at ed Ded ob eet abe ie i ite riage i Ht Lay Pa ul ap, let L Al Wi - ta Ha t HE TBAT tats ites tate | ee iy Hirt aenaneniit titi ae itis batt Hil t Hit ht i iit i i oo Ai Huis i nigitty as + facet HE Haines il pit Hut rae AY | ms ua S 24 it AGI es ot i 2 a a i. stasis ae Te HY a it mite ete abel eater hosat ait} iti ba raat 7 ren Bit i tai ri $i a tite Uitte rt bi 7 i th a a it i tet a i ie i sit ‘4 i, bite! r Het SE. =F. Feaetsters Hn is arte stret, . y eecesceeressecseess Pe ate = Comes Sgparsas : a i ait that atthe Dit 1 ie ‘isi iti bea males it Pit . ta iH i ea H a I Te tb etait: ie Hite cnet Ut : Pett ieetitelntetshaiitets i ee EAR ETL iat iririsrrtetrien betty a Le isha ay tty ce - if ‘ Co 4 ity ; is i t aie 1 ii eae a ty ah aint i a i} et Haat it} sii a tet Hite ee tera he i itty i a a agittst i A 7 rf HT fit aya! ithe Hf i ! pu i thas ‘ ie iftrnth th it | Hi tlie , ee res it eit i se Heh is rei tt att pt remesr Seta! steer: sist Sazee steed B55 = SHEL Settee wate wesetatsess stoe SEESED —s = 3: eserscorrasee aris = ststctes Dyes pepebepecererereneeces st geteierseraes pas 5 oslo pases sects =F pic peuene stews S325 So Tetene—e. THE UNIVERSITY - OF ILLINOIS LIBRARY $512.9 anorectic penance Tags MATHEMATICS LIBRAM Return this book on or before the Latest Date stamped below. University of Illinois Library +- OCT 68 RLY L161—H41 American Mathematical Series E. J. 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TURNEAURE, Dean of the College of Engineering in the University of Wisconsin. xii+266 pp. 12mo. $1.60 Analytic Geometry of Space By Vircit SNYDER, Professor in Cornell University, and C. H. Stsam, Assistant Professor in the University of Illinois. xi+289 pp. 12mo. $2.50. Elementary Geometry By J. W. Youna, Professor of Mathematics in Dartmouth College, and A. J. Scuwarrz, William McKinley High School, St. Louis. School Algebra } BycH: L.-Ruievtz, Professor of Mathematical Statistics, A. R. CRaA- THORNE, Associate in Mathematics, University of Illinois, and E. H. TayLor, Professor of Mathematics in the Eastern Illinois State Nor- mal School. Functions of a Complex Variable By E. J. Townsenp, Professor and Head of the Department of Mathematics in the University of Illinois. HENRY HOM AN D COMPANY NEW YORK CHICAGO SCHOOL ALGEBRA SECOND COURSE BY is hod Rd od 8 ed BVA aa ered DB UNIVERSITY OF ILLINOIS mee CRA THORNE, Pu. D. UNIVERSITY OF ILLINOIS AND Hue, LAY LOR, ‘PH.D; EASTERN ILLINOIS STATE NORMAL SCHOOL NEW YORK HENRY HOLT AND COMPANY 1915 * . “wr at 4 E : , ‘ as ret - ae ‘COPYRIGHT, 1915 | a HENRY HOLT AND COMPAN uy se 1919 MATHEMATICS LIBRARY PREFACE This Algebra is the second volume of a series of two books. It contains material for a half-year course and presupposes a year’s work such as is provided in the authors’ First Course, or any similar text. Since, in many schools, a year or more intervenes between the first and second courses in algebra, it has been deemed advisable to include a review of the First Course before proceeding to new subjects. This part of the book is, however, more than a mere hasty resumé. While the student is reviewing the topics of the first year, he is at the same time making a distinct advance by seeing the subject from new. view-points, which his added maturity and training enable him to appreciate. The selection of new topics in this Second Course is in accord with current practice in the best high schools. While the book contains material for thorough preparation for college, the authors have not had in mind solely the students making such preparation, but rather the great body of students who will go no further in our educational system than the high school. The chapter devoted to the notion of a function is introduced in the belief that the high school student should become somewhat familiar with the idea of the correspondence between two re- lated variables. Along with the notion of a function, the graph is presented with a view to leading the students to picture and to visualize this correspondence. A special effort has been made to obtain interesting problems that have an inherent value even when considered apart from the algebraic principles involved in their solution. Without aiming to teach physics or engineering, some of the problems lv . PREFACE have a physical setting; but, in such cases, care has been taken to assume no more technical knowledge than is possessed by the average high school student. A feature of the numerous review exercises and problems is the inclusion of some typical college entrance examination questions. The authors wish to express their thanks to the teachers who have contributed. to this book by their suggestions. We are especially indebted to Prof. E. J. Townsend and Dr. E. B. Lytle of the University of Illinois and to Miss Jessie D. Brakensiek of Quincy, Ill., High School, for careful and critical reading of the manuscript and for suggestions as to exercises and problems; to Mr. J. L. Dunn of Lewis and Clark High School, Spokane, Wash., Mr. C. H. Fullerton and Mr. W. B. Skimming of East High School, Columbus, O., Mr. E. A. Hook of the Commer- cial High School, Brooklyn, N. Y., Mr. L. C. Irwin. of the Township High School, Joliet, Ill., and Mr. R. L. Modesitt of the Eastern Illinois State Normal School, Charleston, IIl., for reading the proof and seeing the book through the press. * i. eRe A. R. CRATHORNE E. H. TAYLOR CONTENTS CHAPTER I ALGEBRAIC OPERATIONS ARTICLE PAGE ier our vondamental Operations. . . 2.0. . ste ee ew 1 ret Ubtraction . . . . ww woe 8 we lee we 2 MUPIIPROTUreNGHeSeS:. . . kk kk i kh ee we 3 4, Multiplication. .... Poe is codes Labnre as | Se 5 TTR ee we ee 6 6. Multiplication of Rate wninls See ish. kc ease Pe cast, es 6 7. Multiplication of Polynomials by Noemi SS; eR ea . cous ws te et te 12 CHAPTER II LINEAR EQUATIONS IN ONE UNKNOWN Eg SS et Pe er ak eer ee 15 18. Equations and Petes as Pentcnces > 1, ideale eee ae a Peon or Hoot of an Equation. ...... 4... . eo ee MenGmetivmiraieng quatiOnNS . . . . . 1 1 ee tk ww we 16 Derren OT TAQUALIONS .~. . . . - «ss 6 8s 0 eee 1; 22. Linear Equations in One Taino . ew tye 18 23. Verification by Substitution... 2... 6 ee ee ee 18 vl CONTENTS CHAPTER III FACTORING ARTICLE PAGE 24. Rational Integral Expressions .. . 5 « « . si\smin: sen 25 26. Definition of Factoring . . .°. . «2s «© 0s sneene 25 26. Important Special Products . .) ... . 4. » % en eee 26 41.2 be urthersey pe Products 20.5 oo a) sna eee Pe 29 28, Factor Theorem . >. . . 4%. Su) 2) so 30 29. The Sum or Difference of the Same Two Powers. ...... 32 30. Summary of Factoring’ .-°.) 2) 2 34 31. Solution of Equations by Factoriige oes ce ene 38 32. Highest Common Factor . . ...°5. 5. 2) seve 38 33. Lowest Common Multiple. . 1... ... . 1: sts eee 40 CHAPTER IV FRACTIONS $4. Algebraic Fraction’. ... .o 0a 3. %. « cats 6s 4} 36. Reduction of Fractions . . . .-.. 55>. . . eee 41 So, heduction to Lowest Terms:< <=). 05-5 6s ee Ye) 3 42 37; Addition and Subtraction. . . 4%. 5... 2°.°. =.) nn 44 38. Multiplication and Division . . . . . « . . 94 4 46 39. Complex Fractions... . ak sa) a 48 40. Equations Involving F pcre ame 50 CHAPTER V FUNCTIONS 41. Constants and Variables .. . ... . 2. . . ee eure 42. Definition of a Function ....... < « . «nn 59 43. Functional Notation . 2... 9... .. . . . . ne 60 44. Evaluation of Functions 61 45. Formulas Frequently Used “5... . =. . . 13 62 46. System of Coérdinates JF... . . 2.) (on 66 47. Graph ofa Function . ee... dS 68 CONTENTS CHAPTER VI SYSTEMS OF LINEAR EQUATIONS ARTICLE 48. . Graphical Solution of a System of Linear Equations .... . mmeneranear Mquations ~. .-« +s. «1 + et be ft NISSIM Fs Ch yg - ey vem deen ee ge elk cde Ry . Elimination by Addition and Subtraction. ......... Pemesrurnstnstivution 6. ek A wows System of Linear Equations Solved by Determinants... . . Perera Os CNG.) bird Order «<< gel ee . Solution of Three Equations with Three Unknowns ..... MeeerOIenG EATUAGIONS . 2 4 6 ky te te we CHAPTER VII RATIO, PROPORTION AND VARIATION RES a ea TeLarieumVvolved.in Wieasurement . .. . . . 6 65 6 a 8 PMU et ee ee es MM Te een TOOOTIIONS 4... se ke ee ee RE ee sk a aa we a es SS RCO a a CHAPTER VIII EXTENSION OF THE NUMBER CONCEPT eM eUTaCtIONSs Joo 6 8). ew ee ee MEERREMMMTUOOTS © § (05: ys, 5) spe ee (sb wen op te Slows ITE tiers FF ta eo es ee Ieee gg Py SARTO oe Se te i OG sO” PS RP Rr ea aa SEMeetee sche lL tg Oh ae alg es BS . Products of Imaginaries. ... . 53 ew SE a DONTE Sk ew ee . Operations with Complex inne on ee BS Se meenmiueate Compiex Numbers .....: .. +... + Gee . Graph of a Pure Imaginary Number . . .. .’< te . Graph of a Complex Number .....--. +++ e+e ees vil Vill CONTENTS CHAPTER IX QUADRATIC EQUATIONS ARTICLE F . : PAGE 77. Typical'Form . . 1°... 6 « « + # ts) nots 123 78. Solution by Factoring: . ... . . si. 5 + > eee 124 TomeeoulLion py. Hormula, .. . a 8 ee 0 a 125 80. Special or Incomplete Gusdeates reer? ? roe 128 81. Formation of Equations with Given ear . 9 ee Aer ase 82. Nature of the Roots of a Quadratic . . . .. 5 Vj 83. Graph of the Quadratic Function .. 5 . 7g 134 CHAPTER X 84, 85. 86. 87. 88. 89. 90. 91. 92. 93. 94. 95. 96. 97. 98. 99. 100. 101. 102. SYSTEMS OF EQUATIONS INVOLVING QUADRATICS Introduction =e ee 141 One Equation Linear el Gna Quadtratie - 2 ee +e ee Both Equations Quadratic .. . oe + a 145 Both Equations of the Form az? + by? ae c=0. . 2 2 ae 145 Further Special Methods .... .% . . 5. en 146 CHAPTER XI EXPONENTS, RADICALS AND ROOTS Positive Integral Exponents ... . . + ey 2 Extension of the Meaning of an Ranonert: l). ... . 86 To Find from the Table the Copentnn ‘of a ara Nanna a Tre To Find from the Table the Number that Corresponds to a PEPE LERSIITYS She Sar, SMa Var ics sha ts ono dete ss . 189 Computation by Means of Logarithms .......... 190 CHAPTER XIII EQUATIONS INVOLVING RADICALS Solution of Equations Containing Radicals. ........ 200 CHAPTER XIV EQUATIONS IN QUADRATIC FORM i Si err irers ci ceo 204 pameemeeanieous Quadratic Form .......++.+s+«see 207 CONTENTS CHAPTER XV — PROGRESSIONS ARTICLE 126. 127. 128. 129. 130. 131. 132. 133. 134. 135. 136. 137. 138. 139. 140. 141. 142. Arithmetical Progressions ...... ee ee ee ee ee he Elements of an Arithmetical Progression . . Relations among the Elements ... . Arithmetical Means 2-2) a. ee oe Geometrical Progressions .. . sige Elements of a Geometrical Brieresnicn Relations among the Elements ... . Geometrical Means: .. 4.48%, seas oun Number of Terms Infinite ...... Repeating Decimals. ........ Harmonical Progressions ...... . Harmonical’Means™. =") 72 = 4) cae CHAPTER XVI © 8 @€° 0° Se) Gy eee see * «© 6° F £8 “Sel? & 6 eee e 8 © (6 #8 6.58) BAe e 8° 688 (Ste; ete) ae eee eo “ease & (SF ee Te) eee ee, ee a @- @ «= 6! Se 1S) at See eee THE BINOMIAL THEOREM Factorial... Wee coi 2a Powers of -a: Binomial). 330-e aeeee The Expansion of (a+b)” . ¢€ sos © @& Set Oe ae eee Se a Formula for the rth Term of the Henan as (a +b)” Binomial Formula, Any Exponent . . INDEX PAGE 210 210 210 211 212 213 213 213 215 217 217 © 218 221 221 221 224 225 231 SCHOOL ALGEBRA SECOND COURSE A LIST OF SIGNS AND SYMBOLS +, read plus. —, read minus. X, or -, read times. +, read divided by. =, read is equal to. =, read is identical with. ~, read is not equal to. <, read zs less than. >, read is greater than. S, read is less than or equal to. =, read is greater than or equal to. a! or |a, read factorial a. ( ) Parentheses. [ ] Brackets. { } Braces. Signs of aggregation. These signs are used to col- lect together symbols which are to be treated in operations as one symbol. Vinculum. dy, read a subscript r, or a sub r. x’, x’ ..., read x prime, x second, ... respectively. a: b, read a ts to b. lim x, read limit of x. x=o, read x becomes infinite. logan, read logarithm of n to the base a. a”, read a to the nth power, or a exponent n. a, read square root of a. \/a, read nth root of a. f (x), F (a), ete., read “f£’ function of x, “F” function of x. (x, y), or P (a, y) read point whose coordinates are x and y. SCHOOL ALGEBRA SECOND COURSE CHAPTER I ALGEBRAIC OPERATIONS 1. Four fundamental operations. In the first course in algebra, it has been seen that operations with numbers are made up of additions, subtractions, multiplications and divi- sions. These operations are called the four fundamental operations of algebra. EXERCISES 1. If a = 2, name the fundamental operations used in find- 4 4 ing the value of a? + 2a +1; of a — fe + 2. 2. What fundamental operations are used in finding the aoc ot — f value of when x = 2? x+2 3. The area of a trapezoid is $(b + )’)h, where b and Db’ are the bases and h is the alti- 1! tude. What operations are used in finding the area when be = 10, h =“8? 4. The safe load W that may be put on a pile so that ‘the pile will not sink under it is given by the formula, 2 ALGEBRAIC OPERATIONS [Cuap., I. where W is the safe load in pounds, w is the weight of the pile driver hammer, h the distance in feet the hammer falls, and k the number of inches the pile goes in at each stroke of the hammer. What fundamental operations are involved in this formula? 2. Addition and subtraction. The sum of two or more num- bers is the same in whatever order they are added. Thus, a+b=b6+4a4. This is the law * of order or the commutative law, for addition. The sum of two or more numbers is the same in whatever man- ner the numbers are grouped. ‘Thus, a+b+c=(a+b)+c=a+(b+0o). This is the law of grouping or the associative law, for addition. Example 1. Add 5a + 22, 4x — 6a, 8a — x. The sum may be written thus: (5a + 22) + (4% — 6a) + (Sa —2). ee By the associative and commutative laws, we may write (1) in the form 5a +8a —6a+20+4n—2. (2) Again, by the associative law, (2) may be written as 1a + 5a. This example illustrates the fact that polynomials may be added by combining similar terms. For this purpose, it is well to write the polynomials so that similar terms are in the same column. This process applied to Example 1, gives: 5a + 2% —6a + Ax 8a —- 2 7a + 5a * These laws of algebra are not to be proved. They are in the nature of useful assumptions that have a wide application in experience. Anns. 2, 3] USE OF PARENTHESES 3 . Definition of subtraction. To subtract a number a from a num- ver b is to find a number which added to a will give the sum b. Example 2. Subtract 522 —6x + 4ax from 8x2 — 4ax + 2ze. PROCESS: 8x? — 4ax + 2x oz? + dar — 62 3x? — 8axr + 8x 3. Use of parentheses. In order to group terms together, wwe use parentheses. It should be remembered that parentheses may be removed with or without change of sign of each term neluded, according as the sign — or + precedes the parentheses. Thus, a—(b-c)=a-b+4¢e, ' a+(b—c)=a+b-e. In case several parentheses or signs of aggregation are in- rolved together, one within another, they may be removed dy removing the innermost ones in turn, according to the rule ‘ust stated. Thus, a — {3x — A[x — (y — 5) — (2y + 6)]} =a — {3x —4[x —y + 5 — 2y — 6]} — {3u — 4a + 12y + 4} a+ax— 12y —- 4. Add: EXERCISES 1. 4axz, —2az, bax. 2. 4x + dy, 2x — by. 3. 5cx + 6a, 4cx — 8a. 4. 2x + 3y — 62, 3a — 8y +. 72, 8x — Ty + 4a. 5. 4a? + d5ac” — 3c’, 8a? — Vac”, 2ac? + 5c’. 6. Qa? — 382xy, 16x? + 172ry. jai & + 40x + c; | — 3x? — Zax + 3c, Tae — 2c. 4 ALGEBRAIC OPERATIONS [Cuap. I. - Subtract the first expression from the second in the following : 8. 3a +b, 4a — b. 9. 2x? + 3a — 4, 4a? —-4 + 5. 10. 32? + 4ry — by? — 6a, 5x? — zy — y? + 62. 11. 4a? — Gar? + 2ax — 3a + 4, 2a? + ax — 8ax* + a —6. 12. 3(a — b), 5(a — b). 13. Show how you can check a subtraction by means of an addition. 14. From the sum of 6a — 7b + 4c and 8b — 4a + 2c, sub- tract the sum of 2a — 5b + 2c and 6a — 3c + 5b. 15. From the sum of az? + 3y + 42 and 3ax? — 8y + 10z, subtract 2ax? + 7y + 102. 16. From the sum of 8(a—b) + 5(c—d) and 10(a — b) — 6(c — d), subtract 4(a — b) — 10(c — d). Combine coefficients of similar terms: 17. aw + bx 4+ ce. SOLUTION : ax + bx + cx = (2 +6 + 0)2. 18. 2ax — 6bx + 4cz. 19. 4c — abx + cx. 20. m(a+ 6) +n(a+b). SoLuTION: m(a+b) + n(a+ b) = (m+n) (a+ BD). 21. m(a+b—c)+n(a+b-—c)4+l(a+b-c). 22. 3a(5m — 3n) + 5a(—3n + 5m). Simplify by removing parentheses and combining like terms: 23. 4a —5 + (y — 2a) + 3(y — Q). 24. 8« + (82 — 5) — (67 — 11). 25. 6y — 2(y — 4) + 3(y — 5). 26. 2(7 — 5) + 6(x + 1) — 8(@@ — 7). 27. 5x — 2[2 + 3(a — 1) + 5(@ — 2) — 4(a — 5)].* * For names of various symbols of aggregation, or of other symbols, ) see page facing Art. 1. Arts. 3, 4] MULTIPLICATION D Somer tee) — {27 — yy — (2 yy — 2) + Dr) — By]. 29. a — (-2b —c — {a — b}) — (5a —.2c — [—3b + c]). BOs — (—2 — { —32:— [+ — x + 3] — 4} — [22 — x — 3)). 31. 6% — 2[2 — 3(2x — 3 — a) — 5{a — (Qa — 2a) — 4}]. 32. 7a — [(3a — 4) — 6 — (8a — 4c — 2)]. 33. 5x — 4[2(a — 4) + 3(22 — 1) —- (a — 7). Simplify the following by combining like terms: Bae 1) 4 3 4/x? — 1. 35. DV a? — a? + 2/2? — a? ~— 3V/2? — a’. 36. mv/a? — a? + nv 2? — a?. eee tin — 2) + (v7 +1) (n — 1) 38. n(n — 1)(n4+ 2) + n(n — 1)(n+ 1) + n(n — 1). In finding the value of this expression for n = 3, is it simpler to evaluate the three terms separately or to combine them first? 39. State the rule for enclosing terms in parentheses pre- ceded by a minus sign. 4. Multiplication. The factors of a product may be taken in any order. Thus, ab = ba. (1) _ This is the law of order or the commutative law, for mul- tiplication. The factors of a product may be grouped in any manner. Thus, abc = (ab)c = a(be). (2) This is the law of grouping or the associative law, for multiplication. To illustrate the use of the laws of order and grouping, we may write (5ab) (Zab) = (5- a+ b)(3-a-b) =5-3(a-a)(b- b) = 15a7b?. 6 ALGEBRAIC OPERATIONS [Cuap. I. 5. Exponents. We write a-a.... (to m times) = a”, and read a” as “‘ a exponent m”’ or the “ mth power of a.” When m and n are positive integers, a”™- a" =a-:a...(mtimes)-a-a... (nm times) =a-:a-:a-:...(m+n times) ee, or a"a® = art, (3) That is, the exponent of the product of two powers of a number is the sum of the exponents of the powers taken singly. This is often called the first law of exponents. ILLUSTRATIONS: a? - a® = a’, 3?- 34 = 38 5- 5?- 5 = 5%, 6. Multiplication of monomials. The main points involved in the multiplication of monomials are well illustrated by a simple example. Thus, (6a*b*) (—8a?b) (4ab) (6- —3- 4)(a?- a? - a)(b?-b- 6b) [(1) and (2), Art. 4.] —72a*b+. It is to be noted (1) that the coefficient in the product is the product of the coefficients in the given factors, (2) that the exponent of any letter in the product is the sum of the expo- nents of that letter in the given factors. I 7. Multiplication of polynomials by monomials. It is a law of multiplication that abb+c+d+...)=ab+ac+ad+... (1) This is known as the distributive law. According to this law, to multiply a polynomial by a monomial, multiply each term of the polynomial by Ue monomial and add the resulting products. Arts. 8, 9] ZERO ve Thus, 3a7b2 — 2ab? + 3a%b Bab 9a%b? — 6a2b? + 9a3b2. 8. Multiplication of polynomials. To find the product of two polynomials, we take the sum of the products obtained by multiplying each term of one polynomial by each term of the other. To illustrate, multiply 2a? + zy + 4y? +y by x —y + 2y. PROCESS: Qu? +xzy +4y? +y co —-y +2y 203 + xy + 4ry? + xy . — 2x*y — xy? — 473 -— y? xy" + 2xy + xy? + 4ary 223 — xy +4ry? + ry — 4y? — y? + 2a3y + wy? 4+ 42ry’. This process is, in the main, an application of the distributive law of multiplication. 9. Zero. ‘The number zero may be defined by 0=a-a, where a is any number. Then, it follows at once that the product of any number k and 01s 0. For, k-0 = ka — ka (By the distributive law) = (); (By the definition of zero) EXERCISES Perform the multiplications indicated in the following : Pie © 2°); 0°. 2ab? - 3a*b? - 4a°*b. Ary - 30°y - — 2axy?. 3abe - — 4ab’c - 2abc’. On A gl 4 ghd, gn . gn tl : gr —t, Drn+tm+1 F 3r—m—1 5 Dm—n—1 ' gm tn— 1 See aye (1 +x)" 1 ae ek Ee GMAT soa ala a 8 ALGEBRAIC OPERATIONS (Cuap. I. 9. 2a(4a? — 3a). 10. 5a?(4ab + 3a7b — 2a%b?). | 11. (2a? + 3ay) (3a? — 2ay). 12. (a+ b)(a+ b)(a+ 0). 13. (1 —x)(a + ax + az? + az’). 14. (a+6+4¢)?. 15. (a+b — Cc)’. 16. (a+b+c)(a+b-c). b b 3 18. (22 - u) 10. Division. T'o divide any number a by any number b (6 4 0) * is to find a number x such that the product ba = a, Note the condition that 6 # 0. This means that the divisor is not zero. | From this definition of division, a a’ . = da, at a’, a a ¥ a™ and in general — = ne, (1) a” where m and n are positive integers and m > n. That is, the exponent in. the quotient of two powers of a number is the exponent of the dividend minus the exponent of the divisor. 11. Division of monomials. The main points involved in the division of monomials may be brought out by a simple example. ae 4f})2 4 2 Fre. 6a‘b os; a b Sa eee) ab = —2a*b. It is to be noted (1) that the coefficient of the quotient is the quotient of the coefficients with due regard to the rule of signs, * The sign ~ is the sign of inequality and is read “is not equal to.” Arts. 11, 12, 13,14] ZERO IN DIVISION 9 (2) that the exponent of any letter in the quotient is equal to the exponent of that letter in the dividend minus its exponent in the divisor. 12. Division of a polynomial by a monomial. According to the distributive law (Art. 6), a polynomial is divided by a monomial by dividing each term separately. 622y — dary + Yary? Thus, Bry = 2x —a + day. Cueck: Makex =y=a=1. Dividend = 12. Divisor = 3. Quotient = 4. But 12 dl 13. Division of a polynomial by a _ polynomial. The process of dividing one polynomial by another may be easily shown by examples. In general, the dividend and divisor should be arranged according to ascending or descending powers of some letter, called the letter of arrangement. Example. Divide a‘ — 7a? + 2a* — 8a + 12 by 2 + a? — 3a. PROCESS: a4+2a3 — 7a2-— 84+12 | a? —3a4+2 at — 3a° + 2a? a? + 5a+6 5a*— “9a? — 8a+4+12 5a? — 15a? + 10a 6a? — 18a + 12 6a? — 18a + 12 Check this division by finding the product (a? + 5a + 6) (a? — 3a + 2). 14. Zero in division. In the definition of division (Art. 10), it was stated that the divisor cannot be zero. That is to say, division by zero is excluded from our operations in algebra. When the dividend is zero, the quotient is zero ; for, 0 ay since k-0=0. (Art. 9) 10 ALGEBRAIC OPERATIONS [Cuap. I. Where is the fallacy in the following ? Let t= 0. (1) Multiply both sides by z, x? = az.: (2) Subtract a? from both sides, x? — a? = ar — a2. (3) Divide both sides by x — a, x+a =a. (4) But by (1) 2 =a. (5) By (4) and (5) 2a =a. (6) Hence, 2=1. (7) Divides EXERCISES 1.23 = 31. 7oihyD.. Bye | 2. 3a’b’c by abe. Dal eye. 4. qe + 3d + 2 by qe +dadt 2. 5. a" — 5y2n +1 by gnr— syn —1, 6. (x — 1)*(@ + 1)"*! by (@ — 1)*"(@ + 1)". 7 (& — y)(a + y)” by ( — y)®(a + y)>. 8. (a2 — b2)2"+38 by (a2 — b2)2, 9. xr y* — 5x2yt — xy? by —x?y?. 10. vs? — 2gsv — 2s?v by vs. 11. St? — 3g by ig. 12. 12a7¢? + 9xtt® — 1523 by 3x02. 13. 9Yax*w — l5ax?w? — 8artw* by 3ar2w. a iw . Sap*g’ + 24bp?¢? — 32p%q? by 8p2q?. . —ldx?yz — dry’z — 1l5ryz? by 5zryz. 16. a? + 3a*b + 3ab? + b8 by a? + 2ab + b?. Check your result by making a= 1, b= 1. - ao 17. a — ab — ab? + b8 by a? — 2ab + B?. Check your result by making a= 2,b=1. Why ean this result not be checked by making a = 1, b = 1? 18. 2 —-y by x-y. 19. z+ y° by a+ y’. 20. 4y?+ 4xy + 2? — 12yz— 6x2 + 92? by 2y + x — 3z. Arts. 14, 15] POWERS AND ROOTS 1] Check Exercise 20 by making x = 1, y = 2, z= 1. 21. 24+ 32° + 7274+ 824+ 6 by 224248. 22. (a—b)?— 3(a—b64+2 bya—b-1. 23. z*— yt by r—- y. Check the result by making x= 2,y=1. Why can the result not be checked by making x = y = 1? 24. a‘ + 4ab® — 4abe + 3b* + 2b?c — c? by a? — 2ab + 3b? — . 25. A man owns three farms whose values are oat + 5a + 277 +2 +5, 5a* + 2x3 + Ox? + 8x — 7, and Aart + 823 + Tx? + 2x + 2. Find the value of the three together. He divides the land into parts of equal value among his x sons. What is the value of the land that each obtains? | 26. Divide 7R? — mr? by TR + Tr. 15. Powers and roots. In exercises under Art. 4—11, we have performed operations with integral powers by means of the laws of exponents a™a" =amtn. (1) and | a” > gh as (2) We shall soon find it convenient to use a third law of exponents (ann = ann (3) To illustrate the law, (3*)? = 34 - 34 = 38. To derive this law, we note that eye — aja" a” > ...to n times = qrntm+m+-->+-+ton terms =a, The pupil knows something about square root and cube root. We may now well extend the notion of a root by defin- ing the nth root of a number, where n is any positive integer. 12 ALGEBRAIC OPERATIONS [Cuap. I. DEFINITION. An th root of a given number is a number whose nth power is equal to the given number. For a given number z, the nth root is written ~/z. When n 1s 2 the root is called a square root. When vn is 3 the root is called a cube root. The number n is called the index of the root. From (38), we note that the nth root of a” is a”; for, a™ taken n times as a factor gives a". Thus, V28 = Va!-a! = at, Ayia = Wyia 4 = ya, 7/710 = x? W(x? — a?)t = x? — a? In general Van = an, This may be stated in words as follows: The root of a power is found by dividing the exponent of the power by the index of the root. 16. Algebraic operations. An operation that involves only a definite number of the four fundamental operations and of the operations of finding powers and roots is often called an algebraic operation. EXERCISES Extract the roots indicated in the following : 1. V/16a'. 5. ~/—8a'. 9. Vf ain”, 2. \/40”, 6. V 16a. 10. x4. 3. /4a™. 7. v/a. 11. V/a%?. 4, ~/8a3. 8. v/a". 12. Yop. 13. Expand (a — b)? and check the result when a = 2, b = 1. 14. Expand (a — b) and check the result by making a = 8, MISCELLANEOUS EXERCISES Simplify the following by performing indicated operations : 1. 1—[-1 + {25ers 6)} — 1]: 2. 4a — [—-6y — (5 — 2y — 3x) + 4a] — [5 — (4y — 32)). . Arr. 16] MISCELLANEOUS EXERCISES 13 av af a - Se e+ y* —- ry Va+vVa 1 4. eee —: Check result for x = 9, a = 4. Vzi-Va Vrt+vVa 5. (a + b)[a? — (ab — b?)]. 6m* + m*> — 29m* + 27m —9 Check result by making 3. Check result by making x = y = 1. os 3m + 5m? — 7m + 3 m= 1. 7 T3x* — 2.92° + 0.92? + ae E 0.3% — 0.5 6x22" Qar2m “= 1 a3 oa, y® < : ; : 10) 6 eee 2 3 ah Aes FH xe+auyt+y? falta =o, y = 1,2 = 0, and a = 1, find the value of : fee ie — (y+ 2) } — [er —(y —2 — a)]. 12. Ife =m+n-—2p,y =m —2n+p,andz =n+p — 2m, show thatx+y+z2=0. 138. If a=2°4+ 427-1, b=1 —2 — 22”, and c = 273 + 27? + 2-+1, find the value of — a — [b — (2a — c) +c] whenz = 2. 14. Find the value of he ea when a = 4, b = 10, andc = 4. : 15. Find the value of eee oe when a = 4, b = 10, and c = 4. 16. Divide x° — y® by x’ — y’, and check the result by mak- ing « =2 and y=1. Why can not the check be made for ee ey eae Wy 17. Divide 7 — y™ by x” + y™. 18. Divide wR? — mr? by r(R — 7). 19. Divide 47R? — 47r° by R — r. 20. Divide §7D* + ¢7d3 by 7(D + d). - Historical note on the fundamental laws. When the attention of the student is first called to these laws, it is not unlikely that he will regard 14 ALGEBRAIC OPERATIONS | [Cuap. I. them as self evident propositions. That they are not self evident should be clear from the fact that systems of numbers have been devised, that do not follow the same laws of combination as apply to the numbers we use in algebra. In one prominent system, we have the product ij = — ji. While the study of kinds.of numbers that do not obey the commutative, associa- tive and distributive laws, belongs to advanced mathematics, the fact of the existence of such numbers is mentioned here to emphasize the fact that the laws of algebra are assumptions, even if they have a wide applica- tion in experience. Strange as it may seem at first thought, we find, upon examining the history of algebra, that much was known of certain complex processes before it was understood that laws about order and grouping are at the basis of these operations. Sir William Rowan Hamilton (1805-1865) stands out prominently as a man who helped to make clear the nature of algebraic processes by devising numbers that do not obey all of the laws given in this book. CHAPTER II LINEAR EQUATIONS IN ONE UNKNOWN 17. Equalities. A statement that one expression is equal to another expression is called an equality. The two expres- sions are called the members of the equality. There are two kinds of equalities, identical equalities or identities, and conditional equalities or equations. The two members of an identity are equal for all values of the symbols for which the expressions are defined.* Thus, xv? — a? = (4 — a)(x + a), 5a = 10a — 5a, are identities. But in the equality x—5 =4, the two expressions x — 5 and 4 are equal only whenz = 9. An equality of this kind, in which the members are equal only for particular values of the letters involved, that is, are not equal for all values, is sometimes called a conditional equality. In this book, we use the term equation to mean conditional equality. When it seems necessary to emphasize that an equality is an identity and not a conditional equality, we use the sign = instead of the sign = between the members. But the sign = will be used for both identities and equations when this usage can lead to no confusion. * This statement implies that we may not assign values to the lctter 1 =e z 1-2 1-2 excluded when x = 1, since division by zero is excluded. (See Art. 14.) is which will make the members meaningless. Thus, 16 LINEAR EQUATIONS IN ONE UNKNOWN [Cuap. II. EXERCISES Which of the following equalities are identities? 1. (x — a)? = x? — 2ax + a’. 2. sine lth pt Tio 8.27 — 32-4 2es' 0, 4, 27°+6=0. 2 pa ve 5. Top teh ieee 6. 472+ 474+ 1 =0. 18. Equations and identities as sentences. An equation may be regarded as essentially an interrogative sentence, ask- ing for values of the letters that make the members equal. Thus, x-—-5=6 asks for the value of x that makes z — 5 = 6. An identity is essentially a declarative sentence, stating the fact that two members are equal without regard to the values of the letters. Thus, x? + 2ax = 2x? — x? + 8ax — ax for any values that may be assigned to the letters. 19. Solution or root of an equation. To solve an equa- tion in one unknown is to find values of the unknown that reduce the equation to an identity. Any such value is said to satisfy the equation and is called a solution or root of the equation. 20. Equivalent equations. Two equations are said to be equivalent when they have the same roots; that is, when each equation is satisfied by the solutions of the other. Thus, the equations ¢ — Dee and 37 —- 6 = 0 are equivalent. Art. 21] OPERATIONS ON EQUATIONS 17 21. Operations on equations. It is here taken for granted that in the first course in algebra the student has made con- siderable use of the following operations on members of an equation in the process of finding solutions : (1) Adding the same number.to both members (2) Subtracting the same number from both members. (3) Multiplying both members by the same number, other than zero. (4) Dividing both members by the same number, other than zero. These operations are permissible because they lead to equiva- lent equations. The operations (1) and (2) are often replaced by an equivalent operation called transposition. It consists in changing a term from one member of an equation to the other and changing the sign of the term. - The necessity for the restriction in (3) that the multiplier is not zero may be seen from the following examples : Example 1. Given z—1 =3. (1) Multiply by x — 2, x? — 3% +2 =3(a — 2). (2) From (2) x? -—-6x +8 =0. (3) Whence, (x — 2)(a@ — 4) = 0. (4) But, x = 2, and x = 4 satisfy (4) while (1) has the root 4 only. The non-equivalence of (1) and (4) is due to the fact that the multi- plier — 2 is zero when x =2. Such values as x = 2 which satisfy a cer- tain derived equation, but do not satisfy the original equation, are sometimes called extraneous roots. Example 2. Given 2? — 3x +2 = 3x —- 6. (1) Divide by x — 2, z—1 =3, . (2) and x =4. (3) But (1) has solutions x = 2 and x = 4, while (2) has only the solution x = 4. The non-equivalence of (1) and (2) is due to the fact that we have made the error of dividing formally by x —2, which is zero when x = 2. But division by zero is excluded from algebraic operations. (See Art. 14.) 18 LINEAR EQUATIONS IN ONE UNKNOWN [Cuap. II. 22. Linear equations in one unknown. An equation of the form 7 ax+b=0,a40 (1) is called a simple or linear equation in the unknown xz. We have solved many such equations in the first course. The general solution of (1) is as may be verified by substitution, but it is in general better to treat each case separately than to remember the formula. 2%. Verification by substitution. The operations of Art. 21 are useful in finding solutions, but the solution is incom- plete until the values of the unknown are substituted in the - equation to be solved, and are shown to satisfy it. EXERCISES AND PROBLEMS Solve the following equations for x: 1. 32 + 6 = 524 4+ 10. SOLUTION : 3a +6 = 54 + 10. (1) Transpose and collect, —2x¢ = 4. (2) Divide by —2, z= 2. (3) CuEck. Substitute in (1), and 3(-2) + 6 = 5(-2) +10. 0 =0. jae Seer ieee eed oy or re ; SOLUTION : Oy he 7 ema (1) Multiply by 12, 6z +42 = 15 — 122. (2) Transpose and collect, 22x = 15, (3) 15 © = 55 (4) Art. 23] EXERCISES 19 CHECK. From (1) and (4) 15 5 Woy ko 44° 99 4 99? 25 25 44 44 3. (x2 +9 = 2x — 6. 4. 54+7 = 2x7 + 10. §. 54+ 7 = 27 + 9. Gor 27+ 1) = 6r?.—-9. ri 8 2 or ~ d(@ +1) + 2? = x? + 12. moe 2) = 3(c¢ + 1) — 13. me + 1)(z +3) =a(e — 2). 10. ax +a = 6a. iv 3(a — xz) = 9a. 12. 2(a? — 27 + 2) — (a +1) = 22? + 10. 13. ‘(@ + 2)(z@ — 5) = («1 + 4)(2 — 1). ia. 2a + a(x — 1) + 1] = Qe —1)(x + 2). 15. (m+ n)x + (m — n)x = mn. $4 aa, x+b b Tooeeod + o.00 = 2.52 -+- 17.5. 19. 0.1(52 + 20) — 0.4(a — 5) = 25. 5 —8 2(10 — 2) ASA Teak aera ge pe ol. is a 22. p(p —x) = yy +2). 23. mx = mW + MWe — nx.* 24. y—5 = m(x — 10). * The subscript notation is found very convenient both in pure and in applied mathematics. The symbol m is read ‘“‘m sub 1.” 20 LINEAR EQUATIONS IN ONE UNKNOWN [Cuaap. II. The following exercises, 25-32, give relations in the nota- tion of physics. 25. Given s = vt, solve for v. 26. Given s = vt + so, solve for v. 27. Given s = $gt?, solve for g. 28. Given s = 3gt? + so, solve for g. 29. Given EH = 4Mv’, solve for M. I 30. Given D = Ww » solve for W. UW — WW 81. Given F = 2c + 32, solve for C. 82. Given PV = po( 1 + =a) express ¢t in terms of P, V, p, v. 33. What extraneous root is introduced into an equation x = 5 if both members are multiplied by 2? by a2 — 2? by x+3? 34. What extraneous root is introduced into the equation ax +b =0 if both members are multiplied (1) by x? (2) by x—2? (8) byx-—a? 35. What root is lost if the members of x? — 4 = x — 2 are divided by x — 2? 36. If each member of an equation is multiplied or divided — by the same expression that does not contain the unknown, are the roots changed ? 37. Five times a certain number when diminished by 23, is equal to 5 more than the number. What is the number? 38. A father is four times as old as his son and the sum of their ages is 65. What is the age of each? 39. Find three consecutive integers whose sum is 78. 40. What number must be added to each of the numbers 18, 108, and 48 in order that the product of the first two sums shall be equal to the square of the last sum? 41. What number must be added to each of the numbers a, b, and c, in order that the product of the first two sums shall be equal to the square of the last ? Art. 23] PROBLEMS 21 42. How soon after noon will the hands of a clock be to- gether again? SuaceEsrion: Let x = number of minute spaces which the minute hand has traveled from noon until it overtakes the hour hand. Then 7 will be the number of spaces the hour hand covers meanwhile. The difference, xL— 5 is 60 spaces. Why? 43. Denochares has lived a fourth of his life as a boy; a fifth as a youth ; a third as a man; and has spent 13 years in © his dotage. How oldis he? (From the Collection of Problems by Methodorus, 310 4. D.) 44, A man can do a piece of work in 4 days, another in 6 days - and a third in 12 days. How many days will it require all to do it when working together ? 45. A room is 4 feet longer than it is wide, and if the length were decreased by 2 feet and its width increased by 4 feet, the area of the floor would be increased 40 square feet. Find the dimensions of the room. PROBLEMS PERTAINING TO MIXTURES 46. A grocer mixes two grades of coffee which cost him 20 cents and 35 cents per pound respectively. How much of each must he take to make a mixture of 60 pounds which he can sell at 30 cents per pound with a profit of 20%? 47. A dealer has 2000 gallons of alcohol 85% pure. He wishes to add water until it is 75% pure. How much water must he add? 48. How much cream that contains 25% butter fat should be added to 1000 pounds of milk that contains 34 % butter fat to produce a standard milk with 4% of butter fat? | — 49. How many pounds each of water and milk with 4% of butter fat should be taken to give a mixture of 160 pounds with 31 % butter fat? 50. If 24 pounds of iron weigh 22 pounds in water and 20 pounds of lead weigh 19 pounds in water, find the amounts 22 LINEAR EQUATIONS IN ONE UNKNOWN [Caap. II. (weights in air) of iron and lead in a mass which weighs 180 pounds in air and 170 in water. 51. A dealer mixes a pounds of tea worth x cents a pound with 6 pounds of tea worth y cents a pound and with c pounds of tea worth z cents a pound. Find the value, v, of the mix- ture in cents per pound. PROBLEMS PERTAINING TO BUSINESS 52. The gross income of a certain man was $110 more in the second of two years than in the first, but in consequence of an income tax of 1% on the part of the income above $3000, the net income (after paying tax) was the same in the two years. Find his income in each year. 53. A married man has a $4000 exemption from income tax, but pays at the rate of 1% on the rest of his income. He © finds that after deducting income tax, his actual income is $5980. On what amount does he pay the 1% tax? 54. A man made two investments amounting together to $6000. On the first he gained 10%, and on the second he lost 5%. His net gain on the two was $60. What was the amount of each investment ? 55. At a ball game the charge was 75 cents for each re- served seat and 50 cents for each general admission ticket. The ticket seller found he had sold 4320 tickets for $2785. How many people bought reserved seat tickets ? PROBLEMS INVOLVING THE LEVER Two boys, A and B, are balanced on a teeter board as shown in Fig. 2. The board is twelve feet long with the point of support at the middle. The boys find that they must sit at distances from the Fie. 2 support such that the products ob- ; tained by multiplying the weight of each by his distance from the fulcrum or point of support are equal. : Thus, (78) (5) = (65) (6). 6-ft: ArT. 23] PROBLEMS 23 In the general problem of the lever the force F required to lift the weight W depends upon the position of the point of support (fulcrum) and the lengths of the arms of the lever. If f and w represent the distances of the force and weight respectively from the fulcrum, then the law of the lever is expressed by, Fig. 3 56. A and B weigh 70 and 120 pounds respectively. On a teeter board, A is to sit 6 feet from the fulerum and B 5 feet. A stone is placed on A’s side 5 feet from the fulerum so that the sides will balance. How heavy is the stone? © 57. A and B together weigh 320 pounds. They balance when A is 8 feet from the fulcrum and B is 6 feet. Find the weight of each. 58. How heavy a stone can a man, by exerting a force of 200 pounds, lift with a crowbar-6 feet in length if the fulcrum be six inches from the stone (neglect weight of crowbar) ? PROBLEMS INVOLVING UNIFORM MOTION The rate of uniform motion is the distance traversed in each unit of time. The rate of motion is often called the speed or velocity. .It is customary to use s for distance or space traversed, v for velocity, and ¢ for the time. 59. A Zeppelin flies 55 miles per hour and an aeroplane 70 miles per hour. If the aeroplane is 7.5 miles behind the Zeppelin, how long will it require to overtake the Zeppelin? 60. A freight train leaves New York for Chicago making a speed of 30 miles per hour. Five hours later a limited express leaves New York making 54 miles per hour. In what time will the express overtake the freight? : 24 LINEAR EQUATIONS IN ONE UNKNOWN [Cuap. II. 61. In how much time does the minute hand of a clock gain a complete revolution on the hour hand ? 62. If the speed at which sound travels in air is given by vy = 1090 + 1.14(¢ — 32), where v is the speed in feet per second, and ¢ the temperature of the air expressed in Fahrenheit degrees, find the tempera- ture of the air when sound travels 1120 feet per second. CHAPTER III FACTORING 24. Rational integral expressions. By a rational integral expression in certain letters, say in x, y, 2, we mean the sum of terms of the type axmyn2?, where the exponents m, n, and p, are any positive integers. Thus, 523 + 10a? + 42 — 6, Bry +y?V5 — a0? — «v7, and ax’y + 10xy? — cxyz are all rational integral expressions in all the letters involved, while 3x , . : : ‘ 5, are rational integral expressions in x but not in y. Y 323 + avy, and x? + It should be noted that a rational integral expression with respect to a letter x contains no indicated root of that letter, as Va, and does not have that letter in the denominator of a fraction. The degree of a term of a rational integral expression is the sum of the exponents of the letters in the term. The degree of a rational integral expression is defined as that of a term whose degree is equal to or greater than that of any other term in the expression. Thus, the degree of 3x?y + We are especially concern hapter with the factors of rational integral expressions. 25. Definition of factoring. We shall use the term factor to mean a rational integral factor. A rational integral expression is prime when it has no factor besides itself and unity. : 26 FACTORING [Cuap. III. Finding the prime factors of a rational integral expression is called factoring the expression. 26. Important special products. Factoring in elementary algebra depends mainly upon the recognition of certain type products. ‘The following have already been treated in the First Course. They should be reviewed and memorized. a. Common monomial factor. ax +ay =a(x+y). Example. 2am — 4a? = 2a(m — 2a). b. Difference of two squares. a? — b? = (a+ b)(a — b). Example. 812? —1 = (9x + 1)(9a — 1). c. Trinomial square. a? + 2ab + b? = (a+ b)*. Example. 42? — 12x%y + Oy? = (2x — 3y)?. ‘- d. Trinomial of the form 2? + (a + b)a + ab = (a +a)(x+5). Example. 2? + 7x +10 = (4 + 5)(a@ + 2). e. Sum of two cubes. a’? + 6? = (a + b)(a? = ab 4+ B?). Example. 1252° + 8y!5 = (5a? + 2y5) (254 — 10x?y*> + 4y). f. Difference of two cubes. a? — & = (a — b)(a? + ab + B?). Example. my? — 27n® = (my — 3n') (my? + 38mn3y + On‘). g. Factors found by grouping. ax + ay + ba + by = (a TF b)(a + y). Example. 2ax + 4bx — 8ay — 6by = 2x(a + 2b) —3y(a + 2b) = (Qe — 3y) (a + 2b). h. Trinomial of form +c. Certain expressions of this form can be factored ection. Example. Factor 6x? + x — 10. SotuTion. The factors are two binomials whose first terms are 3a and 2x or 6x and x, and whose last two terms are +3 and +5, or +1 and #15, if the coefficients are integers. We must now choose the terms of the bi- nomials so that the algebraic sum of the cross products is +x. By trial _ we find that the factors are 2x — 3 and 3x + 5. _ ART. 26] EXERCISES IN FACTORING EXERCISES Factor the following : a+ ab. 3+ 3x. m* — A, a? + 10a + 25. m* — 9m + 14. a’bt — x4. mp +mq+np + nq. 10am + 8aq + lima + 12qz. 5r3 + 5. k? + 6k — 55. . ct — 12c? + 36. . op? — 9p — 18. . Gar? + 24ax? + 24az. - Zax + 6a? — 4ay. moe l. . re 4+ 22r + 117. i : pepe S. er ain a ere glad) te a ee ee ee ee oOnPr wo DH OS = ~j = ioe) 1 Ve Sale serie te: 4 . 3l? + 161 — 35. . am+bm+a+ob. . a + 20a" +100. Find two factors only. eee? — ye tl He. . 2 — 52" + 42". Fmd three factors only. . a —1. Find two factors only. ae 3. . abt+at+bd+1. . dax — Sby — day + 36a. Aap rd Oe : meowene Fe woe Oo © NNN NY NY WD on on -» Ww 28 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. FACTORING [Cuap. III. mt — ni. (38a — y)(2a + 4) + 5(8a — y). | (2a — 3b)? — 4b?. a” — b? + 2be — c?. x? + 4x? + 42. a? — Tab + 12b?. 8x3y® — 125m*. Find two factors only. 3bmx + 2bm — 3pqu — 2pq. 1 — x. 60a? + 8ax — 32”. a‘b + 5a%b? + 40763. 24miny — 81lmy*. (m + 2)? — 5(m + 2) — 176. 5a? — 8ab — 4b?. 1z° — 12° — 842. x? — 4ax — 4b? + 8ab. (a + b)(c? — d?) — (a? — b?)(c — d). 2a? — 5x — 38. 2ilx? — 232 + 6. (Qx +a —b)?— (x —a+b)?. x — 2° — 647? + 64. Find six factors. x® + 1623 + 64. 12x? + 25a — 7. 152* — 2° — 62. 9a2x* — 30a7x? + 25a72?. (x — 1)? — 9(@ — 1)? + 8(@ — 1). 1 — a — 2ry — y?. a*c? + acd + abe + bd. 2? — 64" +8. Find two factors only. a‘? — 5a?Pht — 146%. Find two factors only. a"b2n+2 — 16a°"b?c%, Find four factors. Arr. 27] TYPE PRODUCTS 29 _ 2%. Further type products. The following type products are also useful and should be mastered. i. Square of a polynomial. The square of a polynomial equals the sum of the squares of the terms plus twice the product of each term by every term that follows. Thus, (a+b+c+d)?= Beeb ce + d* + 2ab + Zac + 2ad + 2be +- 2bd + 2cd. If a polynomial can be put into the form of this product it can be factored. Example. x? + 4y? + 922 + 4ry + 6rz + 12yz = (x)? + (2y)? + (82)? +2-2-2y4+2-4-38242- 2y: 3z = (x + 2y + 32). j. Cube of a binomial. By performing the multiplication, we find that (a + b)® = a? + 3a7b + 8ab? + 0b. If a polynomial can be put into this form it can be factored. Example. 82° + 362*y + 54ry? + 27y' = (2x)' + 3- (2x)? (8y) + 3- (2x)- By)? + (By)? = (22 + 3y)’. The student should find the cube of a — 6. k. Expressions that can be reduced to the form of the difference of two squares. Example 1. Factor x* + xy? + y'. By the addition and subtraction of x?y? we have, = (x2 + y?)? — 2%? = (2? + y? + cy) (2? + y? — ay). Example 2. Factor 36a‘ + 44ab? + 25b*. Find two factors only. If 16a2b? be added to 36a! + 44a2b? + 25b4, the resulting expression is a perfect square. Adding and subtracting 16a7b?, we have 36a + 44a%b? + 25b! = 36a! + 60a2b? + 25b4 — 160%? = (6a? + 5b?)2 — (4ab)? = (6a? + 5b? — 4ab) (6a? + 5b? + 4ab). 30 FACTORING [Cuap. III. EXERCISES Factor the following : x? + y? + 1627 + 2xy + 8x2 + Byz. 4a? + 9b? + c? — 12ab + 4ac — 6be. x? + 6x7y + 12xy? + 8y’. 8m? — 36m2n + 54mn? — 2773. ed ral re ie rhe a zt — 3274+ 1. a* + 4a? + 16. a® — 6a’b + 12ab? — 8b’. x y%z3 + 6aroytz? + 1273y2z + 8. x* — 11lxz? +1. x? + Dy? — bry + By — 2a 4+ 1. ~ d+ 32? + 32%. x§ — 321 + 32? — 1. ~ cP + yt + Qr3y? + Qy2z + 2? + Qrz. iets: fe ee . 2+ 2e3 4+ 32? + 274 4+1. a’z® — 3a*x* + 3a*x? — a’. 4a‘ + 12a°b + 12a7b? + 4ab?. 8as™ — 12a?2"5" + 6a™b2" — 5%. aim +o?" + 1. Find two factors only. — Fe Se ee ls eee bo ee XE ll eal ool oe = 8 SOMABRAR WD . Factor theorem. Divide 3x7 — 10% +7 by 2 - a. cS) @ 32? — 10% +7 |. Dee ox? — 3ax ox + 3a — 10 (3a — 10)a +7 (8a — 10)x — 3a? + 10a 3a? — 10a +7 The remainder is seen to be the same as the dividend when a is substituted for x. This exercise illustrates the principle that. the remainder resulting from dividing a rational integral expression in «x by « — a may be obtained by substituting a for x un the given expression. ArT. 28 ] FACTOR THEOREM dl To find the remainder when the above expression is divided by x — 1, we substitute 1 for vz. This gives o:1l?—10-1+7=0. Since the remainder is zero, the expression is divisible by x — 1; that is, c — 1 is a factor of 3x? — 10a + 7. This exercise illustrates the important principle known as the Factor THEOREM.* If a rational integral expression in x becomes zero when a is substituted for x, then 2 — a isa factor of the expression. The factor theorem gives a simple means of factoring cer- tain expressions. Example 1. Factor 5x? — 13x + 6. We wish to find a factor of the form x —a. In order that the given expression be exactly divisible by x —a, where a is an integer, a must be a factor of 6, that is, a must be +1, +2, +3,or +6. If we substitute 1 for x we have, 5-12-18-146 = -2. If we substitute —1 for x we have, 5 - (-1)? —13- (-1) + 6 = 24. If we substitute 2 for x we have, 5-2? -—-13-2+46 =0. Hence x — 2 is a factor of the given expression. The other factor is found by division to be 5x — 3, and we have 5a? — 134 +6 = (x — 2) (5a — 38). Example 2. Factor 3x’ + 4¢ ~15. By substituting —3 for x we have, | ees ep 4. (3) — 15 - 27 —12 = 15 = 0. Hence x — (—3) =x +3 is one factor. The other factor is found by di- vision to be 3a — 5, and we have 3a? + 44 —15 = (x + 3) (82 — 5). * For the expression az? + bx + c, a proof is given on p. 130. For proof of the general theorem see Rietz and Crathorne, College Algebra, p. 121. 32 FACTORING [Cuap. III. Example 3. Factor x* — 2x7? — 52 + 6. By the use of the factor theorem one factor of the given expression is found to be x —1. The quotient found by dividing 2? — 2a? -— 544+ 6 by «x —1lis x? —x-—6. The factors of this quotient are x + 2 and z— 38. Hence ze — 22? ~ 54 +6 = (& —1) (44+ 2)@—s) EXERCISES Factor the following expressions : 1. x? — 34 4 2. 5. 2° + 3a? — 5x — 15. 2. 377 — 9x + 6. 6. 2? — 12%7 + 27x + 40. 3. 2x3 + 347+ a. 7. 2 + 8a? + 17x + 10. 4. 2° + 627? + 32 — 10. 8.004 = 27a 9. e*+2°+4+ 8 +8. Show without actual division 10. That a‘ — 6! is divisible by a—b and by a+b. 11. That «2° — 5xty + lla*y? — 147?y? + 9ry* — 2y® is divisi- ble by x — y. 12. That a> — 6° is divisible by a — b, but not by a + b. 13. That a® + 6° is not divisible by a + b ora — 6. 29. The sum or difference of the same two powers of two numbers. The type form is a” = Db”. By applying the factor theorem, we can determine whether a+bora-— bis a factor of a" + 6". 1. Divisibility by a — b. Let n be either an even or an odd integer. If we substitute a = b in a” — b", we have 6” — b” = 0;if we substitute a = b in a” + 6" we have b” + b” = 2b”. Hence a — b is a factor of a” — b", but not of a” + b”. Arr. 29] FACTORING OF a" 33 2. Divisibility by a + 6. Let n be odd. Substituting a = —b in a” — 6" we have (—b)" — b” = —2b"; substituting in a* +6" we have (—b)" + b” = -b" + 6" = 0. Therefore if n is odd,a + isa factor of a” + b", but not of a” — 6". Let n be even. Substituting a = —b in a” — b” we have (—b)" — b” = 6” — b" = 0; substituting in a” +6", we have (—b)" + b” = 6" + b* = 26". Therefore of n 1s even, a + 6 is a factor of a” — b”, but not of a” + b*. | Example 1. Factor a’ — b®. Find two factors only. SoLuTion : a> — b5 = (a — b)(at + ad + ab? + ab? + B4). The second factor is found by division. Example 2. Factor a’ +b’. Find two factors only. SoLuTIon: a? + b7 = (a +b) (aS — ad + ath? — a3 + abt — ab + bS), : Example 3. Factor a! — b!. Find four factors. SouuTion : a!° — b! = (a> — b)(a5 + 55). Finish the solution. Example 4. Factor x!° + 243y5. Find two factors only. SOLUTION : gid | 243y° = (x?) 5 4. (3y)> = (x? + 3y)[(@)* — (w)83y + (x?)? (By)? — 27(3y)* + (By)4] = (22 + 3y) (28 — 3a°y + Oaty? — 272?y? + 81y*). The above examples illustrate the following rules which are useful when we wish to find the quotient of a” + b” divided by a+borbya-6. When the divisor is a — 6 all the terms of the ‘quotient are positive. ! - When the divisor is a + b the terms of the quotient are alter- nately positive and negative. The number of terms in the quotient is n, if the divisor is a factor of the dividend. 34 FACTORING [Cuap. III. The exponent of a in the first term is n — 1 and decreases by 1 in each succeeding term. The exponent of b is 1 in the second term and increases by 1 in each succeeding term. EXERCISES ' Factor: 1. “2° + 7°. Zaye Find two factors only. 3. ty", Find. three factors only. 4, x6 + y?, Find two factors only. 5. 32a° — 1. Find two factors only. 6. v4 + y’. Find two factors only Ton Ce Oe Find two factors only. ioerate eg pe Find six factors. 9. 12827 — y#. Find two factors only. a ee 243a" + m>. Find two factors only. . Find the first four terms in the quotient = S ep — = 12. Find the first four terms in the quotient e ges 13. Find the remainder when x* — 7x* + 2° — 547 —247 1s divided by x —1; by x +1. 14. Find the remainder. when 42° — 224 — 773 +249 is divided by « — 2; by x + 38. 30. Summary of factoring. The following suggestions will be found helpful in factoring. I. Take out monomial factors, keeping in mind numerical factors. II. After the monomial factors have been removed, if such factors occur, see if the expression to be factored can be classed under any of the following type forms of this chapter: ‘Arr. 30] SUMMARY OF FACTORING 35 1. Binomials. (a) The difference of two squares, as a? — b?. (b) The sum or difference of the same two powers, as Ge b. 2. Trinomials. (a) Trinomial squares, as a? + 2ab + b?. (b) Trinomials of the form x? + (a + b)a + ab. (c) The general quadratic trinomial, ax? + bx + ¢. 3. Polynomials of four or more terms. (a) Expressions to be factored by grouping terms. (b) Difference of two squares or expressions that can be reduced to that form. (c) Square of a polynomial. _(d) Cube of a binomial. | _ III. See if a rearrangement of terms will bring the expres- sion under any of the above forms. IV. If the above methods fail, try the factor theorem. V. Test each factor to see if it can be further factored. VI. It is convenient to remember that x? + y?, v7 4+ xy + y’, and x? + zy — y? are prime. MISCELLANEOUS EXERCISES Factor the following expressions : 1. x3 + 4a? + 42. 3. 162° + 250y°. 2. x2 +-(a — b)x — ab. 4. mx — nx + pr. 5. 9(2a — d)? — 4(3a — x)? 6. (4a — 3b)(7m — 2p) + (a + 4b) (7m — 2p). 7. 8a? — 2lab — 9b’. 8. ax — be + cx + ay — by + cy. 9. 8m — 36m?q — 27q° + 54mq’. 10. 16a*x* — 8a7z? + 1. 36 uli Be 12. 13. 14. 15. 16. Uy 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. FACTORING (Cuap. III. m? + 4n? + 1+ 4mn — 2m — 4n. a® + 68, m> — b>. Find two factors only. at + +. Gt 7 | 8la? — 16(2a — 8z)?. x — y*, Find two factors only. gi — yy, Find three factors: x? — (c+ 5)x + Se. n(e-y)—-x+y. aba’ +4- 8abx? — abx — 8ab. 30x? + 13x” — 77. a’ + 27a’. l2az%y! — 36a*bx?y? + 27a°b?x. x? + 5a? — 29x — 105. er>+y?+4+ 2ry — 4x — 4y. a‘ + 4. Hint: Add and subtract 4a?. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. x* — 10x? + 9. a” + lla” +18. Find two factors only. mtni+m+n. k10 — k. Find four factors. eetat+ i. p2nt4 = r2, 8m'n — 36m?2n2x + 54mnx? — 2723, vit 8a? +a —- 42. 1224 + 962. 62? + a — 2. 5a? + 2a — 3. a? — 4ac + 4c? — 2. - Art. 30] 41. 42. 43. 44, 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 70. EXERCISES IN FACTORING a’” — 3a" + 3a" — 1. m> +1. Find two factors only. 582 — 1]. Find two factors only. 2% — 7, Find two factors only. xe — 3x4 + 2. 10° + 1000. Find two factors only. at + 464. 3bmx + 2bm — 8anx — 2an. b? — 4b? + 8. a? — b? — 8ab(a + BD). (m +n)? — 3(m + n)*p + 38(m + n)p? — p’. m+7+(m+7)?. a’ + 19a? — 216. 30 30° Pee x? +”. 32° — 1502° + 1472. vt + 203 + 3x7 + 2a 4+ 1. xg? — 54x —171. 2a* + 102? — 25227. or +7 +9"-+1. Find two factors only. x -_ = + 4. px” — 2pgrry” + gy". Find two factors only. 23 — 3%. Find two factors only. 125¢!* + 752 + 1525 + 1. 56x? — 38x + 6. Ibe — 1)? — 2(k — 1) — 8. (7 — 2)§ + 2(r — 2)?. a + yy’. r? — 5°, 1 — (4.5). Find two factors only. 37 38 FACTORING [Cuap. III. 31. Solution of equations by factoring. ‘The method of solving quadratic equations by factoring (Art. 84, First Course) can be extended to equations of higher degree than the second. Example. Solve the equation 32° + 16x? + x — 20 = 0. Factoring the left member, we have (x — 1)(@ + 5) (3% + 4) =0. We next set each of these factors equal to zero and solve for x and obtain x=1,2 = —5, andz=-— = These values check when substituted for x in the original equation. EXERCISES Solve the following equations and check : 1. x? + 8x — 209 = 0. 5. v2 — da79-4+5=0. 2. 36a" + 17 — > = 0; 6. 2° — 77 +6 = 0. 3. 2° +422 +2-6=0. 7. 2° — 8 = 0: 4. 102° — 112? — 62 = 0. 8. 22 +2? = 0. 9. x* — 9277 + 8 = 0. 10. x* — 102? + 35x? — 50x 4+ 24 = 0. 11. 2? — 5ax — 24a? = 0. 14. 22° :— (a? 4 Jee 12. x? = 16. 15. x* — 2a*z7? + a* = 0. 13, 2 — 2? - 2420; 16. 2° —- 422527 17, 32°.— 7° — 32-2 = 0. 18. 2 — 674 lla —6=0. 19. 2? + 5x? — Qe = 10. 20. 3021 + 1323 = 7727. 32. Highest common factor. A number or expression which is a factor of two or more expressions, is called a common factor of those expressions. The product of all the common prime factors of two or more expressions is called their highest common factor (H.C.F.). ArT. 32] HIGHEST COMMON FACTOR 39 Two expressions which have no common factor are said to be prime to each other. The H.C.F. of two or more expressions can be found by resolving eech of the expressions into its prime factors, and then finding the product of the common prime factors. EXERCISES Find the H.C.F. of each of the following sets of expressions : ve eS eh ee Pwd HE oO nth baal Sie tonsa ai dhe” had ax”, 2abx, and 3a*b?. 12abc, 28a7bc?x, and 4a‘b*cy. 60, 144, 84, and 156. 52, 117, and 78. a? — b? and a? — 5ab + 407. x? + Qry + y?, 2? + xy, and x? — Txy — 8y’. 240, 1260, and 3300. 900, 1650, 3150, and 4850. 12a? — 24ab, a? — 3b + 2b?, and 2a? — 3ab — 2b?. m2? — 3m — 4, 2m? — 7m? — 5m + 4, and m3 — 4m? — m +4 4. . 2 — y8, 2? + Qry + y*, and 24 — yt. .r?—6r+9,r? + Sr — 24, and r? — Or + 18. . a 4+ 8a2y + day? + y®, v3 + Qa?y + ry’, and xy + 2ry? + y?. . 4 —2?, and x? — 47 + 4. 15. 1 — 2x, x? + 182 — 14, and 1 — 2’. 16. 22 — (y+ 2z)?, (y — x)” — 2, and y? — (x — 2)?. 17. ma+mb—mc, a?+b? +c? + 2ab — 2ac — 2bc, and (a + 6)? — c?. 18. 23 — y3, x — y®, and 2° — y’. 19. x22" — ryt, gen Qery" ag an and x22" — 1p 20, —1)7, a — 1, anda? +2*-—<2z- Il. AO FACTORING [Cuap. IIT. 33. Lowest common multiple. The lowest common multiple (L.C.M.) of two or more expressions has been defined to be the product of all of their different prime factors, each taken the greatest number of times that it occurs in any of the expres- sions. This definition gives the method of finding the L.C.M. It may be noted that the L.C.M. of two or more expressions is the expression of lowest degree which contains each of the given expressions as a factor. EXERCISES Find the L.C.M. of each of the following sets of expressions : 8 3ax?, a’bx, and 2ab2z. . 6m2n, —4mnz3, and 12mn?z. 240, 225, and 500. 160, 450, 600, and 840. z?24+ ary, 22+ y’, and x27 — 3xry — 4y?. a? — 14a — 32, a® + 2a, and 3a‘b. x+1,x2+2, and zx? + 32 4+ 2. a? + 7a + 10, a? + 6a? + 3a — 10, and a? + 4a — 5. m — 8m? + 16, 5m? + 10m, and 3m. ax — ay + bx — by, a? + 2ab + b?, and x? — zy. . a?,a — 1, (a — 1)?, and (a? — 1)?. . 2m? + 2m — 1, 8m3 — 4m + 1, and 2m — 3m + 1. . a(m — n)(m — p), b(p — m)(m — q), and c(q — m)(n — m). (2 ~ 3)(a — 7), (8 — x)(@ — 2), and (7 —HG =a . a+ 8, 8 — a’, and a? — 4. rad Ng? ire SG AP SE pel led a eS SS CHAPTER IV- FRACTIONS 34. Algebraic fraction. An algebraic fraction is the indi- cated quotient of two expressions. Thus, a b means a divided by 6. 35. Reduction of fractions. The form of a fraction may be modified in various ways without changing the value of the fraction. For different purposes in the treatment of fractions, we make use of the following principles which have already been studied in the First Course : I. The value of a fraction 1s not changed by multiplying or dividing both the numerator and denominator by the same number. That is, II. Changing the sign of either the numerator or the denomina- tor of a fraction is equivalent to changing the sign of the fraction. That is, —— = — III. Adding two fractions having a common denominator gives a fraction whose numerator is the sum of the numerators and whose denominator is the common denominator. That is, ) ee oY cee. C C Likewise, gee ae 2 42 FRACTIONS [Cuap. IV. IV. The sum and the difference of two fractions are expressed b » respectively. a C We can reduce b and F; to a common denominator, since by I, a _ad ge = 2. b> bd ee Wood By III, we can complete the process. V. The product of two fractions is a fraction whose numerator is the product of the numerators and whose denominator is the product of the denominators. That is, Pee mers VI. To divide one fraction by another, invert the divisor and then multiply. That is, The reciprocal of a number is 1 divided by the number. Thus, the reciprocal of a is = aaa i is” “Of ous - Hence, to divide. by a fraction, multiply by its reciprocal. Propositions III and V, stated for two numbers in each ease, are readily extended to three or more numbers. 36. Reduction to lowest terms. To reduce a fraction to its lowest terms separate the numerator and denominator into their prime factors and then cancel common factors by division. We know by Principle I (Art. 35) that this cancellation may be performed. Art. 36] REDUCTION TO LOWEST TERMS Reduce oo to its lowest terms. 7 x -8 (a — 2) (a? + 27 + 2) SOLUTION : (¢ 2) = @ = 2) 4 zc? +27 + ve meget 2)? EXERCISES Reduce to lowest terms: 48 ap 1 aa Tre Ga Bx a® — 6? 2. Seapear a a2 a® + b “ a? + ab 2 a? +B 4 aa 4. ee? 19, Mt = @. a* — ab ny + my — qy a—1 (m + n)3 oe Se ses 5a2b3c gen ani Te acne 2 ee ae ! 25a°b2c : (a + y")? 13 2a7x? — a®x — bat ' 3 — 3az? + 2a%x 14 v + y? +2? + 2ry — 2x2 — 2yz é x? + Qry + y? — 2? 15 ae + 62? -+ lle + 6 ; eT — 6 a+ 28. @2@+i+(a¢e+1) 17 (a — b)(c — d)(b — c) (a — c)(c — b)(a — b) 1g, HWS - yy") "(x3 + 3) (x? + y?) 43 44 | FRACTIONS x — Ses 597 — 13° tu: eee eee gen + 3 a co 20. gin +l _ pany , °1 20a?m2 — 5a’m?n4 * 20a2m — 20a?mn? + 5atmn4 (a? Eb O72 a ae 2nhn n 2n 2npn 23. a2"b"c + abc?" + ab"c abe oa (a +b)? + 4(a +b)? + (aie (a + 5)? [Cuap. IV. 37. Addition and subtraction. The method of addition and subtraction of fractions is given in Principles III and IV (Art. 35). That is, we reduce the fractions to be added or subtracted to a common denominator, and then add or subtract the numerators. 1 2 Example. Add ma ee and ln See ae SOLUTION : 1 y @ —2)@-3) ' @-3)@-4 2-4 24 —4 “@-2)@-3)@-4 "@-2@-3)@-4 3x — 8 ~ (2 —2)(z — 3) (2 — 4) EXERCISES Perform the following additions and subtractions: Comet, “YY aie 3 + 3" i j ee 1. v- “Tia ee die pe Z a u b 5 5 ArT. 37] EXERCISES 45 24. 25. eee te D = =D Cc ape meg Me Sa 1 2 dr z Aaa oe wees ate Peo, or — 3 12 ye a 9 a—b 5 2 ie Tia eee: 1 a 8 a ee aree hy Me Bye ATS 7 les) a mee 14,a+b+7—1. ba — 4. fo oe 17 x—2 z* — 52+ 6 2 a Sha 2£+b 2?+(a+b2+ab a b ab mee cha (acc) 8 5 om — 4 ; eee SO Om? = m8 Coie + 1- a+ il a? — 1 Se aS °a—-b b-a ; __ es Re ee ~@—yy—z). @-z)\e-y) (y-—2)@-2) eee ey ae? — Sry + Gy? sy? — 4zy + 2? x+a on x+6 1 xr+e x?—(b+c)x+be x?-—(a+c)x+ac 2*?—-(a+b)x+ab x+1 - 1 ou Ss ee 46 , FRACTIONS | [Cuap. IV. ant + b™ > a" — pb" 26. ae, bn a qa” + bn 27 3a a a—1 au 2 "“@+ta—20 a*?—15a+44 °- a? — 6a —55 u 20.8 : : 29. 1+-4+54+54+ 5+ all i iy: CR Sy ee 31. l-y-y Peay 32. By how much is the value of the fraction = changed by 3 adding 6 to both terms? By subtracting 6 from both terms? Answer the same question for =. 33. Find the value of the fraction eat 2+ 2 values 1,10, 100, 1000. (6) When z has the values .1, .01, .001. Arrange the results in order of magnitude in each case. (a) when « has the 38. Multiplication and division. The rules for multiplica- tion and division are given in Principles V and VI (Art. 35). x? — 2ar +a? rita Example 1. Find g2 pak a2 x? ea a2. SOLUTION: at aaa Pon) tae eo a? +a? 6? +a? LA -Ha ad x? —2ax +a? x-a Example 2. Divide PY oe savas R en 20k a a Peers e+a +a? rs x? — a? va? +a? er xr—a xz+a Art. 38 ] SIMPLIFICATION OF FRACTIONS 47 EXERCISES Perform the following -multiplications and divisions and simplify the results : b Sab a \? ti {eg 10. ae, + (oe) : 7 6ab Sax 2. 8-57 Leann tt he , &% % ote) 3. 83 12. wot + my EON. ee 4. 8k - 12. 13. G+) +a Sax say. a b EeGhy dbx i b—a ye ieee eee ior. ol " g? ay a? +ab ee 40 | ab ab ts, a oA iJ 5) we m+-1l m—I1 a to+! m (142) + (242) 5 n 19 (a — aes x? — xy + y? "a+ y a? -Qar+ 2° 20 222 wae aie Ig es — 18x + 42. ey Ee xe? + Qa m2 + 2mn i m? — 4n? m?+4n? mn — 2n? oe 2 22. (5 -£41)+(G+241): a a a a 4; et a \2 b a a ies) «(1 -i 48 : FRACTIONS : [Crap ay: 2x pees uy 2 tn 2 A ——————————— aa : : t vend 2D eee - Sipe et me 2 25. ( - a= a “ys (a — b)?. eats 3 AY es —, ate 26. (a? + 3a? — 6a — 8) = Aor An fh ee ey: ee pee ret pyc. — y 28. ay gt yi” 76 — 8 ay arma — a*bsa m3 —bex3 (m — b?x)8 abm? + 2ab’ma + ab>x? 30, 2° — 627 + 362 ie 24+ 2162 zg? — 49 gis 2 A> 39. Complex fractions. A complex fraction is a:fraction that contains a fraction either in its numerator, or in its denomina- tor, or in both. Complex fractions are e simplified by the rules for division of fractions. One of the most useful methods of simplifying complex frac- tions consists in multiplying both numerator and denominator of the complex fraction by the L.C.M. of the denominators of the simple fractions gs make up the terms. Example 1. aoe 12(2 - 5) * Weie Awe 8 —6y 2708 IN: 5y 38 aes 4 12 4D mame bac erway en a ab(a? — b)| = — - — a a-b —b S a(a? — b?) + ab(a — b) ~ b(a? — b?) — ab(a +d) _ a(a? — 2b? + ab). b2(a + b) ArT. 39] COMPLEX FRACTIONS EXERCISES Simplify the following fractions: ‘tema ; 913 a ee 6 2-3 _ 1+4 peel b 3. i: Pe. a he 13. 35 50.0), aa be \ ara ee" * zr) a® + ab + 0° te 1+(“5+ —) 49 c-y +3 aay z J i; Seeman 4 4 sree eee 3 fh dei hy b + As d+- i 00 : FRACTIONS [Cuap. IV. ty 14. ¢+1+ ee ttl te oa 16. sree ; m—-nt+ m+t+tnt m—n += +> OG, Moe E 2 [See 2 DOS. ab abe oe ae eaieed ah 1 1 aly & ae {Lea onl AS Dia os) Pim eb 1a ae 2 b+ ; d + Wea, h ae: Che e = 12 a Le ae 1 20. Po 1, 6 6h To 1 i ee : tr t+1rz 40. Equations involving fractions. In solving an equation that involves fractions, it 1s usually convenient to clear of fractions by multiplying each member by the L.C.D. of the fractions. Art. 40] EQUATIONS INVOLVING FRACTIONS ol If the unknown occurs in any denominator, multiplying by the L.C.D. may or may not introduce new roots that do not satisfy the equation to be solved. Example 1. Solve = 2 ae Z ire 2. SOLUTION : p : Teite: a =e. (1) (1) - @& — 1)(@ — 8) gives 5(@ — 3) + 8(@ — 1) = 2(4 — 1) (a — 8). (2) Simplifying (2), 2x? — 16x + 24 = 0. (3) 2(x — 6)(a — 2) =0. (4) Hence, =D, or Gem 2. The roots of (2) are 6 and 2, and both satisfy (1). It is important to note that if we should multiply (1) by a common denominator other than the L.C.D., roots may be introduced which do not satisfy (1). To illustrate, if the members of (1) are multiplied by (« — 1)(a — 3)(a — 4), the resulting equation, 5(@ — 3) (a — 4) + 38(@ — 1) (@ — 4) = 2(e -— I) (x -— 3) (@ — 4), has the root x = 4 that does not satisfy (1). x —1 Example 2. Solve TORO eee L x -l SOLUTION : Ss eae (1) (1) - (2? — 1) gives ¢ -l=77 -1, (2) or re greene eae UP Hence, gi=Oor x = 1. (3) The roots of (2) areO and 1. Now = 0 satisfies (1), but x = 1 does not satisfy (1) since the left hand member has no meaning when x = 1. Hence, the root « = 1 is introduced in clearing of fractions. The introduction of roots in clearing of fractions when an unknown occurs in the denominator should make clear the importance of checkmg each solution by substitution in the equation to be solved. 52 ’ FRACTIONS [ Cuap. IV. EXERCISES Solve the following equations and check the results: 1. 5 Pay iy 4, 2(7x — 10) — $(50 — x) = 20. C—O DR 1 age ay 3 I 4 a AUR OT CBT: o(22? +3) fp — 5 274 +1 27 — 5 5 4 8 1 2-6" 2 Qin Aer 1G) x—2 x—l De 1. BU Be x—3 1 4x% —1 ae cai tae a ae 21 10 11 11° > 9 een ob Ll eae 2x+3 47+5 = 5r — 6. Or 15. — Art. 40] EXERCISES AND PROBLEMS o3 16. a—-=c x x caesar ade area ig, Mt _ min ie I > 7 ee eee Pp q r We phe Uk Rika br ——i(i«i‘ié‘iak N-x p—-x mp — 2z) a1. ——~ 4 = = —O m+txrm-z2 nt — 7" x Sat ars aril Ain od a x—.5 on oe leh 9.5. ier Atel. Oe BLS Ro Be Sek SO | 1,00 2 8 7 Ban ie a oe EXERCISES AND PROBLEMS 1. What part of a piece of work can A do in 1 day if he can do all of the work (a) in 5 days ; (6) in 17 days ; (c) in 33 days ; (d) in 52 days; (e) in m days; (f) in m+n days? Tell in each case what part he can do in 3 days ; ind days. 2. A can do a piece of work in m days, and B can do the work in n days. What part can each do in 1 day? What part can both together do in 1 day? How many days will be required by A and B working together to do the work ? 3. The marked price of a bill of goods is $600. It is dis- counted 7%. Find the discount and the selling price. 4. Find the discount and the selling price if the marked price is m dollars and the rate of discount 7%. o4 FRACTIONS [Cuap. IV. 5. Find the marked price and selling price if the discount is $40 and the rate of discount 2 %. 6. Find the marked price and selling price if the discount is d dollars and the rate of discount r%. 7. Find the marked price and the discount if the selling price is s dollars and the rate of discount r %. 8. In a division the dividend, divisor, quotient, and re- mainder are represented by D, d, q, and 7, respectively. Express D in terms of d, g, andr. Express g in terms of D, d, and r. 9. If a oranges cost c cents, what is the cost of b oranges? — 10. What is the average selling price for a horse if m horses are sold at a dollars each, and n horses are sold at y dollars each ? 11. What is the average selling price if a articles are sold at p cents each, b at gq cents each, and c at r cents each ? 12. If a train goes m miles in x minutes, how far will it go in 1 hour? 13. At c cents a pound how many pounds of butter can be bought for d dollars? 14. AtS bought for $7? For a dollars? For b dollars? 15. What is the gain per cent on the cost if an article is bought for $120 and sold for $150? If bought for x dollars and sold for y dollars ? 16. What number added: to both the numerator and the denominator of ;°; gives a fraction equal to 2? 17. What number added to both the numerator and the a | 5 18. If for the sum of the fractions 3 and % we take the sum of the numerators divided by the sum of the denominators, by how much does the result differ from the true result ? Answer “and © b d dollars per yard how many yards of muslin can be denominator of = gives a result equal to=? the same question for the fractions f "Arr. 40] EXERCISES AND PROBLEMS 55 19. Divide the number n into two parts such that the first . part equals 2 of the second. 20. State the rules of arithmetic for multiplying and dividing a fraction by an integer. Apply these rules to multiplying and dividing : by c. REVIEW EXERCISES AND PROBLEMS ON CHAPTERS I-IV Insert expressions in the parentheses so as to form identities. 1 oct+y—-z=27+( Te 2.2-y—-z=2x — ( : 8. 2-y+2=2 — ( ,: 4,.2+ 6y —6 =2 — 6( Je 5. a — 25) + 10 =a — 5( a4 6. ax — ay + bx — by = (a + b)( e 7. How do you check the solution of an equation ? Solve the following for x and check the solutions: S42 — Liz = 0.3. 10. L7 (2 — 2) —0.3(2z% +1) = 08. 5(2—2) 2(% —4) © 11. eae 14... = 0. or wae 4+ 5 12: oTal 9 = 22. 1 2 ee et oe a 14. Solve x — (= a a +2) ee +S. (DarTMouTH *) 15. Remove the parentheses from 3a — {3a — [3a — (8a — 3a—- 3a) — 3a] — 3a} — 38a, and simplify the result. (DARTMOt TH) 16. Find the value of — a — {5b — [a — (8c — 3b) + 2c — 38(a — 2b — o)J}, where a = —3, b = 4,c = —5. (Y ALE) * Institution that gave the question in an entrance examination. 56 | FRACTIONS [Cuap. IV. 17. Find the product: (1 —a#)(14+2)(1 +27)(1 + 24)(1 + 7)(1 4+ 2"). (ILLINOIS) 18. Find the greatest common factor and the least common multiple of the three expressions : at + a2? + x1, a? + ax + 2, a? — ax +2. (HARVARD) a? + b? . a 84+ 063 19. Simplify 1 i preety (YALE) GLI6 20. Divide 62°" — 2522" + 272" — 5 by 2a” — 5. (ILLINOIS) Express the following in as few terms as possible : 21. 4.827? — 2x? 4+ 3.5y? — 2.42y — by? + 2a? — (6.42y — y?). 22. 6.40? — [1.672 + (7a — 1 + 1.42?) — 3 4 62]. 23. A person who possesses $15,000 employs part of the money in build- ing a house. He invests one third of the money that remains at 6 per cent, and the other two thirds at 9 per cent, and from these two investments he obtains an annual income of $500. What is the cost of the house? (Mass. INstiTuTE) 24. The admission to an entertainment is 50 cents for adults and 35 cents for children. If the proceeds from the sale of 100 tickets amounted- to $39.50, how many tickets of each kind were sold ? (ILLINOIS) 25. Divide x increased by 3 by 3x decreased by 4. Divide the result by 32? less the binomial 32 — 7, and call the result A. Form the same expression where x — 1 takes the place of x and call it B. Divide A by B and reduce to a simple fraction. (PRINCETON) Factor the following : 26. 9at — 16. 33. x? — 1. 27. Yat — 4. 34. xt +1. 28. (cx —y)? — (x + y)?. 35. 78? — FF 29. x + 4x2 + 4, 36. x? + 6x? + lla +6. 30. 35p? + 4pq — 15q?. . 37. xz? — 62? + Ile — 6. 31. x4 — 162? + 64. 38. m® — né, 32. x(x +1)(@ +3) +27 +4¢ 43. 39. m® — n8, ae Art. 40] REVIEW EXERCISES AND PROBLEMS 57 ord Find the values of expressions in Exercises 40-43 for a = 5, x = 8, y = 7, in the shortest way possible. zy +- a2 x? sae a? 40. x2 2! a2 xz? + a2 x? + a? yz? bee a2 go? —aq* z* 4 a? (= ne a" i. v) (5 ean u) 41. y cia ae BD tee Le y x ip) of Ur u * g(a —1)(x —2)(@ —3) x(x +:1)(z@ — 1) (@ — 2) 43. : : x@+1)(@+2)(@+3) (+I) +2)(@+3)(@ +4) 44. Reduce to a single fraction 1 7 r+3 eee oa A (OHIO STATE) ad ea 45. Simplify : = a ; : - (1 cal =). (Onto Srarn) 46. Find in the shortest way the numerical value of (x? — zy + y”) (x + y), if x = 24 and y = 12, 47. A train running 30 miles an hour requires 21 minutes longer to go a certain distance than does a train running 36 miles an hour. How great is the distance ? (CORNELL) 48. Factor, and find the highest common factor and the lowest com- mon multiple of the expressions 2x4 — 2° — a#?, 277 + 4% — 3,23 —a2? —x +1. 49. Simplify : ae - ay - ay (OHIO STATE) 50. A bicyclist averaging 12 miles per hour is a half mile ahead of an automobile running 40 miles per hour. How many minutes before tke automobile dashes past him? (STANFORD) 51. Given the equation 5x — 10 = 3x —2. What extraneous root is introduced into the equation by multiplying each member by x — 3? 52. Verify that 2 is a root of 3r(x — 2) =z? — 4. Is 2 a root of the equation obtained by dividing each member of the given equation by xz —2? 58 | FRACTIONS (Crap. IV. 53. The difference of the squares of two consecutive integers is 41. What are the integers ? 54. Concrete is to be made of a mixture of cement, sand, and gravel. The weights of sand and gravel used are equal, but the weight of the cement used is $ that of the sand. How many pounds of each are required to make 7800 pounds of concrete ? 55. The formula for converting a temperature reading of F degrees Fahrenheit into its equivalent temperature of C degrees Centigrade is C = $(F —.32). Express F in terms of C, and compute F for C = 25 and for C = 42. 56. The numerator of a fraction is 7 less than the denominator ; if 4 is subtracted from the numerator and 1 is added to the denominator, the resulting fraction equals }. Find the fraction. 67. Simplify the fraction ! a+ Beyrartel (NEBRASKA) 1+ 3 =a 58. A train runs 100 miles in a certain time. If it were to run 5 miles an hour slower it would run 20 miles less in the same time. What is the rate at which the train runs? (Missouri) 59. Divide 54 into two parts such that twice the smaller shall exceed 29 as much as 143 exceeds four times the greater. (WISCONSIN) 60. Solve the equation x+b 2a oa 2e 2a —b es eee ) (HARVARD) oa b-—a its lowest terms. (HARVARD) 62. Which takes the lesser force to lift a weight of 100 pounds with a crowbar 6 feet long ; (a) when the weight is at the end of the bar and the fulerum one foot from the end ; (b) when the fulcrum is at the end of the bar and the weight one foot from the end. Estimate your answer, then calculate the actual difference. 63. The distance from San Francisco to Los Angeles is 475 miles. A train running 26 miles per hour leaves San Francisco for Los Angeles at the same time that a train running 31 miles per hour leaves Los Angeles for San Francisco. In how many hours will they meet? (Caurrornia) 61. Reduce the expression eae (2 = =) to a single fraction in CHAPTER V FUNCTIONS 41. Constants and variables. A constant is a symbol which represents the same number throughout a discussion. Thus, in the formula for the volume V of a cylinder of radius rand height h, Vc= treks the symbol 7 is a constant, whatever values r and h may have. In problems dealing with falling bodies the space s through which a body falls in time ¢ is given by the formula oie agt’, _ where g is a constant. | The Arabic numerals are, of course, the most common cx- amples of constants. A variable is a symbol which may represent diiferent num- bers in the discussion or problem into which it enters. ' Thus, in the foregoing illustrations, V, r, hk, s, t are variables. Many mathematical expressions contain both variables and con- stants. Except in certain geometrical and physical formulas it is customary to use the letters a, b, c --- from the beginning of the alphabet to denote constants and the letters --- x, y, 2 at the end of the alphabet, to denote variables. Exercise. In Fig. 4, let the point P move around the circle, while the points P:, O, and Q are fixed. What lines in the figure represent variables and what represent constants ? Fig. 4 42. Definition of a function. Many discussions and prob- lems of algebra involve two variables which are so related that 60 7 ‘FUNCTIONS [Cuap. V. - when a value of one is given, a corresponding value of the other can be found. Throughout the course in algebra, we have had many problems in which two or more such related varia- bles occur. In the equation 2” — 3y = 4, if a value be given to x, the corresponding value of y can be found. ‘Thus, if x=0, y = —$; ifz =1,.y = —4, and so on. In evaluating the expression 2?+3x—-—1, we find that x? + 3x2 —1= —1 when x=0, 27+ 32% —1=3 when @ =1, and so on. Fixing the value of x in the first illustration fixes the value of y; in the second illustration fixing the value of x fixes the value of x? + 3a — 1. DerinitTion. If two variables are so related that when a value of one is given, a corresponding value of the other is determined, the second variable is called a function of the first. | | Thus, in the equation 2x — 3y = 4, y is a function of xz. The expression x? + 32 — 1, and in general any expression contain- ing z, is a function of x We may therefore and shall speak of a ‘“‘function of x,” instead of an “ expression involving the variable 2.” 43. Functional notation. If we refer to the same function of x a number of times in a discussion, it is convenient to have some suggestive abbreviation to represent it. It has become customary to represent a function of x by the symbol f(z) which is read “f function of z.”’ If another function occurs in the same problem it can be represented by F(a), which is then read ‘“F major function of 2,”’ while the f(x) is read “f minor function of x.” } If in a discussion f(x) is used as an abbreviation for the ex- pression a* + 323 + 22 — 3, then f(a) means the result when a replaces z. Thus, : f(a) =a* + 3a? + 2a — 38, f(h) =h* + 3h3 + 2h — 8, and f(2) = 2443-23? 42-2-3. _ Arts. 43, 44] EVALUATION OF FUNCTIONS 61 meet F(x) = re then F(a) = eres PQ) =" and f(2) = sped These illustrations bring out an important point in this notation, viz.: If the x in the parentheses of the symbol of f(z) is replaced by any other number, the new symbol means the value of f(z) when the number is put in place of z. 44, Evaluation of functions. To evaluate a function of z is to put a given number in place of x and to calculate the corre- sponding value of the function. Thus, if f(z) = 24 + 323 + 27 — 3, we evaluate it for x = 2 by replacing x by 2 and reducing. Thus, f(@2) =2° +3 -2? 42-2 -—3 =41. For xz = 3, we find f(3) = 34+ 3 - 33 +2-3 -—3 = 165. EXERCISES If f(x) = 2a — 3, find f(1), f), f2), ((-D. If f(x) = 3x + 2, find f(0), f(4), f(-4), f(@). If f(x) = «?, find fO), f(2), FG), fm). If f(x) = 20° — 1, find (4), £7), FAD), fp). If f(®) = 16¢?, find f(1), f(2), f(x), f(10). If fy) =y +y +1, find f(2), f2.3), fLOD), f@). 7. If f(s) = <=, find (03), £9), (1.1), FO. 2 Se eee 8. It fin) = +" *-, find f(1), $4), $@), SO. 62 FUNCTIONS [CuHap. V. 9, If f(x) = 42-42 +3 and F(x) = 2? — 1, find f(2) and F(2) ; f(6) and F(6). : 10. If f(x) = a3 +2 and F(a) = 2x? — 4a — 5, find the quo- fC) eZee F(1) F(3) 45. Formulas frequently used. Most of the formulas of mensuration learned in arithmetic and geometry express variables as functions of other variables. Thus, for the area A of a circle of radius r, we have the formula A= re. tients and Fic. 5 which expresses the area as a function of the radius, so that for a given r the area A can be found. The following list of common formulas will be found useful for reference. 1. Area A of a rectangle of sides a and b. A = ab. 2. Area A of a parallelo- gram of base b and altitude h. b A = bh. (Fig. 6.) Fig.6 3. Area A of a triangle of base b and altitude h. A = 3bh. (Fig. 7.) 4. Area A of a triangle in terms of its sides, a, 6 and c. A = ~Vs(s —a)(s — b)(s — oc), at+bte 2 where s = (Fig. 8.) Fie. 7 5. Area A of a circle of radius r and ‘circumference C. A = 4Cr.. 6. Circumference C of a circle of dia- meter d, or of radius r. C=7d,orC =2rr. w =3.1416. A useful value for practical purposes is n= 3.14 a Fic. 8 Art. 45] USEFUL FORMULAS 63 7. Area A of a circle of diameter j be d or radius r. Aj=47d* or A = zr’. 8. Area A of a trapezoid of | bases 6; and bb. and altitude A. | (Fig. 9.) bi A =4(b: + by)h. Fic. 9 9. Area A of a regular hexagon of sidea. (Fig. 10.) A= BV 3 a2 2 a Fig. 10 10. Length ¢ of the hypotenuse of a ; es right triangle of sides a and b. (Fig. 11.) c = Va? +b. b Fig. 11 11. Volume V of a cube of edgea. (Fig. 12.) V =a’, 12. Volume V of a rectangular solid of length l, width w and altitude h. (Fig. 13.) V =lwh. 64 FUNCTIONS [Cuap. V. 13. Volume V of a cylinder of altitude A and radius of base r. (Fig. 14.) V = rh. 14. Volume V of a pyramid of altitude h and area of base B. (Fig. 15.) V =4Bh. 15. Volume V of a cone of altitude A and radius of base r. (Fig. 16.) V= $777k. 16. Surface S of a sphere of radius 7, or dia- meter d. S =47r?, orS = 1d?. Fig. 15 17. Volume V of a sphere of radius r, or dia- meter d. V = g7r,or V = 47d. 18. Volume V of a spherical segment (a slice: of a sphere between two parallel cutting planes), where his the altitude and a and } the radu of the two bases. (Fig. 17.) Fig. 16 19. Surface S of a zone (the portion of the surface of a sphere lying between two parallel cutting planes), where A is the distance between the cutting planes and r the radius of the sphere. eas S = 2arh. Fig. 17 ‘Arr. 45] EXERCISES AND PROBLEMS 65 EXERCISES AND PROBLEMS 1. The length of a rectangle isa. The length is three times the height. Express the area as a function of a. _ 2. The altitude of a triangle is 2 inches less than the base. Express the area in terms of the base. 3. The sum of the two bases of a trapezoid equals three times the altitude h. Express the area as a function of the altitude. 4. Iixpress the area of a circle as a function of the circum- ference c. 5. The sides of a rectangular solid are a, a+ 2, a +3. Express the volume as a function of a. 6. The lengths of the sides of a rectangular solid are con- secutive integers. The shorter side is of length J. Let V be the volume, then V = f(l). Find f(J). 7. The base of a pyramid is a square whose side equals the altitude h. Express the volume of the pyramid as a function of the altitude. 8. Find the volume of a pyramid of altitude 10 and square base of side 4. 9. Find the volume of a cone of altitude 7 and radius of base 4. 10. The altitude of a cone is one greater than three times the radius r of the base. Express the volume as a function of the radius. 11. Express the Ae of a sphere in terms of the circum- ference C of a great circle (the largest circle that can be drawn om the sphere). 12. Let V be the volume of a sphere, and c the circumference wf a great circle. Then V =f(c). Find f(c). 13. Find the volume of a spherical shell. The thickness of she shell is one inch and outside diameter 10 inches. 14. Express the following sentence in functional symbols. [he number N of eggs bought for a given amount of money lepends upon the price c per dozen. 66 FUNCTIONS [Cap V. 15. Express in functional symbols: If the age x of the applicant for life insurance is known, the premium P is fixed. 16. If d is the distance passed over in ¢ hours by an aero- plane going 70 miles an hour, then d = f(t). Find f(é). 46. System of codrdinates. Let X’X and Y’Y (Fig. 18) be two straight lines meeting at right angles. Let them be considered as two number scales with the point of intersection as the zero point of each. Let P be any point in the plane. From it drop perpendiculars to the two lines. Let x represent the perpendicular to Y’Y and y the perpendicular to X’X. | If P lies to the left of Y’Y, x is to be considered negative. If P lies below X’X, then y is negative. It is clear that no matter where P is in the plane, there corresponds to it one and only one pair of perpendiculars, x and y. The lines of reference X’X and Y’Y are called the coérdi- nate axes, and their intersec- tion is called the origin. The first line is called the X-axis, and the second the Y-axis. The perpendicular to the X-axis from a given point in the plane is called the ordinate or y value of the point. The perpendicular to the Y-axis is called the abscissa or x value of the point. i If we have two numbers given we can find one and only one point P which l:as the first number for its abscissa and the second for its orcinate. If, for example, the numbers are 27 and —5, we measure from the origin, in the positive direction a distance 2 on the X-axis and at this point we erect a perpen- dicular and measure downwards a distance 5. We have then located a point whose x is 2 and whose y is —5. This point may : be represented by the symbol (2,—5). The symbol (a, 6b) Art. 46] PLOTTING OF POINTS . 67 denotes a point whose abscissa is a and whose ordinate is b. : The symbol P(a, b) is sometimes used and is read, “the point whose coordinates are a and b.”’ _ When a point is located in the manner described above, it is said to be plotted. In plotting points and obtaining the geometrical pictures we are about to make, it will be convenient to use codrdinate paper (Fig. 19). Then the side of a square may be taken as the unit of length to represent a number. To plot a point, count off from the origin along the X-axis the number of divisions required to represent the abscissa and from the point thus determined count off the number of divisions parallel to the Y-axis required to represent the ordinate. EXERCISES AND PROBLEMS 1. Plot the points (2, 4), (—2, 4), (2, —4), (—2, —A4). 2. Plot the points (8, 7), (7, 3), (0, 0), (0, 1), (2, 0). 3. Plot the points bad by, ook, 0), (3, 3) (3, an Ge eit (-3, —$), (— 3, 0). 4. Draw the triangle whose vertices are (0, 0), (0, 1), (2, 0). 5. Draw the triangle whose vertices are (1, 3), (—1, —1), (2, —2). 68 FUNCTIONS [Cuap. V. 6. Draw the quadrilateral whose vertices are (0, 0), (2, 3), (5, —1), (3, —3). 7. A square of side 2 has one corner at the point (2,1). If the sides of the square are parallel to the codrdinate axes, what are the codrdinates of the points that may be at the other cor- ners of the square ? 8. A line is bisected by the origin. One end of the line is the point (2, —5). What are the codrdinates of the other end ? 9. A circle of radius 2 with its center at the origin intersects the codrdinate axes at what points? At what points does it — intersect the straight line which passes through the points (2, 2), (—2,. —2) : 10. Let the X-axis be considered as an east and west line, and the Y-axis as a north and south line. The following points on a river indicate its general course. Map the river from x«-= —4 to 2=+4. (—4, —22); (— 33, es) La 2} (27; =a) 2 14), Looe at) a i (—3, 3); (0,1), (2,8), (1, 18), G, 8), 2, 1), a, 9), (3, uy (33, ak a 135 47%. Graph of a function. The change in a function can be represented to the eye on codrdinate paper. For example, the change in the area of a square due to the change in the length of the side can be represented to the eye as follows: Let A be the area and a the length of the side. Construct a table showing the area for —-X various values of a. Peed eena., on tee a=-|3/1\| 3/2) Fig. 20 A=|1|1|2;|4 7 Draw coérdinate axes on paper and plot the points (4, 4), (1, 1), (2, 21), and so on (Fig. 20). Connect the points by a smooth — curve. Art. 47 | GRAPH OF A FUNCTION 69 The fact that the area increases more rapidly than the side _is shown in the upward bending of the curve. By this method, which is the same as that used to map a river ) (Exercise 10, Art. 46), any function may be represented on co- ordinate paper. This representation of a function is called the graph of the function. The graph of f(x) contains all the points whose coérdinates are [x, f(x) ] and no other points. Example. Obtain the graph of 5 — 1 for values between —5 and +5. Let f(x) = ; —1. The object is to present a picture which will exhibit the values of f(x) which correspond to different values of x. Any value of x with the corresponding value of f(x) determines a point whose ordinate is f(x). Assuming values of x and computing the corresponding values of f(x), we obtain the following table. emmy 2 | 23 | 4 | 5| -1| -14| -2 | -3.| -4 | 5 f(z) = —1 |)-# 10] 414111281 -$| -12 1 -2 | -3 | -3 | -33 The corresponding points (0, —1), (1, —2),---, are plotted in Fig. 21. It is seen from the figure that all the points lie on a straight line. This shows that the function 5 1 increases uniformly as x increases. 70 | FUNCTIONS [Cuap. V. EXERCISES AND PROBLEMS 1. Draw a curve showing the change in the volume of a cube as the length of the edge x changes from 0 to 5. 2. One side of a rectangle is x, the other side is x + 1. Show by a curve the change in the area as 2 changes from 0 to 10. 3. Show by a graph the change of the circumference as the radius 7 changes. 4. Show by a graph the change in the area of a circle as the diameter d changes. 5. Exhibit graphically the change in the volume of a sphere as the diameter d changes. Graph the following functions on codrdinate paper from x = —4tox = 4, plotting at least 17 points. 6. «+4 2. TO 7 2¢+1. 11,.-2* a 8. 27 — 1. 12. 2742. rye ae 13. 27 +241. 3 14. l+2—-— 27. 15. 1 —2 — x. MISCELLANEOUS EXERCISES AND PROBLEMS Evaluate the following functions : Late) _ at fore =1,2=3,2% = = =e 2. f(s) = s* — 7 fors = —4,8 =1.2,3 =17, . f(m) = m? + 38m +1, for m = —4, m = 13, m = —10. is) 4. If f(a) = 0 2a? + a, find fQ), Te f(0). 6. Tse) = SEER, find f08), f(-3), S@), f0). | Arr. 47] EXERCISES AND PROBLEMS 71 Beate y(n) os 4° find f(2*), f(a + 1). m7, If f(r) = a — 1, and F(x) = 1 — 2x, find F(2) + f(2). 8. The dimensions of a rectangular solid are consecutive integers. The shortest edge is represented by x. Express the volume. Find the volume when x = 7; when z = 11. | 9. Arrange the following numbers in order: f(—38), f(—2), ie (— 1); f(0), £2), F@), F(22), f(3), where fv) = 2? +2 +1. 10. Fill out the table where f(a) = ue ai arch a7 +1 a=|-3|-2/-1/0|1|[2|3|[41|5 f(a) =| | | ge ae et | 11. Fill out the table for f(x) = 10x + 1002?. Mereroyer | 203 | 41.5 | 6] .7 ] 8] 9] 1 RS a Rea cain ce es | Peeeutihe table for f(y) - 5 On codrdinate paper graph the following functions from x= —d tox = Dd: x x—l x+1 13: = ise 17, ZH. 14. : bat? 16. x — 22. 18. (x +1)(x +2). 19. If f(x) = 2* — 223 — 27? + 2x show that: f(2) =f(1) = Bie b) = f(0). 20. For the function in Exercise 19, show that f(3) = f(—2). 21. If f(a) = a? + 1, find f(a + 1). Peeeiiaieer + 7 4 1, find f(2x — 1). 23. If f(a) = 2a + 1, find f(2z + 1). 72 FUNCTIONS [Cuap. V. 24, If f(x) = 22 —2x —1, find i(;). 25. It F(a ee find (3): ate 26, Lo yia)= 1’ find f(# — 2). Bie ee Hrontts the length of an adult’s foot in inches and s the size of the shoe, then s F = 3 + 8. Represent by a diagram the relation between length of foot and size of shoe for sizes 1 to 12. 28. For the 14 days preceding the opening of the European war the daily average price of 20 well-known stocks was as follows: 100.70, 100.63, 100.01, 104.49, 98.30, 98.77, 98.49, 97.95, 97.05, 97.16, 96.58, 93.14, 94.12, 89.41. Show this de- cline in price by a diagram. Hint: To save work in handling these large numbers, 80 may be sub- tracted from each of the above numbers and the differences plotted. 29. The weight of one mile of fine iron wire depends on the diameter of the wire according to the table: Diameter in inches.|.007|.008|.009|.010|.011|.012|.013|.014|.015|.016|.017|.018]|.019|.020 Weight in pounds. |.7 |.9 |1.1|1.4|1.7|2.0|2.4 | 2.7| 3.1|3.6|4.0]| 4.5] 5.0 | 5.6 Exhibit graphically. 30. The weight of water changes slightly with the tem- perature according to the table: Temperature Fahrenheit. | 40 | 50 | 60 | 70 | 80 | 90 | 100] 110] 120 Weight of cubic foot water in lbs. |62.42|62.41162.37|62.31|62.28|62.13]62.02|61.89|61.74 Temperature Fahrenheit. | 130| 140| 150| 160| 170! 180| 190] 200] 210 Weight of cubic foot water in lbs. |61.56|61.37|61.18|60.98|60.77|60.55|60.32|60.07|59.82 Exhibit graphically. Hint: The number 60 may be subtracted from each of the above weights and the resulting graph will be unchanged in shape. Art. 47] PROBLEMS 73 31. The weight which a manila rope will safely bear de- ' pends upon the diameter of the rope according to the table: - Diameter of rope in inches. bie] 1 [1d] 1d1 121 2 | 23 [25122] 3 1|2| | Z| Maximum safe load in tons. | 4 | 34 | 54 | 7} [1041 133 | 174] 21 | 25 | 293 Represent graphically. Hint: Let 2 inches be the horizontal unit and one quarter of an inch the vertical unit. CHAPTER VI SYSTEMS OF LINEAR EQUATIONS 48. Graphs of linear equations. The graphs of some func- tions are found in Chapter V. REVIEW QUESTIONS 1. What is meant by (1) codrdinate axes, (2) coérdinates of a point? 2. Locate points represented by the symbols (2, 5), (—2, 5), (2, —5), (-2, —5). 3. What are the approxi- mate codrdinates of P,Q, R, S, T, U, V, and W in Fig. 22? 4. What is meant by the origin? What are its codérdi- nates ? An equation of the form ax +by+ce=0 (1) is called a linear equation. When 6 = 0, this equation may be put into the form Wine es Ba bib (2) Since in (2), for any given values of a and b, we may assign x and compute corresponding values of y, the equation gives any number of pairs of values to plot as codrdinates of points. In other words, y is given as a function of a in (2), and the graph of this function is called the graph or locus of equation (1). Arr. 48] GRAPHS OF LINEAR EQUATIONS 75 When a or 6 is zero, the graph is a line parallel to the X-axis or to the Y-axis. Thus, the equation j ie =A) _ gives a line parallel to the Y-axis, and 2 units to the left of that AXIS. that the points determined fall on The linear equation is so called because its locus is a straight line. 3 EXERCISES AND PROBLEMS Draw the loci of the following equations: y= 1. 7 SOLuTIoN: Let x = 0, and we have y = -1. Let y =0, and we have Ne *=1. The points (0, —1), (1, 0) being located, we draw a straight line through them. If we are given that the graph is a straight line, why is it sufficient to plot only two points? Assign further values, 1, 2, 3, 4, 5 to x and verify the locus plotted. xu xX 2. y — 2x = 4. Been oy = 2. 4. 3x2 — 2y = 6. 6. 3x + 2y = 6. 6. 7x + dy = 0. 7 (x+ dy = 4. | 8. ies ai= 0. Seale, 5 aici Hint: Show the position of all Ys points for which y = 2. Pia: 23 Breer B=: 0), Hint: Show the position of all points for which x = 8. 10. The relation between temperature in degrees Fahrenheit F and degrees Centigrade C is given by 9 F=,C +32. Construct the locus of this equation. 76 SYSTEMS OF LINEAR EQUATIONS [Cnap. VI. 11. The speed v in feet per second of a falling body at time t is given by » = 32.2t + 10 aeys where 10 denotes the speed at the instant from which time is measured. Construct the locus that shows the relation. 12. Where does the locus 38x — y = 6 cut the X-axis? The Y-axis? Answer the same question for the loci 2% — 4y = 9, and x + 3y = 0. 13. Plot the loci of x —y=5 and of 2x+y=4. Do the two lines have any point in common? What are the codrdinates of this point? Do these codrdinates satisfy both equations ? 14. Plot-x.— y = 5 and x —y = 10. Do the two lines have a point in common ? 15. Plot x — y = 5 and 2x — 2y = 10. Do the graphs have points in common ? 49. Graphical solution of a system of linear equations. As stated in Art. 48, the locus of any linear equation in x and y is a straight line. Any such equation is satisfied by an indefinitely large number of pairs of values of x andy; that is, by the codrdi- nates of all points on the locus. — In the graphical solution of the system of two equations, we seek the codrdinates of points common to the loci of the two equations. As the loci are two straight | [Naa lines, three cases arise: Seale, 2 spaces=1 Unit (1) In general, two lines inter- sect in one and only one point. This situation is illustrated (Fig. 24) by the graphical solution of Eye, y + 2a = 4. Yu Fic. 24 - Arr. 49] GRAPHICAL SOLUTIONS (ii The solution is x = 2, y = 0. Hence, in general, a system of ' two linear equations has one and only one solution, but two exceptions are now to be noted. (2) Two lines may be parallel, and thus have no common point. This situation is illustrated (Fig. 25) by the graphs of the equations of the system xr—y=2, 2x — 2y = 9. ~ When the loci are two parallel | lines, there is no pair of num- xi. —— | aX bers that satisfies both equa- tions, and the equations are said to be incompatible or inconsistent. (3) Two lines may be co- incident, and thus have an indefinitely large number of points in common. ‘Thus, if we plot « — y = 2, and 3a — 3y = 6, we shall find that the graphs coincide. There are an indefinitely large num- ber of pairs of values that satisfy this system. The two equa- tions of the system are in this case equivalent or dependent. Pre Scale, 2 spaces=1 Unit Pia. 25 EXERCISES Find the solutions (if they exist) of the following equations : Piso ~_— hates y=. 19, 4. 3a + 4y = 24, 26 —y = 1. ox — dy = 11. ‘ g: or + 3y = 4, mae g+1.5y = 3. y—«x =4. 3. 2e+y =9, 6. 2x + dy = 10, 6a — dy = 23. 5a + dy = 7. 78 SYSTEMS OF LINEAR EQUATIONS [Cuap. VI. 7. 7x + 8y = 15, 8. dx + 4y = 6, Pe 4flen(, 5a + 4y = 18. 9. Plot the locus of y + 2% = 5, then multiply both members by 3, and plot the locus of the resulting equation. Compare the graphs. 10. Plot the loci of y + Qn = 5, and of 8x — 2y = 4. Then form an equation by adding the mentee of these two equa- tions and plot the locus of the resulting equation. What are the coordinates of the point common to the three loci ? 50. Elimination. The process of deriving from a system of n equations, a new system of n — 1 equations with one fewer unknowns, is called elimination. Thus, when the given sys- tem consists of two equations in two unknowns, the new system consists of a single equation in one unknown. 51. Elimination by addition or subtraction. Elimination by addition or subtraction consists first in multiplying the mem- bers of the two equations by such numbers as will make numer- ically equal the coefficients of the unknown to be eliminated. Then, by addition or subtraction of the members, this oe is eliminated. Example. Solve 6a — 5y = 14, (1) 1% + 2y = 32. (2) Soutution : Eliminate y as follows : (1) - 2* gives 12% — 10y = 28. (3) (2) - 5 gives 30x + 10y = 160.. (4) (3) + (4) gives © 47x = 188, (5) x =4. (6) gave 4 for x in (1) gives 24 — dy = 14, (7) —5y = —10, y= a (8) Cueck: Substituting 4 for x and 2 for y in (1) and (2) gives 24 —10 = 14, (9) 28 + 4 = 32. (10) ~ * (1) - 2 means “ the members of equation (1) multiplied by 2.” ek ot ae, Lee eh 2 a - a ee, ee ae ee Art. 52] ELIMINATION 79 52. Elimination by substitution. Elimination by substitu- tion consists In expressing one unknown in terms of the other in one equation, and substituting this result in the other equa- tion, thus obtaining one equation in one unknown. Example. Solve tz —3y = 26, (1) Qn + lly = 43. (2) SoLuTion: From (2), 2x = 43 — lly, (3) 43 — 11 and C= Teas (4) Substituting 8 ty for x in (1) gives 7 STU _ 3y = 26. (5) (5) - 2 gives 7(43 — 1ly) — 6y = 52. (6) Then —83y = —249, yrs. (7) Substituting 3 for y in (4) gives) a = 5. (8) Cueck: Substituting 5 for x and 3 for y in (1) and (2) gives 35 — 9 = 26, (9) 10 + 33 = 43, (10) EXERCISES Solve the following systems of equations, and check the results : 1. 6x + 5y = 22 y ; 6. ks, 4x —y = 6. gens: 4 2. « + 2Qy = 14, ae 2x + 8y = 23. 7. 5 + 8y = 15, 3. 2a + 5y = 4, y 4a — 10y = 48. + 4 = 387. 4, 87+ 2y+5=0, x 6x + 10y + 16 = 0. 8. 5 + oy +1 =9, 5. 2(2+y) = 16, x 2 = + ¥ + 3(x — y) = 6. eed 80 SYSTEMS OF LINEAR EQUATIONS [Cuap. VI. 9. 0.8¢ + O.ly = 0.19, 11. ax + by = 2ab, 0.62 + 0.9y = 0.39. bx + ay = a? + 07. x+5 10. —5— + Sy = 9, 12, 2¢+y=m+n, mx — ny = m? — n?, psy +2=5 2 dd. Systems of linear equations solved by determinants. Let ax + by = a, (1) art + boy = &, (2) be a system of linear equations in two unknowns. We set the problem of finding values of 2 and y that will satisfy both equations. Multiply the members of (1) by 62: and those of (2) by —b. Adding the members of the resulting equations, we obtain (abe a= dob) x = boty — Dice, >) bocy = bice or t= abs — dob; ) provided ibe — deb, ¥ 0. In a similar manner, by multiplying (1) by —a: and (2) by a, and adding, we obtain 1C2 — Ae : hr ER provided aibe — doh, # 0. We note that the denominators of the above fractions are alike. This denominator may be conveniently denoted by the symbol ay by 2 be which is called a determinant. Since it has two rows and two columns, it is said to be of the second order. The letters a, bi, a2, be, are called the elements of the determinant, and a, bs are said to constitute the principal diagonal. A determinant of the second order then represents the number which is ob- Art. 53] SOLUTIONS BY DETERMINANTS 81 tained by subtracting from the product of the terms in the principal diagonal, the product of the other two terms. Thus, 1 2 o 4 ai Se Gee [= ew — vz, | Although we have solved systems of linear equations by cer- tain methods of elimination (Arts. 51, 52) it is often better to employ determinants. The use of determinants introduces us to an important instrument that is essential in the study of more advanced algebra. Using the determinant notation, we may now write the solu- tions of our equations in the form C1 by mM A C2 be a2 C2 a olay bil dz be az be We note that the numerator of the solution for x is obtained from the denominator by substituting in place of a, a2, which are the coefficients of « in the equations to be solved, the known terms G, ¢. In a similar manner, in the numerator of the solution for y we replace bi, be by c, @, respectively. EXERCISES 1. Solvex+y = 3, 22 + dy = 1. SOLUTION: sya | Leys —] Sg | om a ae a 2 3 2 3 Solve the following pairs of equations, using determinants: 2. Tx + 9y = 41, 3. +5 = 9, oO 82 SYSTEMS OF LINEAR EQUATIONS L[Cuap. VI. 4, 3x -—y+1=0. 5. 52 — 2y = 26, sy —4% +10=0. © 4x + 3y = 7. Ps Iie — dy _ 8a t+y | ; 22 32 8x — by = 1. Hint: First clear of fractions and simplify. 4r —3y—% 32 2y 95 ie 5 = LD etbaet) 3 pre PAS 15 6 vel 8. ax — by = 0, Lie 9f =. C. Y iF 4(z — 1) =3 + 32 — 5 24 — 3y + 2 = 0. 10. a(a + y) + D(a — y) = a? + D?, a(at+y)+b0(y—2) =a — B. 11. (m+ n)x — (m — n)y = 21m, (m + 1)x — (m — l)y = 2mn. Az. 2 — oy — 6, 14. 0.32 + 0.2y = 4, Ax — dy = 24. 0.7x — 0.6y = 4. 13. 37 + 4y =7 — dy, 15. mx + ny = 2mn, z—y = 6 — 2z. nz +my =m? + n?. 54. Determinants of the third order. The square array of nine numbers with bars on the sides ay bi C1 (0b) bs C2 a3 bs C3 is a convenient abbreviation for the expression aibec3 + byc2a3 + Cidebs — asbecr — b3@Q, — C3Q2by, (1) and is called a determinant of the third order. As in the case _ of the determinant of the second order, the letters a, bi, --- Art. 54] SOLUTIONS BY DETERMINANTS 83 are called the elements, and the letters ai, be, c3, form the prin- cipal diagonal. The expression (1) is called the expansion or development of the determinant. It is seen that each term of the expansion consists of the product of three elements, no two of which lie in the same row or in the same column. Any determinant of the third order may be easily expanded as follows. Rewrite the first and second columns to the right of the determinant. From the diagonals Qy by Aly b2 a3 bs QA bi C1 dz be & a3 bs C3 running down from left to right we find the first three terms of (1). From the diagonals running up from left to right we find the last three terms of (1), after prefixing negative signs. EXERCISES Obtain the expansions of the following determinants: 1(13 4 23 Fra When the first and second columns are written to the right of the determinant, we have 13 4)1 3 + od Gas aw | MesyDi| b 3 The diagonals running down from left to right are Pape oo, 1:4, 2; 3. The diagonals running up from left to right are eto: 3, 1; 5, 2, 3. The expansion then is 1:7-543-3-144-2-3 —1-7-4-—3-3-1-—5-2-3 =1. 84 SYSTEMS OF LINEAR EQUATIONS [Cuap. VI. SVL eS 5.0 ieee 3.3 12 Om bh es hag 0 az be Bret earceaeL 6. ld ae 3 2-4 Q) ey See ao u Ok 4, |-1 1 I 7s lta ene alert 20u 2 1 1-1 6 34 Sia ay Lode x4y 55. Solution of three equations with three unknowns. Let the three equations be axe + by + a2 = dh, (1) dex + bey + @Z = de, (2) asx + bsy + c32 = ds. (3) Multiplying (1) and (2) by be and —b; respectively and adding, we get (aibe 2 doby)x + (cibe = biC2)z = bed, ee bid. (4) Eliminating y in a similar manner from (1) and (3), we find (asby — aybs)a +- (bics _ bsc1) 2 = bid3 — bsdy. (5) We now have two equations in two unknowns x and z._ Elimi- | nating 2 from these two, we find [(aibe — agbi) (bcs — bse1) — (asb1 — aibs) (bec, — dice) |x = (dibs _ dob;) (c3by — b3C1) oe (dsbr a dybs) (becy = bice), . which after some simplification gives us ms diboc3 + debscr + dsbic2 — dibsce — dsb. — dob1¢3. 7 ayboc3 + debsc: + asbice — arbsc, — azsbec, — aebic Art. 55] SOLUTIONS BY DETERMINANTS 85 The denominator is the development of the determinant in Art. 54, while the numerator is the same as the denominator with a replaced by d. Hence, we can write the solution for x in the form dy bi C1 do bo C2 ds bs C3 ’ an bi Cy (lg bo C2 a3 bs C3 provided the determinant in the denominator is not zero. In a similar way, we can find the value of y and of 2. a dy C1 a by dy a2 do C2 a2 be dy ag ds C3 a3 bs ds 4 ian bs a) Seariby. ci). d2 be C2 2 bs " a3 b3 C3 a3 b3 C3 The denominators in the expressions for 2, y, and 2 are the same, while the numerators are obtained from the denomina- tors by replacing the coefficients of the unknown in question by the known terms. For example, in the numerator of y, the knowns dj, de, ds replace 61, bs, bs, respectively. Historical note on determinants. It appears that the German philos- opher and mathematician Leibnitz first used determinants in solving equa- tions. In a letter to a friend, written in 1693, he introduced the notion of -determinants in his efforts to simplify expressions arising in the elimina- tion of certain unknowns from systems of equations. It seems, however, that Leibnitz made very little use of the method and that it did not become known to mathematicians ; for, in 1750 Cramer, a professor at Geneva, rediscovered the method, and his work is accepted as forming the begin- ning of the development of the subject. 86 SYSTEMS OF LINEAR EQUATIONS [Cuap. VI. > EXERCISES Solve: 1 #«—y—z= —-6, 2x+y+2=0, ax — dy + 82 = 13. SOLUTION : -§6 -1 -l 0 1 1) 8.225" sl; 273 ae 71 =f ip 29saee 2 1 1 3 —9d 8 1 -6 -l 1 -1 -6 (2 0 1 2 Le 0 _ (3.418 8' 39, «8, 8 =o ie ae Glee. Gel aoa “| [Ii 9 eo 2 1 1 2 1 1 3.5 RS 3) Doe Ee 2. 32 + 2y —2 = 4, 7. it aoe ba — 3y + 22 = 5, y 6x — 4y + 32 = 7. 2+2_. 3. t+y+2=1, Yee ox + 2y + 7z = 1, eatin. 15a — Ay + 82 = 18. Ty ae 4, 2x + 3y + 52 = 2, 8. ax + by =a, ba — y + 42 = 5, by + cz = b, 7a — 2y + 62 = 5. ax + cz = 5. «+ 2y+z2=0, 9. ax + by — cz = 2ab, 2x +y+ 22 =3, Ay — 6y + 82 = 14. sete = als y+z=2, ~e+2=4. 10. by — ax + cz = 2be, ax — by + cz = 2ac. x+y = 3a, u.+ 2 = 4a; y+z2= 5d. Art. 55] EXERCISES AND PROBLEMS 87 Lele One: D eeGurta Cae Sit See ars 973 13. Ton Aero = Or eo, 10A + 2B — 36C = -1. PROBLEMS 1. The sum of two numbers is 95, and their difference is 15. Find the numbers. 2. Thirty-four tons of slate are to be hauled in two-horse and four-horse loads. It is found that this will require 9 two- horse and 4 four-horse loads. Assuming a four-horse load is twice a two-horse load, how much is carried on each kind of load ? 3. A workman is engaged for 30 days on the following terms. He is to receive $3 for each day that he works, and is to be fined $1.50 for each day he is idle. He received $63 for the 30 days. How many days did he work ? 4. Two books cost a cents. The one cost a cents more 5 than the other. Find the cost of each book. 5. A was m times as old as B a years ago, and will be n times as old as B in b years from now. Find the ages of each . in terms of a, b, m, and n. PROBLEMS PERTAINING TO DIGITS 6. Two numbers are written with the same two digits ; the difference of the two numbers is 45 and the sum of the digits is 9. What are the numbers? 88 SYSTEMS OF LINEAR EQUATIONS [Cuap. VI. 7. The sum of the two digits with which a number is writ- ten is 15. The digit in tens’ place is one larger than that in units’ place. Find the number. 8. There is a number of three digits whose sum is 14. The © sum of the digits in tens’ and hundreds’ places is equal to that in units’ place. If the digits were written in reverse order, the resulting number would be 495 greater than the given number. Find the number. 9. There is a number of three digits whose sum is 12. The digits in tens’ and units’ places are equal. That in hundreds’ place equals twice that in units’ place. Find the number. PROBLEMS PERTAINING TO. MIXTURES 10. What quantities of silver 72% pure and 84.8% pure must be mixed together to give 8 ounces of silver 80 % pure ? 11. What quantities of two liquids, one 95% alcohol and the other 15 % alcohol, must be used to give a 10-gallon mixture of 45 % alcohol ? 12. The crown of Hiero of Syracuse was part gold and part silver. It weighed 20 pounds, and lost 1.25 pounds when weighed in water. How much gold and how much silver did it contain if 19.25 pounds of gold and 10.5 pounds of silver each lose a pound when weighed in water ? EXPLANATION OF WEIGHT IN WatTER. A body like a piece of-gold or silver when weighed in water loses an amount of weight equal to the weight of the water displaced. Thus, if x and y denote the weights of gold and silver, © y (oe 105 24°; 13. One bar of metal is 20% pure silver and another is 12% pure silver. How many ounces of each bar must be used, if, when the parts taken are melted together, a bar weighing 40 ounces is obtained, of which 15% is pure silver? eae - + Arr. 55] PROBLEMS 89 PROBLEMS PERTAINING TO BUSINESS 14. A man has $35,000 at interest. For one part he receives 35% and for the other 4%. His income from the money is $1300 per year. How is the money divided ? 15. What is the capital of a person whose income is $1302, when he has + of it invested at 8%, $ at 6%, and the remainder at 5%? 16. The garrison of a certain town consists of 240 men who receive $3850 as monthly pay. The monthly pay of a cavalry- man is $20 and of an infantryman $15. How many men of each class are there in the garrison ? 17. A farm laborer engaged to work 90 days at $1.50 per day and board. For days which he was idle, he was to pay $1 for board. At the end of the time he received $97.50 net. How many days did he work ? 18. A man distributed a dollars among n persons of two classes, the one class receiving b dollars each, and the other c dollars each. Find in terms of a, b, and c the number that received 6 dollars and the number that received c dollars. PROBLEMS PERTAINING TO AVERAGES 19. Four numbers have the property, that when succes- sively the arithmetical average of three of them is added to the fourth, the numbers 29, 23, 21, 17 result. What are the numbers ? 20. A student has three grades which give an average 85. The sum of two extreme grades exceeds the intermediate grade by 75. The highest grade exceeds the intermediate by 3. Find the grades. 21. A high school student received the same grade in solid geometry and in second-year algebra, and 3% less in plane geometry than in first-year algebra. To find the average grade of the student in all his mathematics, the grades in second-year algebra and in solid geometry were each multiplied by 3, and 90 SYSTEMS OF LINEAR EQUATIONS [Cuap. VI. added to the grades in first-year algebra and plane geometry. The sum thus obtained was divided by 3. The average thus computed equalled 87. If the grades in first and second year algebra were interchanged, the average would be decreased by exactly 1%. Find the grades. PROBLEMS PERTAINING TO MENSURATION 22. The three angles of a triangle are together equal to 180°. The largest angle is 4 times as large as the smallest one, and equal to the sum of the two smaller angles. Find the three angles. 23. The perimeter of a rectangle is 330 inches. If the length of the longer side is doubled, the perimeter is 528. What are the dimensions of the rectangle? 24. The sums of the three pairs of adjacent sides of a tri- angle are 180, 200, and 220. Find the sides of the triangle. 25. The area of a trapezoid is equal to one-half the product of the sum of the two bases by the altitude. The altitude is 6, and the area is 54. The lower base is 2 longer than the upper base. Find the upper and the lower base. PROBLEMS PERTAINING TO PHYSICS 26. In Wilson and Gray’s determination of the temperature of the sun the Fahrenheit reading of the temperature is 5552 more than the Centigrade reading. What is the Centigrade reading. See Exercise 10, Art. 48. ) 27. It is required to find the amount of expansion of a brass rod for a rise in temperature of 1 degree Centigrade, also the length of the rod at temperature 0°. If ¢ represents the expan- sion, and bo the length required, it is known that b = ct + bo, where b is the length of the rod at temperature t. Whent = 20°, the length of the rod is 1000.22 ; when ¢ = 60°, the length is 1001.65. 28. If h represents the height in meters above sea level, and b represents the reading of a barometer in millimeters, it is Arts. 55,56] INDETERMINATE EQUATIONS 91 known that b =k-+hm, where k and m are constants to be determined from observation. It is observed that at height 120 meters, the barometer reads 751, at height 769 meters it reads 695. Determine k and m. Use the resulting formula to calculate the barometer reading for a height of 1000 meters. 29. Two boys, Jack and Bill, wish to determine their weights by means of a teeter board. They find the board balances when Jack is 8 feet from the fulcrum and Bill is 6 feet from it. When Jack takes a 5-pound weight in his hand, he has to move 4 foot nearer the fulcrum to maintain the balance. Find the weight of each of the boys. (See p. 22.) 30. A weight of 400 pounds is carried by two men by means of a pole, at a certain point of which the weight is hung. One ~ man holds the pole at a distance 33 feet from the weight and the other at 44 feet. What amount of the weight does each man lift ? : 31. Two unknown weights balance when placed 8 and 12 inches from a fulcrum ; if their positions are reversed, 25 pounds must be added on the side of too small leverage in order to balance. What are the weights ? 56. Indeterminate equations. In this chapter, we have been engaged mainly in solving systems of equations where there are aS many equations as unknowns. However, in Art. 49 attention is called to the fact that a single linear equation in two unknowns has an unlimited number of solutions. In general, when the number of equations in a system is fewer than the number of unknowns, we have what are called inde- terminate equations. In this book, we shall consider only the case of a single equa- tion in two unknowns. ‘Thus, x+2y=4 (1) is satisfied by aio Rimes a = 2, denen y = 2, Y = % = 1, ) 92 SYSTEMS OF LINEAR EQUATIONS [Cuap. VI. It is sometimes desirable to solve such an equation for inte- gral values or for positive integral values of the unknowns. EXERCISES AND PROBLEMS 1. Show that the equation 3x + 6y =5 has no integral solutions. SOLUTION : Since 3x + by = 5, (1) 32 = 5 — by, (2) and z= 3 — 2y. (3) When integral values are assigned to y, we get no integral values for z. 2. A farmer spent $386 buying two kinds of sheep, the one kind costing $3 per head and the other kind $5 per head. How many of each did he buy ? SoLuTion : Let, x = number of sheep at $3 and y the number at $5. Then 3a + Sy = 36, 3x = —dy + 36, 5 To give integral values greater than 0, y can take values 3, and 6. Hence, 7 head at $3 and 3 head at $5, or 2 head at $3 and 6 head at $5. 3. How many whole watermelons at 35 cents and whole cantaloupes at 15 cents can be purchased for $2.10? 4. Separate 37 into two parts one of which is divisible by 5 and the other by 11. 5. How many selections of coins can be made to pay a debt of 60 cents with quarters and dimes ? 6. I desire to weigh a mass of 15 ounces by means of two- ounce and three-ounce weights placed in the same pan. How can this be done? Historical note on indeterminate equations. It may be of interest to note that the early work in algebra centered on indeterminate equations rather than on determinate equations. The writings of the Greek alge- Arr. 56] HISTORICAL NOTE 93 braist Diophantus of Alexandria (about 350 a.p.) were devoted largely to the solution of indeterminate equations. He did not, however, devise general methods for the solution of indeterminate equations. Each prob- lem had its own peculiar method, and it took great ingenuity to devise solutions for the many particular cases. The credit for the invention of general methods of treating indetermi- nate equations belongs to the Hindus. The Hindus set the problems of indeterminate analysis in a different form from that of the Greeks. The object of the former was to find all possible integral solutions while Dio- _ phantus was content with a single rational answer. CHAPTER VII RATIO, PROPORTION, AND VARIATION 57. Ratio. The ratio of a number a to a number 60 is the a b written a: b. It is clear from the above definition that any ratio is a frac- tion and any fraction may be regarded as a ratio. Thus, quotient - obtained by dividingaby 6. The ratio a to b is also 2 C ae 2 9’ and q are ratios. 38. Ratios involved in measurement. It is good usage and often convenient to speak of the ratio of two quantities if they have a common unit of measure. Thus, the ratio of 6 feet to 2 feet is . To measure a quantity is to find its ratio to a given unit of measure. Thus, when we say a bar is 3 yards long, we mean that the ratio of the length of this bar to that of the standard yard is 3. EXERCISES Express the following ratios as fractions and simplify : 1.° 6:: 8. 5. (1-4): (142); 74) 5 haan 6. 10 feet : 6 feet. 3. 22:8. 7. 37°53. 4, (2? — y*): (@—y)? 8. (x? — y*): (u + y). 1 1 ek Pe yt Se 4 2 Arts. 58, 59] Se ePROPORTION 95 10. Separate the number 168 into two parts in the ratio of 3:4, _ 11. Two numbers are in ratio 2 to 5, and if 1 is added to each number, they are in ratio 3 to 7... Find the numbers. 12. Separate the number 168 into three parts that are to each other as 1:2: 4. Hint: Let x, 22, and 4x be the parts. 59. Proportion. A proportion is a statement of the equal- ity of two ratios. Thus, er as he Gee is a proportion and is often written Ore OO: Itis read “ ais to b as cis to d.” The four numbers a, b, c, and d are said to be in proportion, a and d being called the extremes and 6 and ¢ the means of the proportion. EXERCISES Find the value of x in the following proportions : x 7 3 - i; een 10. 4, riemt Ete. a1 2. 10: 5. Ca as Smet i2= 2:3 6. 4:7 =2: 14 a Cc 7. If ad = bc, show that 5 a Meote show that © = ¢. a C Me now that + ore as Ch ae OL a C 96 RATIO, PROPORTION, AND VARIATION [Cuap. VII. 10. He = then xz is said to be a fourth proportional to a,b, andc. Find a fourth proportional to the following sets of numbers : (a) 5, 10, —6. (6) 4,3, —12. (c) 4, —6, —5. Liceit : = ; then x is said to be a mean proportional between aandd. Find the mean proportional between the following sets of numbers: (a) 4 and 9. (b) +3 and +48. (c) —3 and —48. PP kee °, x 1s said.to be a third proportional to a and Db. b Find a third proportional to the following pairs of numbers : (a) e2./83 (b) 2, —3. (CMe aelals (d) —7, —11. 13. Find two numbers in the ratio of 3 to 4 whose sum is 28. 14. Find two numbers in the ratio 3 to —4 whose sum is =aNGY 15. What number must be added to each term of the fraction 1 2 ao a fr ae i0 to give a fraction equal to 5 16. What number must be added to each term of ; to: id pe =e obtain 3! Write as proportions the following : 17. OS = 2 < 28 19..33:5 =a ae 18. 5-8=4 > 1 20.. ab)= 4aas _ Arts. 59, 60] 21. a? —b- =3.- 5. fee — or + 2: = ab. 23. xy = 20. 24. ~ = ab c 25. ab = cz. Solve the following for x: BUree se 0: SS. 31. 7:4 = 28:2. 2 ete 2) = 2: 60. Properties of proportions. 86. 4:2 = at 3° x -10 — 2. PROPERTIES OF PROPORTIONS 26. 4% =6- 7. 27. a = 0b. Soturion: © = 2 Tew 28. 4:= 2% 2 20s Tit 33h Oe =a ar. 34. rit eae x 30S eet = L382: oi. () — =): (1-4-2) = (2 — 2): (2: +:2). 97 I. In any proportion, the product of the means equals the _ product of the extremes. For, in 3 b ) when we multiply each member by bd, we obtain ad = DG: Hl. 1 = ©, show that b Hint: Multiply both members by °. d That is, in a proportion the means may be interchanged with- out destroying the equality. In this case, the second proportion is said to be obtained from the first by alternation. Tiss Hint: Divide the members of 1 = 1 by the members of the equation. b C d » show Pare = 2 Hoes In this case, the second proportion is said to be obtained from the first by inversion. 98 RATIO, PROPORTION, AND VARIATION [Cuap. VII. is ahow that oe C LY, If; 7 5 7 Hint: Add 1 to each member and obtain = +1= at i; In this case, the second proportion is said to be obtained from the first by composition. GeaeG a—b cd V. lf; = ae then ? = d Hint: Subtract 1 from each member. In this case, the second proportion is said to be obtained from the first by division. pn. Ge GD: Cad I]. If- = -— then = . eg HP rire eh a Hint: Divide members of “ ne eb: +s in IV by the members of a0. 6 a pak aa in V. O.--.b° =e ae In this case, the proportion is said to be ob- poe ee a ; a C tained from > = = bad. In this connection, instead of the words “composition,” “division,” “composition and division,’ some authors use the words “addition,” “subtraction,” and “addition and subtraction ”’ respectively. by composition and division. EXERCISES From the following proportions, find the proportions that arise by alternation, inversion, composition, division, and com- ‘position and division. Loe) 4 3's: 4.0: > 2 Pes Seo 5. 2:24 [See 2 2 3 9 8. 7:5 = —2l eae 6. («+ y):(e—Yy) = oem Art. 60] EXERCISES AND PROBLEMS 99 ia@26)=¢:d, prove a ¢ a C BoE Od’ ota pioe sy a@+2b c+2d 10 a+3b c+3d ee 25s co — 2d Sees ODAue Se OF. Lt en QE Ca Oe OAT Oy LT If; 7 UO ree rey, re lis F Hint: Let = 5 = 7 =k, then a = bk, ¢ = dk, ¢ = fk. id d3 a% ile If bi = be = bs = ha show that Atra+rd+d bi tbe +bs +b, by PROBLEMS INVOLVING SIMILAR FIGURES 13. The sides of a triangle are 3, 4, and 5. In a similar triangle the shortest side is 5. What are the other sides? (Fig. 26.) Similar figures are figures of the same shape. (Figs. 26, 27.) In two similar figures any two of the sides of one are proportional to the two corresponding sides of the second. The areas of similar figures have the same ratio as the squares of corresponding sides. 14. The sides of a triangle are 9, 10, and 11. If the shortest side is lengthened one inch, what are the increases in the length of the other two sides in order to make the new triangle similar to the old? 15. In Fig. 27 the sides of the larger figure are 1, 2, 3, 4, 5. The longest side of the smaller figure is 3. What are the lengths of the other sides? 16. The area of a triangle whose base is 12 inches is 60 square 5 Fig. 26 100 RATIO, PROPORTION, AND VARIATION [Cuap. VII. inches. If the area of a second triangle similar to the first is 135 square inches, what is the base of the second triangle ? 17. A post 4 feet high, 20 feet from a street light, casts a shadow 7 feet long. What is the height of the light on the lamp post ? 18. If in a map the distance between two points 450 miles apart is 4 inches, what is the distance between two cities which are 5% inches apart on the map? 19. The perimeter of a triangle is 78 inches, and the longest side is 9 times the shortest, while the third side is a mean pro- portional between the other two. Find the sides of the triangle. Pige- 2 PROBLEMS INVOLVING SIMPLE MACHINES 20. The arms of a lever are 6 and 43 feet. What weight on the end of the short arm will just balance a weight of 100 pounds on the end of the long arm? For principle of the lever, see p. 23. The formula there given may be written as a proportion W ie fw. 21. A pump handle is 4 feet long and works on an axis 4 inches from the pump rod end. What force is required at the end of the handle to overcome a pump rod resistance of 20 pounds ? 22. Where must the support of a seesaw board 14 feet long be placed, in order that two children weighing 60 and 75 pounds may balance? 23. A crowbar 6 feet long is used to lft a stone weighing 800 pounds. The fulerum of the crowbar is 6 inches from the end of the bar. What force must be applied at the end of the bar to lift the stone ? 24. The length of the handles of a wheelbarrow is 44 feet. The center of a load of 200 pounds is 20 inches from the axle of the wheel. What force is required to lift the load ?° A Sila laigerion Papas ain Art. 60] PROBLEMS 101 25. In a wheel and axle, the radius of the wheel is 10 inches and the radius of the axle is 3 inches. What force applied to the rim of the wheel will wind up a weight of 140 pounds? The wheel and axle (Fig. 28) consists of a wheel and cylinder fastened together and turning about the same axis. The force F’ applied at the circumference of the wheel to lift the weight W by a rope wound around the cylinder depends upon. the radii ry r of the cylinder and 7’ of the wheel according to the ae formula, (f) fo yo ort Fe _ 26. The crank to a well windlass is 20 inches F long and the cylinder upon which the rope is wound is 6 inches in diameter. What force is necessary to lift a bucket of water weighing 60 pounds? 27. After the cylinder in Problem 26 has been closely wound with rope and a second layer of rope has begun to be wound on the cylinder, how much is the force increased to lift 60 pounds if the rope is one inch in diameter ? 28. A brakeman pulls with a force of 160 pounds on a brake wheel 18 inches in diameter to wind the brake chain about the axle of the wheel. If the axle is 4 inches in diameter what is the pull on the brake chain ? Fig. 28 29. A safe weighing 900 pounds is pushed up an inclined plane by means of rollers. What force is necessary to keep it from F slipping back if the inclined plane is 10 feet long and raised 2 feet at one end. The force F necessary to hold a weight W on an inclined plane (Fig. Fie. 29 — 29) is given by the proportion F:W =h:l, ; where | is the length of the plane and h the height through which one end is raised. No account is here taken of friction. 102 RATIO, PROPORTION, AND VARIATION [Cuap. VII. 30. Find the force necessary to prevent a barrel weighing 350 pounds from rolling down a plane 21 feet long, one end of which is raised 4 feet. | 31. If a boy can exert a maximum force of 100 pounds, what is the maximum height of the end of the inclined plane in Problem 30, so that the boy can handle the barrel ? 61. Variation. In Chapter V we have seen that if y is a function of x, written ya F(x), then in general y changes when x changes. We may say that y varies when «x varies, but the word ‘“‘varies” has come to have a more restricted meaning when used in this connection. Each of the statements “y varies as 2,” ‘“‘y varies directly as x,” ‘“y is proportional to 2,” ‘“‘y is directly proportional to 2,” means that y equals the product of « by a constant. That is, U= kx. The constant k.is called the constant of variation. The expression “y varies as «”’ is sometimes written Ya L. The area of a circle varies as the square of its radius. That is, ) A = kr?, if A represents the area and r the radius. With our restricted ‘meaning of the word “ varies,” it is not correct to say that the area of a circle varies as the radius, for, in the equality A =k-r, k isnot a constant for different values of r. If a train moves with a uniform speed, the distance s travy- ersed varies as the time ¢t. That is, San Arts. 62, 63, 64] VARIATION 103 62. Inverse variation. Each of the statements ‘“‘y varies inversely as 7,” iz3 = by me ay 2 ) . “y is inversely proportional to 2, means that y is equal to the product of the reciprocal of x and a constant. That is, Y= Thus, the volume of air in the cylinder of a bicycle pump varies inversely as the pressure on the piston. That is, te p if V represents volume and p pressure. 63. Joint variation. The statement ‘“z varies jointly as x and y”’ means that z equals the product of x, y and a con- stant. That is, gered an Tys The distance which a train, moving with a uniform speed, travels varies jointly as the speed and the time, or d= kut, where d is the distance covered, v the speed, and ¢ the time. In this case k = 1, if v and d are measured with the same unit of length. 64. Combined variation. The statement ‘“z varies directly ‘as x and inversely as y”’ means that z varies jointly as 2 and the reciprocal of y. That is, ka g=—: Y If T varies directly as x, directly as the square of y, inversely as w and inversely as the cube of v, we have T =p. w v8 104 RATIO, PROPORTION, AND VARIATION [Cuap. VII. — The attraction F of any two masses m and mg for each other varies as the product of the masses and inversely as the square of the distance r between the two bodies. That is, | kmume — , 9 e PF EXERCISES AND PROBLEMS Reduce each of the following statements, Exercises 1 to 7, to an equation: 1. y varies as x, and y = 34 when z = 2. SOLUTION : y = ke. Since y = 34 when z = 2, we have 34 = 2k, ork = 17. Hence, hy 2 2. T is directly proportional to ¢, and ¢ = 3 when T = 27. 3. n varies inversely as m, and n = 39 when m = 3. 4, S varies as ??, and S = 144 when ¢t = 3. 5. A varies jointly as B and C. When B = 3 and C =4, it is found that A = 14. 6. P varies directly as q and inversely as r?. When gq =1 and r = 3, it is found that P = 27. 7. The surface S of a sphere varies directly as the square of the radius r. The surface of a sphere of radius 1 is 47. 8. If x varies as y, and if y = 1 when zx = 8, find x when ge): SOLUTION : x =ky. Since y = 1 when z = 3, we have eH open Lg aes or t= oy. When y = 5, we find x = 15. 9. Ifa varies as y and if « = 3, when y = 6, find y when 2 = 7. 10. If a varies inversely as b, and if a = 3 when 6b = 1, find a when 6 = 8. 11. If 2 varies jointly as z and y, and if 2 = 120 when a = 2 and y = 3, find z when x = 3 and y = 3. Art. 64] PROBLEMS 105 12. If z varies directly as x and inversely as y, and if z = 120 when x = 2 and y = 3, find z when z = § and y = 3. 13. If x varies directly as the square of y and if x = 4 when y = 1, find x when y = 4. 14. The safe load of a horizontal beam 10 feet long supported at both ends varies jointly as the breadth and square of the depth. If a 2 by 6 white pine joist 10 feet long safely holds up 800 pounds, what is the safe load of a 2 by 8 joist of same length? 15. Write in the form of an equation the law: The safe load w of a horizontal beam supported at both ends varies jointly as the breadth b and the square of the depth d and inversely as the length / between supports. 16. A beam 15 feet long, 3 inches wide, and 6 inches deep ~ when supported at both ends can bear safely a maximum load of 1800 pounds. What is the maximum load for a beam of the same material 10 feet long, 2 inches wide and 4 inches deep ? 17. What is the safe load for the second beam mentioned in Problem 16 if it is turned so that the width is 4 inches and depth 2 inches ? 18. Write in the form of an equation the law: The crushing load of a solid square oak pillar varies directly as the fourth power of its thickness ¢ and inversely as the square of its length l. 19. If a four-inch oak pillar 8 feet high is crushed by a weight of 100 tons, what weight will crush a pillar half as high and 6 inches thick? (See Problem 18.) 20. What weight will crush a four-inch oak pillar 4 feet high ? : 21. The deflection d of a rectangular beam of a fixed length varies inversely as the product of the breadth 6 and the cube of the depth d. Write this statement in the form of an equation. 22. In the formula kbd s = ) 106 RATIO, PROPORTION, AND VARIATION [Cuap. VII. s denotes the strength of a rectangular beam, b, d, and I, the breadth, depth and length respectively, of the beam, and k is a constant. Translate the formula into English using the terms of variation. PROBLEMS INVOLVING MOTION 23. The number of feet a body falls varies directly as the square of the number of seconds occupied in falling. If the body falls 16.1 feet the first second, how many feet will it fall in 6 seconds? 24. How far will a body fall during the sixth second ? | 25. The velocity of a falling body at any time varies directly as the number of seconds occupied in falling. What is the velocity at the end of 6 seconds if the velocity at the end of the first second is 32.2 feet per second ? 26. An object dropped from a balloon strikes the ground in 7 seconds. At what velocity does the object strike the ground and what is the height of the balloon when the object is dropped ? 27. A wrench is dropped from an automobile at a height of 3 feet while the automobile is traveling at the rate of 70 miles an hour. How far does the automobile move while the wrench is falling ? 28. The time for one vibration of a pendulum at a given place varies as the square root of the length of the pendulum. In Chicago a pendulum 4 feet long requires 1.1 seconds for a vibration. What is the time of vibration of a pendulum 1 foot long ? 29. What is the length of a pendulum which vibrates every second at Chicago ? 30. A weight is suspended by a wire 94 feet long. What is the time of one vibration at Chicago ? 31. A pendulum supposed to vibrate every second registers 90,000 vibrations in 24 hours. How much must the pendulum be lengthened ? Art. 64] PROBLEMS 107 PROBLEMS INVOLVING PRESSURE 32. The volume of a gas enclosed in a vessel varies inversely as the pressure upon it. Twenty-four cubic inches of air under a pressure of 100 pounds will have what volume when the pres- sure is decreased to 50 pounds? 33. If a toy balloon contains 150 cubic inches of gas when under a pressure of 15 pounds per square inch, to what size will it shrink if subjected to a pressure of 45 pounds per square inch? (See Problem 32.) 34. The pressure of wind on a sail varies jointly as the area of the sail and the square of the wind’s velocity. When the velocity is 15 miles per hour, the pressure on a square foot is 1 pound. What is the velocity of the wind when the pres- © sure is 10 pounds per square foot ? 35. The pressure of gas in a tank varies jointly as its density and its absolute temperature. When the density is 1 and the temperature 300°, if the pressure is 15 pounds per square inch, what is the pressure when the density is 2 and the temperature 290° ? CHAPTER VIII EXTENSION OF THE NUMBER CONCEPT 65. Integers and fractions. The first numbers studied in school are the numbers called positive integers. With these numbers the pupil is able to add and to multiply two numbers together and to subtract a smaller from a larger. He under- stands the answer, for it is always a positive integer. But division cannot always be performed if we have only positive integers; for example, the division of 4 by 3 is set aside as impossible. A student knowing only the positive integers might learn to solve equations of some. kinds. For example, the equations 4—2z =0, 32 — 6 = 0, would give no difficulty, but the solu- tion of the equation 2x — 5 =0 would be declared impossible, for the student knows no numbers which will satisfy the equa- tion. He might call the result imaginary. With the set of positive integers our elementary mathematics goes little further than counting; even the simple operation of measuring cannot be performed. For this reason early in his school life the number system of the pupil is increased by add- ing to it the numbers which we call, fractions. These new numbers, while having no meaning in counting, have very definite practical applications. | EXERCISES Assuming that we know only positive integers and zero, show which of the following exercises are possible of solution. Solve: 1; ¢-— 3: = 0. 8. 37 + Dae 2. 32 — 15 =e 4. 2xr+/7=0, _ Arts. 65, 66] NEGATIVE NUMBERS 109 6. 52 + 42 = 0. . 8. 27 — 1 = 0: beer — 0 = (0. Geer 4 re i). fe oe + 2 = 0. 10. 2? — 32 = 0. Discuss the results of the following problems : 11. Three boys out hunting shot altogether 21 rabbits. Two of the boys shot an equal number, but the third boy shot 2 more than either. How many rabbits did each shoot ? 12. Three boys gathering nuts get altogether 21 bushels. Two of the boys gather the same amount, but the third boy gathers 2 bushels more than either of the others. How many bushels did each gather ? 66. Negative numbers. The numbers consisting of posi- tive integers and fractions are sometimes called the numbers of arithmetic. With this set of numbers, the operations of addition, subtraction, multiplication, and division, can be per- formed with but one exception. This exception is the sub- traction of a larger number from a smaller, which is declared impossible. For example, there is no number in the set which answers the question, ‘What number added to 4 gives 3.” If astudent, knowing only this set of numbers, knows how to solve an equation, he would declare the solution of the equation 272 + 3 =0 to be impossible, or he might say the answer is imaginary. In reading a thermometer, in debits and credits, in latitude and longitude, a physical meaning is found for the subtraction of a larger number from a smaller. The introduction of new numbers called negative numbers is one of the purposes of the study of algebra. With positive and negative integers and fractions the student is enabled to solve any problem which reduces to an equation of the first degree. The equation ax +6b=0 of Art. 22 has always a solution among these numbers. 110 EXTENSION OF NUMBER CONCEPT [Cuap. VIII. EXERCISES Assuming that we know only positive numbers, show which of the following exercises are possible of solution. 1. 22 —1 =. 6. x? — 54 +6 = 0. 2. BLS DF, 7 v—-x-2=0. $3. 2.42 = 4. 8. 2? — 32 4+2=0. 4. 4¢4+9=1. 9. 2277 + 37 — 2 =O: 6.52 = 2a: 10. 2? 4+ 1-="0: How many answers are there to each of the following prob- lems? 11. Two thermometers differ by 3 degrees in reading. If . the reading of one thermometer is doubled it equals the reading of the other. What are the readings of the two thermometers ? 12. The ages of two children differ by 3 years. The age of - one child is double the age of the other. What are the ages of the children ? 67%. Irrational numbers. The quotient of two integers is called a rational number. Since an integer may be written as a fraction with 1 as a denominator, the numbers discussed in Arts. 65 and 66 are rational numbers. For example, 3, —6, 4+, .67, —} are rational numbers. In operations with radicals we often deal with irrational numbers. Irrational numbers cannot be expressed as the quotient of two integers. For example, V2, Vi, V4 are irrational. A student having only rational numbers at his command would find it impossible to obtain the length of a diagonal of a unit square. In other words, he would not be able to solve the equation zx? = 2. A further extension of the number sys- tem to include irrationals is then necessary. _ Arts. 67, 68,69] REAL AND IMAGINARY NUMBERS II11 EXERCISES Represent by points on a straight line the following numbers : 1. 4, —3, —3, §, 0. 2. V2. SoLuTion: Construct a square of side 1 (Fig. 30). The length of its diagonal is V2. Laying off the length of this diago- nal from Owe find a point on the line representing ~/2. 3. V8. 4.2+V2, 2-v2, -2- v2, -24+ v2. Bree + V/8, —3 + V8. Arrange the following sets of numbers in ascending order. 6. V2, -4,2 — V2, -V8, 3, 2vV2. Pert 4) —/2, —2, 4, —3V2, 68. Real numbers. The four classes of numbers which we have discussed, that is, positive integers, fractions, irrationals and negative numbers, together with zero are classed together under the name real numbers. These numbers have the im- portant property that they may all be represented by points on the same straight line. 69. Imaginaries. We have seen that man found it desirable to invent fractions, irrationals and negative numbers. In just the same way it is found convenient to invent still another kind of number. The square of any real number, positive or nega- tive, is a positive number. The square root of a negative number cannot then.be a real number and is given the name imaginary number. It is not represented on the same line with the real numbers. The term “ imaginary numbers ”’ is here used in a technical sense. The numbers are imaginary in the same sense that a y] 112 EXTENSION OF NUMBER.CONCEPT ([Cuap. VIII. fraction, a negative number or an irrational number is imagi- nary for a person knowing only the positive integers. It is desirable to include these imaginary numbers in our number system; for otherwise we could not solve such an equation as 24+1-=0. The number 1/—1 bears somewhat the same relation to the system of imaginary numbers that the number 1 does to the real _ numbers. Any real number may be written with 1 as a coeffi- — cient; forexample, 4 = 1-4. Any imaginary number involves the product of a real number and ~/ —1. Thus, we may write Af HA DA: / 5 SAD ee EVE va Daf SO) ee) RAY The number V —1 occurs so often that a special symbol is used for it; that is, EXERCISES Raa, i. Write the following imaginary numbers in terms of 7 1. /—5. Sotution: V-5 = V5- -1=V75-vV -1 =iv5. ave 6. V—ae S20 A 0! 7. Voie : ae ar A/S 8. \/ re. 5. =. 9. V-07- oe 10. W-1 — 42. Arts. 70, 71] IMAGINARY NUMBERS 113 70. Powers of i. Since any imaginary may be written in terms of 7, it is important to consider the powers of 7.. We operate with 7 as with any other letter and remember that 2? is to be replaced by —1. Hence, for powers of 7, we have er |) aa a pee (ay) "1: SAS 0 hl ee v= ts 742 = —), and so on, repeating the numbers —1, —7, +1, +7. 71. Products of imaginaries. If imaginary numbers are written in terms of 2, the product of any number of imaginaries is easily found. Example. Find the product of —3, /—5, and /—7. SoLUTION: V/-3 =iv3, V—-5 =iv5, V-7 =iv’7. V8 V5 VAT =iv8 1 V5 iT = OVS 5 «V7 = —iV/105. In general if ~/—a and —6 are two imaginaries we have Va: V—b= iva: ivb =?Va- Vb = -Va- Vb = —Vab, ye ae /-a-/—b = — Vab. | EXERCISES 1. Continue the table of Art. 70 for 27, 28, 79, 7, a. Find the following products : Pewee... 4/ —3. 1. VSL een Bape a. 8. (/—2)3. ee 4/ —10: . 9. (V5) 5. s/he n/— 16. 10; (\/ 4) Bn cay nt 4/ — 11; Et (/ —3) Vas 12. (V—-4)'V—§. 114 EXTENSION OF NUMBER CONCEPT [Cuap. VIII. 72. Complex numbers. The sum of a real number and an imaginary number is called a complex number. Thus, 1 +7, 2 —\/— 3, 5 + 37 are complex numbers. In general, a complex number has the form a + 2b, where a and b are real numbers. The expression “ imaginary numbers ”’ is often used to mean complex numbers, and the term pure imaginary is applied to numbers of the form 7b. We shall find such numbers in solving quadratic equations. ‘ For example, 2 + ~/—38 satisfies the equation xv? —4244+7 =0, as shown by actual substitution. Thus, (24/7 —3)? -42 + J7-3) +7 =444/-3 —38 = See 73. Operations with complex numbers. The ordinary op- erations of algebra may be performed with complex numbers by operating with 7 as with any other letter and remembering that 7? = —1. Example 1. Add 6 + —3 and 3 -— W-2. In the 7 notation we are to find (6 +7173) + (8 —iv2) =6 +1V3 43 -iv?2 =9 +7(vV3 —- V2). Example 2. Multiply 6 + /—3 by 3 — V—2. In the 7 notation (6 +iV3) (8 — iV2) = 18 - 6iV2 + 3iV3 — 23-2 = 18 —1(6/2 — 3V3) — (-1)V6 = 18 + V6 —71(6V2 ae 74. Conjugate complex numbers. Complex numbers which differ only in the sign of the imaginary parts are called con- jugate numbers. Thus, 3 + 22 and 3 — 21, a+ 7b and a — 1b are conjugate. Since (a + 2b) + (a — 2b) = 2a, (a + 1b) — (a — 2b) = 20d, (a + 2b) (a — 1b) = a? + B?, we see that the sum and product of two conjugate complex — numbers are real numbers, but the difference is an imaginary number. Art. 76] REVIEW EXERCISES OF PROBLEMS 119 REVIEW EXERCISES AND PROBLEMS ON CHAPTERS V-VIII 1. If x varies inversely as y, and when x = 6, y = 2, what is the value of x for y = 3? 2. If x varies as y, show that x + y varies as y. 3. Determine x and y from 0.52— sy =0.15, 4¢ +0.2y = 2.41. (STANFORD) * 4. Solve a i = 4, ay i ra = 6. | (YALE) ey), 5. Find a mean proportional between (@ + y)? and (a — y)?. 6. Find a third proportional to - + : and - 7. Evaluate li 2ar3 4 5 6 (Pan 8. Solve the equation La 2B EDs pe ee | See: 9. The weight of an object above the surface of the earth varies in- versely as the square of its distance from the center of the earth. An ob- ject weighs one pound at the surface of the earth. What would it weigh 1500 miles above the surface? (Take 4000 miles for the radius of the earth.) 10. The weight of an object below the surface of the earth varies directly as its distance from the center of the earth. An object weighs one pound at the surface of the earth. What would it weigh at a point 500 miles below the surface ? 11. Simplify (a — ib)? — (a + 2b)?. 12. Represent graphically on the same diagram the numbers 2 — 31, ya | Be Aseely tae S, sl gma V/2— in/2. 13. Solve . 3ax — 2by =, a®*z + b’y = 5hbe. (SHEFFIELD) 14. If the base of a triangle is constant, show that the area varies directly as the altitude. * Institution that gave the question in an entrance examination. 120 EXTENSION OF NUMBER CONCEPT [Cuap. VIII. 15. Show that the radius of a circular cylinder varies inversely as the square root of the altitude if the volume is constant. 16. The price of sugar for the 21 days immediately following the open- ing of the European war was as follows: 4.40, 4.40, 4.40, 4.40, 5.00, 5.00, 5.00, 6.00, 6.50, 6.75, 7.50, 7.50, 7.50, 7.50, 7.50, 7.50, 7.25, 7.25, 7.00, 7.00, 7.00. Show the change in price by a graph. 17. On the same sheet of codrdinate paper, represent graphically the : 2 three equations y=27?+24+1,y=1l-2,y= ach 1. What can be told from the points where the graphs intersect each other? 18. Give an example of two linear equations in two unknowns whose - graphs are parallel. 19. Give an example of two linear equations whose graphs coincide. 20. If a:b =c:d, prove that ma +b: ma —b = mc +d:me —d. 21. If 7x — 42:82 — 32 = 4y — 7z: 38y — 8z, prove that z is a mean proportional between x and y. 22. The following scores were made by city school children and by rural school children in grades from the third to the eighth inclusive, in a speed test in the fundamental operations in arithmetic. Grade 3rd 4th 5th 6th 7th 8th iby 5.4 | 6.6 | 9.0 | 10.3 | Tb 13.1 Rural | 2:5 6.5 6.9 8.0 10.0 11.1 Make graphs of these scores to show the comparative speeds. Lay it 23. Solve: — pat, Vay ee pide ae i ea Ps + mt v5 (Mass. INstiTuTE) 24. If one of two numbers be multiplied by m and the other by n, the sum of the products is P, but if the first be multiplied by m’ and the second : by n’, the sum of the products is P’. Find the numbers. (DarrmMoutH) 25. Represent graphically, y = x? — 2z. (WISCONSIN) 26. Compare the cubes of the following numbers, Seis) 1 1/3 oe 2 27. By actual substitution show that 3 + 27 and 3 — 27 de the equa- tion x? — 62? + 13x = 0. 1, Art. 76] REVIEW EXERCISES AND PROBLEMS 121 28. Given that z varies as the sum of two quantities one of which varies directly as x and the other as x?. If when x = 1, 2, 3, z = 5, 16, 33 respectively, what is z in terms of x? 29. Simplify (@® + 21° +21 + 72)’, if 7 = —1. 30. Find the value of y by determinants from the equations : 3a — 4y + 22 =1, 2x + 3y — 32 = -1 oa — Sy + 42 = 7. (COLLEGE ENTRANCE EXAMINATION BOARD) d mae — meh —,, V—-2 + 37-1. 31. Simplify 8V —2 + V —4)(2V-3 -— V-2) ; es ea (ILLINOIS) 32. The force P necessary to lift a weight W by means of a certain machine is given by the formula P = a + bW, where a and Bb are constants depending on the amount of friction in the machine. If a force of 7 pounds will raise a weight of 20 pounds, and a force of 13 pounds will raise a weight of 50 pounds, find the force necessary to raise a weight of 40 pounds. (First determine the constants a and 6.) (HARVARD) 33. Solve the simultaneous equations ee Loh we Peele y. isa), —-1 1 -1 i ae eS Eat =i 2 20 ea -* =-, prove that ax +by +cz is a mean _ proportional between x? + y? + 2? and a? + b? + c?. 35. Find a number such that by adding it to each of the numbers, a, b, c, d, we obtain four numbers in proportion. (Mass. INstTiTuTE) 36. A glass of milk and cream is sold for the same price as a glass of milk of twice the size. If milk costs 7 cents a quart and cream 32 cents a quart, what is the percentage of cream in the smaller glass? (HarvARD) 37. Solve: “4x — 8y. = —3; ? llx + 5y = -15. Draw a graph and verify your answer. (Missourt) 38. If a:b =c:d, prove that ab + cd is a mean proportional between a? +c? and b?+d. — (Mass. INSTITUTE) 39. A man invested $2720 in railroad stock, a part at 95 yielding 2 per cent, and the balance at 82 yielding 3 per cent; his income from both in- vestments is $70. Find the amount invested in each kind of stock. (ILLINOIS) 122 EXTENSION OF NUMBER CONCEPT ([Cuap. VIII. 40. Suppose the earth to be a smooth sphere 25,000 miles in circumfer- ence and that an iron band 1s fitted closely around the earth at the equator. Suppose this band to be expanded by the heat of the sun until its inner circumference is 25,000 miles and 1 foot in length, and that the distance between the surface of the earth and the band is the same at all points. Is it possible to insert a knife blade ;4 inch thick between the surface of the earth and the band? Estimate your answer and then check it. 41. A concrete pavement is 28 feet wide. A cross section gives an arc of a circle with an 8-inch rise at the center. (Fig. 34.) It is necessary to determine the radius of this circle in order —___-———+,———__-— to construct the form (templet) by which dork the concrete is laid. Find the radius. Fic. 3¢ 42. Solve the above problem if the width of the pavement is a and the rise at the center is D. 43. The lengths of the sides of a triangle are consecutive integers. If - -a represents the side of middle length and A the area of the triangle, then A =f(a). Find f(a). 44. If 8 yards of silk and 12 yards of woolen cost $27, and 12 yards of silk and 8 yards of woolen cost $28, find the price per yard of the silk and of the woolen. (CALIFORNIA) 45. The radius r of one base of a segment of a sphere is one-half the radius of the other base. The altitude of the segment is equal to the radius of the smaller base. The volume of the segment and the surface of the zone are then functions of r._ Find these functions. (See Formulas 18 and 19, Art. 45.) CHAPTER IX QUADRATIC EQUATIONS 77. Typical form. A quadratic equation in one unknown zis an equation that can, by transposing and collecting terms, be written in the typical form ax? + br +c =0, where a, 6, c do not involve x, and have any values with the one exception that a is not zero. Since the result of multiply- ing the members of an equation in this typical form by any given number is an equation in the typical form, the a, b, c can be selected in an indefinitely large number of ways. A quadratic equation is said to be of the second degree. The function az? + bx + ¢ (a # 0) is called the typical quad- ratic function. EXERCISES Arrange the following equations in the typical form and select a, b, and c from the resulting equations : 2 1. Br +44 5-5 $241. SoLuTIoNn : By transposing and collecting terms, 8 Z ea AD hee en pa at 5 +3 0, , 8 1 so that Segiei 5) & = 9. 2.93 +27 — (27? = 24 + 9 — 9x7. Sere Wi)? = x + 2. 1 es Rae 4 124 QUADRATIC EQUATIONS [CHar. IX. ] 5. 4y* + oy +6 = 4 — 8y + dy’. 6. 22432-2744 oe ier 22 x ~ sats 4_—_ = —* 7 a eR + 2 i I+ i 2 8. 32? —xa +k =a" + 2. SouuTIon : By transposing and collecting terms, 7 at? — x + hein = (), so that | a= 9. 4? + (22-4 n)? = 1) 11. (mx +n)? + 2? = n?. (@ ++-1)? (eae 12. Tg 2 138. (k — x)? — (n+ 2)? = (m — @)?. 10. 274+ 2nr+t=k4+ te. 14. d*y? + y? ae (y+d)? =0. 15. (¢+n)? — ( — nj = ni’. 78. Solution by factoring. In solving a quadratic equation, it is generally best first to reduce the equation to the typical form. After such a reduction, if the left-hand member of the equation can be factored readily, the solution is very simple. Example. Solve (x —.3)?.= 6.— 2a. . Arranged in the typical form this equation becomes xv?—4r +3 =0. The factors of the left-hand member are easily found. They are x —3 ~ and x — 1, and we may write the equation in the form (¢ —3)(4 — 1) = 0. Any value of x which makes either factor zero will satisfy the equation. If x = 3, we have (3 —3)(8 —1) =0-2 =0. Again, if x = 1, we have (1 — 3)(1 — 1) = -2-0=0. Hence, 3 and 1 are solutions of the given quadratic equation. _ Arts. 78, 79] SOLUTION BY FORMULA 125 EXERCISES Solve the following by factoring: ih a +62 + 5 = 0. 9. 377+ 4r+4+1 =0. Bee 3)? = 1. 10. 32? — 17x + 10 = 0. Sat = 025. 11. 6y? — Sy + 2 = 1. ace = 12. 122 287 SS 9. 5B. (x +1)\(a@-—1)-8=0. 13. 4+ 2(42 —17) =0. 6. x2? — nx = mn — mx. 14. 2x? — ax — a’? = 0. To 2? — 2nz+ n? = 0. 15. (@ +1)? = 2(2? — 67 — 3). 8. 277+ 54 — 3 = 0. 16: a? ='Sa? — 2a: 79. Solution by formula. Any quadratic equation may be solved by the process of “‘ completing the square.” For example, to solve 2%? — 4% — 7 = 0, write the equation in the form x? — 2x = ni Adding 1 to both members to make the left-hand member 2 7 9 a perfect square, 22? —22 +1 = mee 1 = 5 or (x —1)? = . 9 3 Extract the square root, 2-—1=+ as ate BV 2; 3 = etae and | x =1+5V2, orl -5v2. Both of these values of x satisfy the original equation, as we see on substi- tuting them forz. Thus, x a ; = ee _ 9 3 2\1 rea —4\ 1 45/2 ) —7 =2\ 14+3V72+,-2) -4 1422 —7. 2 2 4 2 =2+6/7249 -4 -6vV2 -7 =0. In the same way, substituting 1 — SV 2 for x, we find aie ae ee 2(1 - 5v2)? - 4(1 -5v2) -7 =0. 126 QUADRATIC EQUATIONS [Cuap. IX. If we apply this method to the typical quadratic equation ax? + bx +c = 0, we obtain a formula by means of which we can write down the roots of an equation without factoring or conn Cue the square. Transposing c and dividing by a, the general equation be- comes 5 EO c ee +-% = —-: ca a Add the square of one half the coefficient of x to each member to make the left-hand member a perfect square, i oi, (2)}. 2: (2 - 90s aha 2a) aaa +5) uF O5 ee tO * pee he eee Te. Extracting the square root, bees Mb = 4ac eGo 2a —btivb? — 4ac or Qs) a 2a If we let x, and 22 represent the two solutions of the quadratic — equation az +ba+e= 0, ‘ —b +2/b?— 4ac _ we may write Ly = 2 =) a 2 cae ee; —+/b?— 4ae 4ac 2a These expressions may be used as formulas for the solution of any quadratic equation. — Arr. 79] EXERCISES AND PROBLEMS 7 Thus, to solve the equation 2x? — 4x.— 7 =0, we substitute in the formulas, a = 2,6 = —4, c = —7, and find a Ee + Vv 16 SEE Cah) af SEA oe 1 +33, ee 2 7) es 3g. EXERCISES Solve the following equations by use of the formulas, and verify by substitution: 1. v7 -— 4% +3 = 0. 9. 677 +2 - 12 =0. Zor — (x + 10 = 0. 10. 9x? = 13 — 42. 3. 224+ 7r +12 = 0. 11. 97? + 147 +3 =0. 4.327 + 10x = 82. 12. 3a? — 187 + 5 = 0. 5. 2r? — 3a — 14 = 0. 13. 1077 + 92 + 8 = 0. 6. 477+ 7744+ 3 = 0. 14. x7 + 6.51 — 5.27 = 0. toon — 2 —6 = 0. 15. (1 —2z)(x —2) +3 =0. 1 3 8. 67? + 54 +1 = 0. 16. xi + 5t —7— = 0. 17. y? + 0.9y = 1.2. - Solve by any method : 18. 16(1 — x)(1 +2) — 92? = 0. 19. 22° +974+14=0. © 22. x(x — 2) =5 4+ 32. 20. 22 +7x —-9 =0. 23. (2x — 5)? — (a — 6)? = 80. 21. 32? + 1.03¢ + .1 = 0. 24. 3x7 = 332 — 90. 25. (x —6)(x —5) + (& - 7)(a — 4) = 10. 26. Divide the number 29 into two parts such that. their product is 198. 27. A square is cut from a rectangular piece of paper. The remainder of the rectangle has the shape and dimensions shown 128 QUADRATIC EQUATIONS [Cmap. IX, in Fig. 35. What is the side of the square if the area of the origi- nal rectangle was 100 square inches. 28. The product of two consecutive numbers is 182. Find the numbers. 29. Three times the product of two con- secutive even integers exceeds ten times their sum by 60. Find the numbers. 30. The sum of two adjacent sides of a rectangle is 28 inches, and the area of the rectangle is 195 square inches. What are the dimensions of the rectangle ? 31. One side of a rectangle is 3.5 inches longer than the other. The area of the rectangle is 181 square inches. What are its dimensions ? 32. A picture whose dimensions are 56 x 60 inches has a frame of uniform width whose area equals that of the picture. Find the width of the frame. 33. Within a rectangle whose sides are 30 and 40 inches, a second rectangle is drawn so that its sides are everywhere equally distant from the sides of the given rectangle. The area of the second rectangle is one-half that of the first. Find the perimeter of the second rectangle. , y PHIG?H 85 8@. Special or incomplete quadratics. If b orc is zero in the quadratic equation ax? + bx +c = 0, the equation is said to be incomplete. I. When c = 0, ax? + bx = 0 is the typical form of the equa- tion. We can always write this equation in the form : z(ax + 6) = 0. =p : Hence, the roots are 0 and ee Conversely, if 0 is a root of a quadratic equation ax? + bu +c = 0, Arr. 80] SPECIAL QUADRACTICS 129 then a-0+6-0+c=0. That is, c= 0. Therefore, a quadratic equation has a root equal to zero when and only when the equation has no known term. II. When 6 = 0, ax? + c = 0 is the typical form. In this case, Case ya Conversely, if the roots of a quadratic equation are arithmet- ically equal, but opposite in sign, there is no term containing x in the first degree ; for if +7 and —r are both roots of ax? + br +c = 0, Cl) we have ar? + br +c = 0, G2) and ar? — br +c = 0. (3) Subtracting (3) from (2), 2Qbr = 0. Since it is given that r is not zero, it follows that b = 0. Hence, a quadratic equation has two roots arithmetically equal but opposite.in sign, when and only when the term in x vanishes. III. When b = 0, c = 0, the typical form is ax? = 0. Both roots of a quadratic equation are equal to zero when and only when the known term and the term in « vanish. Solve : EXERCISES 1. (a — 6)? = 72 — 122. Santi er*s d. x? rs (x + 3)( — 3), 2 ne 2. 9 = 3 4, x? — 32 = 0. 5B. (@ +1)? + (& + 2)(@ + 8) = 7. 6. (5 — x)(a — 6) + 30 = 0. Determine k so that each of the following equations shall have one root equal to zero: 1. of oe + 3k —.15 = 0. 8. 2? +72 1 ee: 9. 2y? — 5 +k — 4k? = 0. 130 QUADRATIC EQUATIONS [Cuap. IX, Determine k and m so that each of the following equations shall have two roots equal to zero: 10. 5x2 + 3ka — Q2mz — 5a +k+m—5=0. 11. 247? 4+ 4k?a + 8ka — 38m —1 = 0. Determine k so that the roots of the following equations may be arithmetically equal but opposite in sign: 12. 277 ++ 4+ 2kx +1 =0. 13. k*x.— 9x — 3x? = 14. 14. 2? +4 7k*z — 2kz — 8 = 0. 81. Formation of equations with given roots. It is some- times necessary to form a quadratic equation with given roots. The following theorem and its converse enable us to do this very easily. TuHEorEM. If r is a root of the equation ax? + bx +c = 0, then «a — r is a factor of the left-hand member. If r is a root of the equation, we have ar? +br+c=0. Then aav*?+bxa+c = az? + bx+ec-— (ar? +br+c) = a(z? — r?) + b(@ — r) (x —r)(ax + ar +b). Hence, x — r is a factor of ax? + bx + €. ConVERSE THEOREM. If #—vr is a factor of the left-hand member, then r is a root of the equation. If x — r is a factor of ax? + bx + c, then the substitution of r for x makes the factor x — r vanish, and 7 is a root of az? + br4+c=0. Arts. 81, 82] NATURE OF ROOTS 131 EXERCISES Form quadratic equations of which the following are roots: eee Sotution : When the right-hand member of the equation to be formed is 0, the left-hand member has factors x — 3 and x — 3. Hence fo 3 at Petre ee er os (v — 3)( — 9) =2? -ar +5 = 0 is a quadratic equation with roots 3 and 3. Since we may multiply both members of the equation by 2 we may write the equation freed from frac- tional coefficients in the form 2a" = 12. +5 =, OU. 2. 2, >. Ter b= Bo 1 -eNeas 3. 3, 8. ce mon 4. —4, —3. GD hia Fae: 5. +/3,.-vV3> 10, OTN oe Aes 6. a, 6. . 11. SS ee expe ae 7 m 12. Form an equation whose roots are 2 and * such that the coefficient of x? is 6. | 2 3 such that the coefficient of x is 5; such that the known term is 7. 13. Form equations whose roots are 2 and 82. Nature of the roots of a quadratic. In Art. 79, we found the two roots of the quadratic equation, ax? + br +c =0, | 35 b? — 4a a mp 4/07 — 4ac and Lo = 2a In case a, b, c are real numbers, the numerical character of these roots depends upon the number b? — 4ac under the radical 132 QUADRATIC EQUATIONS [Cuap. IX, sign. An examination of x and 2 leads at once to the follow- ing conclusions : | (1) If 6? — 4ac>0, the roots are real and unequal. Thus, in the equation ov? — 1llz +10 =0, we find b? — 4ac = 112 -4-3.-10=1>0, hence the roots of this equa- tion are real and unequal. Upon solution they are found to be 2 and >. (2) If 6? — 4ac<0, the roots are imaginary and unequal. Thus, in the equation x? — 62 +58 =0, we find b? — 4ac = 6? — 4- 58 = —196<0, hence the roots of this equa- tion are imaginary and unequal. Upon solution they are found to be 3 — 7i and 3 + 7. (3) If b? — 4ac = 0, the roots are real and equal. Thus, in the equation 4a? + 4 +1 =0, we find b? — 4ac = 42 -4- 4-1 =0, and the roots are real and equal. Upon solution they are found to be —2, —3. It should be observed that if the coefficients are real and one root is imaginary, then both roots are imaginary. | The expression b? — 4ac is called the discriminant of the equation ax? + bx +c = 0. If we add together the two roots of the equation ax? + bx +c = 0, we have -b. ++/b2 tue . — b —\7b eae b ay +t = —— 5 + — = - 2a 2a a Hence: I. The sum of the roots of a quadratic equation in x is equal to the coefficient of «2 with its sign changed, divided by the coefficient Of a7. Art. 82] NATURE OF ROOTS 133 Thus in the equation dav? — 444+ 5 =0, we know immediately that the sum of the two roots is . If we multiply the two roots together, we have —b ++/b? — 4ac\f —b a ie! C ae a SY rad Hence: Il. The product of the roots of a quadratic equation vn x vs equal to the known term divided by the coefficient of x. Thus, in the equation 3x? — 4x + 5 =0, we know immediately that the product of the two roots is bs 3 EXERCISES Determine the nature of the roots of the following equations : Trtig? — 237+ 2 = 0. 3. 27? + 6x + 199 = O. 2. 27— 2c + 2 = 0. 4. 4977+ 147 +1 =0. & ytyt+1 =. Determine the real values of k so that the roots of the follow- ing equations may be equal : 6. w+r+k =0. SotutTion: In order that the roots of this equation may be equal we must have 6? — 4ac = 1 —4k =0. Hence k is }. Substituting this value in the equation we have wP+e+s = (© +7)? =0. 7 v@tke+1=0. 9. v2 +3kr+k+7 =0. 8. 2a? 4+ 824 +k = 0. 10. kx? —- 4x4 +3 =0. Determine by inspection the sum and the product of the roots of the following equations : di ae dae 1 = OQ. 13. 2+ 5x — 2a? = 0. 12. y+ 8y — 11 =0. 14. 2? — 274 + 3k = 0. 15. 2 +pr+q=Q0. 134 QUADRATIC EQUATIONS [Cuap. IX. Determine the value of k in the following equations : 16. 2? + kx + 3 = 0, where one root is. 2. Sotution : Let 2; be the second root. The product of the roots of this equation is 3. Hence, Pies 1 = 13; or 2; -3. aet er ta AH a i The sum of the roots is —k. -—— Be iad Hence, 2 +5 ee —k eet pea per i: | HH are and k=—>5 ) | | 17. x? + 5¢ 4+ 2k = UF aes! | Se E where one root is 2. | | | i 18. 227 + ke + 14 = O, PCE TW, ae eae where the difference between | ° | Y the roots is 9. HEEH | b | LI 19. 2x? — 6kx + 9k = 0, Se pana me eS where one root is 6 times | \ / See ee the other. U HH 20. 627 + kz — 7Fo=10; where the quotient of the / eo ee n ° =F 0 ig Ripert trie nae two roots is——- Fig. 36 2 83. Graph of the quadratic function. To graph the quad- ratic function x? + « — 2, compute a table as follows: r= Fe sede kee ae hee e+2—-2=| 63] 4] 18| 0|—13|—2|—22|/—2|—12/0/92/4 - Plotting the points (—3%, 63), (—3, 4), - - - - from the table and drawing a smooth curve through them we have the curve in Fig. 36. The graph crosses the X-axis at —2 and 1, hence for these values of x the function x? + 2 — 2is zero. In other words, —2 and 1 are the solutions of the equation r+ 2 — 2'= 0, Arr. 83] GRAPHS OF QUADRATIC FUNCTIONS 135 They are represented graphically by the abscissas of the points where the graph crosses the X-axis. It can be shown, if a is positive and different from zero, that the graph of the function az? + bx +c has the same general characteristics as the curve in Fig. 37. This curve is called a parabola. The real roots of the equation az” + bz +c = 0 are given by the abscissas of the points where the curve crosses the X-axis. If the curve has no point in common with the X-axis, then the roots of the equation are imaginary. If the curve touches the X-axis, both roots of the. equation are real and equal. These three cases are shown in Y Fig. 38, where the graphs of 7? +2—-—2, +244, x? + a+ 2 are shown. Kiev 3/ EXERCISES Construct the graphs of the functions in the fol- lowing equations, and by measuring determine the roots if they are real : Lea to), a? 22 —3 =0. x?—2x4+5=0. 7—-2744+1=0. 2a? + 3x —2 =0. 2x? + 34 +2 = 0. Av? — 122 +9 =0. 8. 1+ 27-2? = 0. | 9) 2? 5 Sea Teor — a2 = (). ~ | Scale, 2 spaces=1 Unit ‘77 Fic. 38 TS eS ee 8 b 136 QUADRATIC EQUATIONS [Cuap. IX. MISCELLANEOUS EXERCISES AND PROBLEMS Solve: | 1. a? — 97 + 14 = 0. 3. 1277 —2z —~1=0) 2.) 2? bor + li 0: 4. 77+ 54-1 =0. 5. 5x2? + 247 — 5 = 0. 6 («+1)?4+3¢4+1) +2 =0. 2 3 3 11. ——— 4+ —— =], ie BD ty ters 243 Lae: x . n m 8. A 9. $2: A ——— =], x+2 — + —+n x 1 Lema x—-a 2%a—2z)—b rosette Oi: tet ba iad: Wome “2, 10. = a. ££) °a ae CF: see — + — a. ££ 15. Find all of the roots of the equation x? — 9x = 0. Solve the following equations: 16. x? — 7.35% + 2.45 = 0. Wy 277 + O27 — 8.04 = 0. What are! the values of & if the following equations have equal roots? 18. 2? — 2ka + 3k? = 9. 19. kx? + 2kx — 32 42 = 0. 20. Find by the graphical method the approximate values of the roots of the equation x? — 4x — 13 = 0. Find the quadratic equations whose roots are : 21. 22. 23. Tata's (a — b), (a + 5b). n(1 +n), n(1 — 7). Art. 83] | PROBLEMS 137 24. For what value of k will the equation a 2 ke 1. ied Oty ea | have roots arithmetically equal but opposite in sign ? 25. Find two numbers whose sum is 12 and the sum of whose squares is 74. 26. What numbers differing by 9 have a product 198 ? 27. Find two consecutive integers whose product is 506. 28. Find two consecutive even integers whose product is 288. 29. Separate 42 into two parts such that the first is the square of the second. 30. Separate the number a into two parts such that the first: is the square of the second. 31. One leg of a right triangle is twice the length of the other. The number of feet in the perimeter of the triangle is the same as the number of square feet in the area. Find the sides of the triangle. 32. One leg of a right triangle is n times the length of the other. The measure of the perimeter in feet is the same as the measure of the area in square feet. Find the sides of the triangle. 33. One leg of a right triangle is 5 feet and the hypotenuse is 5 feet longer than one half the other side. What is the area of the triangle ? 34. The length of one leg of a right triangle is a and the hypotenuse is a feet longer than one half the other side. What is the area of the triangle ? 35. Find the radius of a circle such that the number of feet in its circumference equals the number of square feet in its area. | 36. By increasing the radius of a sphere by 1 inch, its volume is increased 40 cubic inches. Find the radius of the original sphere. (See 17, p. 64.) 138 QUADRATIC EQUATIONS [Cuap. IX. 37. The area of a triangle is 16 square feet and the base is 12 feet more than twice the altitude. Find the base and altitude. 38. The area of a triangle is b square feet and the base is n feet more than twice the altitude. Find the base and altitude. 39. Show that the equation x? — ax — b? = 0 has one posi- tive and one negative root. 40. Around a rectangular flower bed which is 3 yards by 4 yards, there extends a border of turf which is everywhere of equal breadth and whose area is 10 times the area of the bed. How wide is the border ? 41. Graph on the same sheet of codrdinate paper the function a7 +4+6¢ where c takes on the values —3, 0, 1, 10. What effect does changing the constant term in a quadratic function have on — the graph ? 42. Graph on the same sheet the function ax? +241 where a takes on the values 10, 1, 3, 75. Decreasing the coeffi- — cient of x? has what effect on the graph ? 43. The edges of a cube are each increased in length 1 inch. It is found that the volume is thereby increased 331 cubic inches. What was the length of the edge of the cube? 44. A cistern is filled by two pipes in 40 minutes and the larger pipe can fill it in one hour less time than the smaller one. In how many minutes can the smaller pipe fill the cistern ? 45. A rectangular piece of tin is three times as long as it is ' wide. From each corner a 3-inch square is cut out and the ends turned up so as to make a box whose cubic contents are 312 cubic inches. What were the dimensions of the piece of tin ? 46. A man sold some railroad shares for $4050. 2 Ot Wie and x =5 — > 2, Se ee. emer A; and, y= 5. x JE, where 7? = —1. Check these solutions by sub- stitution in (1) and (2). The graph of equation (1) is the circle shown in Fig. 41, and the graph of equation (2) is the straight line there Fig, 41 shown. Itis to be noted that these graphs do not intersect. This fact means that there exists no pair of real numbers that satisfies both equations (1) and (2). 144 EQUATIONS INVOLVING QUADRATICS [Cuap. X. EXERCISES AND PROBLEMS Solve the following systems, verify each set of roots by substitution in the given equations, and draw the graph for at least Exercises 2, 3, 5, to show the graphic interpretation of the results. j Rag era Os The Fn Bs 9. 7 — reeds x+y =d. ry =a?+a. 2. «7? — y? = 48, 10. 2* + y? = 260, x — ly = 0. LY ae Pp Gre oa het MO 11. xy + 150 = 0, 2x + 3y = ls. x—y+31=0. 4. 5x? —3y?+7 =0, 12. 2x? — 2ry + y? = 10, 2e+y—-—7=0. Cys TT 5. ry + 54 = 0, 13. x—y = 4, rt+y+3=0. _ dx”? + Qry + 4y? = 25. CS ee re A BP 14. 7+y+2y = —4, xy +15 = 0. r—y=4., ie 2) 1 1. +71, a ide OP ha) el a ij 8. 44 — 3y + 16 = 0, 16. 2 ee Dee se cee 47 + dy = 3. 17. The perimeter of a rectangular athletic field is 1124 yards, and the area is 16 acres. What are the dimensions? 18. The difference of the two legs of a right triangle is 7, _ the hypotenuse is 17. Find the sides of the triangle. 19. Find two consecutive integers the difference of whose squares is 31. 20. The difference of the areas of two squares is 700 square feet and the difference of their perimeters is 40 feet. Find a side of each square. Arts. 86, 87] SIMULTANEOUS QUADRATICS 145 86. Both equations quadratic. When both equations of a system are quadratic, the problem is often so difficult that the system cannot be solved by methods at present at our disposal. There are, however, some forms of such equations for which we may easily obtain solutions. In Arts. 87 and 88 we shall consider a few such equations. 87. Both equations of the form aa’?+by?+c=90. If, instead of considering x and y as the unknowns, we consider — ee ee Scale, 2 spaces=1 Unit Fig. 42 first x2 and y? as the unknowns, the method of solution is that for linear equations. Example. Solve 9x? + 16y? = 288, (1) | x2 4 y? = 25. (2) SoLvuTIon : Solving for x? and y?, we have x? = 16, (3) y? =9. (4) From (3) and (4), g=4+4,y = +8. 146 EQUATIONS INVOLVING QUADRATICS [Caap. X. Iorming all possible pairs, we have the four solutions ly =3, ly = -3, 7 aoe ly = -3. The equation (1) has for its locus an oval shaped figure called an ellipse (Fig. 42.) The equation (2) has a cirele for its locus. The four points of intersection represent graphically the four solutions. EXERCISES Solve the following systems of equations and give the graphical representation for 5, 6, 7, and 8. 1. 9x? + 25y? = 225, 5. 9x? + 16y” = 288, i oh OP) SY x? + y? = 9, 2. O08 Lge ee G0, 6. 9x? + 16y? = 288, 7 yr Lay, e* 4p eas 3. 49? —9y? ‘=.19, 7. 9x? + 16y? = 288, x? + y? = 34. x? + y? = 32. 4. 4074+ y? = 61, 8. 9x 7+ 16y? = 288, 2x? + 3y? = 93. x? +y? = 64. 88. Further special methods. Many systems may be solved by special devices in which the aim is to find values for any two of the expressions, x + y, x — y, and xy, from which the values — of x and y may be obtained. Other devices, such as substitu- tion, are suggested in the exercises which follow. Example 1. Solve the system x? +2y = 12, (1) y? + ay =4. (2) SOLUTION : Adding (1) and (2), x? + 2ry +y? = 16. (3) Hence, each Oye: clera (4) Subtracting (2) from (1), Co dat) bart (5) (5) + (4) gives, —-y=+2o0r—-2. © (6) From (4) and (6), Zee oor —3; y = 1 or —1. ArT. 88 | EXERCISES By substitution in (1) and (2), we find that only the two pairs eee 3 : ey i satisfy (1) and (2). Example 2. Solve the system x+y’ = 16, a? = OF. SoLuTION: Substituting 6x for y? in (1), we have x? + 6x = 16, or x? + 6x — 16 = 0. By formula (Art. 79), 2 = aes | If x = 2, we have y = +V/12 = + 2V3. 147 (1) (2) (3) If x = —8, we get y = +V—48 = +4i\/3, which are imaginary numbers. The solutions are (2 =2 “i ~=2 c= —-8 x= -8 Peat y 2/8 Ny = a5) Check the results by substitution. EXERCISES Solve the: following systems: 1.y* = 62, 6. 27 + zy = 36; x—y+8=0. ry + y* = 45. meee ey? = 8), Tee ri 10) x? — y? = 77. ry + y? = 6. 3. 27+ y? = 25, 8. p?—q? =9, cy = 12. 4p? = 25g. 4, 5x7 — 9y? + 121 = 0, 9. +7 = 34, Ty? — 3x7 — 105 = 0. Sie to; 6. 9aee a = 50} 10. s* + st = 45, ry = —7. s? — st = 5. y = —4ivV/3. EQUATIONS INVOLVING QUADRATICS [Cuap. X. MISCELLANEOUS SIMULTANEOUS EQUATIONS INVOLVING 148 QUADRATICS Solve : 1)? = 77 = 100, 14, ox + 4y = 50. 2.2—y = 9, 15. xy = —6. : Si 97? 2-254? = 225, 16. Ox? — 25y? = 225. 4. 477 = 977, ti 4x? + Oy? = 1. hee rie ee) lean 8 Hint: Subtract second from first. 6. x?—xzyt+y’? = 3. x—y =0, Cy? = 2a) — 3¢Y en bx + y = 3. LY — By = 5, 2y + ry = 6. xv? +y? + 3x + dy = 28, xy —6 = 0. Hint: Multiply second by 2, add to first and solve for x + y. 18. 19. eaters xy —10 = 0. 2(z — y)? +2(4 + y)? = 85, Qa —y) + 4(a + y) = 25. ey 12 =0; x? + y? — 25 = 0. x? — y* = 3, C4 = Ole Hint: Divide first by second. 10. 11. 12. 13. 4x? — y? = 16, 2 Ay = se r+ y = ary,’ etyt+ar+y? = 49. x? — Qey — 2y? = 32, x — by = 2. x? + 4y? = 24, CY nO, 20. 21. e+y+art+y =2, xy —2=0. ty = 2(r + y), x? + 2ry + y? = 1. 22? + 5y? = 125, 5a + 2y = 10. 2x* — zy = 10, Ot" = Ate Hint: Multiply first by 11, second by 10, and subtract to. eliminate the constant terms. Factor the resulting expression into 22. 23. 24. (x + 2y) (8a — 5y) =0. x? + xy + 2y? = 44, 2x? — ay + y? = 16. x? +y? = 16, a? y” Og. (eu x? + y? = 25, ox — 2y = 6 Art. 88] EXERCISES AND PROBLEMS 149 25. 27 + 4y? = 17, 31. 6x? + dry — by? = 0, wipe 2: 2a? — y? + 5x = 9. 26. v?+y’?=9, 32. 274+ sry + y? = 41, ot — 2. ety tety = 32. aT ee OY 4, 33. 2? + dry + 2y? = 3, Po y= 18, Qr? + y? = 6. 2 va le 28. 3x” + dxy + 2y” = 8, 34. (x + y)2—3(x + y) = 10, x? — xy — 4y? = 2. g? — y? = Hint: Let y = vx. Substitute in both equations and solve for v and z. = 5, S0 ety 10, UY ie Te 36. 2? +y?+x—-—y-—12=0, tomes oY ced OF 29. x7 + 32y — 2y? = 2, 2x? — dry + by? = 3. 30. x77 + zy — by”? = 0, a? + 4x + 3y = 69. 37. x? + 2ry + y? + 2x + 2y Hint: Solve the first equation . te bes, : for x in terms of y, and substitute v* — dry + y* — 2x + 2y in second. +1=0. 38. The sum of two numbers is 12 and their product is 35. Find the numbers. 39. The sum of the squares of two numbers is 100, and their difference is 2. Find the numbers. 40. Find two numbers whose product is 150 and whose err quotient 1s 5 41. The area of a rectangle is 120 square yards and the diagonal is 17 yards. Find the sides. 42. A rope 48 feet long exactly surrounds an enclosure in the form of a right triangle of hypotenuse 20 feet. Find the other sides of the enclosure. 43. A loop of twine 34 inches long is to be put around four pegs set in the form of a rectangle of area 60 square inches. Find the sides of the rectangle. 150 EQUATIONS INVOLVING QUADRATICS [Cuap. X. 44. A farmer raised broom corn and pressed 6120 pounds into bales. If he had made each bale 20 pounds heavier, he would have had one bale less. How many bales did he press and what was the weight of each ? 45. After a mowing machine had made the circuit of a 10-acre rectangular field 33 times, cutting a swath 5 feet wide each time, 23 acres of grass were still standing. Find the dimensions of the field. 46. A farmer bought 5 cows and 30 sheep for $525. He bought one more sheep for $25 than he bought cows for $300. Find the price of each per head. , 47. A field contains 9 acres. If its length were decreased by 20 rods and its width by 4 rods, its area would be less by 4 acres. Find the length and width. 48. If the length of a diagonal of a rectangular field of 30 acres is 100 rods, how many rods of fence will be required to enclose the field ? 49. It took a number of men as many days to pave a street as there were men, but had there been five more workmen em- ployed, the work would have been done 4 days sooner. How many men were employed ? 50. The sum of the squares of two consecutive integers is 1301. Find the numbers. 51. A rectangular piece of tin containing 400 square inches is made into an open box, containing 384 cubic inches, by cut- ting out a 6-inch square from each corner of the tin and then folding up the sides. Find the dimensions of the original’ piece of tin. 52. If the product of two numbers is increased by their sum, _ the result is 79. If their product is diminished by their sum, the result is 47. Find the numbers. 53. For two numbers a and 6 it is found that the expressions a (a — b), a? — b give the same number. What is this number? Art. 88] PROBLEMS 151 54. The area of a field is 367,500 square feet. The length of the diagonal is 875 feet. What are the dimensions ? 55. The difference of the squares of two numbers is divided by the smaller number. The quotient is 4, and the remainder is 4. If the difference of the squares of the numbers is divided by the greater number the quotient is 3 and the remainder 3. What are the numbers? 56. A certain floor having an area of 96 square feet can be covered with 432 rectangular tiles. If the workman use a tile one inch longer but one inch less in width, it takes 512 tiles to cover the floor. What are the dimensions of the two sizes of tiles? | 57. A club of boys bought a motor boat for $192. Four boys failed to pay their share as agreed, so each of the others was compelled to pay $4 more than he had promised. How many boys were in the club? 58. A father divided $1000 between his two sons and kept it for them at simple interest until called for. At the end of 24 years, one son called for all the money due him and received $632.50. At the end of 3 years the other son received $531 as his share. How was the money originally divided and what interest did the father pay ? 59. A few days after the outbreak of the European war the cost of 100 pounds of sugar in the United States was $3.10 more than it had been just before the outbreak. For $330 a grocer received 3100 pounds less sugar after the outbreak than he would have received before for the same amount. What was the price before and after the outbreak of the war ? CHAPTER XI EXPONENTS, RADICALS, AND ROOTS 89. Positive integral exponents. The expression @”, when n 1s a positive integer, means the product a:-a-a... ton factors. In operating with positive integral powers we make use of the following laws: 1k OO = 1 Example. os ON a 1H fs a” +a" =a”, wherem > n. Example. 47 + 43 = 44, ITI. (7) a Example. (O4)* sess, IV. (a°bic" is -)n = arrpirmen... Example. (52a5c7)4 = 58q!2¢28, Vv. (=) a le: a” a”"P Example. (Fe) = oo EXERCISES Perform the indicated operations and simplify the results - when possible. Fractions should be reduced to lowest terms. 2 a’ GS aN Ee Te tare 5. hea a a’ het 2. ath? « aids, 1g aibadle 6. = Art. 89] EXERCISES 153 6min®p? — ay : ; . . (=—])- Lies) i 2mn*p a (= ee 3 12a*bc 1SapUe a 182 (ee, E Sab*c? 14. (—3)? - (—3)8. 19. (rrs)4, Be on0"C)”. azt4h§ 10. (—3m2n’)%. 15. nebo PAVE Chal acl 4ab gergntt ainb™ wie (=) 16. aALat PA at qzt2 (24 ei y')? ys _ ys 22. qaz-2 , 26. ee ae iets ; hey Re ys)? (a <3 b)3 6aq22-8y b3y—-5 23. RD 27. (2az2u by-9)3 6a\?2 /b\8 b \2 (a — 1)3” (F) (23) (se) ae a2\" c2\n xy n (6a2x3y)? Be) (pe) (or) aw CR 31. Find the value of 1 +2 ae : gto : + ;. for x.= 10. 32. 33. 34. 35. exis otc 4h 4442 +5 ; for x = 10. Find the value of 22% + x +3 . 2 + 7a 2 ; for x = 10. Find the value of each of fi ae sums for z = 4: (a) 1+2. (b) l+2a2+4 2%. (c) L+aua+a2? 4+ 2%. (7) l+a+274+ 2° 4+ 24, Find the sum of 2, of 3, of 4, and of 5 terms of ee fore = 10; So a 154 EXPONENTS, RADICALS, AND ROOTS [Cuap. Ub 36. Find the value of 5é + 6#2+9t+3 for t=10. Also Brdiee sae enees eet ae | 37. Write in terms of powers of ¢ = 10, the numbers 8648, 48609, and 100101. 38. Expressed in terms of powers of ¢ = 10, 35.762 = 31+ 5 +i +4 Ba Write in terms of powers of t = 10, the numbers 3.1416, 1.4142, and .06041. 90. Extension of the meaning of an exponent. Thus far we have dealt entirely with positive integral exponents. No meaning has thus far been given to a” if n is zero, fractional, or negative. It is found useful to give a meaning to such expressions as a°, 8?, and 2°. It is convenient to define them so that the laws for positive integral exponents hold for these new kinds of exponents. Such definitions will now be given. 91. Meaning of a. In order that Law I, Art. 89, may hold for an exponent zero, it is necessary that CE. (UT el Ch at eset or a =)1) 11.a = 0 That is, for all values of a, except 0, ao =1, 92. Meaning of @’. Assuming that Law I, Art. 89, is to hold for fractional exponents, . at - at. at = aititt =a, and a? is one ‘of the three equal factors whose product is a. But this is what has been defined as the cube root of a. Hence we say that he a= Va. Arts. 92, 93, 94] NEGATIVE EXPONENTS 155 In general, if g is a positive integer, eer ne... to g factors = av eta tT ** a terms = 4, and a’ 1s one of the q equal factors of a, or a qth root of a. That is, ai = Va Examples. gi = V8 =2, 50? = 50 = 5V2. QIK 93. Meaning of nee Let p and q be positive integers. Then applying Law I, Art. 89, we have Pp , +ott+...topterms _q@ ) at?-a’-at...to p factors = a! p and a% means the pth power of the gth root of a. That is, | P &y ae aa = (s/a)? aa Va". Examples. 273 = (0/27)? = 9, 9? = (V/9)5 = 243. We have now found meanings for zero and _ fractional exponents. 94. Meaning of a” when mm is negative. Let m = —n, where n is a positive number. Applying Law I, Art. 89, we have a-a*=ar"=@ =1. i ee Hence, Cech cer. Urdu 0: 1 a™ - and | a” = Thus, a” and a~” are reciprocals of each other. Historical note on exponents. It appears that Nicole Oresme, a bishop of Normandy, in the fourteenth century conceived the idea of fractional exponents, and gave some rules for operating with them. But his nota- tion was very different from ours. To Simon Stevin of Bruges in Belgium (1548-1620) belongs the honor of inventing our present method of de- 156 EXPONENTS, RADICALS, AND ROOTS [Cuap XI. Therefore in a fraction, any factor of the numerator may be changed to the denominator, or any factor of the denominator to the numerator, if the sign of the exponent of the factor is ° changed. 27a-*b = 22-5 b 5243 2a Example. We have now found meanings for zero, fractional, and negative exponents. EXERCISES Give in simplest form the values of the following : Depots oO walt es eee 2.502 e14e ee Sree oars Dy | as (ae aig ey 16. (5y5)?. 5 17. (3)° ie 5. 5-1 187-1250 | 1\-2 ; Die! (Gy 19. 648. . ~20. 8-1(3)° - 363. ee (3) en —8. (x + 2)°. 21. 14 34 9; (=o) oe. —2 «2 oo, 7am 10. 16: lis 11. 83. —23. (394,)9 + 84 — Gj. 12.0 2ote Ace 24, 2-37-67. 25. 8 - 273 — 27 - 83 + 8-3 - 27-4 — 8 . 273. 26, 2-5 — 5% + (§)?— noting powers. It ‘remained, however, for John Wallis (1616-1703), an Englishman, to show the advantages of fractional and negative exponents to such an extent as to assure for them a permanent place in algebra. _ Art. 94] EXERCISES ; 157 yo) ieee ES ie 29. 3 - 32% — 3 - 32! i> > i 2 pe ie . i ten Je (—1)4 fee 2 + 374), at ( 14° — Qu ): Change the following into identical expressions without zero or negative exponents; and simplify when possible: Pain O27". a‘ 45. ees 32. (x? + y”)°. a*+a oe Gets te LO, SotutTion: By Art. 94, : 47. (a —b)- (a2? — Bb). Ba2e-2y-8 = 5a? - 1 i. = 5a? mys oy =48. (1 a) (a a) 0 ok. 49. 4x fy + 2). Borg 2 2. ie aie 3-2 a?bm She a on 51. . m2n—4 aib—ic8 St. ns 52. (22 + y)* (2? + y)!. SoLuTION: By Art. 94, 63. a-! + 2a? + 3a-3 mn-* min? m4 , ; m ns ni n- 54 Z 38 12ab-3 a7) + 2a-* +. 3a-° 8a 55. a! + (2a) + (3a). 3 39. 32" — —- bree ee. % "72 y? B02 4a) (—4).°. j Al. aty t: ~ 57. ae Pot. 7 1 ke a ‘be? =58. 2a% — dq-2 1 | ———: 2 =8 “er 2 (y-" —2 —2 , +P © a gy? 60. 8° + 81 + 8 — 84. 158 EXPONENTS, RADICALS, AND ROOTS ([Cuap. XI. Change the following into identical expressions without — fractional or negative exponents: Slee. 62. ai. 63. 8 - 53, 64. 65. 24. n 66. ym. 67. te 68. 69. e€ TO 71. 72. 73. 74. 75. 76. ads 78. 79. 80. 81. (m + n)?. (8xy°)*. amen’, 12a7?. a*b-*c8, (72 + 4)—4, (k-) 4 1-8), Tx Ty!, [(a))-1 82. 83.0 84. 85. 86. yi+y?+y. - y? + y? + y?. 3 gel ete gel, . 8a® \3 89. (E) 90. fo) Os 3 225 Change the following into identical expressions without radical signs or negative exponents: 91. 92. 93. 94. SOLUTION : Va*b'ct = (ab5c!)2, by definition of a fractional exponent, = a*bc2, by Law IV, Art. 89. 95. Wa: Vb. 96. v/a"bce, 97. v/a%b’c. 98. Wa + b. 99. v/(a + b)*. 100. v/a® + B®. 101. 102. 103. 104. 105. 106. 107. 108. 109. 110. 5 gly mM 5930 4 \/ ab?c3 | rye! Vath » Wx. SOO, Va 5, Vat. v/16a-8c-3. V8.8, V/36 + 64. V9 5. Arts. 94, 95] IRRATIONAL NUMBERS 159 Perform the indicated operations. By the use of nega- tive exponents if necessary, leave the results in forms without denominators. oii. ~ O°. 123. (=). a2\3 tiered o*. 124. (Ss) Plame f= x, *y?. 125. (—a*)-?. a4, 7. 126. (—m')-8, TAD yt 27°, 127. (GE): aoecae” + at. 128. m* + m-*. 117. ki - ki. 129, rt}. 72, Pipeie ~ 73. 130. (=). 119. (2a°b-%) + (a-%b?). 131. (rH)tH, 120. (a?)*. TS 2H te ON, 121. len). LS be (rot Reema 122. (a-")~2, 134. (Qh . Qn-1)3, 135. (a')5. Be ty fey) (st — a tyd + 41). hy (m-? + n~?) (on? —m? n-4 ve ia) n Dp 138. (a ™ + ba)’, 139; (oa? + bx-! + c)?. Ue Tale 95. Rational and irrational numbers. A rational number is defined (Art. 67) as one that can be expressed as the quotient 160 EXPONENTS, RADICALS, AND ROOTS (Cuap. XI. of two integers. An irrational number is a real number that cannot be thus expressed. Thus, 16, 4, and 12 are rational numbers; V2,* 72, V3, V5,.1 + V5, and 9? are irrational numbers. Any irrational number can be enclosed between two rational numbers that differ from one another by as small a number as we please. Thus, we may write, 1i Two radicals which, when reduced to their simplest forms, have the same order and the same radicand, are said to be similar. Thus, 5\/2 and —2v/2 are similar; so also are 7/81 and 2bv/3a' since 4/81 = 3/3 and 2bW3a? = 2abw/3. On the other hand, V2 and V3 are dissimilar ; so are V2 and w/2. The algebraic sum of similar radicals equals the common radical factor multiplied by the sum of its coefficients. Example 1. Example 2. Example 3. 3Va@-a+2 bo 8/2 +3/2 — 4/2 = (8 +3 —40 eee /75 +3712 = 5727 = 25 See ee 5V3 + 6V3 -— 15V3 = Aaa oe ae 2 1/4 2aba\ _ 3b 25a? — 3bvV/25a‘b2a = 5va + eva — 15a*b?V/a ( a 2 2b? 2a es 242 te 15a2b va 3 -5V9-3 ArT. 101. ] EXERCISES EXERCISES Collect terms and simplify : le Ss a ae bo a is) 14. 15. 16. 1G Re ee eek ere oe ee 38V2 + 8/2. V3 — 2/3 a L173: /20 + 8/45 — V5. 3/28 — 63 + 4/175. W/81 + 5W/24 — W375. 16 + 9/250 — V3. 6\/§ — /24 — V2 + 8V6. Va + 6Va + 2Va. 3/3 + 4V/ abc! + /4b%c?. Var — 4V/8a*bic3 + 3/6400". Vatbc® + 8v/b8c? — 5W/a®bec?. : V5 + Vash? — ~/27b5. . V{a + 6)? — Va? + ab — Vad? + 5B. Va+\/. + Ver Fa, V2 be ve Yy wv 3a%b + 4a2c + 5aid. Qatyt — Saity? + Qaby}. 165 . , be | SoLuTion: Taking out the common monomial factor x7y?, we have 2x*y? - Baty? + Qaxty? = (2 — 5xy? + Qy)a*y?. 18. 19. 3aby? — 2aby3. 2'a2b + 8a2b? — 18%a2b*. 20. + 22 + 38. 166 EXPONENTS, RADICALS, AND ROOTS ([Cuap. XI. 21. 3 + 12%a — 274d. 22. /a%b? + 3»/4a%b! + 5av/9ab? — 4abv/ab?. 23, 3/8 — 4/72 + 6V/48 — V108. 24. VW(at+y) — Vat + xy — V8ry> + By". 25. vF + Ways + A/a" 102. Radicals reduced to the same order. Two radicals are of the same order if they have the same index. Radicals of different orders may be reduced to the same order.. This is most easily done by the use of fractional ex- ponents reduced to a common denominator. Example 1. Reduce Va and Wa to the same order. SoLuTion: Ja =a =a = Vai. AVG = Gt =e A/G. Example 2. Reduce 2\V/xy and 5W/xy?z? to the same order. SoLuTION: Q/xy = 2x*y? = Qx*y*? = 2V/x3y3. | Saye = 5xdyizi = 5aby'z? = 5/x%yizt, EXERCISES Reduce to the same order: Ley 2,44, 6. 2/m, Vxy?. PREM AE W od 7. 2/2, 3V 3) 3. /xy, 20/mn. 8. wW/ay2z, 2W/axypr2?. 4.4/5, V7. 9. 1V/ab3, 4v/abee. 5. V/ab, W/x?y. 10. 14\/am?, 9V¥/a3m. 103. Multiplication of radicals. We may well illustrate the process of finding the product of two radicals by examples. Che 103] EXERCISES 167 Example 1. Multiply V3 by V5. Sotution: V3-V5 =3?- 5? = 15? (Law IV, Art. 89.) = 15. Example 2. Multiply 2¥/ab2m by 3~/mn. Sotution: 2v/ab?m - 3W/mn = 2(ab2m)* - 3(mn)? = 6(ab?m2n)* (Law IV, Art. 89.) = 6Vab?mn. Radicals of different orders may be reduced to the same order by the methods of the last article. Example 3. Multiply 2\/ab by 5W/a®b. SotuTion: 2Vab = 2V/a%b', 5a = 5Va'b?, 2W/ aid’ - 5V/ a4? = 10W/aib® = 10av/ab®. The principle used in finding the product of radicals of the same order may be stated as follows: The product of two or more radicals of the same order is equal to the product of their coefficients times a radical of the same order whose radicand is the product of the given radicands. EXERCISES Perform indicated multiplications and simplify as far as possible : 1. 3\/m - 5V/an. 6. 10 -w/2. 2. A/abe av/be. 7. 2a - 5a Wate, 3. 2/5 -v/10 - 40/35. 8. V/4 V8. 4, Sx -wWy?. 9. 6 -V/36 -W/5. 5. av/b -V/ab?. 10. 5 Ve 168 EXPONENTS, RADICALS, AND ROOTS [Caap. XI. The multiplication of radicals is often much more easily performed by the use of fractional exponents. That method should be used in the next ten exercises. © sb a/m -/mn + WY min?. av evenee ‘aoe Ls 20 eae SoLution : W/m +» Wmn +» oW min? =m'min'dmin? 2 * § 2 =5min® = 5m + min oo So =5mvV/ mn", 5 6 9 190 = 5m + m™*ni? 12. —/xy - 2/2". 17. V/x4y324 - Vx yz. 13. Va: VJd- Wa’b>. 18. 3a%c} - 5a%be. 14.-N/ 3165/3: 15. 375 - W/2. 19. —S8a‘b - —5a*%be?. 20. 2a%b?ch - 9akc?. 16. ~/128 -~/500. 21. —risi - 3r2sp. 22. SOLUTION : 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. (Va + 2v/b)(3Va — 5v/b). Va +2Vb 3Va —5Vb 3a + 6Vab — 5/ab — 10b 3a + Vab — 10b. (V3 -V5)(V3 +V5). (Vm +/n)?. 2/3(V3 +5 — 4/6). 2av/b(abv/b — av/ab + 5vV/a). (5 — 2v/5)(3 —V/5). (V7 +VID(V3 -V5). (Bey 8 — 3/5) (atb? — c*)?. (3a? + 2b*)?, (at + B)3, Arts. 103, 104] DIVISION OF RADICALS 169 Booey fe 8/28 —-/63)* 34. (a? + b*)(a? — 5). 35. (5 + 4V6 — 2v/3)?. 36. Find the value of xz? — 47 + 1 if 2 = 2 4+. 37; Find the value of 32? + 4x — 2 if x = Se = b2 — dac 38. Find the value of az? + br + cif x = ae 104. Division of radicals. When the dividend and divisor are monomials of the same order the division may be performed by the use of the principle Va 4 Vb Vb 66 6 = Example 1. oes 34/9 = 3/2. : aVaN NG } Example 2. es a ae Vi eel. /3I. V12 12 pees For purposes of computation it is usually desirable that the denominator of the quotient be made rational. A A AA) Example 3. ee A Mx, vi V7 V7 7 14 1-6 RA Example 4. ee VO ee ae oes Yo VV V6 6 Example 5. 8V5+ V2_ BV5+V2) 2V543V2) _ De eye (2\/5 - 32) (2/5 + 30 30 + 2/10 + 9/10 +6 _ 36 +11V10, 20 — 18 2 170 EXPONENTS, RADICALS, AND ROOTS [Cnap. XI. EXERCISES Perform the following divisions, obtaining results with rational denominators ° Ln Dian 8.. 6s/hiae DEN S80 i ND 9. 6/5 + 25. BON AD wb. 10. Vie 4, ~/35 + W/5. 11. Vab + Wcd. 5. W135 + W5. 12. Wa'm + Wam’. 6. V/ab + Va. 13. Wax + Vy. 7. VWabe? + Vac’. 14. 6\/150 + 5/45. 15. (4712 — An7 Gee. 16. (\/10 + 3/15 — 7V/35) + V/10. 17 eo 4) Te 7 20. (/a+ Vb) + Vab. 18.070 2415: 21. (/a +476) Se 19. ; + ab. 22. (3 +./6) + (26 -1). 23, 1 (28 — dao). 24. (5\/7 — 8/11) 4/7 ey 25. 7/2 + (2/5 — 9/2). ies og, Vt tvy. 3 +V6 Va -vVy 27. = 29. ov Oey em I YEeEN \/5 Seu PROBLEMS (Reduce the answers to the simplest radical form.) 1. What is the diagonal of a square whose side is s? 2. What is the diagonal of a square whose area is 280 Square inches ? Arts. 104] PROBLEMS 171 3. Express the volume V of a cube in terms of d, the diagonal of one face. Also express d in terms of V. 4. Find the altitude and the area of an equilateral triangle whose side is s. 5. Find the side s of an equilateral triangle whose area is A. 6. Find the side of an equilateral triangle whose area is 420 square inches. 7. If S is the surface and r the radius of s 8 a sphere, then S = 47r?. Find r if S = 2000 square inches. ; 8. A regular hexagon can be divided into 6 ante, S equilateral triangles. Find the area of a regu- lar hexagon whose side is 10 inches. (Fig. 45.) 9. The volume V of a pyramid of base area b and altitude A is given by the formula V = is A pyramid with a volume of 300 cubic feet and an altitude of 9 feet, has a square base. Find the length of the side of the base. BES 10. Find the altitude and the volume of a regular pyramid with a square base, if the side of the base is s, and the lateral . edge 3s. In a regular pyramid, the perpendicular from C the vertex to the base meets the base at its 4. - y center. In Fig. 46, OA is a lateral edge. 11. Find the altitude and volume in the last problem if s = 10. O 12. Show that the total surface of the pyramid in Problem 10 is s*(1 + 2/2). 13. From the formula of Problem 12 find A B the total surface if s = 10. 1D 14. The pyramid in Fig. 47 is a regular c tetrahedron. If each edge is 10, find AD, Fie. 47 AP, and lastly the altitude of the pyramid OP. Fic. 46 172 EXPONENTS, RADICALS, AND ROOTS ([Cuaap. XI. The faces of a regular tetrahedron are equilateral triangles. In Fig. 47, AP =2 AD, where AD is an altitude of the triangle ABC. 15. Find the volume of a regular tetrahedron whose edge is 10. | 105. Square roots of polynomials. The following solution will illustrate the process of square root. Find the square root of 9a + 2a° + 22x? + 12% + 9. SOLUTION : Oat + 120% + 2247 +122 +9 | 32? 4+ 2043 9x4 6x? + 2x 122° + 222%? | 1223 + 42? 6a? + 4a +3 Pe 18x? + 12% +9 EXPLANATION. (1.) The first term of the root is 9x! = 322. (2) The trial divisor is 2(3x?) = 62. (3) The second term of the root is 1223 + 6x? = 2z. (4) The complete divisor is 6x? + 22. (5) The second trial divisor is 2(8x? + 2x) = 6x? + 4a. (6) The third term of the root is 182? + 6x? = 3. (7) The second complete divisor is 6x? + 4x + 3. EXERCISES Find the square roots of the following : 1. 25x? + 60xy + 36y’. 2.. 4a* — 20a7b + 2507. 3. at — 2a® + 3a? — 2a + 1. 4. 4x* + 20x? + 292? + 102 + 1. 5. a? + 4b? + 9c? — 4ab + bac — 12be. 5 wy be ee 9 5 Bite ic 15 7. 28 — 27> Seay 22° + 5x? + 12a ee 8. al? — 4a9 + 4a6 — a® + 2a? + 3. Arts. 106, 107] SQUARE ROOT 173 106. Square roots of numbers expressed in Arabic sym- bols. The following solution will illustrate the process of extracting the square root of a number expressed in the Arabic notation. Find the square root of 419904. SOLUTION : 41’99’04 [648 360000 Trial divisor = 2 x 600 = 1200 59904 Complete divisor = 1200 + 40 = 1240 | 49600 Trial divisor = 2 x 640 = 1280 Mone Complete divisor = 1280 + 8 = 1288 10304 EXERCISES Find the square roots of the following. In each of the Ex- ercises 5—12, find the root correct to two decimal places. 1. 2116. 5. 85. 9. 3.1416. 25 2. 6084. 6. 3. 10. 2. 5 3. 524176. 7. 1. st 4. 822649, 8. 016. 12. . 107. Square roots of radical expressions of the form a+4/b, The square of /xtVyisa+y+ 2V/xy. Hence if a+/6 can be put into the form of x +y + 2vV/zy its square root can be found. Example. Find the square root of 7 + V/48. SOLUTION : mee ae = 7 2/12 = 443 42/4" 3S. Hence, te 4g a 2 + 1/3. 174. EXPONENTS, RADICALS, AND ROOTS [Cuap. XI. EXERCISES Find the square root of each of the following : hen are Oe 4. 10 ++/96. OSG n/p: 6. 151 + 24/7. 3. 11 + 6/2. 6. 27 =0ye * 108. Cube roots of polynomials. In a first course in algebra the process of finding the square root is explained by the aid of the formula (a + 6)? = a? + 2ab + O?. Similarly the method of finding the cube root is obtained by studying the formula (a + b)? = a? + 3a*b + 38ab? + B?. (1) Example 1. Find the cube root of 8x* + 362?y + 54ary? + 27y'. SoLtuTion: The work may be put in the following form. 8x5 + 3627y + 54xy? + 27y? | Qe + dy. 823 Trial divisor = 3(2x)2 = 122? 36x2y + 54ay? + 27y8 ‘ Complete divisor = 12x? + 3 (22) (8y) + (8y)? = 127? + 18zy + 9y? B6x22y + 54ary? + 27y3 EXPLANATION. In (1) the first term of the cube root is Ya? =a. Simi- larly in the present example, the first term of the cube root is 78x = 2z. In (1) the second term of the cube root, b, may be obtained by dividing 3a’b by 3 times the square of the root already found, which is 3a?, and which is called the trial divisor. In the present example, the trial divisor is 3 (2x)? = 122%, From 3627y + 122? = 3y, we find the second term of the root. The remainder in (1), after a3 is subtracted, may be written as the product of two factors, or 38a7b + 3ab? + b? = b(3a? + 3ab + b?). In the present example we need to find an expression of the form 3a? + 3ab + 6? where ‘ a = 22, and 6b = 3y. This gives 3 (2x)? + 3 (22x) (8y) + (By)? = 122? + 182y + Oy’, which is the complete divisor. Then 3y (122? + 18%y + Oy?) = -- = 36r?y + 54zy? + Q7y\. * Articles 108 and 109 may well be regarded as supplementary work. Art. 108] CUBE ROOT 175 When this product is subtracted the remainder is zero. The process is then complete and the cube root is 2% + 3y. It should be noticed that the complete divisor is formed by adding to the trial divisor an expression of the form 3ab + b?. Example 2. Jind the cube root of 8x° — 36275 + 6621 — 6323 + 332? — Ox +1. SOLUTION: 82° — 36z° + 6624 — 632° + 332? —- 9x7 41 | 27? —- 327 +1 a ad 1st trial divisor, 3 (2x7)? = 1224 | — 3625 + 6624 — 6323 1st complete divisor, | 1224 — 1843 + 92? | — 3625 + 54a4 — 2723 2nd trial divisor, 3 (2a? — 3x)? = 1224 — 3623 + 272? | 1204 — 3623 + 332? — 97 +1 2nd complete divisor, 1224 — 362° + 332? —- 9x +1 1224 — 362° + 332? — 9x +1 The first two terms of the root are found as in Example 1. To find the third term we proceed as in finding the second term. Three times the square of the part of the root already found is 3 (2x? — 3x)? = 1224 — 3623 4 272?, which is the trial divisor. Dividing, we find the next term in the root to be 1. To find the complete divisor we must add to the trial divisor 3(22? — 3x)-1 + 1? = 62? — Ox +1. The complete divisor is then 1224 — 362° + 332? —- 97 + 1. Multiplying this by the last term of the root and subtracting, the remainder is zero. The process is then complete and the cube root is 2a? — 3x + 1. If the root contains more than three terms, the succeeding terms can be found by continuing the process used in finding the second and third terms. EXERCISES Find the cube roots of the following expressions : 1. 273 + 272? + 9x 4+-:(1. 2. x? + 6a?y + 12ry? + 8y’. 3. 8a° — 36a2b + 54ab? — 276%. 176 EXPONENTS, RADICALS, AND ROOTS ([Cuap. XI. p> — 3pty® + 3p2y® — . 64a°b? + 144a7b? + 108ab + 27. 1 — 15m?2x +. 75m4*x? — 125m 523. - 216x%y> + 108m2rty2 + 18m*x2y +m. m> + 3m? + 3m + §. 1: — 624+ Qla? — 449*:-+ 632+ — bat ease 10. b® — 3b + 5b? — 36 — 1. 11. 8z!? — 36z!° + 662° — 632° + 33z4 — 9a? + 1. 12. 1252° + 1502y + 2252? + 60zry” + 180zry + 135” + 8y? + 36y? + 54y + 27. * 109. Cube roots of numbers expressed in Arabic symbols. The process used in finding the cube roots of polynomials can be applied to numbers expressed in Arabic symbols. This will be illustrated by an example. Example. Find the cube root of 185193. ret bacspe wil ahh lho SOLUTION : 185/193 |57 t = 50, 125000 = @ u=7. Trial divisor = 3t? = 3 - 50? = 7500 60193 contains 3t?u + 3tu? + wu. situ =3-50-7 = 1050 u? = 7% = 49 Complete divisor = 3? + 3tu+u? = 8599 60193 = 3f?u + 3tu? + u3. EXPLANATION. (1) The given number is divided into periods of 3 figures each to determine the number of orders in the cube root. (2) Since the number contains 2 such periods, the root contains two orders, tens and units, which may be represented by ¢ and u, respectively. Hence the cube is in the form | (¢ + u)? = #8 + 3éu + 3tu? + Ww, (3) The largest cube of tens contained in the number is # = 50? = 125000. (4) After @ is subtracted the remainder contains 3f2u + 3tu? + u3. * Finding cube roots of numbers may well be delayed until the next chapter where it is done much more simply and easily by the aid of logarithms. Art. 109] = EXERCISES AND PROBLEMS 177 (5) The trial divisor is 3 times the square of the root already found, which is 3¢? = 3 - 50? = 7500. (6) The next figure in the root is found by division to be 7. (7) The complete divisor is 3t? + 38tu + uw? =3 - 50? +3 .- 50-7 +7? = 8599. (8) Multiplying the complete divisor by the new figure in the root we have 7 x 8599 = 60193. Subtracting, the remainder is zero. Hence the process is complete, and the cube root is 57. (9) If the cube root contains more than two orders, the remaining orders may be found by repeating steps 5, 6, 7, and 8. EXERCISES Find the cube roots of the following : 1. 13824. 5. 636056. 2. 110592. 6. 21952. 8. 421875. (fm F334 0559 be 4. 804357. 8. 4913. MISCELLANEOUS EXERCISES AND PROBLEMS In Exercises 1-6 rationalize the denominator and find the values of the fractions correct to two decimal places. x Be. ‘ 2/5 —V7 1472 V5 +377 ; ae Rae VJ/3 +/5 37 3. 5+V3 | 6. aS FA Od eae Oh 24/3 5/2 — 4/3 7. Find the value of x? — 62 + 2 when 2 = 3 + 2v2. 8. Find the value of x? — 4a — 1 when x = 2 -vV/5. V19 oe 10. The area of a square is 14 + 6\/5. Find the side correct to two decimal places. 9. Find the value of 3x? — 27 — 6 when z = 1 + 178 EXPONENTS, RADICALS, AND ROOTS [Cuap. XI. 11. Find the side of a square whose area is 998001 square feet. 12. Find correct to two decimal places the radius of a circle whose area is 50 square inches. 13. In Fig. 48, T is a right triangle, and 71, T:, and 73 are equilateral. Prove that 7; = T, + 73. 14. The circumference of a circle equals the perimeter of a square which is 4s. Find the difference in their areas in terms of s. e 15. Assume a given unit of length and Fic. 48 construct geometrically /5. ‘SuccEstion. Use codrdinate paper. Construct a right triangle with legs 1 and 2. Then the hypotenuse equals V1 +4 =-+V/5. By actual measurement the value of 5 can be determined correct to tenths, or even hundredths, if the scale is taken sufficiently large. 16. Construct the following square roots: 7/2, 10, 17, x/3, V 20. Suacestion. In constructing 3 construct a triangle with base 1 and altitude 2. 17. Plot carefully the graph of y? = x for values of y from 1 to 10. Solve the equation for y. This gives y =z. If now for given values of x we read off from the curve the values of y, we thus obtain the square roots of those values of z. By the aid of this curve find the approximate value of the square root of 2, 3, 10, 16, 38, and.75. 18. Similarly from the curve y = 2° plotted from z=1 to x = 5 obtain the cube root of 2, 3, 27, 40, 65. CHAPTER, XII LOGARITHMS 110. Definition of a logarithm. In the expression 103 = 1000, the exponent 3 is called the logarithm of 1000 to the base 10, and is written 3 = logi1000. In general, if a* = y, (a>0, a #1), then 2 is said to be the logarithm of y to the base a, which is written x = logyy. The two equations . iy; and t= l0g,y are thus equivalent. That is, the logarithm is an ex- ponent. Without using symbols, the logarithm of a given number may be defined as the exponent of the power to which a number called the base must be raised to equal the given number. In what follows, we shall assume that the laws of exponents (Arts. 89-94), which apply to rational exponents, are valid for any set of exponents with which we are concerned. 111. Numbers expressed as powers of given numbers. We know how to express many numbers as powers of a given number. These same relations may be expressed by means of logarithms. | Thus, 16 = 42, and 2 = log,16; 8 = 4°, and : = log, 8 ; 4 = 4% and 1 1 = 1 —3 = logiea ; 100 = 102, and 2 = log100 ; ioe 10“1, and —1 = logw7o: 180 LOGARITHMS [Cuap. XII. EXERCISES 1. logs36 = ? logiwl00 = ? logs = ? logel28 = ? logs81 = ? 2. Fill out the following table : Base Number | Logarithm 10 10 | a 81 4 3 8 8 : 10 —3 10 1 —E7 3. Write the following in the notation of logarithms: (a) 452 =.10'5!- = (6) 5.1 = 10°: (6) 6470 DADO Tey SOT 4. Write the following using exponents: (a) logi2 = .38010 ; (b) logi99 = 1.9956; (c) logw:34.2 = 1.53840; (d) logis7000 = 3.8451. 112. Characteristic and mantissa. The integral part of a logarithm is called the characteristic and the decimal part the mantissa. Thus, in log:9245 = 2.38892, the characteristic is 2 and the mantissa is 3892. The characteristic may be negative. Thus, logi.01 = —2. It is usually convenient to keep the mantissa positive even though the logarithm is negative. Thus, logio+ = —.6990. But this may be written 1.3010, which has a positive mantissa, and a characteristic —1. The minus sign is placed over the 1 to indicate that the characteristic alone is negative. Arts. 113, 114, 115] PROPERTIES OF LOGARITHMS 181 113. Logarithm of a product. The logarithm of a product equals the sum of the logarithms of its factors. . Proor: Let logau = x and logav = y, then a® =u,a¥ =v, (Definition of logarithm) and uv = arty, (Art. 89) Hence, loggw =x+y, that is, log,uv = logau + logy. Similarly, loga(ww) = logvu + log + logaw, and so on for any number of factors. Example. logio(79 x 642) = logiw79 + logi.642. 114. Logarithm of a quotient. The logarithm of a quotient is equal to the logarithm of the dividend minus the logarithm of the divisor. Proor: As above, let logaw = x and logy = y. Then a* =-u, and a¥ = 2, and Sergent v Hence, log =“—Y, ; U that is, loge, = logau — logav. Example. loging; = logw245 — logy912. 115. Logarithm of a power. The logarithm of w’ is equal to v times the logarithm of w. PROGR eLCa et — 102,0, OF a*.= u. Then Pe (tt) a"? (Art. 89) Hence, logaw’ = vx = v logyu. Example. logic (455)? = 3 log0455. 182 LOGARITHMS [Cuap. XII. Making v = n, and then v = we have respectively, (a) The logarithm of the nth power of a number is n times the logarithm of the number. (b) The logarithm of the real positive nth root of a number is the logarithm of the number divided by n. EXERCISES Express the logarithms* of the following expressions in terms of the logarithms of integers: LV 2 oa lt ar cipky SOLUTION : ete log 17 + log W/12 — log 7? — log 11? (Arts. 113, 114, 115) = log 17 +4log12 — 2log7 — tlog11 (Art. 115). 48 3V5 2. log 05 6. log WF 2/180 54. ¥/1.9 Sal Teel . 19 °° 6 te 12? - 325 945 4. lo ae =e ea 6 8. logy sii (58)8 a 5. | : Suit 11/2 el: (V3 : (vs ()) * When in a problem the same base is used throughout, it is customary not to write the base. Arts. 115, 116] COMMON LOGA Express the logarithms of the foll of the logarithms of prime factors : 3 11. lo =" 2 cVd SoLuTIon : log od, loga + 8logb — cVd 12. log le Bt fe vn 16 “iG i a : n(m — 2) its 14. log Va 18. RITHMS 183 owing expressions in terms loge — $logd. log(a? — b?). ° log od oe Vc(ax + b) ta? vy log(a? + 6). log Express each of the following sums and differences as the logarithm of a single expression: 19. log m —1 log n + 2 log z. . 1 “'/ mx? Soxution : log m —tlogn +2logz =logm —logv n + log x? = log Wa n _ 20. ri i 22. 23. 24. 25. 5 log 3 + 2 log 7 — 4 log 95. log 48 + 3 log a — 2 log b — 3 log 15. a log x — b log y. log(a — 6) + log(a + 6). log 1 — log 15. 2 log(a? + b?) — 2 loga — 2 log b. 116. Common logarithms. As 10 is the base of our number system, it is convenient to take 10 as the base of the logarithms used in ordinary computation. These logarithms are known as the common logarithms, or the Briggs system of logarithms. In the following discussion of common logarithms, log x is written as an abbreviation of logioz. 184 LOGARITHMS [Cuap. XII. 117. Determination of the characteristic of a common logarithm. Since 10? = 1000, log 1000 = 3; ” 107 = 100, log JOU3eree * 10' = ©10;ior \ oe BAG Oot te 1, log 1303 7 107 =. 1 log 2) ae ” 10°? = (Ol, log. 0 ae ? 107° = 001, loe-.00I== re and so on. So far as these powers of 10 are concerned it may be observed that when a number increases, its logarithm increases. As- suming this to be true for all positive numbers we have, for example, log 100< log 400< log 1000, or : 2< log 400<3. Hence the characteristic of log 400 is 2. Similarly, log.01 1). eis z—n 15 r—n re: a 4/ ne = 30 = 4b := ——__——. Solve V = ; wr for r. 1/38? 4 for s. Solve A = _ mgl Solve s = ay. for r Solve d = .02758VD -1- WP for P. Solve V = é (B+6+~/Bb) for B. Solve A = P(1 +7) for r. 203 CHAPTER XIV EQUATIONS IN QUADRATIC FORM 124. Quadratic form. Any equation is in quadratic form with respect to an expression containing the unknown, if we may replace such expression by a new letter, and obtain a quadratic in that new letter. ‘Thus, in the equation, | | to/¢ =3. =D = which may be written | 2-3-vVz-3-2=0, if we let z = -/x — 3, wehave 2? —z-2 =0. Similarly, in 223 +4741 =0, if we let z = 2~*, we have 222 4+2+1=0. Again, in ax” + br” +c = 0, if we let z = x”, we have az? + bz +c =0. EXERCISES Solve the following equations and check results by substi- tution: 1. 4—-38Vr%4+2=0. (1) SoLution : Let Vz = 2. Then GL = 2") Substituting in (1), 22 —32+2 +0. Factoring, (z — 1) — 2) =0. The roots are 2 .= 1, 2 = 2. Thus, Vze=1, vax =2, or oe 1,z =4. CHECK: 1 Sai 0;4-3-242 20. Art. 124] EXERCISES 205 2,2." — t127-7 + 30 =.0. (1) SotuTIon : Let a =2. (2) Then from (1) and (2), 22 —11z + 30 =0. Factoring, (2 — 5) (z — 6) =0. Then Zim), Zim G, Hence, : =+¥V5, : =+V6. ~ (3) and + ya AIS V/5 5 V6 6 The four numbers + ied + v6 ~ wy are solutions of (1). The phe eee 6. a G student should check by substitution in (1). 3. z+ + 27 — 20 = 0. (1) SoLuTion : Let a? = 2Z, (2) Then from (1) and (2), 2274+2z—-—20 =0, (3) (2 —4)(2 +5) =0. (4) Then g-= 4,2 = —5. (5) Hence, g=+2,2 = t+iV5. (6) Note that two of the four roots are imaginary. The student should check by substitution in (1). fo 2 — 56 = 0. (1) SoLuTion: Let BM = 2; (2) Then from (1) and (2), 22 —z— 56 =0. Solving by the formula, z = ee = 8or -7. Hence, ; z= V8 or V—7. Curck : (W8)2" — (x/8)" — 56 = 8? — 8 — 56 =0. ogee (4/ — 7)" — 56 = (— 7)? +7 = 5650. 5. 2* — 182? + 36 = 0. 8. 2 = Ore te = 0. 6. 2? +3r71+2=0. 9. Vz +10+ Vx + 10 = 2. 7 2—2zt =6. 10. 2+V72x+6= 14. 206 EA EQUATIONS IN QUADRATIC FORM ([Caap. XIV. = 15 eh ar ae, t : (s+) +4047 = 12, x = . 23 — 972 +8=0. » (2? + 52)? + ale tb = 42: (4a? — 3)? + (8a? — 6)? = 80. . A/ 2a a ea . 3V2? 4+ 6VWe —4=0. . 2 + 3la-4 — 32 = 0. . tab Se +897 ob oo 4 ee 92? = 4n8V/ a = 22-6 ie Ze A Dy eer fie ee Hint: Write the equation in the form 2! + 223 + 2? — 2(z? +2) —3 =0 and make z?+2 =z. 22. 23. 24. 25. Solve the following equations for log x: 29. 30. vt — 843 + 23x? — 28x — 8 = 0. One Tra 20 0): 26. 3(p* —2]) =p a = /a72? —8a* = 0. oT. K(k? 1) p* — 10p? — 11 = 0. 28. 1627 (1 — 2) = 3. (log x)? — 5logx+4=0. 3(log x)? — 11 log x —4 = 0. Solve the following equations for 10*. 31. 32. 10% — 7.107 +6 = 0. 10% — 11.107 + 10 = 0. 33. If the square of a certain number is multiplied by this square increased by 6, the result is 40. Find the number. Are there any integers that satisfy this condition ? numbers? Check results by substitution. Any imaginary Arts. 124, 125] EXERCISES 207 34. If the square of a certain number is multiplied by this square decreased by 4, the product is 192. Find numbers that satisfy this condition, and check results by substitution. Are the results integral? Fractional? Real? Imaginary ? 125. Simultaneous quadratic form. Any system of equa- tions is in simultaneous quadratic form with respect to ex- pressions containing the unknowns, if we may replace such expressions by new letters, and obtain a system of simultaneous ~ quadratics in such new letters. Thus, in the system, eg? —y =6, gt — yl =2, if weletz = 271, w = y~, we have 22 — w* = 6, z2—w =2, EXERCISES Solve the following equations and check by substitution : Lae = yo? = 6, (1) gt=-yt = 2. SoLuTion : Let Po Bex. 2, Yt =a, Then system (1) becomes 22 — pw? = 6, (2) 2—-w =2. (3) Solve for z in (3) and substitute in (1), w? +4w +4 -— w* = 6, (4) 4w = 2, (5) ; w =}. (6) From (6) and (3), z=3. Hence, 2=4,y =2. CHECK : (4)? — (2)? = (5)? — (a)? =6, (Q)?-Q)t=5 4 =2. 1 1 1 1 Beas, +> Ses 1 Loin — = §, ne 3 208 4. = Bt 7 e£+y = 18, cogil Va-Vy =1. eee ——-—=2, ais 8 2+ Vy =7 x = 7, cis Oeste : Sal “Ot? un ty aed) Sia) Ma Vy —-Vai =1, 1 1 preemie 1 Mis VL 2 2 as » (e349 Git : 6 ; i er 229 uae ry = 1. ee pe EXERCISES AND PROBLEMS Solve the following equations for x: 1. 424 — 52? +1 =.0, 2. 324 — 7x7 -6 = 0. 3. vz — 13x? + 40 = 0. 4, (x? — 1)? — 38(7? —1) +2 =0. 3 x—5 a se ye 5.2 2 2. Lem x — 9d. pA: ae 8. 2 — 1laz? + 24 = 0. oxo ees 12: 9. 10zt — 4x? — 32 = 0. 10. (32 + 4)? + 5(82 + 4)4 —6=0. 11. x* — 4(a + b)a? + 4(a + OD)? = O. 12. xt — 4(a + b)x? + 16(a + Bb)? = O. 18.. 35257 -— 12a tel = 0: 14. (a? — 5x + 6)? — 8(2? — 54 + 6) +: 2 = 0. 15. (2? — 8)(a? — 3) = 374. 16. (47? — 5¢ + 2)? = 122? — 1dr + 4. 17. c— 15V/x + 56 = 0. as 18. =r — 4q% er = 0. 2 EQUATIONS IN QUADRATIC FORM ([Cuap. XIV. Arr. 125] EXERCISES | 209 Solve the following equations for x and y: i gl as ee O et 5 cee = 3° ) ha gy? 4 23. (x — 2y)? — 2 + y = 6, 82 — 5y = 11. Veg 20. -+- = 4, x Yy my BA ee 8; fe 17, ants x? y2 2 ce agai § sro 21.x2-y=2, . 25. x? + dy? = 7, Vi + Vy =2. ca a hh 26. (x — 4)? — (y — 6)? = 0, (x —4)?+y =8. 27. George Washington was born a.p. 1732. In the year A.D. «, he was x — 10 years old. Find the year a.p. zx. 28. Find a number whose square multiplied by its square plus four gives 1440. 29. Find two numbers the sum of whose reciprocals is 5, and the sum of the squares of whose reciprocals is 13. 30. Find two numbers whose difference is 36, and the differ- ence of whose square roots is 2. 31. The difference of the reciprocals of two numbers is 5, and the product of the numbers is .05. Find the numbers. CHAPTER XV PROGRESSIONS 126. Arithmetical progressions. An arithmetical progres- sion 1s a succession of numbers each term of which, after the first, differs from the next preceding term by a fixed number called the common difference. Thus, 2 AAG AR is an arithmetical progression with the common difference 2. In the arith- metical progression 10, 8, 6, 4, 2,-- > the common difference is —2. 12%. Elements of an arithmetical progression. Let a repre- sent the first term, d the common difference, n the number of terms considered, / the nth, or last, term, and s the sum of the terms. The five numbers a, d,n,l, and s are called the elements of the arithmetical progression. 128. Relations among the elements. Since a is the first term, we have, by definition of an arithmetical progression, a +d = second term, a + 2d = third term, a + 3d = fourth term, a+ (n — 1)d = nth term. That is, L=at+(n-A)d. (1) The sum of an arithmetical progression may be written in each of the following forms : a+(a+d)+(a+2d)+---+(l-2d) + (l-—d) +], 1+(l—d)+(l—2d)+---+(a+42d)+(@+4+d) +a. S s ‘ we have 8 =3 + 7d, ord = Arts. 128, 129] ARITHMETICAL MEANS 211 By addition 2 =(a@+)+(a+l)+(@t+)+--:: +(a4+)) =n(a+l). Therefore, sa Sa +1): (2) Whenever any three of the five elements are given, equations (1) and (2) make it possible to find the remaining two elements. 129. Arithmetical means. The first and last terms of an arithmetical progression are called the extremes, while the re- maining terms are called the arithmetical means. ‘To insert a given number of arithmetical means between two numbers it is necessary only to determine d by the use of equation (1) and to write down the terms by the repeated addition of d. Example. Insert 6 arithmetical means between 3 and 8. SoLtuTIon: We have to find d, when a = 3,1 = 8, andn =6 42 =8. Since Ll =a + (n —1)d, 5 = Hence, the 6 arithmetical means between 3 and 8 are 26 31 36 41 46 SL (EG NS ay IY SR | EXERCISES Find / and s for the following five exercises: 1.2+11+20+--:-: to 10 terms. SOLUTION : Ll=a-+(n —1)d. Here, a =2,d =9,n = 10. ~=2+9-9 =88. n s = 5 (a +1), = §(2 + 83) = 425 ZLe PROGRESSIONS [CHap. XV. 2. —-4-—8-—12-—--- to 19 terms. D se tied 3: 3 Aegan ee a - ++ to 21 terms. | 1 840e Fol 4. = O51 Ae - to 17 terms. bed pals : 5. riba ri} ost fit Be - + + to 25 terms. 6. Given d= 3,7 = 25,1 = 74 findiamuae 7. Givenl=—4,n=14,d =4; finda ands. 8. Given a = 3, t = 108, s = 1221 > Tita eee 9. Given a = 2,1=9,d =4; find n and s. 10. Given a = —4, | = —64, n = 21; find d and s. 1 74 1633 11. Given d = 5) l= 7 Doreen y bs find a and n. 12. Giver | = 41, n = 41;.8 = 8613 finda 13. Given a = —2,d = o s = 0.; find mance 14, Given a =7,n=7,8 =7; find d and 15. Given d = 10, n = 10, s = 10; find qand?®. 16. Insert 5 arithmetical means between 3 and 21. 17. Insert 8 arithmetical means between 9 and 81. 18. Find the arithmetical mean between 84 and 162. 19. Insert 11 arithmetical means between 2 and 20. 130. Geometrical progressions. A succession of terms in which the same quotient is obtained by dividing any term by the preceding term is called a geometrical progression. This quotient is called the ratio. Thus, Brame, 24, <.° is a geometrical progression with a ratio 2. | . 4 Arts. 131, 132, 133] GEOMETRICAL PROGRESSIONS 213 131. Elements of a geometrical progression. The elements are the same as those for an arithmetical progression with one exception. Instead of the common difference of an arithmet- ical progression, we have here a ratio represented by r. 132. Relations among the elements. If a represents the first term, then second term, Q — I ar? = third term, a be ds | = fourth term, ar’ = nth term. That is, if / represents the nth term, we have t=ar!, (1) By definition, s=a+t+art+ar+ar4+:-++ar, (2) Then, sr=artar+ar+-+++ar-t-+ar’. (3) Subtracting members of (2) from members of (3), we have sr —s=ar —a. Since 1 = ar”, (4) may be written in the form ri a. pa ess (5) Here, as in an arithmetical progression, whenever any three of the five elements are given, relations (1) and (5) make it possible to find the other two. 133. Geometrical means. The first and last terms of a geometrical progression are called the extremes, while the remaining terms are called the geometrical means. To insert n geometrical means between two given numbers is to find a 214 PROGRESSIONS [Cuap. XV. geometrical progression of n + 2 terms having the two given numbers for extremes. Example. Insert four geometrical means between 2 and 64. SoututTion: In this case a= 2, lL = 64, and n=. Then, by (1), Art. 132 64 = 27°, 32 = 7", and | pes ps Hence, the four geometrical means are 4, 8, 16, 32. | EXERCISES 1. Given a = 5,r =3,n=10; findland s. 2. Given a = —3,r = 2,n =8; findl ands. 8. Given a = 4,r = 4,n = 10; findl and s. 4. Given.a = 5,71 = —2, ee find J and s. 5. Givens — 50a = 670 = 10+ ind eee 6. The third term of a geometrical progression is 3, and the sixth term is 81. What is the tenth term ? 7. What is the fifth term of a geometrical progression whose first term is 2 and third term 4? 8. What is the sum of the first five terms of a geometrical progression whose first term is 2 and third term 8 ? | 9. The first term of a geometrical progression is 4, and the last term 256. If there are four terms in the geometrical pro- gression, find the common ratio and the sum of the series. 10. Insert one geometrical mean between 6 and 150. 11. Insert two geometrical means between 2 and 250. 12. Insert three geometrical means between 12 and 2 13. What is the eighth term of the series 3, $, yy,...? 14. If each term of a geometrical progression is multiplied by the same number, show that the products form a geometrical progression. Art. 134] INFINITE NUMBER OF TERMS 215 134. Number of terms infinite. Consider the geometrical progression Z) zy % ie boas It may at first thought appear that the sum of the first n terms could be made to exceed any finite number previously assigned by making n large enough. That this is not the case and that the sum can never exceed unity, will be seen from the following illustration. Conceive a particle moving in a straight line toward a point one unit distant in such a way as to describe 4 the distance in the first second, $ the remaining distance in the second second, 4 the remaining distance in the third second, and so on indefinitely. This is represented in Fig. 52. The distance AB represents one unit of distance. In the first second the particle moves from A to P;. In the second second it moves from P; to Ps, and so on. The total distance traversed by the particle in n seconds is given by the sum ,+4+4+-°::: ton terms, which sum cannot exceed nor equal 1, no matter how many terms we take, but can be made to differ from 1 by as small a positive number as we please by making the number of terms large enough. Thus, when n = 10, the sum is ae (Exercise 3, 1024 Art. 133). In this illustration, 1 is said to be the limiting value Ys P, P3; Pi Ps Se eT TT le eet A B Fig. 52 of the sum of the first n terms of the progression. If s, repre- sents the sum of the first n terms, we write lim 1 = 00 amt which reads, “‘ the limit of s, as n increases indefinitely is 1.” The sum s of the infinite progression is defined as this limit. For any geometrical progression in which the ratio is less than 1, the above argument can be repeated, and it can be 216 7 PROGRESSIONS [Cuap. XV. shown that there is a limiting value to the sum of the first n terms of such a progression. In Art. 132, we have shown that the sum, a+ ar + ar? +.--+*-- are, is given by Sn = a Sn) We then write lim » shim a. ligne eee hi 2 sn = We shall consider only the case where r < 1. In this case, as n increases, the value of r” becomes smaller and smaller. In arr —r value than any number that has been assigned and left fixed, or is smaller in absolute | fact, we may take n so large that i a nm l-r Hence, the sum of the infinite geometrical progression as we say, approaches the limit 0 as n becomes infinite. a, At, OF: <= , where 1. is s = lim s, = ° n=o 1—r EXERCISES Find the sum of the following infinite progressions : 1. Ota we ar SoLuTION: Here ai=6, 7 =A. a 6 desk. Shree | jess ten | Tel 2. 1 OY 4 Rie 5. 27 30 7 Rag be | 6. .5,'.125) 03125 . Be Q’ 27” Pe 1 1 4229-16, Radere 1. V2, 1 a oe 8. 2, 25, 03125 0am 9. .3, .03,c00aaaee Arts. 135, 1836] HARMONICAL PROGRESSIONS 217 135. Repeating decimals. Repeating decimals furnish good illustrations of infinite geometrical progressions. For example, .66666 - - - may be written as 6 + .06 + .006 + .0006 + .00006 +: - - where a = .6 and r =.1. The limiting value of the sum of n terms as the number n increases indefinitely is 3. Again, .4919191 - - - may be written A+ .091 + .00091 +--- where the terms after the first form a geometrical progression in which a = .091 and r = .01. EXERCISES Find the limiting value of each of the following repeating decimals : eit >. 6. .83333 --. 11. .404040--. 20800 - hy ea Pa Pee 12. .242424 --. oo 2LbG66 >. 8. .005050 - -. 13.. 1.181818 --. 4. .363636 --. 9. 234234 --. 14. 2.272727 --. fereuulU*:. 202 OUSOU ee. 15. 3.363636 - -. 136. Harmonical progressions. ‘Three or more numbers are said to form a harmonical progression if their reciprocals form an arithmetical progression. The term ‘ harmonical”’ as here used comes from a property of musical sounds. If a set of strings of uniform tension whose lengths are proportional to 1, 4, 4, +, $, , - - - be sounded together, the effect is harmoni- ous to the ear. The succession of numbers i. se Vege). 5). 6.) is a harmonical progression since the reciprocals form the arithmetical progression Bee’ <. ly 2, 3, 4, v) 6, 218 PROGRESSIONS [CHap. XV. 13%. Harmonical means. To find n harmonical means between two numbers, find n arithmetical means (Art. 129) between the reciprocals of these numbers. The reciprocals of the arithmetical means are the harmonical means. EXERCISES Insert a harmonical mean between 4 and 3. Insert two harmonical means between 1 and 4. What is the harmonical mean between x and y? 4. Show that 3, 2, 3 are in harmonical progression and continue the series two terms in each direction. SOF PSs MISCELLANEOUS EXERCISES AND PROBLEMS Find the seventh term of the progression 18, —6, 2, -- Find the fifth term of the progression a, a + d,a + 2d, -- 1 n+ n+2n + Tee What is the arithmetical mean ee. a and 6? Insert 5 arithmetical means between —4 and 20. Insert 3 geometrical means between 5 and 405. What is the geometrical mean between a and 6? 8. The fourth term of a geometrical progression is 108, the sixth is 972. Find the ninth term. 9. The sum of three numbers in arithmetical progression is 15; if 1, 4, 19 are added to the numbers, the results are in geometrical progression. Find the numbers. 10. Show that the sum of the first n odd numbers 1, 3, 5, 7, -is n?. Find the tenth term of the progression ~ : oN LLE aa hd pete edt A ade pe PROBLEMS PERTAINING TO MOTION 11. If a body falls 16 feet the first second, 48 the next, 80 the next and so on, how far does it fall in the twelfth second ? How far has it fallen at the end of 12 seconds? Art. 137] PROBLEMS 219 12. If a body falls 16 feet the first second, 48 the next, 80 the next and so on, how far does it fall in the tth second? How far has it fallen at the end of the tth second ? 13. A ball rolling down an incline of 30° goes 8 feet the first second, and in each second thereafter 16 feet more than in the preceding second. How far will it roll in 10 seconds? 14. A marble rolls down an inclined plane, passing over distances 3 feet, 9 feet, 15 feet, in successive seconds. How long will it take it to pass over 108 feet ? 15. Assume that a ball falls 16 feet the first second, 48 the next, 80 the next, and so on. A baseball was dropped from the top of Washington Monument, 550 feet high, and caught by an American League catcher. About how fast was the ball falling when caught ? 16. In a potato race there are placed 20 potatoes, at the distances 5 feet, 8 feet, 11 feet, and so on, from a bag. A man starting from the bag is required to pick up the potatoes and carry them back to the bag one by one. What is the total dis- tance that he goes in thus collecting the potatoes ? 17. A wheel of perimeter 6 feet, free to rotate on an axis, is started to rotate 20 revolutions per second, and then is to be retarded by friction. If it makes 95% as many revolutions each second thereafter as it did the preceding second, how far will a point on the rim have moved when the wheel comes to a standstill ? Hint: These distances are assumed to form an infinite progression. 18. If a particle moves in a straight line from a given posi- tion, with such a speed that during any given second it moves 60% as far as it did during the preceding second, and if it moved 25 feet during the first second, what is the limit of the distance the particle will move? 19. If a ball should fall 9 feet and bound back 6 feet, then fall 6 feet and bound back 4 feet and so on indefinitely, what is the limit of the distance through which the ball would move? 220 PROGRESSIONS [Cuap. XV. 20. A travels uniformly 25 miles a day; B starts at the same — time, and travels 5 miles the first day, 10 miles the second, 15 miles the third, and soon. In how many days will B over- take A ? | PROBLEMS PERTAINING TO BUSINESS 21. Find the present value of twenty annual payments of $1000 made at end of each of the next twenty years, if money is worth 5% compounded annually. $1000 Hint: The present value of $1000 due in n years is (1.05)"" 22. Find the present value of $125 per year paid at the end of each year for forty years if interest is at 4% compounded annually. 23. I owe a debt of $4675, to be paid in instalments, the first payment to be $800, the second $725, and decreasing by a common difference, until the last payment which is $50. Find the number of instalments. 24. A gentleman being importuned to sell a fine horse, said he would sell him on the condition of receiving 1 cent for the first nail in four shoes, 2 cents for the second, and so on, doub- ling the price of every nail; the number of nails in each shoe being 8, how much would he receive for his horse ? CHAPTER XVI THE BINOMIAL THEOREM 138. Factorial. The symbols r! and |r are read “ factorial 7,”’ and are used to indicate the product 1-2-3-...r. The symbol r! is easier to set up in type than the symbol |r, and is used more in print than |r. Thus, 3! =1-2-3=6; 7!=1-2-3-4-5-6-7 = 5040. 139. Powers of a binomial. By actual multiplication, (a + b)? = a? + 2ab + B?, (1) (a + b)? = a3 + 3a%b + 3ab? + B3, (2) (a + b)* = at + 4a%b + 6a7b? + 4ab3 + Dt. (3) Exercise. Expand (a + b)* by actual multiplication of (3) by a + b. 140. The expansion of (a +b)”. The expressions (1), (2), and (3) of Art. 139 are special cases of the general formula PA sees | (a+b)" =a" +na"" b+ Bee a” 2 if n(n — 1) (n —2) 31 a"—3$3 as oe, (4) This is called the binomial formula or binomial theorem. The following properties of this formula should be noted: (1) The first term is a”. (2) The second term is a”b. (3) The exponents of a decrease by unity from term to term while the exponents of b increase by unity. 222 THE BINOMIAL THEOREM [Cuap. XVI. (4) If in any term the coefficient be multiplied by the ex- ponent of a and divided by the exponent of 6 increased by 1, the result is the coefficient of the next term. (5) The factorial number in the denominator of the coeffi- cient of any term is equal to the exponent of b in that term. (6) The numerator of the coefficient of any term is the prod- uct of factors beginning with n, decreasing successively by 1, and ending with n — r + 1, where r is the exponent of Db. (7) When n is any positive integer, the expression contains n + 1 terms of which the last is 6”. The binomial theorem holds for any positive integral value of n, and, under certain restrictions on a and b, for negative and fractional values of n. (See Art. 142.) The theorem is proved in more advanced algebra, but here it is merely stated without proof. EXERCISES 1. Expand (a — 6)”. SoLuTion: Put —b for b in (4), Art. 140. The even powers of —b are positive and the odd powers are negative. We may therefore write (n — 1) (a — 6)" =a" — na™b + a oy a"—2b2 = ne asb) 45s oa The last term is either —b" or +b" depending upon whether v is odd or even. 2. Expand (8y + 52)5. SoLuTION : Substitute 3y for a, 5% for 6, and 5 for n in (4), Art. 140. | 4 cis eae (By + 52)> = (By)® + 5@y)! 5x +2~5 (8y)* (5x)? + 2-2-8 (By)? Gn) 5-4-3-2 | + [Sey Gy) (6x)! + (Gx)® = 243y5 + 2025 yx + 6750y%x* + 11250y%x? + 9375ya* + 312525. Art. 140] EXERCISES 223 6 3. Expand (5 _ = x SoLuTION: Substitute ; for a, 2 for b, and 6 for n in (4) Art. 140. Then ges 2p \s e\* cy 2b EN? 2b\? c\3 2b\3 pee) = (a) +6(5) (- 7) +¥5(5) (- 2) +20(5) (- 2) ¢\?/ 2b\4 c 2b\° 2b\e 8 bc +15 (5) (-2) +6(5)(-2) +(-2] E7200 Sie 20b%c! 160b%c?_ 80b'c? = 64b°c =~ 640° 2722 273 ing 1) tan? afte Pe ine 13. (v2 - =) oR *) eA) 22 \ve + 4 +4 e 14. (2 + 2vV/x)4. 23. (x! + 2y*)4. 224 THE BINOMIAL THEOREM [Cuap. XVI. Use the binomial formula to compute the following, correct to three significant figures : DAS Lok) e Hint: Write (1.1) = (1 +.1)% and expand to a few terms. 25. (.99). | Hint: Write (.99)2? = (1 — .01)”° and expand. 26. Se) ae 27. (.999)!° 141. Formula for the 7th term of the expansion of (a + b)”. We have seen by Art. 140 that the third term is n(n —1) eye 2b? the fourth term is eee ar h*; and hence that the rth term is n(n —1)(n—2) ... m—7r+4+2) toon (r — 1)! EXERCISES y\10 1. Find the sixth term of (21 — y) . - Sotution: Here a = 22, b = —3y, n = 10, r = 6. x _10-9:8-7:6 .\.f y\5 __ 896 The sixth term is [2a g.4 05 (22) ( u) =e ee 2. In (a — y)", find the sixth term. 8. In (a? — y?)™, find the middle term. 4. In (x? — y*), find the middle term. 5. In (3a — 4a?)!5, find the sixth term. 5S ai¢ : 6. In (2 i *) » find the middle term. 7. In (.5\/a — .1v/y)", find the fifth term. Arr. 142] EXERCISES | | 225 142. Binomial formula, any exponent. The binomial for- mula (a Mie b)” = qr + nab ae ae a”2h2 a8 n(n 2% 1) (n a 2) qr 3h3 a 3! holds when n is a negative or fractional number, provided the number 6 lies between —a and a. If n is not a positive integer, the formula for the development of (a + 6)” contains an unlimited number of terms. If b is small compared to a, the first few terms of the expansion give a useful approximation for certain purposes. Thus, the first few terms in the binomial expansion of (27 + 1)} gives the ap- proximate value of the cube root of 28. (See Example 3 below). EXERCISES 1. Expand to four terms (a + y). SOLUTION : Substitute in the expansion for (a + 6)”. This gives ug 1) Lilia Vi dee cess 1 4-1 ma ag oe ne a 3- 1% 1 oa 8 ae, ae y- 5 $2 + 352 ys 2. Expand to three terms (32 — 4y)~”. So.tuTion: By the formula for (a + b)", (3e - 5) . = (37)~? — 2(82)- (5 v) ee pe (@x)*{ - su) t+ il y ly “2 wad Oy? Ai 2723 T 082! +. 3. Find approximately 7/28 by the binomial formula. SoLuTION : Write 35-1) 0/28 = (27 +1)? = 278 +3 F o7-8 4 3(3 ~ See O74 Bee ical t 97 — 9187 7 = 3.0366 —. 226 THE BINOMIAL THEOREM [Cuap. XVI. Expand the following expressions to four terms: 4, (1 — x)=. 10. (a?y-? + 1). Ba — 2) 11. (8a* + 1). 6..(2a;+ 1)>. 12. (844 = poner i es 13. (9x? + 27) Al 8 (1 — 2). 14, (« _ *) ; QA eine 15. 7/9 = (84 1)% Find the roots indicated, correct to three significant figures : 16. +/10. 18. (79)}. 20. /730. 17. (26)3. 19. 1/65. 21. (25)}. Art. 142] REVIEW EXERCISES AND PROBLEMS 227 REVIEW EXERCISES AND PROBLEMS ON CHAPTERS XIII-XVI 1. Find the arithmetic mean, the geometric mean, and the harmonic mean between a and b. 2. In a Christmas savings club there are three classes of depositors. One class deposits 1 cent the first week, 2 cents the second week, 3 cents the third week, and so on, increasing the weekly deposit 1 cent each week for 50 weeks ; the second class deposits 2 cents the first week and increases the weekly deposit 2 cents each week for 50 weeks ; the third class de- - posits 5 cents the first week, and increases the weekly deposit 5 cents each week for 50 weeks. Find how much is deposited by a member of each class in 50 weeks. 3. The population of a town is P at a certain time. Annually it loses 2% by deaths and gains y% by births. What is the population at the end of n years? 4. What is the outer diameter of a spherical shell that is 1 inch thick and contains 50 cubic inches ? 5. Solve for x (a) x? + 2a? = daz. (b) a? +327 = 4. (PRINCETON) uff 2n : 6. What term in the developmént of (a =) does not contain a. (SHEFFIELD) 7. Leta, b, c (Fig. 53) be three verticals erected so that a line from the foot of a to the top of c, or from the foot of c to the top of a, bisects the intermediate vertical b. Show that b isthe harmonic mean between a and c; that is, show that G43) 3 Fic. 53 8. In Fig 54 the lines a, b, c, d,+-- are drawn so that the angles marked 1, 2, 3, ° ++ are equal. Show that the lengths of a, b, c, d,--- form a geo- metrical progression. Hint: Show that ; = ° by use of similar triangles, p. 99. 9. Solve (a +11)'+ a? = 552-3. (ILLINOIS) 10. Solve Ve+-1+2 =11. 228 THE BINOMIAL THEOREM [Cuar. XVI. Fiac. 54 11. Find to two places decimal by the binomial theorem the value of (120). (CALIFORNIA) 12. A piece of land is in the form of a trapezoid, with parallel sides 20 rods and 30 rods, and altitude 20 rods. It is desired to divide this piece of land into two equal parts by a line parallel to the parallel sides. Find the distance between the shorter parallel side and this dividing line. 13. Find the value of log;49 + 4log, 64 + logs 216. Fig. 55 14. Insert two arithmetical means, two geometrical means, and two harmonical means between 3 and 24. 15. Find the fifth term of (2 +=3) x and reduce to the simplest form. (DARTMOUTH) 16. Solve for z, (log x)? — 11 log x + 10 =0. 17. The following quaint problem was found in an old Hindu manuscript: The square root of half the number of bees in a swarm have flown out upon a jessamine bush ; $ of the whole swarm have remained behind ; one female . bee flies about a male that is buzzing within a lotus flower into which he was allured in the night by its sweet odor, but is now imprisoned in it. Tell me the number of bees in the swarm. 18. Solve for x and y (x +2)? — (2y - 3)} = 2, x—2y+3 = 14. Art. 142] REVIEW EXERCISES AND PROBLEMS 229 19. Find the sum of n terms of the series, ed i Sen He + + n n 1+ -f- ee (Mass. INSTITUTE) 20. A rectangle has the base x and altitude y, and a constant area 12. Construct the rectangles, on coédrdinate paper when x has the values 24, 12, 8, 6, 4, 3, 2, 1, .5, .4, .3. In each case take’ the origin as ohe vertex and the codrdinate axes as sides. After all the rectangles are constructed draw a curve through the vertices that are opposite the origin. 21. Gerbert (1003 a.p.), who was later: Pope Sylvester, found the alti- tude of an equilateral triangle by multiplying the side by # ; Heron (about 100 B.c.) used the factor +3 ; the factor 7 is now used to make a rough ap- proximation. Which value is nearest the correct one? 22. Continue the progression 3, 1, —4 two more terms and find the sum of the first 13 terms. | (ILLINOIS) 23. Find the value to four decimal places of ae by the binomial theorem. (Mass. INstrTuTE) 24. Insert an arithmetical mean between 97 and 47, also a geo- metrical mean. How do they compare ? (DARTMOUTH) 25. Solve for x 20 (log x)? — 23 log « +6 = 0. 26. Solve for x 6(log x)? +17 log x +7 =0. 27. A number is 12 greater than 4} times its square root. Compute its value to two places of decimals. Verify the smaller value. (COLLEGE ENTRANCE EXAMINATION BOARD) 28. (a) Of a geometric progression, the sum of the first and second terms is 16, the sum of the third and fourth terms is 36. Find the ratio, and the sum of the first 6 terms. (b) Write the seventh term of the expansion of (1 + x)® by the binomial theorem. (CoLLEGE ENTRANCE EXAMINATION Boarp) 29. Find the roots of 492+ — 70x? + 12 Fig. 56 = 0 to three decimal places. 30. The sum of an infinite geometric series is 4, and the first term is 6. Find the ratio and the sum of 4 terms. (YALE) 31. In Fig. 56, the vertical lines are equally spaced. Show that their lengths form an arithmetical progression. 32. In Fig. 57, the vertical lines are so spaced, that the lines drawn from the foot of 230 THE BINOMIAL THEOREM [Cuap. XVI. each to the top of the next are all parallel. Show that the lengths of the verticals form a geometrical progression. 33. A strip of carpet one-half inch thick and 29% feet long is rolled on a roller 4 inches in diameter. Find how many turns there will be, remem- bering that each turn increases the diameter by one inch, and taking as the length of the circumference */ of the diameter. (HARVARD) 34. Solve Vx — Vy =2, (Vx —- Vy)Vxy = 30. (YALE) 35. Solver —y —-Va —y =2,23 —y' = 2044. (YALE) 36. Find the middle term of the expansion of (x — y)®. Express this Ue bt 4 2 10 15 term in its simplest form when z = es , and y = ve _* (HARVARD) 3 5 37. kixtract the square root of 38 — 12/10. (REGENTS) 38. Solve Vora +1 =384+ V2 +2. 39. Jind the sum to infinity of the progression 2 1 —8, - 5) BOS (Mass. INsTITUTE) 40. Solve for x 102% — 107 —2 =0. 41. Find the value of the repeating decimal .3727272 * - - in fractional form. 42. Solve 3(x + 1)2 +. 23 + 322 + 32 +3 =0. INDEX [The numbers refer to pages.] Abscissa, 66 Cube root, Addition, 2 of numbers in Arabic figures, associative law of, 2 176 commutative law of, 2 of polynomials, 174 of fractions, 44 Algebraic operations, 12 Degree, Argand, 116 of an expression, 25 Arithmetical means, 211 of a term, 25 Arithmetical progression, 210 Determinants, 80 elements of an, 210 expansion of, 83 Associative law, historical note on, 85 of addition, 2 of the second order, 80 of multiplication, 5 of the third order, 82 Axes, solution of equations by, 80, 85 © codrdinate, 66 Diophantus, 93 Discriminant, 132 Binomial theorem, 221 Distributive law, Briggs, 198 of multiplication, 6 Division, 8 Cardan, 116 of fractions, 46 Characteristic, 180 _ of monomials, 8 determination of, 184 of polynomials, 9 Commutative law, zero in, 9 of addition, 2 of multiplication 5 Elimination, 78 Complex numbers 114 by addition and subtraction, 78 conjugate, 114 by substitution, 79 graph of, 117 Equalities, 15 operations with, 114 Equations, 15 Constant, 59 as sentences, 16 Coérdinate axes, 66 equivalent, 16, 77 Coérdinates, 66 graphs of linear, 74 Cramer, 85 incompatible, 77 232 Equations (continued) inconsistent, 77 indeterminate, 91 in quadratic form, 204 in simultaneous quadratic form, 207 involving fractions, 50 involving radicals, 200 linear in one unknown, 18 operations on, 17 quadratic, 1238, 128 solution by factoring, 38, 124 solved by determinants, 80, 85 systems of, involving quad- ratics, 141 roots of, 16 with given roots, 130 Evaluation of functions, 61 Exponents, 6, 152 fractional, 154, 155 historical note on, 155 laws of, 152 negative, 155 positive integral, 152 zero, 154 Extremes, of a proportion, 95 Factor, 25 highest common, 38 Factorial, 221 Factoring, 25 solution of equations by, 38, 124 summary of, 34 Factor theorem, 30 Formulas, frequently used, 62 Fractions, 41, 108 addition and subtraction of, 44 algebraic, 41 clearing of, 50 complex, 48 INDEX equations involving, 50 multiplication and division of, 46 reduction of, 41, 42 Function, 59 evaluation of, 61 _ graph of a, 68 Functional notation, 60 Fundamental laws, note on, 13 Fundamental operations, 1 Gauss, 116 Geometrical means, 213 Geometrical progression, 212 elements of, 213 infinite, 215 Girard, 116 Graph, of a complex number, 117 of a function, 68 of a linear equation, 74 of logax, 185 ; of a pure imaginary number, 116 of a quadratic function, 134 Graphical solution, of a system involving quad- ratics, 142 ; of a system of linear equations, 76 Hamilton, Sir William Rowan, 14 Harmonical means, 218 Harmonical progression, 217 Highest common factor, 38 Historical note, on determinants, 85 on exponents, 155 on fundamental laws, 13 on imaginary numbers, 116 on indeterminate equations, 92 INDEX Historical note (continued) on logarithms, 193 Hyperbola, 143 Identities, 15 as sentences, 16 Imaginary numbers, 111 graph of, 116 historical note on, 116 products of, 118 pure, 114 Indeterminate equations, 91 historical note on, 92 Integers, 108 Integral expressions, 25 Irrational numbers, 110, 159 Leibnitz, 85 Linear equations, graphical solution of a system of, 76 graph of, in two unknowns, 74 in one unknown, 18 | solved by determinants, 80, 85 Logarithm, 179 | characteristic of a, 180 common or Briggs, 183 computation by, 190 historical note on, 193 mantissa of a, 180, 185 of a power, 181 of a product, 181 of a quotient, 181 table, 186, 187 Mantissa, 180, 185 Mean proportional, 96 Means, of a proportion, 95 Members, of an equality, 15 233 Monomials, division of, 8 division of polynomial by, 9 multiplication of, 6 Multiple, lowest common, 40 Multiplication, associative and commutative laws of, 5 by zero, 7 distributive law of, 6 of fractions, 46 of monomials, 6 of polynomials, 6, 7 Napier, Baron John, 198 Number concept, extension of, 108 Numbers, complex, 114 conjugate complex, 114 imaginary, 111 integers and fractions, 108 irrational, 110, 159 negative, 109 rational, 110, 159 real, 111 Operations, algebraic, 12 historical note on fundamental, 13 on equations, 17 Ordinate, 66 Oresme, Nicole, 155 Origin of codrdinates, 66 Parentheses, use of, 3 Polynomials, division of, 9 multiplication of, 6, 7 square roots of, 172 234 Powers, 11 of a binomial, 221 of 7, 113 Prime, expressions, 25 expressions prime to each other, 39 Products, important special, 26, 29 of imaginaries, 113 Progressions, arithmetical, 210 geometrical, 212 harmonical, 217 Proportion, 95 by alternation, and by inver- sion, 97 by composition, by division, and by composition and division, 98 Proportional, fourth, mean, third, 96 Quadratic equation, 123 “nature of roots of, 131 simultaneous, 141 solution by factoring, 124 solution by formula, 125 special or incomplete, 128 typical form of, 123 Quadratic function, 123 graph of, 134 Quadratic surd, 161 Radicals, 160 addition and subtraction of, 164 . division of, 169 equations involving, 200 index of, 161 multiplication of, 166 order of, 161 INDEX reduction to same order, 166 similar, 164 simplest form, 163 simplification of, 162 Radicand, 160 Ratio, 94 Rational integral expression, 25 degree of, 25 Rational numbers, 110, 159 Real numbers, 111 Reciprocal, 42 Repeating decimals, 217 Roots, 11 extraneous, 17 index of, 161 of equations, 16 order of, 161 Simultaneous equations (See systems of equations.) Solution, 16 graphical, 76, 142 Square root, of numbers in Arabic symbols, 173 of polynomials, 172 of radical expressions, 173 Stevin, Simon, 155 Subtraction, 2 definition, 3 of fractions, 44 Surd, 161 quadratic, 161° Systems of equations, both equations of ax? + by? ++c=0, 145 both equations quadratic, 145 elimination in, 78, 79 graphical solution of, 76 incompatible, inconsistent, de- pendent, equivalent, 77 involving quadraties, 141 form INDEX Systems of equations (continued) one linear and one quadratic, 141 solved by determinants, 80, 85 Transposition, 17 Variable, 59 Variation, 102. combined, 103 constant of, 102 inverse, 103 joint, 103 Verification, by substitution, 18 Wallace, John, 156 Wessel, 116 Zero, 7 as an exponent, 154 in division, 9 235 URBANA C001 vo02 FIRST-[SECOND] COURSE N UNIVERSITY OF ILLINOIS- 9R44S <= c [= lu iéo} ond = — oS So 4 o ” 512 + behint