aj=9a, or — a, 10, Given 3a:*— a:=140 to find x. Am, x=1^ or — 6|. 1x 30* Given 5a^+-— =1x^—51 to find x. Am. x=6, or —4^. 4/|. 4 21. Given 2a;' — =1x to find x. Am, a;=4, or i, 3 22, Given 3a:'-l'7a;=2ii;'' + 84 to find x. Am, a;=21, or -4. 2l3t Given re'— a; + 3=45 to find x. Am, a;=7, or —6. 24« Given 4a; 7= 14 to find x. Am. a;=4, or — li. a! + l ^^ ^. 10 14— 2a: 22 ^ _ ^ . 25« Given = — =— - to find x. Am, x=3. or l\^, a: a:' 9 26. Given 6a:'— 4a;+ 3 = 159 to find x. Am. x=6, or — 5|. J221 4a; 27i Given 3a: =2 to find x, Ans, a:=19, or — 19|. X 28* Given =— -— to find x. Am, x=6. or I, 2 a:— 3 6 Qa. g Q/j. g 29t Given 5x =2x-\ to find x. Am, a:=4, or —1. a;— 3 2 -I OQ __ 0„ 30i Given 3a: =29 to find x. Ans, a:=13, or —4^. x . 2a:' 4a: 31* Given 16 = h V| to find x. Am, x=S, or — 4i. 3 5 * 7 5 3ic— -4 a:— 2 32. Given - + 1 = 10 -— to find x. Am, a;=12, or 6. a:— 4 2 33. Given — — -= — -- — to find x, 5 x—6 10 «. ^. 3a:— 10 „ 6a;'-40 ^ ^ 34. Given 3a: — - — -—=2-}-— — to find x 9— 2a: 2a:— 1 Ans, a: =36, or 12. V, Am, a?=ll^, or 4. AFFECTED QUADEATICS. 295 ^w«. a;=l, or — a«. 90 2Y 90 S6« Given = to find a:. -4?^. ir=4, or — 1|. ar a; + 2 x-\-\ 1 19 37. Given -^ — -- +-^ — - =7— to find ic. Am, a?=4, or — 3f. ic'— 3a; a;' + 4a; 8a; y>* 10a;' + 1 38* Given — - — — =aj— 3 to find ar. Am. a;=l, or --2a a;''— 6a; + 9 39t Given t-^ — y-^^^—l^-^ to find a?. Am. x—b, or 2. 7— a; a; JA fv '7~12a; a; 8a; + 110 ^ „ , 40* Given — =^r =r — to find x. xli Vx Vx' Ans. a;=9, or —13. -- ^. a—x . a; 6 ^ „ - . 2ac 41. Given i =- to find a;. Am. x=. X a—x c 1c + h±VV'—A:C^ -_ __ 2j/a;' + 60a;' + 9a; + 540 + 89 ^ , 42« Giveni/a; + 60+1/a;' + 9= _1_ to find a?. Vx-irQO^-Vx' + d Am, a;=4, or —5. .- _. 123 4-41^^^ "lOVx + Ax 2x^ ^ . , 43* Given = = = ~ to find x. bVx'-x S-^Vx (5Vx—x){S — Vx) Ans, a;=20|^, or 3. X X b 44* Given -^^ H — := =-= to find x, Vx-\-Va—x Vx—^a—x Vx b±Vb'-2c^ Ans. x= , 2 45* Given hl0a;=i95 to find jc Am. x=1, or —6, X 46. Given — — 14-20^=:42| to find x. Am. x—7, or — 6|. .„ ^. 2X+VX ^, ^2X—VX^ rt :i A .0, 47. Given ^=3y^j— 3. = to find x. Am. x=4^ or 3 Jg-, 2x-^x 2x-\-Vx Ao ^. 54-9i/^ 2Sx—4:6Vx 1x*-Sx + A- ^ ^ 48. Given =:= = 1 ir nr to find X. x-^2Sfx 6 + 4/aj {x^2^x){^^-S/x) Ans. a;=5, or — 2^^. 296 AFFECTED QUADRATICS. 49. Given ^ = ,r— + =- to find x, 8-34/a; 4+4/jB {8-SVx){4 + t/a;) -4w5. x=93j or Y. /:a n- i^—Vx+1 5 ^ _ , DO. Given =r^=—. to find a;. ^ws. a;=8, or — f. 51. Given {V4x + 5)(V1x+l)=80 to find x. Ans. x=5, or —JaV-- ^- ^. 15a;— 5 2 - 52. Given =4-— =31/a; to find x, Ans. a;=4, or I, l + 5Vx Vx 53. Given 9a*5V-6a^6''a:=6'' to find ar. Ans, x=z^^^^^^^. 54. Given3^112— 8a;=19+V3a; + 7 tofindar. An^. .r=6, or VA'-' 55. Given V2x+1 +V8x—\8=V1x + \ to find a;. Ans. a;=9, or — 3f. 5o« Given -y^ 1 — 5=0 to find x. Ans. x=- k c c* a c 57. Given x+ Vx : x-~Vx::3Vx + e : 2Vx to find x. Ans. ar=9, or 4. 58. Given ♦/(4 + ar)(5--ar)=:2ar— 10 to find x. Ans. x=5, or 3^. 59. Given a;'— a;— 40 = 170 to find x. Ans. a;=15, or —14. ^3. 3 60. Given x-] — =8 to find x, Ans. a:='7, or 9}. 61. Given 2V^+3V^=2 to find x. Ans. x=l, or -8. 62. Given 4a;= f-46 to find x. Ans. a;=12, or — #, X ' *' 63. Given 1-^— — =— - to find x. Ans, x=*l. or —3. 7— a; *l-\-x 10 M*^ r,' 3ar+6 3a;— 5 135 ^ , 65. Given ♦/(a;— l)(a;-2) + t/(a;-3)(a;— 4)=|/2 to find a;. Ans. a;=3, or 2. AFFECTED QUADRATICS. 297 66. Given , , ^._ +—^=——-^ to find x. Ans. x=6, or — f . 67. Given '^\/^-^-\/i +45=1^10^+56 to find x. Ans. ^=20, or -VsTVeV/- 68. Given aJV + (l+c)6f;|/c + c6V=[6'(?t/c + (a6 + c)^(l+c)]a; to 6c?1/c 1+c find X, Ans, x= or -^^. ao + c a; Y 69. Given r:r=r to find x. Ans. z=14, or —10. a; + 60 3a;— 5 ' 70. Given f-— =13 to find x. Ans. x=9, or lyV x—5 X Sx 20 71. Given -— 6=— to find x. Ans. a;=10, or — |. 3/ + 2 oiC 72. Given -= tt: — ^ to ^^ ^' ^^*- ^=^i or 5. a; ~f" o X -p xU 73. Given — =:*— — r- + l to find x. Ans. x=6l}, or 4A. 0/p_i_ 3 2a; 74. Given — — -—— — ^ — 6^ to find x. Ans. a;=13|a., or 8. lU — X JiO — oX ' n- 25a; + 180 40a; 3^ ^ , 75. Given -— r^=^ r— ;r to find x. 10a;— 81 5a;— 8 5 Ans. a;=14|, or /^. ^- _. 18 + a; 20a; + 9 65 ^ _ , 76. Given -j- r=r-^ — ^"tt;^ x to ^^^ ^• 6(3— a;) 19 — 7.r 4(3— a;) Ans. x=1^^j, or 2|. _ ^,. 5a + 10a6\ /5i/M^^(l + 26Vv/c\ cc^ , « , . (3-an|/a + 6 Sb'cdVc to find a;. ^ri«. x= ^ .. ' ^„. , or . a6(l + 26') ' 5a »»»■♦..» AFFECTED QUADRATICS INVOLVING TWO UNKNOWN QUANTITIES. (326.) In solving equations of this character the usual plans of elimination may be employed. The student should adopt that plan which seems to be best for the example under consideration. 298 AFFECTED QUADRATICS. EXAMPLES. 1, Given I •'*^y_ 8 / to find the values of a; and y. .] yr=3, or -f . 2. Given \ xy ' > to find the values of x and y. t 9y-9a;=:18, ) ( y=4, or f . 3« Given 1 « , ' ^/ * * ' ' {• to find the values of x and y, ly=4, or — V- 4i Given J "~ ~ i r to find the values of x and y. \ 4—x =y~y% ) , Ans, \ "=^ "^ ^ (2/=:l, or|. 5. Given i i^^~[ ^' f to find the values of a? and y. ( Sarfr— yi=14, ) ^71* i ^=(14)', or 8, 6. Given j i , ^Z f to find tlie values of x and y. ( y=8, or 0. 7. Giren ^+J:+''2'=2y^+*'^' La^dthevaluesof .andy. ( Vx+Vy=5, ) ( y=4, or ^jS-. 8. Given \y^ y ~ 9 ' )- to find the values of x and y. ( a;— y =2, ) AFFECTED QUADRATICS. 299 9« Given ^ Vx ^x Vy > to find X and y, x-\,ox\. Ans. (y=4,orJ/. o . ^ ' ., }• to find the values of a; and y. y'' + 4x=2y-\-llS (aj=— 46, or2, ( y=15, or 3. 11, Given j h aZ^qoo f *^ ^^^ *^® values of a? and y. Ans J ^=35, or -if^iL, (y=:16,orif4A '2a; + 7y ^ 61+2a; ^ = 2y— — T:r— , ► to find the values of a: and y, i y=4:, or Iff, 12. Given ^ 4a; 4:X + Sy =y-2, 13. Given -( 16 4:Xy+Sy-S 4y-^3x—2 18— a; 5a: 5 3x-{-y_Sx—5y T~~~"3 to find a; and y- Ans i^=^'^^-2l6, ^^^- 1y=3,or|Hf 14. Given j ^ + 4V^^+4y=21 + 8f^y + 4i/i^ j to find ^ and y. ( Vx-{yy=6, ) \ ♦ >' »■ 327. QUESTIONS PRODUCING AFFECTED QUADRATICS. QU ESTIO N. A and B sold 130 yards of calico (of which 40 yards were -4*8, and 90 ^'s), for $42. Now, A sold for $1, i of a yard more than B did. How many yards did each sell for $1 ? 800 AFFECTED QUADRATICS. Let X = the No. of yards B sold for $1. .«. aj+» = « « ^ « " - = what B got for 1 yard. ' ' 3 90 X 40 ii u u A " " B " 90 yards. x+- ^ « 40 " ... 90 , 40 42=— + - X x-\-\ ^. 45 , 20" 21= h- 21a;' + 7ic=45a; + 15 + 20aj 21a;'— 58cc=l5 a;= — ~ — =3, or—— No. of yards B sold for |1. a;+i=3i " " A " " QUESTIONS. 1. A merchant sold a quantity of cloth for |39, and gained as much per cent, as the cloth cost him. What was the price of the cloth? Ans. $30. 2. There are two numbers whose difference is 9, and their sura multiplied by the greater produces 266. What are the numbers ? Ans. 14 and 5, or— 9i and — 18i. 3. It is required to find two numbers, the first of which may be to the second as the second is to 16, ; and the sum of the squares of the numbers may be equal to 225. Ans. 9 and 12. 4. Bought two sorts of linen for 6 crowns. An ell of the finer cost as many shillings as there were ells of the finer. Also, 28 ells of the coarser (which was the whole quantity) were at such a price that 8 ells cost as many shillings as 1 ell of the finer. How many ells were there of the finer, and what was the value of each piece ? Ans. 4 ells of the finer ; the value of the finer 16 shillings, and of the coarser 14 shillings. AFFECTED QUADRATICS. .301 6. Two partners, A and B, gained |18 by trade, ^'s money was in trade 12 months, and he received for his principal and gain $26. J5's money, which was |30, was in trade 16 months. How much did A put in trade ? Ans. |20. 6. A person bought some sheep for $72, and found that if he had bought 6 more for the same money, he would have paid $1 less for each. How many did he buy, and what was the price of each ? Ans. 18 sheep, at $4 a piece. I. The plate of a looking-glass is 12 inches by 18, and is to be framed with a frame of equal width, whose area is to be equal to that of the glass. What is the width of the frame ? Ans. 3 inches. 8. There are two square buildings, that are paved with stones, a foot square each. The side of one building exceeds that of the other by 12 feet, and both their pavements taken together contain 2120 stones. What are the lengths of them separately ? Ans. 26, and 38 feet, respectively. 9. A laborer dug two trenches, one of which was 6 yards longer than the other, for £l1 16s., and the digging of each of them cost as many shillings a yard as there were yards in length. What was the length of each ? Ans. 10, and 16 yards. 10. A company at a tavern had £8 15s. to pay ; but before the bill was paid, 2 of them sneaked off, in consequence of which those that remained had each 10 shillings more to pay. How many were in the company at first ? Ans. 7. II. A grazier bought as many sheep as cost him £60; out of which he reserved 15, and sold the remainder for £54, gaining 2 shillings a head on those he sold. How many sheep did he buy, and what was the price of each ? Ans. 75 sheep, at 16 shillings each. 12. What two numbers are those whose sura is 19, and whose dif- ference multiplied by the greater is 60 ? Ans. 12 and 7. 13. If the square of a certain number be taken from 40, and the square root of this difference be increased by 10, and the sum multi- plied by 2, and the product divided by the number itself the quotient will be 4. What is the number ? Ans. 6. 14. A person being asked his age, answered, if you add the square root of it to ^ of it, and subtract 12, there will remain nothing. How old was he? Ans. 16, or 36. (Prove the last answer.) 16. A and JB set out fr<^m two towns which were at the distance of 302 AFFECTED QUADRATICS. 247 miles, and traveled the direct road till they met. A went 9 miles a day ; and the number of days, at the end of which they met, was greater by 3 than the number of miles which B went in a day. How many miles did each go? Am. ^117, and B 130. « 16.-4 set out from C toward i>, and traveled 7 miles a day. After he had gone 32 miles, B set out from JD toward (7, and went every day -jJg of the whole journey ; and after he had traveled as many days as he went miles in one day, he met A. What is the distance from C toD? Ans. 152, or 76. 17. Three merchants, A^ B, and C, made a joint stock, by which they gained a sum less than that by $80. ^'s share of the gain was $60 ; and his contribution to the stock was $17 more than ^'s. Also, B and C together contributed $325. How much did each contrib- ute? Ans. A $75, ^$58, (7 $267. 18. The joint stock of two partners, A and J5, was $416. ^'s money was in trade 9 months, and ^'s 6 months ; when they shared stock and gain, A received $228, and 5 $252. How much was each man's stock ? Ans. A's $192, and i5's $224. 19. A body of men were formed into a hollow square 3 deep, when it was observed, that with an addition of 25 to their number, a solid square might be formed, of which the number of men in each side would be greater by 22 than the square root of the number of men in each side of the hollow square. What was the number of men in the hollow square ? Ans.' 936. 20. A merchant bought a number of pieces of two different kinds of silk for £92 3s. There were as many pieces bought of each kind, and as many shillings paid per yard for them, as a piece of that kind contained yards. Now 2 pieces, one of each kind, together measured 19 yards. How many yards were there in each ? Ans. 11 and 8. 21. A vintner sold 7 dozen sherry and 12 claret, for £50. He sold 3 dozen more of sherry for £10 than he did of claret for £6. What was the price of each per dozen ? Ans. Sherry £2, and claret £3. 22. ^ and B hired a pasture, into which A put 4 horses, and B as many as cost him 18 shillings a week. Afterward B put in 2 addi- tional horses, and found that he must pay 20 shillings a week. At what rate was the pasture hired ? Ans. 30 shiUings per week. 23. A merchant bought 54 gallons Cognac brandy, a,nd ^ certain AFFECTED QUADRATICS. 803 qtrantity of British. For the former he gave ^ as many shillings per gallon as there were gallons of British, and for the latter 4 shillings per gallon less. He sold the mixture at 10 shillings per gallon, and ^ lost £28 16s. by his bargain. What was the price of the Cognac, and the number of gallons of British ? Ans. Cognac 18s. per gallon, and 36 gallons of British. ( 24. What number is that, which being divided by the product of its two digits, the quotient is 2, and if 2*7 be added to it, the digits will be inverted ? Ans» 36. 25. I have a certain number in my mind ; this I multiply by 2i, add 7 to the product, and multiply this sum by 8 times the number ; I now divide by 14, and subtract from the quotient 4 times the num- ber, and obtain 2352. What number is it ? Ans. 42. 26. What two numbers are those whose sum is 100, and the sum of whose square roots is 14 ? Ans, 64 and 36. 27. What niunber is that which if you subtract from 10 and mul- tiply the remainder by the number itself, the product shall be 21 ? . Ans. 7, or 3. 28. What are the two parts of 24 whose product is equal to 36 times their difference? Ans. 10 and 14. 29. What two numbers are those whose sum is 8, and the sum of whose cubes is 152 ? Ans. 3 and 6. 30. Two partners, A and B, gained $140 by trade ; ^'s money was 3 months in trade, and his gain was $60 less than his stock, and ^'s money, which was $50 more than ^'s, was in trade 5 months. What was ^'s stock ? Ans. $100. 31. What two numbers are those, the difference of whose squares is q^, and which being multiplied, respectively, by a and 6, the differ- ence of the products is s^ ? , as'±:bVs*-(a'-b')q^ Ans. r — )-^ ^^. , bs''dtzaVs'-ia^-b'')q'' and 5 — ^^ '—. a —b 32. Into what two parts can a and b each be divided, such that the 304 AFFECTED QUADRATICS. product of one part of a by one part of b shall be p, and the product of the remaining parts ^ ? ^li_ ab-(a-p)±:Vjab~-(g-p)i'-4ahp Ans, -( ^ab + {^-p)^)/\ab-{g-p)Y-iabp ii^rjsr^z^ _ ab-(a p)dzV\ab-(A-p)Y-4aAp ^ ab-\-{h-p)±:V{ab-{§-^'-4.abp 33. During the time that the shadow of a sundial, which shows true time, moves from one o'clock to five, a clock which is too fast a certain number of hours and minutes, strikes a number of strokes = that number of hours and minutes, and it is observed that the number of minutes is less by 41 than the square of the number which the clock strikes at the last time of striking. The clock does not strike twelve during the time. How much is it too fast ? Ans, 3 hours and 23 minutes. 34. A and B engage to reap a field for £4 10s. ; and as A^ alone, could reap it in 9 days, they promise to complete it in 5 days. They found, however, that they were obliged to call in (7, an inferior work- man, to assist them for the 2 last days, in consequence of which B received 3^. 9c?. less than he otherwise would have done. In what time could Bj or (7, alone, reap the field ? Ans. B in 15, and (7 in 18 days. 35. There are three numbers, the difference of whose differences is 8 ; their sum is 41 ; and the sum of their squares 699. Wliat are the numbers? Ans. 7, 11, and 23. 36. There are three numbers, the difference of whose differences is 5 ; their sum is 44 ; and their continued product is 1950. What are the numbers? Ans. 6, 13, and 26. 3*7. A grocer sold 80 pounds of mace, and 100 pounds of cloves for $65 ; but he sold 60 pounds more of cloves for $20 than he did of mace for $10. What was the price of a pound of each ? Ans. Mace 50, and cloves 25 cents a pound. 38. The fore-wheel of a carriage makes 6 revolution more than the hind- wheel in going 120 yards; but if the periphery of each wheel AFFECTED QUADRATICS. 805 be increased one yard, it will make only 4 revolutions more than the hind-wheel in the same space. What is the circumference of each ? Ans. 4 and 5 yards. 39. ^ and B were going to market, the first with cucumbers, and the second with 3 times as many eggs ; and they find that if B gives all his eggs for the cucumbers, A would lose 10 pence, according to the rate at which they were selling. A, therefore, reserves | of his cucumbers, by which B would lose sixpence, according to the same rate. But B^ selling the cucumbers at sixpence apiece, gains upon the whole the price of 6 eggs. What was the number of eggs, and cucumbers, and the price of one of each ? '~Ans. 30 eggs at 1 penny a piece, and 10 cucumbers at 4 pence _ apiece. 40. A person bought a certain number of larks and sparrows for 6 shillings. He gave as many pence per dozen for larks as there were sparrows, and as many pence per score for sparrows as there were larks. If he had bought 10 more of each (the price of the larks remaining the same), and had given as much per dozen for sparrows as he gave per score for larks, they would have cost £l 5s. 5d. What was the number of each ? Ans, 15 larks, and 36 sparrows. 41. A poulterer bought a certain number of ducks and 18 turkeys for $110 : each turkey costing within 1 dollar as much as three ducks. He afterwards bought as many ducks and 5 more, and 20 turkeys, giving one dollar a piece more for each duck and turkey than before ; and found that the value of his former purchase was to the value of the latter one : : 2:3. What was the number of ducks, and the prices of the ducks and turkeys at the first purchase ? Ans. 10 ducks, and the price of a duck $2, and of a turkey |5. 42. A and B put out difierent sums at interest, amounting together to $200. B^s rate of interest was 1 per cent, more than ^'s. At the end of 5 years, B^s accumulated simple interest wanted but $4 to be double of ^*s. At the end of 10 years, A^& principal and interest was-to ^'s as 5:8. What was the sums put out by each, and the rate per cent. ? Ans. A put out $80 at 5 per cent., and B $120 at 6 per cent. 43. When the price of brandy was 3 times the price of British spirit, a merchant made two mixtures of brandy and British spirit, and the prices per gallon were in the ratio of 9 to 10. He afterwards mixed twice as much brandy with the same quantity of British spirit 20 806 AFFECTED QUADRATICS. in each case, and the relative prices. were the same as before. "What was the ratio of the quantities mixed ? Ans. The first mixtm*es were as 3 to 1, and 2 to 1, and the second as 3 to 2, and 1 to 1. 44. A and B traveled on the same road and at the same rate from Huntingdon to London. At the 50th milestone from London, A over- took a drove of geese which were proceeding at the rate of 3 miles in 2 hours, and 2 hours afterwards met a stage-wagon, which was moving at the rate of 9 miles in 4 hours. B overtook the same drove of geese at the 45th milestone, and met the same stage-wagon ex- actly 40 minutes before he came to the 31st milestone. Where was B when A reached London ? Am. 25 miles from London, 45. The difference between the hypothenuse and base of a right- angled triangle is 6, and the difference between the hypothenuse and the perpendicular is 3. What are the sides? Ans. 15, 12, and 9. 46. In digging among some ruins, the workmen found 9 urns, together containing 60 gold coins ; the 2d and 8th containing 8 and 4 respectively. They secreted a certain number of these, greater than the number they left ; which being afterwards recovered, it was found that the number of urns secreted was to the number left as the number of coins secreted was to the number remaining. Now if in- stead of taking the 2d urn, they had carried off the 8th, then the number of coins taken away would have been to the number remain- ing as the square of the number of urns secreted to the difference between that square and 20 times the number of urns remaining. What number of each was secreted ? A71S. 6 urns and 40 coins. 47. Bacchus caught Silenus asleep by the side of a full cask, and seized the opportunity of drinking, which he continued for | of the time that Silenus would have taken to empty the whole cask. , After that Silenus awoke, and drank what Bacchus had left. Had they drank both together, it would have been emptied 2 hours sooner, and Bacchus would have drank only ^ what he left Silenus. How long would it have taken each to have emptied the cask separately ? Ans. Silenus in 3 hours, and Bacchus in 6. CHAPTER XII. / CUBIC EQUATIONS. (328«) Cubic Equations may be divided into Pure Cuhics^ Incomplete Cuhics, and Affected Cubics, (3 3 9.) A pure cubic equation is one in which the unknown quantity is contained in but one term, and whose exponent is 3, or a fraction which when reduced to its lowest terms the numerator is 3 ; as, 3. S. ax^=dbb; ax^=:±:b* ax^=^b,&G, (330*) The solution of a pure cubic presents no difficulty. PROBLEM. Given 3a;' =24 to find x. SOLUTION. 3ic''=24, a;''— 8=0, (a;~2)(a;' + 2a:4-4)=0. Whence we get ar — 2 = 0, or ic" 4- 2a; + 4=0, whose solutions give a;=2, or — Izfc 1^— 3. If but one value of the unknown quantity were required, we might in the equation a;' =8 merely take the cube root of both sides, and obtain a:=2. EXAMPLES. 1. Given a;' = 27 to find x. Ans, a:=3, or . 2 2. Given ia;''=32 to find x. Ans, a;=4, or — 2±2|/— 3. 3* Given a;' =4 to find x. Ans. a;=V4, or -iV4±|/~fV2. 308 CUBIC EQUATIONS. 4. Given ) ^ K'^Z^" '^\ to find the values of a; and y. { X — y =00 ) Ans, i ~ ' , , . (y=2, or — l±f/—3 5. !• Given -j , ' '/ ' \ to find the values of a? and v. ( iry''=384 ) ^ , U=24, ^715. '-=24,or-12±12i/-3 or — 2db2K— 3 (331.) An incomplete cubic equation is one in which the uaiknown quantity is contained in but two terms : the exponent of one being 3, and the other 1 or 2 ; or, the exponent of one being 3 times a proper fipaction whose numerator is 1, and which is either the exponent or half the exponent of the other term ; as, ax^-±.hx=iizc ; ax^:hbx^=±:c; ax±:bxi=dcc ; ax±bx^=d:c ; SsG. Remaek. — 1^0 general practicable method of obtaining the values of the un- known quantity in incomplete or affected cubics has as yet been discovered, and we are, therefore, compelled to resort to what are called specicU, tentativef or approximative methods. Some of these methods are here set forth. PROBLEM. Given a;'-i-3a;=14 to find the values of w, SOLUTION. a:' + 3ar=14 (1). X* + 3a;'' = 1 4a; (2) = (1) x a;. [members. a;* + '7a;''=4a;'' + 14a; ^ (3)=(2) with 4a;'' added to boHi a.*4.'ra;'' + (l)'=4a;''+14a; + (f)'. (4)=(3) square completed, a;" + f = 2a; + 1, or -2a;-|, (5) = Vjl). .'. a;=:2 ; or a;''4-2a;= — *? a;=— 1±1/— 6. EXAMPLES. 1. Given x^—^x=-~Q to find x, Ans. a;=l, 2, or —3. 2. Given a;'— 32a;=: —24 to find x. Ans. x=z—Q^ or 3 ±4/6. 3. Given a;'— 22a;=24 to find x. Ans. a;=— 4, or 2±V^10, 4. Given a;' + 6a;=:88 to find x. Ans. a;=4, or —2dtsV'^. 5» Given a;'+6a;=45 to find x. Ans. x=3, or . CUBIC EQUATIONS. 309 6* Given ic'— 13a;=— 12 to find x. Ans. x=l, 3, or —4. 7. Givena;' + 48a;=:104 to find x. Ans, a;=2,or —ldzV—51. 8. Given x'—6x=9 to find x. Ans, x=3^ or . 9. Given a; + Tari-=22 to find x, Ans, a:=8, or 29rh'7'/— 10. 10. Given a;'+3a;=14 to find x, Ans, x—2, or —iztV^, 11. Given x^ =14 to find x. Ans. x= — , or . 3x ^ 3' 8 12. Given a;"— 2a;=: — 4 to find x, Ans, a;=:— 2, or Idbt^ — 1. 13. Given 5a;''4-2a;=44 to find x, Ans. x=2, or — . 6 14. Given a;'+108a;=665 to find x. Ans, x=5, or^^^^^~^ 2 15. Given x^—S9x= — 10 to find x, Ans, x=2, 5, or —7. 16. Given a;"— 49a;=120 to find x, Ans. x=S, —3, or —5. 17. Given a;'+12ar=-63 to find x, Ans, a;=— 3, or^'^^^~^ 2 18. Given a;''— 21a;=— 344 to find x, Ans. a;=— 8, or 4±3f"^. 19. Given a;'— 6a;=40 to find x. Ans, a;=4, or —2=1:^—6. 20. Given x'-1^x=-^2Q0^ to find x, Ans, .=^7,or-^^ ' 2 (332.) Sometimes the factors of an equation are very apparent, and then the root may be obtained as in the solution to the following PROBLEM. Given a;"— 3a;— 2=0 to find the value of a?. SOUJTION. «'— 3a;— 2=0 x^~x=2x+2 a;(a;'-l)=2(a;4-l) 810 CUBIC EQUATIOKS. Since a; 4- 1 is a factor of both members of the equation, the equatioD will be satisfied by putting a;+ 1 =0 Whence, x= — l. To get the other values of x divide both members by a;+l. And then, x^—x=2 x=2, or — 1. KXAMPtES. I, Given a;'— 5a;=— 4 to find x. Ans, a;=l, or 2. Given a;'— 2a;=l to find x, Ans. a;= — 1, or 3* Given re"— 2a;= — 1 to find x. Ans. x=l, or — 2 2 ' l±V/5 2 4* Given ic^— 8aj= — 8 to find x. Ans, x=2, or — Irhi^S. 2 5* Given a;--l=2H — j- to find a?. -4/is. a;=l, 1, or 4. arg- x' -{-Sx 7 6t Given 7^=1 ^ ^^ ^' ^^' ^=3, —2, or —3. a; + 2+ — a; 7. Given 3x^—^x^=1x^3 to find a:. Ans. x=^, 3, or —1. 8. Givena;' + a;' + a; + l = 10a; + 10 tofinda;. Ans. x=S, —1, —3. 9. Given a;'— a;' --2a; =—2 to find x. Ans. a?=l, or ±4^2. 10. Given (a:''4-2ar)(a;+4)=2— (a; + 4) to find x. Ans. x=—2y or — 2±i^3. i3x-—=v'-v ) 11, Given •< y ^ ^' S to find a; and y. ( y+a;=4. ) . j a;=3, or 1. (333*) It sometimes happens that the coefficients of the unknown quantity have such relations that the equation may be reduced ac- cording to the method adopted in the solution of the following PROBLEM. Givena;'— 6aj' + lla;=r6 to find the values of x. CUBIC EQUATIONS. 311 SOLUTION. aj'-6a;''4-lla;-6=0 x'-6x' + Ux'-ex=0 x'—6x'-^9x'+2x''—6x=0 (x'-'SxY-{-2(x'-Sx)=0 (a) (a5*-3a;)' + 2(a5'-3a;) + l=l x''-Bx+l = ±l x^—Sx=0, or ■ -2, .-. x'=Sx; or aj^— 3aj= 2 05=3 05=— ^r— =2,orl. In equation (a) put y=zx^—Sx, and we shall have y'» + 2y=0 y'» + 2y + l=l 2/=0, or -2, •. a;'— 3£c=0 ; or —2, the same as before. Remark. — Judicious substitution often saves figures, and also often reveala relations which might not otherwise he so apparent Since a:"— 3a? is a factor of (a), we have immediately a;'— 3a;=0, and by dividing by this, a;"— 3a:+2=0, the same as before. EXAMPLES. 1. Given x''—6x' 4- 6a; + 12=0 to find x. Ans, a;= — 1, 3, or 4. 2. Given ar^— 2a;''— a; + 2-=0 to find x. Am. x=l, 2, or —1. 3. Given aj'— Ga^' -f 12a:— 9=0 to find x. . „ SdtV-3 Ans. a;=3, or , 4. Given a;' — 2a;''— 5a; + 6=0 to find ar. ^w«. a;=l, 3,or — 2. 5« Given x^ + 2a;''— 6a;— 6=0 to find x, Ans. a;=2, —1, or —3. 6* Given a;'' + 6a;'' + lla;-!-6=0 to find x. Ans. a;= — 1, —2, or —8. 7t Given a;'— 8a;'' + 19a;— 12=0 to find x. Ans. a;=l, 3, or 4. 8. Given a;' + 2aa;'' + 5a''a;+4a' =0 to find x. —a±aV— 15 Ans. x=—aj or . 2 (334.) The following examples may be solved according to the artifices employed in quadratics. 812 CUBIC EQUATIONS. PROBLEM . Given a;'— 3a;' + 3a; =9 to find the values of x. \ SOLUTION. a;'— 3a;*4-3a;=9 (1). a.«_3a;a 4- Sx— 1=8 (2) =(1) with — 1 added to both members. a;-l=2 (3)=V(2). a;=3 To obtain the other values find the other factor of a;'— 3aj''4-3a;— 9, SB— 3 being one. That factor is ar* + 3 ; hence, we have a;'' + 3=0. 3;^^ =— 3. a;=±4/^. The best method of solving this equation is by factoring, for its factors are very apparent. a;'— 3a;'' + 3a;— 9=0. a;Xa;— 3)+3(a;— 3)=0. whence, (a;''+3)(a;-3)=0. a;— 3=0. and a;' + 3=0. whence, a;=3, or ±4/— 3. Remaek. — ^The examples which follow are not all of the character of the one whose solution has just been given, but embody different principles with which the student is supposed to be familiar. EXAMPLES. 1, Given a;' + 6a;'+12a;=19 to find x. . , -7 + 3*^3 Ans. x=lj or . 2, Given >|/a;'' + 8=*/125— 6a;' — 12a; tofinda;. J Q —9± 51^-3 Ans. a;=3, or . 3. Given Va;"— a'=N/3aa;'— 3a»a; + 86tofinda;. Ans, a;=a+2V6. J ^. Va + x Va+x Vx ^ ^ , . ■ acl 4i Given 1 = — to find x, Ans. x: 5. Given a;'— 6a;' + 12a; =8 to find x. Ans. a;=2, 2, 2. 6* Given a;^— 16a;' + 76a;= 12 6 to find x. Ans. x=z5j 5, 6. CUBIC EQUATIONS. BIB HE NKLE'S METHOD.* (335.) The following tentative process for solving affected and in- jomplete cubics, has never before been given in a work upon algebra. PROBLEM. 1. Given a;"— 8«'' + lla; + 20=0 to find the values of a;. SOLUTION. a;3_8a;3^]l:c_^20 = 0. a;*_8a;' + lla;'' + 20a;=0. {x^—4xy—5x^ + 20x=0. {A), (x''-4:Xy-4:{x'-4:X)+4=X'-4:X-^i, (B). x^-'4x—2=:x—2, or 2—x, x^=5x\ or x^—Bx= =4. x=5 x- =-— =4,01-1, This solution vsdll need some explanation. Every affected cubic equation may be put into the form expressed by (A) that is, made to cjonsist of the square of a binomial followed by two terms. If these two terms contain as a factor the first power of the binomial, the equation may be solved as in (B) ; but if not, we may proceed as in the above example. Let us resume the equation {x''—4xY—bx^ + 20a;=0. Let us see whether x"^ is not the first term of a binomial square, which being added- to both members of this equation will render them perfect squares. If x"^ is added to both members, we see that the coefficient of the second term in {B) is — 4 ; considering {x^—4xy as the first term ; and this coeflScient shows that the third term in the first member of (B) must be 4, and therefore, the third term in the second member must also be 4 ; but if the first and the last term of a binomial square are x^ and 4, the middle term must be either + 4ar, or —4x. From which we see that the square of the binomial of which x^ is the first term must, in this case, be x'^zh4x + 4. Since, 4 is in both members, it follows that if a;'' ± 4a: be added to — 5a;'' + 20a;, and the result should be — 4(a;^— 4a;), or —4a;'' + 16^, that a;' ± 4a; 4- 4 is the proper square to add. As we have already addei a.-*, it only remains for us to see whether 4a; added to or subtracted from + 20aJ * This method of solvinp^ Cubic Equations, which is also apphed to the solu- tion of Biquadratics was discovered by Prof. W. D. Henkle. 1, therefore, have assumed the respousibilitj of calling it the "Henkle Method." — J. F. S. 314 CUBIC EQUATIONS. makes + 16a;. It will be seen that 4x subtracted, or what is the same —4x added makes +20a; become -f 16 a;, therefore, a;'— 4a; + 4 is the proper square to add. The student in passing from (A) to (B) should put his trial work one side. 4 I— 1 ^ I— 1 ex\. 1 S- 5i 1 O QD l-H J II 1 II d + l-H 1 « S 54 1> 1 i-T I— 1 + o 30« Givena;*— iVa;'— 20a;-6=0tofinda;. _ ^ ., Ans. a;=2db|/7, or -2^1^-2. 31. Givena;*— 65a;'— 30a;+604=0tofinda;. Ans. a:=3, Y, —4, or —6. 32. Givena;*— 3a;'— 4a;— 3=0tofinda;. _ 1±V13 -l±i/-3 Ans. X— — , or . , 33. Givena;*— 27a;' + 14a; + 120=:0 tofinda:. Ans. a;=3, 4, —2, or —6. 34. Given a;*+6a;'— 24a;— 16=0 to find x. Ans. a;=2, —2, or — 3±|/5. 35. Given a;* - 45a;' - 40a; + 84 = to find x. Ans. a;=l, 1, —2, or —6. 6 rT 3 /^ 3 v/a;' 36. Given 4/ -, + 4/ -= to find x. f X* ^ X X 21 Ans. a;=itl, or ^-5-- 8 37. Given a;*/l +^) _(3a;' + a;) = 70 to find x. — IdbV— 261 Ans. a;=3, —3^, or , ^2 BIQUADRATIC EQUATIONS. 88« Given ^ H = =— - to find x. Vx* -9x' ^^ 2a; Ans. a;=±6, or ±^^'16661. 39. Given x^—2x+ 4=2^x^ — 1 to find «?. _ Ans. x=4:±: ^/6, or ± 1^—2. 40. Given aj— 2V'a; + 2=l+Vic'— 3£c4-2 to find a;. ^ Ans, aj=9±4r7, or — - — . (346.) An equation of the form ax*zthx^dzcx'±:bx + a=0, is called a recurring equation of the fourth degree, or a biquadratic recurring equation, PBOBLEM. (347.) Given ax* + bx''+c3^ + bx-{-a=z0 (1) to find the values of a?. SOL UTION. By multiplying by 4a we get 4a^x* + 4abx^ + 4acx'' + 4ahx+W=zO (^)=(1) X 4a. {2ax'+bxY + (4ac- b')x^ 4- 4abx + 4a''=0 (-B) (2aa;''+6a;)'* + 4a(2aa;'' + 6a;) +4a''=(8a'' + 6'-4ac)a;» (0) 2ax^ + bx+2a=dixV8a^ + b^—'iac 2aa;' + (6:f j/8a"+6'— 4ac)rc=:— 2a ifSa'+S— 4ac— 6±i/— 8a' + 26''— 4ac±26V'8a'+6'— 4atf a?= :; . 4a It may be perceived that the coefficient of £c' in the term that is added to both members of (A), and which makes the first member a perfect square, is a function of known terms. This coefficient is equal to the coefficient of «* in (B) subtracted from twice the coefficient x* in (A), or confining the explanation lo the primitive equation, is equal to 8 times the square of the coefficient of x*j plus the square of the coefficient of a;', minus 4 times the pro- duct of the coefficient of x* by the coefficient of a;^ If the primitive equation had been multiplied by a instead of 4a, the coefficient of x^ in the term added would have been just J as much, or BIQUADRATIC EQUATIONS. 333 Node. — This mode of treating recurring equations of the fourth degree is original. The method, it will be perceived, is a general one. The discovery was afterward independently made by M. C. Stevens, who> at our request, has furnished the following concisely written rule. In applying it, the original equation* must not be multiplied by 4, as was done in our solution, in order to avoid fractions. RULE. Divide by the coefficient of c»* and transpose the term containing a?» ; then a^dd to each memlefr 2x'^+ the square of half the term containing x, or —27?+ the same, according as the second and the fourth terms have like or unlike signs. Extract the square root and the equation reduces to the second degree. (348.) The following is the plan usually given for the solution of a recurring equation of the fourth degree : PROBLEM. Given ax*-\-hx^-{-cx^-\-hX'\-a=iO to find x, SOLUTION, oaj' + Ja^ + c-l— + ^=0 X a?" « a , h aar*+-5 + 6a;-|--+c=0 x^ X Now put x-\ — =iy X Then x^+-=y^—2 X Substituting a(y'* — 2)+6y4-c=0 yz=. ! X'\ — = — _ X 2« 2aar' + (Jq^l^Sa'' + 5'— 4ac)a:= — 2a ±|/8a' + 6'— 4ac - 6 ± |/— 8a" 4- 26» - 4ac ± 264/8a'' + 6»— 4ac 4a 384 BIQUADRATIC EQUATIONS. ' PROBLEM. Given 10a;*— 9a;'— 20a:''— 9a; + 10 = to find the values of «. SOLUTION 1. 10a;*— 9a;'— 203;"— 9a; 4- 10=0, 400a;* -SeOa;" — 800a;'— 3 60a; + 400=0, (20a;''-9a;)''— 881a;'— 360a; + 400=0, (20a;'— 9a;)'+40(20a;'— 9a;)+400=1681a;'. 20a;'— 9a;+20=±41a;, .-. 20a;'-50ar= — 20; or 20a;' + 32a;=— 20, 2a;'— 5a;=— 2 5a;' + 8a;=— 6, 5±3 ^ , -4±3f^i:T a;=:'-^=2, or ^ x= . SOLUTION 2. 10a;*— 9a;'— 20a;' — 9a;+ 10=0, 10a;'-9a;'-20-- + i^=0, X aj' 10(.'+J)-9(x+l)=20, put «+-=y, then a;«+-=:y'— 2, a;' ^ ' substituting lOy'— 20— 9y=20, 10y'-9y=40, 9±41 6 8 ^=-20r=T'^^-'5' .15 8 .• . x-i — =-, or — , a; 2' 5' and 2a;'— 5a;=— 2; or 6a;' + 8a; =—6, -5^3_ -4±3i^"=T — _2,or^ x= . x= EXAMPLES. 1, Given a;* + 24a;'— 114a;'— 24a; + 1=0 to find x. Ans. x=2:±zV5, or —14± 1/197. 2. Given a;* + 6a;' + 2a;' + 6a; + 1 =0 to find x. Ans, a;=±|/-l, or ~^'^^^\ BIQUADRATIC EQUATIONS. 385 3* Given x*+x'-{-x^+x-\-l =0 to find x. . - -h - -f ^ 4» Given 2a;*— 4a;'— 6ar'— 4a; + 2=0 to find ar. Ans, x=: . 5. Given 3a;* + 2a;' + 43;' + 2a; + 3 = to find x. Ans, x= . 6. Given 4a;* + 3a;' — 8a;' — 3a; + 4 = to find x. ■Z±V1% Ans. a;=r±l, or 7. Given 7-- — r.=* to find x. (1 + a;)* Ans. a;=l=fcf/3dby3|/i/3db2. (349*) There are other biquadratics that are not recurring which are susceptible of a similar solution, but the coefficient of a;' must be decided by trial. EXAMPLE^. 1, Givena;*— 2a;'— Ya;*— 8a;+16=0lofinda;. . ^ ^ -3dti/3Y Ans. a;=l, 4, or . 2, Given a;*-f »'— a;'-t-2a;4-4=0 to find x, A -1^ Ans. x= 3, Given 4a;*4-8a;»-89a;'4-28a;+49=0 to find x. -ldt:f/21±^-10qz2V'21 Ans. x= -2- , 4 Ans. x=l, 3^, or — . 4. Given 2a;* + 24a;'-315a;' + 216a; + 162=0 to find a;. Ans. x=—3dz^V94±:W^^=p^V94, 5, Given a;*— 12a;' + 4'7a;'— 72a;+36=0 to find x. Ans. a;=l, 2, 3, or 6. 6. Given a;*— 9a;' + 15a;'— 2 'ra; + 9=0 to find x. , 9doSV5±^1S±5W5 Ans, x= . 886 BIQtJADEATiC EQUATIONS. 7. Given x*i-B6x^—400x^—Sl68x + 1l44=0 to find x. Ans. a;= — 9±V/137±i/306q=18f/137. 8. Given f^«--= -:= to find x. Ans. a;= 1, 16, or '^^^-'^^ ^ f^aj-2 2 (350.) There is a class of problems which may be solved after the manner given in the solution to the following PRO B LE M. Given a;* + 2a;»— Yar*— 8a;=--12 to find the value of a?. SOLUTION. x*-}-2x' — 1x''—Sx=—12, (x^-\-xY—8x'—8x=^12, {x' + xy-8(x' + x)=^12, {x'-\-xy-8{x^ + x) + 16=4, x^ + x—4=±2, x^-{-x=6^0T 2j a;=2, —3, 1, or —2. EXAMPLES. !• Given x*-\-2x^—3x^—4x-{-4=0 to find x. Ans. a?=l, 1, —2, or —2. 2. Given x*—l 2x^ + 60a;' —84a; + 49 = to find x. Ans. a;=3±j/2, or 8±V'2. 3. Given a;*— 10a;* + 3 6a;'— 60a; +24=0 to find x. Ans. x=l, 2, 3, or 4. 4. Givena;* + 2a;'— 13a;»— 14a;4-24=0 tofinda;. Ans. ar=l, 3, —2, or —4. 5. Given x* + 12a;' + 64a;' + 108a; + 81 =0 to find x. Ans. x=—S, —3, —3, or —3. 6. Given x* -{-2qx'' + 3q^x^ + 2q'x=zr* to find x. Ans. .- -9±V-Sg^±^Vr*+q' 2 7. Given a;*— 14a;' + 6 la;'— 84a; + 36=0 to find x. Ans. a;=l, 1, 6, or 6. 8. Given 4a;*+|=4a;» + 33 to find x. A o ^, l±V^-43 Ans. x=2, —1^, or . BIQUADRATIC EQUATIONS. 887 9, Given a:*— 2a;'— 2a;' + 3a;=108 to find x. Ans, x=4:, —3, or . 10. Givena;*— 2a;'+a;=30 tofindar. Ans, x=S, —2, or . 11. Given a;*— 6a?» + 6a;' + 12a;=:60 to find x. Ans. a;=6, ~2, or . ' 2 12. Given a:*— 8a:* + 10a;' + 24a; + 6=0 to find x, Ans. x=5, —1, or ^zt^B, 13. Given a;*— 2a;'4-a;=132 to find x. Ans. a; =4, —3, or ■ . 2 II. Given a;*— 2aa;»+(a*— 2)a;' + 2aa;=a' to find x. Ans. xz=~db^^ + l±Vl 2^4 15. Given a?*— Boa;' + 8a V 4-3 2a'a;= 9a* to find x. Ans. a;=2azhat/3, or 2aitaV'13. -» ^. 18 81-a;» a;'-65 ^ , 16. Given — + — = — to find x. x^ 9x 12 Ans. a;=9, —9, —4, or —4. 17. Given a;*— 2a;'— 25a;' + 26a; +120=0 to find x. Ans. a;=3, 5, —2, or —4. 18. Given a;*— 12a;' + 44a;'— 48a;=9009 to find x. Ans. x=lS, —7, or 3±3f^— 10. 19. Given ar'--2a;2 +^a;— |/.^z=6 to find x. Ans. x=l, 4, or . 2 20. Givena;*— 6a;' + 13a;»— i2a;=5 tofinda;. . 3±4/l3 3±i/^iri Ans. x= — - — , or . 21. Given a;*— 8aa;' + 8aV + 32a»a;=fl? to find x. Ans. a;=2a±4/8a'±t/l6a*+2,or3±m. ( y=2, 4, or 3qpV21. \ ic=5, 1 or , Ans. j I7rp64/::^ Cy=3, —15, or — 6±i/— 2. ] a;y=:2, ) ^ (y=l, 2, -1, or -2. ^ _ a f t<^ fijid ^ and y- icy: •s ^ ic + y '^ 3aj '> to find a: and y. |aJ2^-(a;+y)=54, ) ^,... ■i^= 6, -4X6, or -4J, (y=12, -9, 12, or -9. (6a:'4-2y'=5ary+12, ) ^ ^ , . 1 « . o a o 3 ^ M- to find re and y. Ans i^=±2,&c., ^^^- |y=i3,&c C a;V+«5y'=30, ^ < 1 1 _ V to find .r and y. i ^^v -" S ^"*-U=2, 3. -e.orl. 842 BIQUADRATIC EQUATIONS. 17. Given } f'^x^y-''^ x^' to find X and y. j a:=4, -2, or lil^A^ 18. Oive. j (;;;,)|;r^^-;^ [ to and . and y, j a:=ll, -1, or eii^^-STld, ^** ( y=l, -11, or _61±4/-3Y16. 19. Given J-StVUo^-Oy-ie.^., ? ^fi,^,,,,^. ( 6a?=4 + 25y^ ) j a;=l, 1, -/o, or -^^, 20. Given j "•"^■'^^Zjg^ f to find . and ,. Ans. a;=5, 4, or 8dbif/-Ai«, or 8TJ4/^^^^. j ^=5, 4, (^=4,6, y+ 21. Given "a: X x^ X __64 ^3 2»/y y\ to find a; and y. An, i ^=*' -WS N> O' W. ^ ^ , -97± 1/6045 aj=6, J, or ^^^ , Ans, ^ y=6, 160, or 58 1682 9'7=pf6045 ^'<^^^Ay'-Z=Zr'^\-^''-'y ( a;=400, 225, or 12i(— l-5f^— 23, ^^** \ .r.r. OH. 125 ±25*/^=^ i y=500, — 3*75, or . BIQUADRATIC EQUATIONS. 343 24* Given \ . * , L f to find x and y. Ans. ^=2, 1, ^— , 25. Given ] 'ft'^'rf '.-^f;' ^,, [ to find. and,. -4»5. 171 65=F^1114 a?=3,.— -— -, or 133' 26 34 — 9±3'^/lll4 ^'^''m'"^ 26 (354*) Biquadratic equations which do not admit of solution by any of the previous methods may frequently be solved as cubics by the addition of a binomial squared to both members. PROBLEM. Given a;* + 4a;'— a;''--16a;=12 to find the values of x, SOLUTION. a;* + 4a;'-a;''-16a?=:12 (a;'' + 2a;)''-6a;'-16a;=12 (aj»4-2a:)'— 4(a;' + 2a!)+4=a?' + 8a;4-16. a?" + 2a;— 2=a: + 4, or — a;--4, .*. a;^4-a;=:6 ; or 3;" + 3a; =—2 --1='=5 ^ „ -3±1 , ^ x= — - — =2,-3 x=: — - — =— 1, — 2 EXAMPLES. 1, Given a;*— 6aj' + 12ar'— 10a; + 3=0 to find x. Ans, x—ly 1, 1, or 3. 2, Given a;*-~4a;' — 19a;' + 46a; + 120=0 to find x. Ans. a;=4, 5, —2, or —3. 3. Given a;* + 3a;' + a;" --3a; =2 to find x. Ans. «•=!, —1, —1, or —2. ip Given x* — Qx^ + 5x^ + 2a;' = 1 to find x. Ans. x=:5, —1, or 1± V^Tl, 344 BIQUADRATIC EQUATIONS. 5* Given x*—4:X^—8x+S2=0 to find x. Ans, x=2, 4, or —ldtV^» 6. Given a;*— 9a;' + 30a;''— 46a? + 24=0 to find x. Ans. x=l, 4, or 2±/^. 7. Given a;*— a;' + 2a;'' + a; =3 to find x. Ans, a;=±r, or ^^^'^^ 8. Given 6a;*— 43a;» + 10'7a;»— 108a; + 36=0 to find x. Ans. a;=|, li 2, or 3. 9. Given a?*+a;'— 16a;'' — 4a; +48=0 to find x. Ans. a;=2, 3, —2, or —4. 10. Given a;*— a;'— lla;'' + 9a;+18=0 to find x. Ans. xz=z2^ 3, —1, or —3. 11. Given a;*— 8a;» + 14a;' + 4a;=8 tofinda;. Ans. a;=3±V6, or litV3. 12. Given a;*-12a;' + 48a;'— 68a;+16=0tofind x. Ans. a?=3, 5, or 2=fcV'3, 13. Given 2a;*— 2a;'— 2a;' H + -=0 to find x. 2 8 1 ±i/2 Ans. a;-±i|/3, or— -— . 14. Given a;* + a;'— 29a;'— 9a;+180 = to find x. Ans. a;=3, 4, —3, or —5. 15. Given a;*— 4a;'— 29a;' + 156a;=180 to find x. Ans. a;=2, 3, 5, or —6. 16. Given a;* + 29a;' + 287a;' + 1147a;+1560=0 to find a;. Ans. a;= — 3, —6, —8, or —13. t<^ n- 4 n 3 . ^Sa;' 27^; 81 ^ , 17. Given a;*— 9a;' + — — +-— =— to find x. 4 2 4 Ans. a;=li li, or 3 ±31^2. 18. Givena;*— 8a;' + 23a;'— 64a;+120=0 tofinda:. Ans. a;=3, 5, or ±2V^2. 19. Given a;*— 16a;' + '79a;'— 140a; + 68=0 to find x. Ans. xz=2±V\ or 6 iV'Y, BIQUADRATIC EQUATIONS. 345 20. Given x*-5x'—5x^ + 45x=SQ to find x. Ans. x=l, 3, 4, or —3. 21. Given a;*-3a;'— 15a;' + 49a;=12 to find x. Ans. x=S, — 4, or2±V3. 22. Given x*-\-x^—x''—5x+4:—0 to find x. -3±1/-'7 Ans. ar=l, 1, or . 23. Given x* -{-x^—x* + 1 Oa; + 4 = to find x. Ans. X— , orlif"— 3. 24. Given a;* — 7a;' + Qx" + 27a;=:54 to find x. Ans, x=Z, 3, 3, or —2. 25. Givenar* + 3ar'— 7a;'— 27a;=18tofindar. Ans. a;=3, —1, —2, or —3. 26. Given 6a;*— 25a;' + 26a;' 4- 4a; =8 to find x, Ans. a;=|, 2, 2, or -J. 27. Given 8a;*— 38a;» + 49a;'— 22a; + 3=0 to find x. Ans. x=z\^ ^,1, or 3. 28. Given a;* — 9a;' + 1 7a;' + 27a;= 60 to find x. Ans. a; =4, 6, or ±VS, 29. Given a;* + a;' — 24a?' + 43a; = 2 1 to find x. _ ^ , o -5±|/53 Ans. a;=I, 3, or , i 30. Given a;* -fa;' 4- a;' — 120a; =100 to find a;. -5±5f/-5 Ans. a;=2±2V^2, or . 2 3L Given x^ -{■ x^ -\- x" -\- 141a;= 100 to find x. __ 5 4-4^41 Ans. x= f-^— , or 2±4/-21. 32. Given a;*-f-a;' — 19a;' — 49a;=30 to find x, Ans. x=5, —1, —2, or —3. 33. Givena;*— a;'— 19a;' + 49a;=30 to find ar. Ans. a;=l, 2, 3, or —3. 34. Given x*—^^x'-\-^x^—\^x + ^ — to find x. Ans. a;=|^, i, 1, or 3. 35. Given a;*-38a;' + 210a;'4-5.S8a; + 289=0 to find x. Ans. x=-l, -l,or 20±V'lll. 846 BIQUADRATIC EQUATIONS. 36. Given4fl?*— 14a;'— 6a;» + 31ar+6=0tofinda?. Ans. x=2j 8, or . 4 87. Given a;*— 6a;'~68a;'~114a;=ll to find x. Ans. x=^±^VS±j/l1±si^VS, (355*) This method of solution is applicable to all afiected biquadratic equations, as is shown by the solution of the following literal equation. But it is not always practicable, as the quantity to be added is frequently of such a character that it can not be easily found. PROBLEM. Given x*'-{a+h + c+d)x' + (ab + ac + ad-\-bc + bd-^cd)x^-'{abc-^ abd + acd + hcd)x + dbcd=0 to find x. SOLUTION. 4a;*— 4 (a + 6 + c + d)x'' + 4(a6 + ac -\- ad + bc-^ bd + cd)x^— 4 (abc + abd 4- cicd + bcd)x + 4a6cc?= 0, [2x^—(a'{-b-{-c-\-d)xy—(a''—2ab—2ac—2ad + b^—2bc—2bd + c' — 2cd + d')x'' —4:{abc + abd + acd + bcd)x + 4a6cc? = 0, [2a;'— (a + 6 + c+c?)a;]'+ 2{ab + cd) {2x^ — {a ■^b + c+d)x^ -f a'i' + 2abcd-{-c^d''={a-\-b—c^dYx''—2{a + b^c—d)(ab—cd)x-\- a''b^—2abcd + c^d\ .-. 2a;'— (a+b-\-c+d)x + ab-\-cd = (a-{-b—c—d)x— {ab—cd)j or ab-'Cd—(a + b—c—d)x, 23^—2{a + b)xz=—2ab; or 2x^—2{c + d)x=-'2cd, a-\-b±(a—b) , c-^d±(c—d) _ x= -^ ^ =a, or X— -> ^—Cy or a. A close inspection of this solution will enable the student to see what relation the coefl5cient of a;', in the added square, bears to the values of x, as finally ascertained. MISCELLANEOUS QUESTIONS.* 1. A vintner draws a certain quantity of wine out of a full vessel that holds 256 gallons ; and then filling the vessel with water, draws * These questions should be solved without resorting to the method just given. BIQUADEATIC EQUATIONS. 847 off the same quantity of liquor as before, and so on, for four draughts, when there were only 81 gallons of pure wine left. How much wine did he draw each time ? Ans, 64, 48, 36, and 2*7 gallons. 2. An upholsterer has 2 square carpets divided into square yards by the lines of the pattern. Now, he observes that if he subtracts from the number of squares in the smaller carpet, the number of yards in the side of the other, the square of the remainder will exceed the difference of the number of squares in the smaller carpet, and the number of yards in its side, by 88. Also, the difference of the lengths of the sides of the carpets is 6 feet What is the size of each carpet ? Ans. 16 and 36 square yards. 3. A man, playing at hazard, won at the first throw as much money as he had in his pocket ; at the second throw he won 5 shil- lings more than the square root of what he then had ; at the third throT^ he won the square of all he then had, and then he had £112 I6s» How much had he at first ? Ans. 18, or 24^ shillings. CHAPTER Xiy. HIGHER EdUATIONS. (356*) Equations of the fifth degree^ formerly called sur solid equations and equations of higher degrees, have not as yet been found to be susceptible of any general solution. Particular examples, how- ever, frequently occur that may be reduced by known methods. It is the object of this chapter to present some of them. PROBLEM 1. Given «'=«' to find the values of x, SOLFTION I. (a:«+a«)(a:'-a')=0. (a?+a)(ar'— aa; + a") =0, a:'— aa: + a'=0, x=—- . x+a=zOj x=—a, x'-a'=0, {x—a){a^+ax-\-a')=0, x= . a?— a=0. x=a. a±aV—3 —a^aV—i xz=zay —a, , or - 2 HIGHEB EQUATIONS. 849 SOLUTION II. —a + aVl+aV- -10— 2f6 "" 4 » -10+21^5 "" 4 — a+af5— at/- -10-21^6 "^ 4 _ — a— af'S— af- » -10 + 21/6 EXAMPLE S- 1. Given af'=l to find a?. ^Tis. x=l, -1, ^ — -, or . 2* Given a?'=l to find a;. -ldb|/5±V-10=F2*/6 Ans. a;=l, or » 4 3* Given x*=.a* to find x. Am, xzzzdca, ±aV—i, ±aV~l, or ±aV—V'^. . 4* Given jr*=l to find x. Ans. x=±l, ±V^, ±V^,or ±V-V'^. 5, Given a!"+a'=:0 to find x. a±aV5±aV—10±2V5 Ans. «=— a, or . 4 6t Given a;' + 1 =0 to find x. ^ . , l±f^6±l/-10±2f^-5 Ans. a;= — 1, or . HIGHEB EQUATIONS. 7« Given aj*=a' to find x. tn Atis. " x=a, ah, ab*y ah^, ab^^ ah*, ah', aV, or ah^, in which 8. Given aj"=:a" to find a;. Am, X=z- ■a±aV6±aV—lQ^zV6 a±aV5±aV-\0±2V6 (357.) The following examples may be solved by a combination of the principles already learned. Some of them are inserted for the first time in an American work, and will be found to be the most diffi- cult algebraic problems that have ever been published in any work upon this subject. Many of them, however, will be found to be easy of solution. Some of the values in some of the examples are omitted. PROS LEM. Given 2i/y''' + 6y— 2y''~-46y*— 6y +6'=0 to find four values of «. SOLFTIO 2f^y"4-6y-2y'-46/-6y + 6'=0, 2yWfW-\-h'=2y' + 4fty* + ^V, y' + 2fVfTh'-¥ f + h'= 4.f 4-46/ + 6y, y'+V 7Tl^ =±{2f + hy) V'f + h'^f + hy, y' + b'=y'-\-2hy* + bY, 2hy' + hY=h\ 2y' + hf=h\ 4y'=26, or — 46, 2y=±i/26, ±W^, (^), 362 HIGHER EQUATIONS. By taking the minus value in (A) we should obtain An equation of the sixth degree, therefore, the given equation is of the tenth degree. EXAMPLES. 1. Given 2x^{x^-\-a^)^=2x\x-\-2a)-\-a\x—a) to &id two of the five values of a;. Ans. x=^aj or —a. 2. Given a:'— 4a;' =621 to find all the values of a?. A o z,7^ --3±3f^II^ V23±l/2aK'^ Ans. x=3y — V23, , or ^ . 3. Given a;"— 6a:'=16 to find all the values of x, A o s/o -2=F2>/^ V2:FV2V/^ Ans. x=2, - V2, , or ^ . 4. Given x^ +x^ =756 to find all the values of x. Ans, a;=243, -28V'l8l, 243/ ""^'^ V or 28Vl8lO^^-], 5f Given a:'— a;2=66 to find all the values of a;. Ans. a;=4, V49, 1±V^, or V^izl^^-Y 6* Given aa;8 + 6a;* =c to find two values of x. Ans. ^=±ME±^f. \ 2a / 7. Given 3a;" + 42a;' =3321 to find all the values of a;. A o »/7T -S±sy^ V41±'v/4U/^ Ans. a;=3, -'v/41, , or . 40 8* Given Vx^ ==3a; to find all the values of a;. Ans. a;=4, V25, -2 ±21^^, or V25(~^'^^~^^Y iren (a;— 5)'— 3(a;— 5)2— 40 to find all the values of a;. Ans. x=9, 5 + V25, 3±2^/— 3, or 54-V25[ ~^ ), 17 f 2=-— to find all the values of x. ^ Ans. a;=4, '^I, -2 ±2/^, or Vi( ~^'^^^~^ ). 8 17 lOi Given -^- + 2=-— to find all the values of a; X a;f HIGHER EQUATIONS. 363 IL Given x^-\ =— :=r+a;6 to find all the values of x, Ans. a:=4, 1/49, -lipV'^, or l/W "^^ ). a;* ic' 1 12« Given — — =— — - to find all the values of x. 2 4 32 Ans. .=Vl VI Vi(=^^, or Vi(=^^). 13t Given a;'' + 2'7a:'=2224 + 9a;* to find all the values of a?. Ans..= ±4,or±i/^l^^. 14. Given (a;" + l)(a;' + l)(a;+l)=30a;' to find all the values of a?. Am, x=^~^, or -l±x/ir6±|/-liq:f^36 15. Given (a;-i)»-— = ^^"^ ^ to find all the values ^ '^ 9 2(a?-i) +l/a:(a;-J) _ „ , 4±2f/l3 . 2 .-—- of ar. Ans, a;=3, — i, , or ±-v— 3. 16. Given (l-a:)i/a/l+i]-2=V'a;+l+V'3a;-l to find the five values of a?. Ans. a;=l, or ==:. ji A 5aj^ 17i Given Zx^ =—592 to find the eight values of a;. 2i Am. x=±8, ±8V^, ±V^^^(W, or ±|/-V^=^(^. aj» _j +(«" -\ =— to find the eight values of a?. ^/is. a;=±ar — - — , or ±aY — - — » 19* Given 2a;^l— a;*=a(l+a;*) to find the values of a:. Am. ar=±-|/-l±Vl-a*±|/2(lTV'l-a*. 854 HIGHER EQUATIONS. 20. Given Sx^ +af« =3104 to find the ten values of x. Ans. a;=64,(Y)M6( — 1±1/6±|/— 10qF2V^5,) or 21. Given ^ ^-f ^ ^—=z — to find x. a X c n acn+T Ans. X: an+i — Cn+1 22. Given "^ '='^' ' to find x. Ans. x=(J^\ —"' n 8 \ms ' 28. Given ic^»-maf =p to find x. Ans. ^=\ ra±Vm^ + Ap '^i ^ 24. Given a;"— 2aa:2=6 to find x. Ans. x={a±Va'-\-b)'^. 25. Given 3a;*'*~2a^=25 to find x. Ans, a:=(l±^?^j". — 4af 26. Given So;" Vo(^ ==4: to find x. Ans. a;=(8)T^, or {—^^)~^, 27. Givena;*"— 2ic"* + fl;"=6 tofinda;. . i/l±1/13 \/l±V-1 Ans. x=y -- g— , or V ^ • 28. Given (X^—2x^ + x=za to find x. l±:V3zt2V4aTi Ans. x= , 2 JL 3. *"+" 1 29. Given a'6'a?»—4(a6)2 a; 2 »»'»=(«— 6)'fl:w to find «. .„.„{ x'+y'= EX A M P LE S. 13 ^ xt/= X—1/ 6 x—y " to find the six values of x and y. Ans, " a;=3, —2, 3a, or —2a, y=2, —3, 2a, or —Za, m wnicn a= . 2, Given | ^^J'^^^H^' | to find the six values of x and Ans. x=5, 4, Sttf or 4a, y=4, 5, 4a, or 5ay in which a= '"' , 8» Given a!'y--4=4ariy— |-, to find two of the six values js 111 I f of X and y. fy2(a.2_y2)^ Am, = 1. -1:f2/-2, ja:=l. -1 ( y=4, -2 4* Given 16aJ-y^=62^Ja^, a;* 12 a? to find the values of x and y, L y ^' V^' J ^w^ i^==±4, ±16, db2V^, or ±8|/^, • (y=256, (256)^ -192, or -3(64)'. r 2a;+y=26— 7t'2a;+y + 4, 5. Given -I 2^-H^_16 2x-y^ I to find the values of x [ 2x—Vi/ A5 23;+^^^' — lTi/321 a:=2, 10, 5y , 16, —24, or ^ ^ , and y. — 1^1^6145 Ans. 64 64 , o^ 161 ±1^321 ^, ,,, 3073 ±^6143 y=l, 26, —32—, 64, 144, or ^^ . 6« Given SIMULTANEOUS EQUATIONS. ^2t/^ — SVx 357 Vx ■\-V^f-16Vx=^Vx, Vx+V8{y-Vx)-4=y + l, to find the values of X and y. Ans. — 4T16I/-39 788 ±24 1^644 •^=4, i g , A» 4, or — , „ , 3±2V^=^ ,, , 87±|/644 y=3, If , If, ~1, or . 7, Given \ ^1~^''X"^^V^^J a nn !• to ^^^ ^"^^ sixteen values of x and y. ^W5. a;=±3, ±f6, ±' ,or /_-13±|/— 11 2 ^ , l±V''^=l7 1±3^5 i±t/-Tr. y=2, -1, , — , or 2 ' 2 ' 2 x-\-y -{-xy -\-iii?y -\-xf -\r x^y 4- 2a;'y' + ajy" + «y +a:y=ll, 8. Given ^ x'y + 3a;y + 3a;y + 2a;y + 4a;y + 2a;y + 4:xY + 4a;V* + xy*^ + a^y + «y + 2a;y [ +a;y4-a:y=30 to find the sixteen values of x and y. ix-\-y=^±\V^,\±\V~^,2orl,ov\ ± V^, ^"^^ -j a:y=|:F^V/21,|:fi^-19, lor2,orl iFl^-2. Note. — ^The solution of these eight simultaneous equations will give the six* teen values of x and y. 9. Given x^yVxy—a^ xz^^xz—h^ y^zVzy=c to find x^ y, and z. _ 3 f to find the values of x and y. xy — c \ I Ans. x=:(a^±Va"'-c^'')'^, y= (a"± ♦/«""— c"')»* CHAPTER XV. ARITHMETICAL FBOGRESSION. (359.) An Arithmetical Progression is a series of quantities in which the difference of the consecutive quantities is constant, as -j-o • a±d ' a±2d • a±Sd • a±4c?^ &c. PROBLEM. (360.) To find a general expression for any term of an arith- metical progression, SOLUTION. \»U 2d, 3<^. 4 (3)=(l) + (2). ARITHMETICAL PROGRESSION. 359 Since, 2S equals (a + l) taken n times, the expression becomes 2S={a + l)n, 5=(»-±i)„or(a+0| which is the expression required. Remark. — By the aid of the two formulas l=a±(n—l)dj and S=l — ^ )w, we are enabled to find any two of the terms a, d, n, I, 3, when three of them are given since we shall have two equations and two unknown quantities. We append a few simple propositions for the student to demonstrate. PROPOSITION 1. In an arithmetical progression consisting of three terms^ the sum of the first and the third term is twice the second, PROPOSITION 2. In an arithmetical progression consisting of four terms^ the sum of the first and the fourth is equal to the sum of the secoTid and the third, PROPOSITION 8. In an arithmetical progression consisting of any number of terms^ the sum of any two terms equally distant from the extremes is equal to the sum of the extremes. PROPOSITION 4. In an arithmetical progression consisting of an odd number of terms, twice the middle term is equal to the sum of the extremes. Remark. — ^In the following examples the known terms should be substituted instead of the letters representing them, and there will thus arise one or two equations, according as one or both of the formula, L—a±{n— l)d and /S^=l — — k, are involved. QUESTION s. 1. The first term of an arithmetical progression is 2, the common difference 3, and the number of terms 8. What is the last term ? Ans. 23. 2. The first term of an arithmetical progression is 3, the common difference 2, and the last term 99. What is the number of terms ? Ans. 49. 860 ARITHMETICAL PROGRESSION. 3. The last term of an arithmetical progression is 100, the common difference 4, and the number of terms 30. What is the first term ? Ans. —16. 4. The first term of an arithmetical progression is —20, the num- ber of terms 61, and the last term 230. What is the common differ- ence ? Ans. 5. 5. The first term of an arithmetical progression is —12, the com- mon difference —7, and the number of terms 101. What is the sum of the series? Ans. —36057. 6. Insert 8 arithmetical means between 3 and 21. Ans. -T-5 • 7 • 9 • 11 • 13 • 15 • 17 • 19 . 7. Insert 3 arithmetical means between i and |. Ans. ^f.f^y.H. 8. The sura of an arithmetical series is 108, the first term 3, and the common difference 2. What is the number of terms ? Ans. 10. 9. What is the sum of w terms of the progression -i- 1 • 2 • 3 • 4 • 5'*» 6«A. In the series -^^a : ar : ar^ : ar^ : ar* : ar^ : (fee, it may be seen that any term is equal to the first multiplied by the ratio afiected by an exponent which is one less than the number of the term. Therefore, the nth term =ar''~\ S66 GEOMETBICAL PEOGRESSION. In a series which tenninates, if we represent the number of terms by n, and the last term by ^, we have "problem. (365.) To find an expression for the sum of the terms of a geo- metrical progression. SOLUTION. Representing the sum by S^ we have >S^=:a+ar+ar''-far'-far* ar'^'+ar^'-Hir'-', (1). rS=ar-\ar^^r^^r^\ar^ ar"-'+ar"-'+ar* (2)=(1) x r. then rS-S=ar^—a (3)=(2)-(l). Since ar^=zar^^ xr, and ar'^^=l, we have ar'^z^lr; .-. rS—S=ar"—a, becomes rS—S=lr—a (r-l)S=lr-a r—1 which is the expression required. When r is less than 1 it is best to put S= , although the same result will be obtained from both forms. PROB LEM. (366.) To find an expression for the sum of the terms of a de- creasing geometrical progression when the number of terms is infinite. SOLUTION. It may be seen from the formula S= that the sum of the series depends upon the first term, last term, and ratio. In a decreasing geometrical series the terms must continually approach zero as a limit. Therefore, when the number of terms is infinite, we are compelled to consider zero as the last term ; since there is no quantity, however small, greater than zero that may not be reached or passed by a finite number of terms. If, then, ^=0, Ir must also = 0, and the above for- mula becomes, for a decreasing geometrical progression having an in- finite number of terms, S=- , which is the basis of the following GEOMETRICAL PROGRESSION. 367 RULE. Divide the first term of an infinitely decreasing geometrical pro- gression by the difference between unity and the ratio, and the result will be the sum of the series. The above formula may also be deduced in the following manner : Let S=ar\^r-\^r^+ar^, &c., ad infinitum, r being less than 1 (1) r;S^= -f«r+a7-'-H»r^&c., " " " "(2)=(l)xr rS-S=a (3) = (2)-(l) S= -, the same as before. r—r A few simple propositions are here appended for the student to demonstrate. PROPOSITION (367.) 1. In a geometrical progression consisting of three terms the product of the extremes is equal to the square of the mean. PROPOSITION (368.) 2. In a geometrical progression consisting of four terms the product of the extremes is equal to the product of the means. PROPOSITION (369») 3. In a geometrical progression consisting of any number of terms the product of the extremes is equal to the product of any two terms equally distant from them. EXAMPLES. 1, Find the 11th term of ^3 : 6 : 12 : &c. Ans. 3072. 2. Find the sum of 9 terms of ^1 : 2 : 4 : + 2p*, and x^-^-y^zzz s" -^5s*p + 5sp\ Remark. — ^It would be a good exercise for the student to ascertain how these results are obtained. QUESTION. (37 2») What six numbers in geometrical progression are those of which the sum of the extremes is 99 and the sum of the other four terms 90 ? SOLUTION. The conditions show that the sum of the six numbers is 189. Let ar, ary, xy\ xy*, xy*, xy^, represent the numbers. * The formula S= — becomes by substitution ,«« xy'—x x(y*^l) 189= ^ , = ^ , S y—l y—\ 24 870 GEOMETRICAL PROGRESSION. But xi/^-hx=99, 99 ^■^ + 1' 189(y--l) _ 99 21 11 y* + f + l-y^^f+f-y + V 21y*-21y'+21y''~21y + 21 = lly* + lly^ + ll, 10y + 10y'' + 10=:21y^+21y (a), 10(y* + y'« + l)=21y(y'^ + l), Putting y' + 1=0, we have 10(2'— y'')=21y2, IO2'— 21y2=10y^ 21y±292/ 6y 20 / + 1='^ 2' 2y''-5y=-2, 6±3 ^ y=— =2, _ 99 99_ ''-yTi^ss-^' Therefore, the progression is -H-3 : 6 : 12 : 24 : 48 : 96. Note. — Equation (C) is recurring, and might be solved according to either of the methods given in biquadratics. Equation (C) might have been obtained without using the general formula for the sum of the series. ANOTHER SOLUTION. .8 ^ re' ar' y' y' , , Let — 5, — , Xy y, -^, -^ represent the numbers. ••• 4+^=99 (1), y' x^ ^ '^ x^ v' and — + a; + y +— =90 (2), y ^ a;'+y^=99ary (3)=(l) x xY, Putting a; + y=« and xy=:py we have s^ — Gs*/? + dsp'^=99p'' (4), x rcy x^ + a:y (ar + y) + y" = 90a:y (5) ■= (2), .T^ 4- y^ = 90iry — .ry(a: + y) (6) == (5) transposed. «'— 35^=90/?— sp, [forward GEOMETEICAL PEOGEESSION. 871 58' 5s _ 99s' i+2s"^(90+ 2sY~(90 + i 5s 65' 995 • 2^- + -^f——---^^l-— (10)=(9)substitute(iin(4). 90+ 25 "^(90+ 25)='" (90 +25)'^' 8100 + 3605 + 45''— 4505— 10s''+65*'=995, s' + 1895=8100, — 189db261 =36, 5' 36-36-36 _ , , ,^^, p= = =18-4-4=12*24, ^ 90+25 2-9-9 ' a; + y=:36, iry=: 12-24. Whence, it is obvious without solution, that a:=12, and 2/=24. Therefore, the series is — 3 : 6 : 12 : 24 : 48 : 96. - Eemark. — ^This problem is one of the most difficult of those generally pro- posed in geometrical progression, and the solutions given should be carefully studied by the student that he may be able to solve others of like character. QUESTIONS. 1. The sum of the first and third of four numbers in geometrical progression is 148, and the sum of the second and fourth is 888. What are the numbers ? Ans. 4, 24, 144, and 864. 2. There are three numbers in geometrical progression, the sum of the first and second is 15, and the dift'erences of the second and third is 36. What are the numbers ? Ans. 3, 12, and 48. 3. What three numbers are there in geometrical progression whose sum is 14, and the sum of whose squares is 84 ? Ans. 2, 4, and 8. 4. What three numbers are tho§e in geometrical progression, whose sum is 52, and the sum of whose extremes is to the mean as 10 to 3 ? Ans. 4, 12, and 36. 6. What three numbers are those in geometrical progression, whose sum is 13, and the sum of whose extremes multiplied by the me^n is 80 ? Ans. 1, 3, and 9. 6. The sun; of the first and second of four numbers in geometrical 372 GEOMETRICAL PROGRESSION. progression is 15, and the sum of the third and fourth is 60. What are the numbers ? Ans. 5, 10, 20, and 40 ; or —15, 30, —60, and 120. 7. The sum of four numbers in geometrical progression is equal to the common ratio +1 ; and the first term =xy* What are the num- bers ? Ans. y'y, y\, |f , uud ^. 8. A gentleman divided |210 among three servants; the sums received were in geometrical progression, and the first received |90 more than the last. How many dollars did each receive ? Ans. 1120, 160, and $30. 9. The sum of three numbers in geometrical progression is 35, and the mean term is to the difierence of the extremes as 2 to 3. What are the numbers ? Ans. 5, 10, and 20. 10. There are three numbers in geometrical progression, the greatest of which exceeds the least by 16. Also, the difierence of the squares of the greatest and the least, is to the sura of the squares of all the three numbers as 5 : 7. What are the numbers ? Ans. 5, 10, and 20. 11. The sum of three numbers in geometrical progression is 13, and the product of the mean and sum of the extremes is 30. What are the numbers ? Ans. 1, 3, and 9. 12. The diagonals of 4 squares are in an increasing geometrical progression, and the product of the squares of the diagonals of the extremes is to the product of the diagonals of the means as a side of the third is to the square root of the common ratio divided by 4^2. What is the diagonal of the third square, and the common ratio, sup- posing their difierence equal to 45 ? Ans. 81 the ratio, and 36 the diagonal of the 3d square. 13. The difference between the first and second of four numbers in geometrical progression is 36, and the difference between the third and fourth is 4. What are the numbers ? Ans. 54, 18, 6, and 2. 14. There are three numbers in geomstrical progression, the sum of the first and second of which is 9, and the sum of the first and third is 15. What are the numbers ? Ans. 3, 6, and 12. 15. There are three numbers in geometrical progression, whose sum is 14 ; and the sum of the first and second is to the sum of the second and third as 1 to 2. What are the numbers ? uins. 2, 4, and 8. GEOMETRICAL PROGRESSION. SIB IG. There are three numbers in geometrical progression, whose con- tinued product is 64, and the sum of their cubes is 584. What are the numbers ? Ans. 2, 4, and 8. 17. There are four numbers in geometrical progression, the second of which is less than the fourth by 24 ; and the sum of the extremes is to the sum of the means as 7 to 3. What are the numbers ? Ans. 1, 3, 9, and 21. 18. The sum of $700 was divided among four persons, whose shares were in geometrical progression ; and the difference between the greatest and least was to the difference between the means as 37 to 12. What were their respective shares ? Ans. 1108, 1144, $192, and $256. 19. A company of merchants fitted out a privateer, each merchant subscribing $100. The captain subscribed nothing, but was entitled to a $100 share, at the end of every certain number of months. In the course of 25 months he captured three prizes, which were in geometrical progression, the middle term being i the cost of the equipment, the common ratio the number of mouths which entitled the captain to his $100 share, and their sum $1375 more than the cost of the equipmept. After deducting $875 for prize money to the crew, the captain's share of the remainder amounted to } of that of the company. What was the number of merchants, and the captain's pay ? Ans. 25 merchants, and captain's pay $100 at the end of every 6 months. 20. There are four numbers in geometrical progression, the differ- ence of whose means is 3, and the difference of whose extremes is 10^. What are the numbers ? Ans. 1^, 3, 6, and 12. 21. The sum of three numbers in geometrical progression is 31, and the sum of their square is 651. What are the numbers ? Ans. 1, 5, and 25. 22. The sum of four numbers in geometrical progression is 16, and the sum of their squares is 85. What are the numbei-s ? Ans. 1, 2, 4, and 8. 23. The sum of five numbers in geometrical progression is 31, and the sum of their squares is 341. What are the numbers ? Ans. 1, 2, 4, 8, and 16. 24. The sum of six numbers in geometrical progression is 94, and the sum of the second and fifth is 27. Wliat are the numbers ? Ans. 1, 3, 6, 12, 24, and 48. 874 GEOMETRICAL PROGRESSION. 25. The sum of six numbers in geometrical progression is 63, and the sum of the means is 12. What are the numbers ? Ans. 1, 2, 4, 8, 16, and 32. 26. The sum of six numbers in geometrical progression is 1365, and the sum of the extremes is 1025. What are the numbers? Ans. 1, 4, 16, 64, 266, and 1024. 27. Whiit six numbers are those in geometrical progression whose sum is 815, and the sum of whose extremes is 165'^ Ans, 6, 10, 20, 40, 80, and 160. 28. What number is that which being severally added to 3, 19, and 51, shall make the results in geometrical progression ? Ans. 13. 29. $120 are divided between four persons, in such a way, that their shares may be in arithmetical progression ; but if the second and third had received $12 less each, and the fourth |24 more, the shares would have been in geometrical progression. What was the share of each ? Ans. |3, $21, $39, and $57 respectively. 30. The sum of three numbers in geometrical progression is 7, and the difference of whose difference is 1. What are the numbers? Ans. 1, 2, and 4. CHAPTER XVII. PROPORTION. (373.) Proportion is an equality of ratios. (37 4») If the ratio of a to 6 is equal to the ratio of c to rf, these four terms constitute a proportion which is usually written a\h\:c\d, and is read a is to 6 as c is to d. Sometimes the sign of equality is used instead of the four dots, as a : 6=c re?, which may be read, the ratio of a to 6 is equal to the ratio of c to d. We may consider the symbol : as an abbreviation of the sign -f- ; whence, we infer that a\h:\c:d\s only another mode of writing a c a-7-6=c-i-rf, which is the same as r =-. This shows that every pro- portion is essentially an equation. (375.) The four quantities of a proportion are called its terms. (376.) The first and the fourth term are called the extremes, and the second and the third term, the means. (377.) The first two terms of a proportion are the first couplet, and the other two, the second couplet. (378.) The first term of a couplet is called the antecedent, and the second term the consequent. (379.) Three quantities are in proportion when the ratio of the first to the second is equal to the ratio of the second to the third. (380.) The second quantity is called a mean proportional be- tween the other two, and the third quantity a tki7'd proportional to the other two. Thus, in the proportion a:b::b:c,bis the mean proportional, and c the third proportional. (381.) The equality of more than two ratios maybe thus written, a:h:ic:d::e:f::g:h, &c., which may be read a is to 6 as c is to d, as eis to/, as ^ is to h, &c. Remark. — The student should observe that the demonstrations of the follow- ing propositions in regard to proportion are based upon the fact that every proportion is essentially an equation. 376 PROPORTION. PRO POSITION (382.) 1. In every proportion the product of the extremes is equal to -the product of the means. DEMONSTRATION. Let a'.b'.:c:d represent any proportion. We are to prove that €td=bc. This proportion expressed as an equation is -=- '(1). b d ^ ^ ad=bc (2) = (l) xbd. Q,E.D, Or, Put a=rb, then by the nature of a proportion c=ird. The propor- tion will then stand rb:b::rd:d. Multiplying the extremes together, we have rbd. Multiplying the means together, we have rbd. These results are identical, therefore, the proposition is true. Remark. — ^This proposition furnishes the test of a proportion. QUESTION. Are 2, 4, 3, 7, in proportion ? SOLUTION. Multiplying 2 by 7, we get 14, and 4 by 3, we get 12, which are not equal, therefore, by the foregoing proposition they are not in pro- portion. QUESTIONS. 1. Are 3, 7, 8, 11 in proportion? 2. Are 8, 16, 4, 2 in proportion ? 3. Are 2a?, 3a?, 4a?, 6x in proportion ? 4. Are i, -J, yV» i ^^ proportion ? • 5. Are i, i, i, jV iii proportion ? Remark. — If any term of a proportion is unknown, put it equal to x, and form an equation by placing the product of the extremes equal to the product of the means, and then solve the equation to obtain the value of x. When one of the means is unknown, it is most convenient to put the product of the means equal to the product of the extremes. PROPOSITION (383.) 2. When three numbers are in proportion, the product of the extremes is equal to the square of the mean. PROPORTION. 877 QUESTIONS. It Are 3, 4, 5 in proportion ? 2» Are 3, 6, 12 in proportion? 3* Are x, Vxt/, y in proportion ? 4. Are ax^ ahx, hx in proportion ? 5* Are oar, xVab, hx in proportion? PEG POSITION (384i) 3. When the product of two quantities is equal to thepro^ duct of two other quantities, the four quantities may he expressed in the form of three different proportions, DEM O NSTR ATI O N. Let ad=zhc. We are then to prove that all of the following pro- portions are true, a',hi\c\d, a \ c \\ h \ d, b:a::d:c, The equation ad=zhc may be put in the following forms : a c h^d' a b ~c^d^ h_d a~ c* and these three equations respectively give a\h:\c:d, a\c\\h'.d, h:a::d:c, Q, E. D. Corollary. — Since, on the supposition that ad— he, we get the proportions a\c wh'.d h'.a w d\c, we infer that these proportions are also true on the supposition that a\h w c'.d, because this proportion gives ad^=hc. From this fact we obtain the two following propositions : 378 PROPORTION. PROPOSITION • (385.) 4. When four quantities are in proportion^ the first is to the third as the second is to the fourth. If a : 6 : : c : c?, then a : c : : b : d. • This is called proportion by AltevTiation, PROPOSITION (386,) 5. When four quantities are in proportion^ the second is to the first as the fourth is to the third. 1£ a:b :: c: d, then b:a :: d :c. This is called proportion by Inversion. PROPOSITION (387.) 6. When a couplet is common to two proportions, the other two couplets will constitute a proportion. If a : 6 : : m : w, and c:d :: miUj then a:b : : c : d. Let the student prove this. PROPOSITION (388.) 1. If a : 6 : : c : c?, then are the following proportions true : ma : mb ::mc: md ma '.mb'.'. c :d. a', b ::mc '.md ma : h wmc '.d a:mb:: c '.md ma '.mb'.'.nc '.nd ma '.nb'.'. mc '.nd a ^ b ^^ c . ^ m' m" m ' m a b — : — :; c m m :d m ' m m m '.d b d m PKOPORTION. 879 Let the student prove these proportions to be true by an applica- tion of Prop. 1, (382.) PROPOSITION (389.) 8. When four quantities are in proportion, the sum of the first and second is to the second as the sum of the third and fourth is to the fourth. DEMONSTRATION. Let a',h\'.c :d (1). We are to prove that u+h : h :\c+d\ d ad=hc (2)=(1) by Prop. 1, (382.) ad-\-hd=hc-\- bd (3) = (2) with bd added to both members. (a + b)d=b{c + d) (4) = (3) factored. BjTro^,3,(3S4:)a + b:b::c + d:d. Q. K D. This and the derivative proportion in the following proposition is called proportion by Composition, PROPOSITION (390.) 9. When four quantities are in proportion, the sum of the first and second is to the first as the sum of the third and fourth is to the third. Let the student prove this. PROPOSITION (391.) 10. When four quantities are in proportion, the difference between the first and second is to the second as the difference between the third and fourth is to the fourth. DEMON STRATI O N. Let a:b'.'.c : d (1). We are to prove that a — b '.b '.'. c — d '. d ad=:bc (2) = (1) by Prop. 1, (382.) ad—bd=bd—bc (3) = (2) with bd subtracted from both members. {a—b)d=:b{c—d) (4) = (3) factored. By Prop. 3, (384) a—b :b:: c—d : d. Q. E. D. This and the derivative proportion in the following proposition is called proportion by Division, 880 PEOPORTION. PROPOSITION (392.) 11. When four quantities are in proportion^ the difference between the first and second is to the first as the difference between the third and fourth is to the third. Let the student prove this. PROPOSITION (393.) 12. When four quantities are in proportion, the sum of the first and second is to their difference as the sum of the third and fourth is to their difference, DEMONSTRATION, Let a:b :: c:d. We are to prove that a + b :a—b : : c-\-d : c—d^ By Prop. 8, (389) and Alternation, a-\-b: c-{-d :: b : d By Prop. 9, (390) « " a-b:c-d::b:d then by Prop. 6, (387) " " a + b:a—b::c+d:c—d Q, E. D. PR OPOSITION (394.) 13. In a continued proportion, any antecedent is to its consequent as the sum of all the antecedents is to the sum of all the consequents, DEMONSTRATION. Let a :b :: c : d :: e :f:: ff : h :: &c. We are to prove that a: b::a-\-c+e-\-ffj t »i » , »h HARMONICAL PROGRESSION. (406.) An Harmonical Progression is a series of quantities, any consecutive three of which are in harmonical proportion. PROPOSITION (407 •) 1. The reciprocals of a series of quantities in harmonical progression are in arithmetical progression. DEMONSTRATION. Let a, 6, c, d, e,f &c., be an harmonical progression. We are to prove that -, r, -, 3, -, -;,, &c. is an arithmetical progression. ^ a c a e f 384 HABMONICAL PROGRESSION. If we prove that * 1 1_2 1 1_2 1 1_2 1 1 2 (fee, &c., we shall establish the truth of the proposition. ,f^ , , 2ac b ac 2 a + c 1 1 We have b=- — , or -= — --, or -= ==-^ — - a+G 2 a+c b ac c a We nave b= , which gives - + -=--, c= 26c? d= b+d' 2ce 1 1_2 b~^d~~c' 1 1_2 6= 1 1_2 C.^.i). We are to prove that d-\-f ANOTHER DEMONSTRATION. 1_1_1_1 a b~ b c' 1_1_1_1 6 c~ c d* 1111 By the nature of the progression, we have Whence, we get c d d e 1__1_1_1 d e'~ e f* &c., &c. a\c'. \a—b : 6— c, b'.d'.', b—c :c—dy c:e : : c — d: d — e, d:f: :d—e :e—f, ab-^ac=ac—bcj bc—bd=:bd—cd, cd—ce = ce—de, de—df^df—efj HARMONIOAL PROGRESSION". 885 Dividing respectively by < def, gives These equations by transposition, (1 1 __!__! a 6~" b c' 1_1_1 1 b e~~ c (T c~~d'~d i 1-1=1-1. . »N THE EXPANSION OF SERIES. (412.) Quantities or algebraic expressions may be expanded or developed in four ways ; namely, The Method of Division, " " " Undetermined Coefficients, " " " Involution, and " " " Evolution. THE METHOD OF DIVISION. PROBLEM 1. Expand into a series. SOLUTION. By dividing 1 by 1 -I- a, we obtain 1— a 4- a'— a' -J-, 'a:'"~" + , &c. Since this equation is true for any value of x, we have a right to suppose X equal to zero. Making this supposition the first member reduces to A. If we suppose all the exponents in the second member to be finite, it would reduce to zero when x equals zero, and we should have A—0, which is impossible, as ^ is supposed to seme finite '^^ quantity. Therefore, it would be improper to consider all the ex- ponents in the second member as finite, and hence, one or more of them must be zero. Since, a', b', c', d', &c., are all different, it is evi- dent that a can not equal more than one of these quantitjps, and, therefore, of the exponents a'— a; 6'— a; c'—a-, d'^a-, &c., but one can equal zero. It now remains for us to decide which exponent it is that equals zero. Suppose it to be b'—a, or the exponent of x in the second term, and then for x equal zero all the terms after the second must vanish, and there would result the equation But since a:°= 1, A=A'xf'-^ 4- B. Since a =6' and a' is less than 6', a'—a—& negative quantity which put equal —n, and the equation becomes AzzzA'x-'^-^rB, A=- + B, A' Supposing x=0 A=-—-\-B, A—B^oD. But A—B must be equal to a positive or negative quantity, or zero, therefore ^—J5= GO is impossible, which shows that it is im- proper to make b'—a=:0. As the same reasoning applies to all the exponents of x in the 396 THE METHOD OF UNDETERMINED COEFFICIENTS. second member but the first, or a'^a, it follows that a'— a must equal zero, and the equation becomes A=A'x\ A=A', Suppressing the equal terms A=A'af''~* the equation becomes ^ar^«+ Cx'-^-\-Dx^ + , &c.,=^V-«+ (7V-«+i>V'-« + , &c. Dividing by a:*"*, we have B+ (7a:'-* + i>a;^-* + , &c.,=i5V'-*+C7V-* + i>V-*+, &c. Then for a; = zero, we have when we make b'—b~ zero, or b'=b B=B'x\ £=B'. Thus we may go on and prove that c=c' and C=zC' ; d=d' and J)==:D\ (fee, which proves the theorem. THEOREM II. (41 5.) If Aaf-\-Bx^+ Cx' + Dxf^-h, :x^-^ + , &c. =0. Know we make x = zero, we obtain ^ ^ = 0. Suppressing ^=0, and dividing by a^, we obtain B 4- Cx' -* + Dx^^ + , &c. =0, in which if a; is put equal to jsero the equation reduces to In the same way, we can get (7=0, i)=0, a;' + 3^ar* + 3i^a;' + =H 3E=2D ^=2% 3F=i2E F=^%\ &c. &c. By substituting these values in the assumed series, we get 2 4 8 16a; 32.tr' 64a;'' , 8a;'— 2a;« Sx^ 9a; 27 81 243 729 2.2 2' 2' 2*a; , 2V 2V » , .i. i i? 3^^=2?=3-?+3V+§^+ 3^+T^-»-T«-+ ^^-^ ^^^"^^ '^^ ^'^ '^ the terms is apparent. 2 Since assuming it equal to ^ + ^a; + Cx^ + Dx^ + &c., and then multiply- 2 2 ing the result by — ^ for the development of — -^ — --|. a; oaj ~^JiX o PROBLEM 3. Expand (1 +a;)i into a series. SOLUTION. Assume (H-a;)i=^ + 5a;+(7a;' + i>a;' + ^a;*+ Squaring l+a;+0a;'-K)a;''+0a;*+&c. =-4'+2^J?a;+2^Ca;' + 2^i>a;' + 2^^a;*+ ^V + 2BCx' + 2^i>a;*+ + (7V+..:.. Equating like coefficients ^'=1 ^ 2^J?=1 2AC+B''=0 2AD + 2BC=0 2AE-^2BD-hC^=0 &c. >•, whence - A=l B=i we have " Substituting these values in the assumed series, we get 0=-i &c. ♦/l+a;=14-ia;-ia-' + yV^«-yf^a;* + DECOMPOSITION OF RATIONAL FRACTIONS. 899 But we had a right to assume Vl+x=-A-Bx~Cx''—I)x'—JEx*— By squaring we should obtain the same equation as before, and hence the same values for A, B, (7, i>, E, =r. ° r—v r—v But since, when vz=r^u must equal s, we have L t (Zq—Uq\ _ Z—U /^~ 'rP- -?n r- -V 3IMF1 nam 7-7 _ -V^ r— -V ^- =^-rP~^^^ qr^-' q —I. 404 THE METHOD OF EVOLUTION. Butsiiicer'=2, wehaver=2j andr^^=2~y =2* '* Substituting — ^1 = z—u /«^, ^ y which is the formula ffiven in the lemma, when »=-, and hence proves the lemma to be true in the second case. CASE III. When w is a negative integer. PROOF. Suppose n=—m. Then since z~"*—u~^=^g~^u~^(z'*—u^)j we liave = —zr^u-^{ 1. z—u \ z—u / j r=mj^\ which 1 = _^-»^-«i,yjg«-i___^2-m-i^ which is the formula given in the lemma when «= — wi, and, therefore, proves the lenuna to be true in the third case. CASE IV. When w is a negative fraction. PROOF. P The proof in this case is found by putting — for —m in the last, and referring to Case 11. instead of Case I. Therefore, the truth of the lemma is fully established. We are now ready to proceed with the general demonstration of the binomial theorem. We are required to ascertain the development of(a;4-y)^ Since, (aJ+y)"=a^"(l+-) it is only necessary in order to find the development of {x-\-yY to find that of (1 +-) and multiply it by a?*. For convenience put 2=-, and we then have(l +zY, whose develop- ment will now be sought. THE METHOD OF EVOLUTIOlf. 405 Assume (1-\-zY=A-^JBz-{-Cz^ + I)z^ + jE:z*+ in which the assumed coefficients A, B, C, B, JS, &c., are independent of 2, and depend entirely for their values on 1 and n. Let us now endeavor to ascertain the values of these coefficients in terms of 1 and n. To find the value of ^ make 2=0 in the assumed equation, which we have a right to do, since A is independent of z, and we get Substituting this value of A in the assumed equation, we have {l+zy=l+£z+Cz'-{-Dz'-{-M*-^ {A). Since z may be any value whatever without affecting the coefficients Aj B, C, &c., let us make z=u^ and we have (l+uy=l+Bu-hCu''-{-Du' + I!u*-{- (B) Subtracting (B) from (A)^ we get {l+zy-{l+uY=B{z-u)+C(z'-u') + D(z'^u') + i;(z*-u*)-{. .... Dividing the first member of this equation by (1 +2)— (1 +u), and the second by its equal z—u, we obtain (i-\-z)-(i+u) ^ \z-ur \z-ur \z-ur ^ ' Putting l+s=r and l+w=v (C7) becomes rlzr^B^ c(^+i>(^^)+i.f-«:)H. (2,) r—v \z—uj \z^u1 \z—uf But by the lemma we have nr^ Lemma But when ?;=r, we also have w=2, and by the iZ^ — U^ \ \ Z — U /«^, Equation (D) will become by substitution wr*^* = ^ + 2 Cfe + 3i>2' + 4^*. Multiplying this equation by r= 1+2, we get wr'»=^ + 2(72 + 3i)2'' + 4^2' + . -vBz-V2Cz^^-ZDz^-\- Since 14-2=r, we have 7i(l +2)''=^ + (^ + 2(7)2+(2C7+3i)) 2'' + (3i> + 4J')2'-f. . . . Multiplying (A) by n^ we get w(l+2)"=ri + w^0 + »6fe= + /ii)«" + . . . . 406 THE METHOD OF EVOLUTlOlT* Whence, ^ + (^ + 2 (7)2 +(2(7+ 3i))2».f(3i> + 4^)2' + . . . . =n-hnBz + nCz'' + nBz^ + Equating the coefficients of the like powers, f B=n, B=n > we have < B + 2C=nB, 3i> + 4^=wi>, (fee. ► , whence < C= D= n[n—l n(n—l){n — 2) 3 n(n-l){n-2)(n-S) 2.3.4 ' &c. Substituting these values of B, (7, 2), &c., in (A)^ and restoring the value of 2=-, we get w(»— 1)(%— 2)(w— 3) y* ■^ 2.3.4 ^"^^ • • Therefore, since a;"( 1 +-J =:(ar + y)'', we get by multiplying the ex- pression of 1 1 + - 1 by a;", ^(^—1) o o w(w— l)(w— 2) , . . (ar+y)"^.g" + naf-V+ ^ ^ ^ ^"^y+ 2 3 "^ n{n-\){n-2){n-3) ^ 2.3.4'' ^"^ which is the formula given in the theorem. For the purpose of reference, we append the following formulas which have been encountered in the above investigation. Which formula it is best to use will depend on the nature of the binomial to be expanded. (a; + y)"=ar'* + Tiaf^^y + ^^^^— -Wy ,(n-\){n-2)^_, af-y + . . (1). THE METHOD OF EVOLUTION. - * 407 EXAMPLES 1. Expand (a^+x)-k into a series. X x^ , 3a;' 3-5a;* Ans, a-\-—-————L'r 2a 2-40" 2-4-6a' 2-4-6-8a* 2. Expand (a+a:)i into a series. X x" . 3a:' S'bx" . ain ^715. ai(H- 2a 2'4a' 2-4-6a' 2-4-6-8a* 3« Expand 4/2 = (1 + l)i into a series. ^^^- l + 2"2^4 + 2^"T4^+ • ^ 4» Expand (a— 6)i into a series. p . J^ b 35'' 3-76' 3-'7-ll6* bi«. ail r 4a 4-8a' 4-8-12a' 4-8-12-16a* / 5« Expand (a + y)"~* into a series. 1 4y 10y» 20/ 36y* ^''*- "^ "^^"^^ a"'""^ "^ 6* Expand 7 into a series. c ^ be ^b\ b'c ,b*c a (T a' a a° 7. Expand (a'+^'H into a series. 6» 6* 5- Ans. a + — -—--, + — —-7 2a 8a' 16a' 8. Expand — • into a senes. n^a^^x^ a/ x^ , 3a;* , bx" , 5.7a;" \ ^^- 6('-2^^-2V ■^-2V + 2V • • • •; 9. Expand (c'— a;'')f into a series. 4/, 3a;' 3a;* 6a;« \ ^^^- n^-2V~2V-2V ; lOt Find the cube root of — — ^3. a -f" " ^ 3a' ' 9a" 81a» 11. Expand 4/5 = = 1/4 + 1 into Ans. a series. '4- -1 + i- 2" 2» 5 2" 12. Expand V6= =V8— 2 into a series. -4ri^. -ji- 2^3'"^" 6 5 2'-3* 2"-3* • • iOQ THE METHOD OV EVOLtTTIOIir. PROPOSITION (420») 1. The sum of the coefficients of the odd terms of the eX' pansion of {a-\- hy is equal to the sum of the coefficients of the even terms* PROPOSITION (421.) 2. The sum of the coefficients in the expansimi of {a + hy is equal to nth power of 2. PROP OSITION (422i) 3. The sum of the coefficients in the expansion of {a—hy is equal to zero. (423.) We have seen that the binomial theorem furnishes the means of expanding polynomials, since all polynomials may become binomials by substitution. But a more direct method of expanding polynomials is by the MULTINOMIAL THEOREM. {x+ay + hy^^-cy^ . . )''=t-^nx'^^ay + \^-^ DEMONSTRATION. r Assume {x-\-ay-\-hy^^cy^ .... )Tz=A-{- By •{- Cy^ -\- Dy^ ■{■ . . (-4). To determine A make y=0. Whence r A=x~*^ Making z—y^YfQ gei {x-\-az + bz'-\-cz^ )l=:A-\-JBz-{-Cz^ + Dz*-}- {B), Subtracting (B) from (A) there results {z-\-ay + by^-\-cy* )'^'-{x + az + bz^ + cz'' . . . )T=^(y— g) + C{f-z')-hD{y'-z')+ ...((7). Putting P* = (.r + ay 4- by' + cy' ), r whence P''=(x-^ay-{-by'-\-cy^ . . . .)7, and Q*={x^az+bz'-\-cz* ) r whence Q'=(x-^az-\'bz'-{-cz^ » • • ')~** wegetR-Q'=B{y-z)+C{f-z') + D{y'-z')-h But P'-^Q'=a{y-z)-hb(f-z')-\-c{y'^z')-^ .... Tfifi MEtHOD OF EVOLUTION. 409 "Whence, by division, we obtain I P^-^Q' ^ B{y-z)+ C( f-z^) + D{f-z')+ .... P'-Q' a{y-z)-\-b{f^z^)+c{f-z')+ Dividing the numerator and denominator of the first member by P—Q and the numerator and denominator of the second member by y—z,we get B+C{y + z) + I){y' + yz-\-z')-^ . a-{-b(y+z)+C(y'' + yz + z') + W Now, if we make y=z, since P will then equal §, Eq. (jD) becomes j.pr-1 y.pr B-h2Cy-{-SDy'' + 4£Jy'-\- .... sF'-' sF* a + 26y + Sc?/" + 4dy^ + . . . . Substituting the values of P' and P', we get r(x + ay + by'' + cy^ . . . . )T_ ^ + 2(7y + 3i)y'+4^/+ . . . . s(x + ay + by'-{'cy^ . . . . ) ~ a+2by-{- dcy"" + 4rfy' + . . . . Clearing of fractions, we have -(x+ay + by^+cy^ + dy^ . . . y(a-^2by-\-Scy^-\-4dy' + 5ey* . . . ) = (a; + ay4.6y»+cy'-f rf/ . . . ^B + ^Cy + SDy' + iEy'-hSFy' . . .) By substitution from Eq. (-4), we get -(A+JBy-{-Cy'' + Dy'-i-Ey\..){a-h2by + 3cy'' + 4:dy'-\-5ey\,,)=: s ^x + ay + by' + cy' + dy* . . . ){B + 2Cy + Sl)y'-{-4:Ey'-{-5Fy* , . .) Performing the multiplication indicated, we have -aA + aB 8 -y-uiC y^aD y^B y.aF -y'+aG s "^f-vaH + 2bA +2bB +2bC -^2bD -\-2bF +2bF +2bG +3cA +3cB +3c(7 +ScD +ScB +3cF ndA\ ^4:dB +4dC -\-4dJ) +4:dF +5eA -\-5eB ^heC ■¥5 el) +6/4 WO +ShA [Forward 410 THE METHOD OF EVOLUTION. BxMiB y-\-h B f+cB f+dB y*+eB f^fB f+gB +2x0 +2aa +2bC +2cO -v2dC +2eC +2/0 ^3xD +3al) ■V3hD ^ZcD +SdI) +3eD +4xB +4.aE •^^hE +4cE +4dE +5xF +baF +5bF +5cF +6xG +6aO +6bG -¥lxH +naH +SxI Equating the coeflficients of the like powers of y, we have T Bx^-aA s aB-{-2xC=aB--{-2hJ!^ 8 S bB-{-2aO+Sx£>=aC - + 2bB-+3cA- S S 8 cB-^2bC-^ 3aD + AxE=aD- + 2bC~ + 3cB- + 4dA- 8 S S S dB + 2cC-h3bD + 4aE+5xF=aE- + 2bD^ + 3cC--{-4dB-+5eA^ s s s s s eB+2dC+3cI)+4tbE+5aF+6xG=aF^2bE-+3cJ)-+idC'^5eB-+efA- S S S S S 8 There is a symmetry about these equations which would enable us to fonn successive ones without resorting to the equation from which these have been derived. r . . L Putting 71=-, and instead of A write its value x>=afy and solving the above equations, we have A=af JB=nax'^^ nin-l){n---2) 3 'a^x'^^-\-n{n—l)ahar-^-{-ncaf^^'' a^bx^'+'!^\2cu;+b')xr-2 „^ n{n-l){nr-2){nr-3) n{n~l){n-2) 2 . 3 » 4 2 +ndar-\ It may be seen from these values we are not able to write all the terms in the value of F, although some are apparent. But, by using A, jB, (7, i), &c., instead of their values, the law becomes evident. Thus, we may obtain the following general formula : THE METHOD OF EVOLUTION. 411. + + f I « 00 I ? + I OS + ? I en I f i: ? ? ? I f CO f + col- I + 05 'oo I Or ? ^ + b ^ f CO ^ to to + 00 + + to 3 -i^ ^ CO I CO b ^ to b Q + 4- 5S b os , I: b *«s, i 1 Si I CO b + 'co I to b en i CO + 'fco" I bO + 'co b + i + wi- ts to b 09 i ^1 + to + + + + + 412 EKVERSION OF SERIES. MeIemabk. — This theorem applies, of course, to both involution and evolution. The student should use the multinomial theorem in solving the following EXAMPLES. 1. Expand {l-h^ + x' + x^ + x* . . .y into a series. Ans. 1 + 3^ + 6a;' 4- 10a;' 4-1 5a;* + 2« Expand (2a; + 3a;' + 4a;'* ....)' into a series. Ans. 8a;' + 36a;* + 102a;'^ + 231a;'+ 3t Expand (1-1- a; 4- a;' + a;' + a;* . . . .)i into a series. Ans. l + Xa;-}-fa;» + y\a;' + -i?jLa^*-h 4* Eicpand (l-f-^a; + ^a;''+ia;' . . .)i into a series. Ans. l+aa; + J^^'-f-eV8«' + «\¥7^*+ REVERSION OF SERIES. (424.) The reversion of a series is finding another series which is equal to the unknown quantity in the given series. PROBLEM (425.) 1. Revert the series aa; -I- 6a;' 4- ca;'H-6?a;* SOLUTION. Assume that x=Ai/-{-By^+ Ci/^ + Dy* , y being equal to ax-\-bx^-hcx^-{-dx*' .... Substituting the value x in the equation yzzzax + bx' + cx^ + dx* . . . ., we have y^Of + 0i/* + Oy*=aAy+aB -' hA ■\-2hAB cA' Equating the. like coeflScients of y, we have aA^V 6^'4a^=0 cA^ + 1hAB+aG=0 dA*+ScA'B+hB'+2bAC+aD=:0 . whence y'4- aD 4- 2b AG + bJB' -\-3cA'JB dA ^4 a or ^^ 26'-ac , y*+ .... I B=- bh'—babc+a^d 1 b , 26'— ac 3 bb''—5abc-\-a^d , Therefore, a;=-y jy' -I g — y CL a a REVERSION OF SERIES. m PROBLEM 2. 'ReveTti/=ax+hx^+cx^+dz'' . . . , SOLUTION. Assume x—Ay + By^ + Cy^ + JOy'' . . Substituting x for its value, we have y+Oy' + Oy' + 0''=aAy-\-aB f-\- aO y'+ aD /+..•. hA' + dhA'B + 3M'(7 cA^ ■\-hcA^B dA' + 3bAJS' Equating like coefficients, we obtain \a==\ - a' aA=l ' Cb ■' bA' + ajB=0 4) cA' + ShA'J3 + aC=0 SlAB'+dA'+5cA'B+BM^C+aD=0 J ■ , whence • a -ac 7 > jy__ I2h-8dbc+cfid a" PROBLEM 3. "Revert ax +hx'*+cx^-\-dx* .... =py-\-qy*-{-ry*+sy* SOLUTIO N. Assume y= Ax -{■ Bx^ + Cx^-\-Dx* .... Substituting this value of y, we have ax + bx^+cx^+dx* =pAx+pB |a;'4- pC qA""] +2qAB rA' Equating the like coefficients of ar, there results a a;»+ pl> + 2qAC + qB' -hSrA'B sA' pA=a' qA^+pB=b rA^+2qAB+pC=c 8A*+3rA^B+q&+2qA C+pD=d , whence ■ P' j^_ bp^-a*q ri_cp*—2abp'^q + a^q + 2cfipr J' • rj_cp*—a*pf + 2aq(a^q—hpV Therefore, y^g^ + ^X:zg!g,.+ '^'-"> + y g: ±l)^.+ . .. P P P 414 REVERSION OF SERIES. PROBLEM 4. Revert ^+aa; 4- 6a? + car* + c?a;* . . . =y. SOL0TIO N. Transposing jt?, we have ax + bx-\-cx^-\-dx* . . . z=y—p, = This series is the same as the one in the first problem except that y — p stands in the place of y. Therefore, its reversion must be. 5b'—5abc-\-a'd, — ^ — ^-^y EXAMPLES. L Reverty=a;+a;' + a;' + a;* . . . Ans, x—y—y*+f^x*'\- 2. Reverty=a;— Ja;' + ia;'— la;* .... Arts. a;=y+iy' + iy' + |/ 3. Revert y=a;—ia;Hia;^—|a;' .... Ans. x=y + if + ^jy' + ^\\f !• Revert y = 2a; + 3ic' + 4a;' 4- 5a;' .... Ans. x^^JZ-T^y' + j'^y'-^WTf 5. Revert y=l+a;+|a;'+J-a;'' 4- aV^* ... Ans. x=y-l-±{y-iy + ^{y-lY^^tZ^ . . ((• Revert y=a:4-2a;*4-4a;^ .... Ans. a;=y— 2y'— 4y' 7. Revert y=aJ—^aj^-f-ia;'—J-a5* + ia;' ... 4ns, x=y^t + ^ +JL4.^L^ . . . ^2 2-3 2-3-4 2-3-4-5 « -^ a;". x' x' 8. Revert ^=0.--+-^:^^-^:^:^:^:^ . . . x' , l-3a;' , l-3-5a;' ^^2-3 2-4-5 2-4-6-7^ 9, Reverty=:a; + 3a;'' + 5a;' + 7a;* + 9a;' Ans. a;=y— 3y' + 13y''— 53/ *-. T> a;' arV a:* , a;' 10. Reverty=a;+— +— + —+— .... Z O 'k O Ans. x=y--—-\- — —-^— ^ 2 2-3 2-8-4 SUMMATION OF SEEIES. 416 SUMMATION OF SERIES. The summation of a series depends upon its nature. ARITHMETICAL SERIES. (426.) The summation of a finite arithmetical series, whether in- creasing or decreasing, we have ah-eady seen is embraced in the for- mula ^=(— ^-Jw, in which a is the first term, I the last term, and n the number of terms. aEOMBTRICAL SERIES. The summation of a finite geometrical series, whether increasing or 1.1/. 1 rt ir—a a—lr decreasmg, is embraced m the formula S= ~ or _ . When the series is infinite and decreasing, the formula is fS= -. RECURRING SERIES. A recurring series is one in which a certain number of consecutive terms, taken in any part of the series, has a fixed relation to the term which immediately follows. Thus, in the series l-{-3x + 4:X* + 1x'-{-llx* + 18x'' .... the sum of the coeflScients of any two consecutive terms is equal to the coefl^- cient of the following term. In the series l-^2x+3x^ + 4x^ + 5x*-\-Qx^ each term after the second is equal to the product of 2x into the first preceding term plus the product of —x^ into the second preceding term. Thus 5x^=2xx4x^—x^x3x'^. The coeflScients of x and a;', or 2—1, is called the scale of relation. Intheseriesl+4a; + 6a;'-f lla;'+28.r*4-63a:' ,2 — 14-3 is the scale of relation. Thus 63aj'— 2a: x 28a:' —a;' x 11a:' + 3a:' x 6x^. (427.) The fraction -^ produces the series ^ ^ +CX ^ a OCX ac^x^ ac^x* h W^'~Y w ex in which each term after the first is equal to the product of — j- into 44§ SUMMATION OF SEEIES. the term that precedes it. In this series, — — is the scale of relation f> of the terms, and - is the scale of relations of the coefficients. CL -I- l)X (428.) The fraction - — ; r produces the series. ^ ^ c + dx +ex^ ^ a Ih ad\ /bd ad^ ae\ in which each term, commencing at the third, is equal to the two cc dx immediately preceding multiplied respectively by — '—, , which c c is the scale of relation of the terms. ft _L J/j« _1_ /»'>•' (429.) The fraction -— -. will produce a series in d + ex -irfx^-\-gx^ ^ which each term commencing at the fourth is equal to the three pre- yy/V* "foi^ 6X ceding terms multiplied respectively by — ^, — ^^» ~T> ^^^^^ ^^ therefore the scale of relation of the terms. (430.) The fraction „ '-^- t^t; will produce a q-r-rx + sx^ . . . ux''-\-vx''-^^ ^ series in which each term commencing at the n + 2th will be equal to the n + \ preceding terms multiplied respectively by vx"^^ u:f sx^ rx T' ~T' ■ ■ • • • ~'V ~1' which is the scale of relation of the terms. When ^=1 the scale of relation is — vaf+\ —war", . . . —sx^^ —rx, which are the terms of the denominator taken in reverse order and with contrary signs, the first term being omitted. (431.) A recurring series is said to be of the first order when each term commencing with the second depends upon the one that precedes it. (432.) A recurring series is said to be of the second order when each term commencing with the third depends on the two preceding terms. In like manner we have recurring series of the third order, fcmrth order, &c. PiJOBLEM (433.) 1. To find the scale of relation in a recurring series of the first order. SUMMATION OF SERIES. 417 SOLUTION. Let A, B, Cy 2), &c., ... be the coefficients of a recurring series of the first order. Now, some relation exists between each two consecutive terms. Let r be the relation, and we have B—rA\ 0=rB ; D—rC^ &c. whence, r=— -=-^=:-;^, &c. PROBLEM (434.) 2. To find the scale of relation in a recurring series of the second order. SOL UTION. Let A, B, (7, D, &c., be the coefficients of the series. Whence by the conditions must arise the following equations. C=zmA-\-nB, D=mB+nC, &c., &(5., in which m and n are unknown quantities. The solution of these two equations must give their values. BC=mAB-\-nB\ AB—m AB + nAC, AB-BC=(AC-B')n, _ AD-BC ""- AC~B'' In the same way m may be found, whose value is seen in the equation C'-BB m=z AC-B'' PROBLEM (435t) 3. To find the scale of relation in a recurring series of the third order. SOL UTION. Let A, B, (7, i). By F^ &c., be the coefficients. By the conditions, we must have B = mA -\-nB-{-q C, E=mB-\-nC+qB, F=mC-\-nB + qB, &c., &c. The solution of these equations gives the values of m, n, and q, 27 418 SUMMATION OF SERIES^ PROBLEM (436.) 4. To find whether the law of the series depends on one, two, three, or more terms. SOLUTION See whether the scale of relation established by problem 1st agrees with the given series ; if not, try problem 2d ; and if that does not agree, try problem 3d, and so on until the scale is found. Or we assume that the series is of a higher order than necessary, in which case one or more of the values found by resolving the resulting equations will be found to be zero, and the remaining one will give the true scale of relation. PROBLEM (437,) 5. To find the sum of a recurring series of the first order. SOLUTION. It is evident that when the series is of the first order (that is, a geometrical series), the sum of the series may be obtained from the formulas, S= -, S=- , and S=- . The first formula must be used when the series is increasing and finite ; the second, when it is decreasing and finite ; and the third, when it is decreasing and infinite. PROBLEM (438.) 6. To find the sum of an infinite recurring series of the second order. SOLUTION. Let -4 + J?+ (7+2> .... be a series of the second order. Then C=mAx''+nBx D=mBx'-^nCx Ii=mPx^-\-nQx • Adding these equations together, and putting A + JB-h C-\-D H . . . .^S, we shall have SUMMATION OF SERIES. 419 {\—nx—mx'^)S=A-{-B—nAx A + B—nAx S=- \—nx—mx* PROBLEM (439.) *1. To find the sum of an infinite recurring series of the third order, SOLUTION. ' Let A + B+ C+JD, &c. be the series. According to the nature of a recurring series of the third order there must result the following equation : D = mAx^ + nBx^ -\-qCx E=mBx^-\-nCx''-\-qDx Adding these equations, and putting S=A-^B-\- C+D-\-E, &c., we have S—A—B—C=mSx'+n{S—A)x^-\-q{S—A'-B)x ^A + B+C—{A-\-B)qx—nAx* wnence. o — ~ — „ , 1—qx—nx—mx PROBLEM (440.) 8. To find the sum of a given number of terms of a re- curring series of the second order. SOLUTION. Let A-^B+ C+J) .... 4-i2 be a finite recurring series of the second order. We have C=mAx' + nBx J)=mBx^+nCx T=mBx^-{-nSx. Let us find the sum of the recurring series of the second order, U+ V+ W, &c., to infinity. We have, according to Prob. 6, o =— 5- 1—nx—mx 420 SUMMATION OF SERIES. Supposing A + JB+Cj (fee, to be an infinite series, we have already found its sum to be A + B—nAx 1—nx—mx^ Subtracting from this the sura of U-h V+ TT, &c., to infinity, and tbeie must remain the sum of the finite series, A-]-£+0 . . . . T. Putting this sum equal to S, we have ^ A + B—U—V-nAx+nUx l—nx — mx Note. — ^In the same way we might obtain the sum of a finite re- cuning series of higher orders. PROBLEM (441,) 9. To find the general term of a recurring series. SOLUTION. «^ , . - . a-}-bx-\-cx* . . . . paf* , The general generatmg fraction, ^_^_^^^^^.^ _ _^^^+v may be thus represented : (a + bx + cx'' , . . .px'')(q-{-rx + sx'* .... +vx'^^)-^ The second parenthesis may be expanded by the multinomial theorem^ and the result multiplied by the quantity in the first pa- renthesis. Then, if we take in this product the part which contains x to any power whatsoever, we shall obtain the general term of the recurring series. We present another method. Take the generating fraction, a-\-bx+cx^ . . , psf" q-{-rx+sx'^ . . . vrr''+* which is supposed to be reduced to its lowest terms. Dividing both numerator and denominator by v, and changing the order of their terms, we have V c „ b a . . . -x^-\.-x-h- V V V a;"+\ . . s „ r a V V V Separate the denominator into binomial factors. Then the fraction may be considered as made up of fractious which have these binomial SUMMATION OF SERIES. 421 fractions for their denominators. Determine these partial fractions according to the method given in undetermined coeflScients, and then find the general term of development of each of these partial fractions, and the sum of these general terms will be the general term of the recurring series. Note. — In the decomposition into partial fractions, when a factor of the denominator is of the form (x-{-a){x-{-b)"'^ assume the fraction to be equal : A B C JI_ x + a {x-^-b)"^ {x + hy*-' ' ' * x + i) Each partial fraction may be put under the form P^p+x)'' in which r is some positive integral number. The expansion of this by the binomial theorem gives for the term which contains x"" the following -r{-r-l)(- r~ 2) .... (-r-(« -l) r-2-3 . . . . « ^^ "^ • It is the sum of the terms containing a:" resulting from the different partial fractions, that composes the general term in the recurring series. (442.) A simpler mode of finding the general term of a recurring series will be found in the following discussion. We have already seen that a recurring series of the first order is a geometrical series, and, therefore, its general term is easily found. Let us suppose that a + ar + ar^ + ar^ + ar* . . . , a geometrical series, is also a recurring series of the second order. When this is the case, we must then have the following equations : ar^=:ma'\-nar, ar^=mar-\-nar^, ar* = mar^ + war', ar*: in which m, n is the scale of relation. - By division each of the above equations reduces to n±V^m-\-n^ x whence r = . 2 If these two values of r are not equal, we see that there may be two geometrical series which will also be recurring series of the second order, having m, n for their scale of relation. 422 SUMMATION OF SERIES. Let the two values of r be c, and h. Then since the first terms of "these two geometrical recurring series of the second order, may be any quantities whatever ; let the variables x and y represent them. We shall then have x+xc + xc^-\-xc^ ...... xc'^\ yj^yh+yh'+yh' yh^^-K Each of these series are recurring and of the secohd order, and the scale of relation in each is m, n. By adding them, we have {x-\-y) + {cx + hy) + (c''x-\-b^y)+{c'x+b'y) {xc^'+yh-% This series is not geometrical, although formed by the addition of two series which are geometrical. But these two series are also re- curring series of the second order. Have we a right then to conclude because two geometrical series added together, do not produce a geometrical series, that these two geometrical recurring series have not when added produced a recurring series ? Let us examine. If the series {x + y) + {cx + hy)-\-{fx-\-h^y) .... is a recurring series of the second order, whose scale of relation is m, n, we must have c^x ■\-h'^y=m{x + y)-\- n(cx + by) SinceaJ + caj-f-c'ic+c^ &c., and y-\-by + ¥y-\-b''y, &c., are recur- ring series whose scale of relation is m, %, we have the following equations : c'^a; = mx + ncx b^yz=my + 7iby. By addition c'x + b^y=m(x-^y)-\-n{cx-\-by) which proves that {x + y)-\-{ax+by) + {c'x+b'y) .... {xa^-'+yb^') is a recurring series of the second order, whose scale of relation is m, n. Let this series be represented by A-^B+C-^I) ...... T; 7" standing for the generaL term. The following equations must then subsist : x-^y=A, cx + by=:Bf B-bA ' whence, x= — , ' c—b B-cA 2^="-^:=6- We also have T =:a?c^' + ^6" ^ B-bA , B-cA' T=- — j-C^' j-6*-*. c—b c—o SUMMATION OF SERIES, 423 By assuming a geometrical series to be a recurring series of the third order, we should find r by solving a cubic equation. Then pro- ceeding in a similar manner, we should find that a recurring series of the third order is equal to the sum of three geometrical series, having the same scale of relation, whence the general term for a series of the third order might be obtained. • EXAMPLE S 1. Find the sum of 1 + 2a: + Sa;" 4- 28a;' + 100a;' + 356a;',S, in which ^rep- resents the true sum of the whole series. Now let a—h-\-c—d m—n-^p—q-\-r be the series to n-\-\ terms, n-\-\ being an odd number. The terms after r are —s + t—u + v These terms may be thus represented, — (s— <)—(%— v)— .... The quantities in the parenthesis are positive, and since each parenthesis is negative, it follows that the series which terminates in r is followed by a series of negative quantities, whence the following inequation, a—h-\-c^d m—n-\-p—q-\-r'^S. From this, we see that the quantity necessary to be added to the first member of the first inequation to make it equal to S must be less than r. CASE II. When the number of terms taken is odd. Let a—h-\-c—d m— w+^ be the series to n terms, n being odd. Reasoning, as in the last case, we have the following inequations, a—h + c—d m—n+p'^S, and a — h + c — d ..... m — n-^p—q<^S^ 426 THE DIFFERENTIAL METHOD. which show that the quantity to be ^btracted from the first member of the first inequation to make it equal to *S^ must be less than q. The solution of this problem establishes the following THEOREM. (4:4:T,) The error committed by taking n terms of a decreasing converging series^ whose terms are alternately positive and negative^ for the sum of the whole series, is numerically less than the following term. Note. — In applying this theorem to certain examples, the series should be considered as commencing at the second term of the ex- pansion. i^ »> ♦ •> »■ THE DIFFERENTIAL METHOD. (448.) The differential method of summing a series is a method of finding the sum of a series by ascertaining the successive differences of its terms. PROBLEM. (449.) 1. To find the first term of any order of differences. SOLUTION. Let a, 6, c, d, e,f ,=6— a, i>2=c-26 + a, B^^d—'Sc + Sb—a, i>4=:e— 4c? + 6c— 46 + a, i>5=r/~5e + 10(]?— 10<;+66— a, &c., &c. Whence may be obtained b=a + D^, c=:a + 22)^+2>2, &c., &c. We see that in the {n-\-\)i\i term of the series a, 6, c, c?, e, 4 +, &c., from whicli we see tiiat the nth term 2 • 3 . 4 428 THE DIFFERENTIAL METHOD , , « . ^ 7i(w» — 1)^ w(w— l)(n— 2)^ a, 6, c, d, , + -^—^^IB^ +-^^ — ^'^-^-^s + «(w— l)(7i — 2)(?i— 3) 2 ; 3 . 4 must bea + (7i-l)l>,+-^^ ^^^ -■Di + - — ^r-^"^ — ^^^ It 2 . o (w-l)(n-2)(7i--3)(w-4) _, , .. 1. . vx • J V 4-^-— — ^^^-- — '^ '^ I^^t <»c., which is obtained by writing n— 1 for n, PROBLEM (451*) 3. To find the sum of n terms of a series. SOT^UTION. Let a, 6, c, ,, D^^ -Dg, D^^ &c., we have by the last problem for the (w + l)th term of the series 0, o, a + 6, a-\-h-\-c^ a-\-h-\-c-\-d^ &c., or the sum of n terms of the series , « -^ ^(^— l)-r^ 7i(« — l)(w — 2) ^ . a, 6, c, (^, &c., -0 + nB, + ^ ^ ^ A + \ % ^ A + ^^_l)(^_2)(^-3) 2.3.4^* +, ,, J^g* Z>3, &c., of the assumed series are respectively equal to a, Z>i, D^^ &c. of the proposed senes) 8=na^—^-- — -D^^ — ^-- — '-^^ — ^-^a + 2.3.4 ^' ' EXAMPLES. 1, What is the first term of the fourth order of difierences of the series 1, 8, 27, 64, 125, &c.? Ans. 0. 2. What is the first term of the eighth order of difierences of the series 1, 3, 9, 27, 81, &c. ? Ans, 256. THE DIFFEKENTIAL METHOD. 429 3. What is the first term of the fifth order of diflferences of the series 1, i, i, }, j\, 3V, eV, &c. ? Ans, - 3V 4. What is the first term of the sixth order of differences of the series, 3, 6, 11, 17, 24, 36, 50, 72, &c.? Ans. —14. 5. What is the first term of the third order of differences of the series 1, 2*, 3*, 4*, &cJ. Am. 60. 6. What is the 10th term of the series 1, 4, 8, 13, 19, &c. « Ans. 64. 7. What is the 15th term of the series 1', 2\ 3\ 4^ &c. ? Ans. 225. 8» What is the 20th term of the series 1, 3, 6, 7, &c. ? Ans. 39. 9. What is the wth term of the series «, a+c?, a+2c?, a + 3c?, ±1)\ 480 APPLICATION OF THE DIFPEHENTIAL METHOD. 18. What is the sum of n terms of the series 1, 3, 6, 10, 15, &o. 1 7i(%+l)(w + 2) Ans, -\ ^ ■ ;^ ' , 1.2.3 19. What is the sum of n terms of the series l(w + 1), 2(w+2), 8(m4-3), 4(m + 4), &c*? Ans, '^ ; o ^-^—^- 1 • 2 . 3 APPLICATION OF THE DIFFERENTIAL METHOD. PROBLEM. (452.) To find the number of balls or shells in a triangular pile. SOLUTION. By an inspection of a triangular pile of balls, or shells, we see that the first tier, commencing at the top, has one ball ; the second, 1+ 2 ; the third, 1 + 2 + 3 ; the 4th, 1 + 2 + 3 + 4; and the nth, 1+2 + 3+4 + 5 . . . +w. The number of tiers equals the number of balls, or shells, in one side of the base. Then, to find the number of balls, or shells, in a triangular pile of n tiers, we must find the sum of the series 1, 3, 6, 10, .... to » terms, which is _ n{n+\){n + 2 ) 1.2.3 PROBLEM. (453.) To find the number of balls or shells in a square pile. SOLUTION. By an inspection of a square pile of balls, or shells, we see that the first tier, com- mencing at the top, has I'' ; the second, 2^* ; the third, 3^^ ; the fourth, 4^* ; and the nth, w' balls, or shells. Then, to find the number of balls, or shells, in a square pile of n tiers, we must find the sum of the series l^ 2', 3", 4', 5', to n terms, which is n{n^l)(inj^) 1.2.3 APPLICATION OF THE DIFFERENTIAL METHOD. 431 PROBLEM. (45 4«) To find the number of balls or shells, in an oblong pile, SOLUTION. By an inspection of an oblong pile of balls, or shells, we see that if m + 1 represents the number in the first tier, commencing at the top, the number in the second is 2(w+2); intheSd, 3(w^ + 3)• in the 4th, 4(m + 4) ; and in the nth, n(m + n). Then, to find the number of balls, or shells, in an oblong pile of n tiers, we must find the sum of l(m + I), 2(//i4-2), 3(m + 3), , . . . to n terms, which is w(% + l)(l+2w4-3m) 1.2.3 (455«) These three formulas may be written as follows : In the triangular pile, 3=^ . -^^— — • . (w + 1 + 1) " " square " 5=i . ^^^^\ (w + w + l) " " oblong ** 5=:i.^?i!!±D i(ri+m) + (w+m)+(m+l)} Since -~- — - represents the number of balls, or shells, in the ii triangular face of each pile, and the other factor the number in the longest base line, plus the number in the line parallel to it, plus the number in the top tier, we have the following GENERAL RULE. Multiply one-third the number in the triangular face hy the num' her in the longest base line^ increased by the number in the opposite parallel line and by the number in the top tier, and the product will be the whole number of balls in the pile, EXAMPLES. 1* How many balls in a triangular pile of 16 courses ? Ans, 680* 432 SPECIAL SERIES. 2. A complete square pile of shells has 14 courses : how many shells are in the pile, and how many remain after the removal of 5 courses ? Ans. 1015 in the pile, and 960 remain. 3* In an incomplete oblong pile of balls, the length and breadth at the bottom are respectively 46 and 20, and the length and breadth at the top are 35 and 9. How many does it contain ? Ans. 7190. 4. The number of balls in an incomplete square pile is equal to 6 times the number removed, and the number of courses left is equal to the number of courses taken away. How many balls were in the complete pile ? Ans. 385. 5* Let h and k denote the length and breadth at the top of an oblong truncated pile, and n the number of balls in each of the slant- ing edges. How many in this truncated pile ? Ans. ^J2n^ + 3n{h+k) + 6hk—S{k-[-k+n)+l^, 6« How many shot in a triangular pile whose bottom row contains 8 shot? ' Ans. 120. 7. How many shot in a square pile whose bottom row contains 8 shot? Ans. 204. 8. How many shot in an oblong pile whose length and breadth at the bottom contains respectively 16 and T ? Ans. 392. 9. How many shot in a triangular pile whose bottom row contains 30 shot ? Ans. 4960. 10. How many shot in a square pile whose bottom row contains 30 shot-? Ans. 9455. 1 1 . How many shot in an oblong pile whose number of courses is 30, and top row 31 ? Ans. 23405. 12. How many shot in an incomplete oblong pile whose length and breadth at the bottom are respectively 46 and 20, and at the top 85 and 9? • Ans, 1190, i» ■■ » t. »■ SPECIAL SERIES. THEOREM. 1 (456.) Any fraction of the form ^ is equal to - of tha q q difference between the fractions - and . SPECIAL SEKIES. 488 DEMONSTBATION. g q ^ qn-hpg-nq ^ pq ^ q _l/g_ 9 \ n n+p n(n+p) n(n-\-p)^' ' n{n-\-p) p\n n-\-p)' Q. K D. OoROLLABY. — If the difference between - and — — is Imown, the •^ n n+p value of —r^ — c will be known, whether - and — — are known or n(n +p) n n -\-p not. Hence, we obtain the following PRINCIPLE. In any series of fractions having the form — ^-^ — r the sum of the n{n-\-p) ^ series is equal to -th of the difference between a series of fractions of the form -, and a series of fractions of the form QUESTION . (45Ti) What is the sum of the series t:^+-^ + — +... to in- finity? SOUTTION. Here q n{n-{-p) 1*2 q - 1 n{n-\-p) 2*3 q _ 1 •. §'=1 ; n=\ ; w+jt>=2, or^=l. '. ^=1; 71=2; w+^=3, or^=l. •. 3'=1; w=3; w+^=4, or^=l. n{n 4-j9) 3*4 From which, we see that q constantly equals 1 and p constantly equals 1, and n successively equals 1, 2, 3, ^^'^ *^ ^^' I'o'o o'o'i 5*7'9 finity. Ans, |. 2. What is the sum of the series -±-+^+ J^ + -^+, &c., to infinity ? Ans. -^j. 3 9 15 3* What is the sum of the series — — --+ (..-..,.-,. + n.iAo^ "^^ al to the of the difference between a '' ^ mp series of fractions of the form —. .-7 ~"^ / , i\ \' •^ -^ J J n{n^p)(n-\-2p) .... [n + {m—l)pj and a series of fractions of thefwm -. tt — — — r 7 — ; ^. Again reasoning as in (456), we obtain the following SPECIAL SERIES. 487 PRINCIPLE. (46 It) In any series of fractions of the form a(a-{-b) .... (a+pb) n(n + b) . . . , (n+pby the sum of the series is equal to of the difference between a j: J- *' j^ *-L X ' «(« + &) .... {a-\-pb) series of fractions of the form —7 r^ f ^7-^ — tt^* «^» a '' '^ J J n{n + b) .... [n-h{p—l)by series of fractions of the form —} y! ' ' ' ' ; it- — —^* •^ "^ ^ ^ n(n + b) .... (n+pb) QUESTIONS. 1. What is the sum of r terms of the series - + —— + — —- + 1-3-5-7 ^ , . l'3-6-'7 . . . . (2r+l) ^ — — — — + ,&o.? Ans. — — ^ ^— 1. 2-4-6-8 * 2-4-6 .... 27* 2 2*4 2'4'6 2. What is the smn of r terms of the series -4--—+ + Z'o 0*0*7 2-4-6-8 ^ . , . 2-4-6-8 . . . . (2r+2) ?5W + '*"-' ^'"- 3-5-T-9 ■ ■ ■ ■ (2r +!) -'• 9 2*3 2*3*4 3. What is the smn of the series rrr + ^-^77: + t^-^ttt + » <^c*» ^ 0*0 o*D'7 o*D*7*8 infinity? Ans. ^. 4* What IS the sum of r terms ot the senes - + • ' n n(n + b) a(a + b){a + 2b) ^ n(n + b)(n-h2b) ^^'^'"^ ^^^ 1 / a(a + b){a+2b) (a + rb) \ ' n—a-b\ n{n+b){n + 2b) .... [n + {r—l)b])' Remark. — If 7i=a + 26, whence ni-b=a + Sb and n + 2b=a-{-4by &c., and n + (r—2)b=a + rb, the fraction in the parenthesis becomes a(a + b) a + {l^r)b' which vanishes when r is infinite, and we therefore have for the sum of the series to infinity a n—a^h' 438 SPECIAL SERIES. When 71= a 4- ft, the fraction in the parenthesis becomes a, and the sum of the series -(a--a)=-(0)=-, an expression of no defi- n — a — nite signification in its present form. It may be observed that this is obtained without reference to the value of r ; hence, whether r is infinite or finite, the result is the same. (46!2#) Some series may be very beautifully summed as follows. QUESTION 1. What is the sum of the infinite series x + x*-}-x'-\-x*-\-y &c. ? 1*^*3 ^*o*4 o'4'o 3 4 5 4* What is the sum of the infinite series — — + + 3 + , (fee. ? 1*2*2 2*3*2 0*4*2 Ans. 1. tj ft 5* What is the sum of the infinite series ,:o.Q.oa + o.Q.^.oa + 1*2*3*2 2*3*4*2 ^+,&c? Ans.i. 6. What is the sum of the infinite series + in.or + ToTo"^~ + 8*18 10*21 12*24 1 3 + ,