LIBRARY 
 
 OF THE 
 
 University of California. 
 
 GIFT OF 
 
 Class 
 

 
 
 (C\ 
 
TWENTIETH CENTURY TEXT-BOOKS 
 
TWENTIETH CENTURY TEXT-BOOKS 
 
 THE YOUNG AND JACKSON ALGEBRAS 
 
 Elementary Algebra, adapted to one 
 and a half or two years' work. It com- 
 pletely fulfills college entrance require- 
 ments in the subject, xii, 442 pages, 
 
 $1.12. 
 
 A First Course in Elementary Alge- 
 bra, adapted to one year's work. It 
 includes the maximum number of sub- 
 jects contained in high-school first-year 
 courses in algebra, ix, 294 pages, 95 cents. 
 
 A Second Course in Elementary 
 Algebra, designed for pupils who have 
 had one year of algebra, and contains 
 no advanced algebra, vii, 206 pages, 
 70 cents. 
 
 D. APPLETON AND COMPANY 
 NEW YORK CHICAGO 
 
 163 
 
TWENTIETH CENTURY TEXT-BOOKS 
 
 A SECOND COURSE IN 
 
 ELEMENTARY ALGEBRA 
 
 BY 
 
 J. W. A. YOUNG, Ph.D. 
 
 ASSOCIATE PROFESSOR OF THE PEDAGOGY OF MATHEMATICS 
 THE UNIVERSITY OF CHICAGO 
 
 AND 
 
 LAMBERT L. JACKSON, Ph.D. 
 
 FORMERLY PROFESSOR OF MATHEMATICS, STATE NORMAL 
 SCHOOL, BROCKPORT, NEW YORK 
 
 NEW YORK 
 
 D. APPLETON AND COMPANY 
 
 1910 
 
Copyright, 1908, 1910, by 
 D. APPLETON AND COMPANY. 
 
PREFACE 
 
 The material in this volume is prepared for the use of pupils 
 who have done a year's work in Elementary Algebra. Many 
 high schools divide their Algebra work into two courses sepa- 
 rated by some work in Geometry and elementary science. 
 These schools often find it more convenient and economical 
 to use two books in Algebra, one for the first course and another 
 for the second. The authors' First Course in Elementary 
 Algebra is planned for first-year work and this book is 
 planned for the second-year work. 
 
 The first six chapters are a review and extension of the 
 topics of the First Course. The chapter on logarithms is 
 new in name only, because in theory it is an extension of the 
 subject of exponents. The remainder of the volume treats 
 the usual topics, Equations, Proportion, Variation and Series, 
 supplemented by problems applying Algebra to Geometry. 
 
 Throughout the treatment the authors have constantly kept 
 in mind both the logical value and the practical utility of the 
 subject. 
 
 The logical value of Algebra is of prime importance ; hence, 
 the proofs of processes are based upon reasons, both correct 
 and satisfying to the mind of the pupil. On the other hand, 
 subtle distinctions and arguments savoring of higher mathe- 
 matical methods without their true rigor have been avoided. 
 
 The utility of Algebra is given the emphasis which it so 
 richly deserves. This is done by making the equation promi- 
 nent, by introducing simple formulas of Geometry and Physics, 
 and by applying Algebra to modern industrial, commercial, 
 and scientific problems whose content can readily be undcr- 
 
 iii 
 
iv • PREFACE 
 
 stood by the pupil. Useless puzzles and problems relating 
 
 to past conditions have been excluded, with the exception of 
 
 a few supplementary problems retained on account of their 
 
 historical interest. 
 
 The Summaries and Reviews at the ends of chapters furnish 
 
 systematically, and in small compass, the essentials of Algebra. 
 
 By reference to these the pupil can best review and unify his 
 
 knowledge of the subject. 
 
 THE AUTHORS. 
 
CONTENTS 
 
 CHAP. 
 
 I. — Fundamental Processes .... Pp. 1-24 
 
 Definitions, 1 ; Addition, 1 ; Graphical Representation of Addition, 
 2, 3 ; Commutative Law of Addition, 2 ; Associative Law,' 3 ; Sub- 
 traction, 4 ; Relative Numbers, 5 ; Graphical Representation of 
 Relative Numbers, 5 ; Rules for Signs of Operation, 6 ; Absolute 
 Value, 6 ; Addition of Relative Numbers, 7 ; Subtraction of Relative 
 Numbers, 8; Multiplication, 11 ; Commutative Law of Multiplication, 
 11 ; Factor, 11 ; Associative and Distributive Laws, 12; Multiplica- 
 tion of Relative Numbers, 13 ; Division, 14 ; Fractions, 15 ; Graphical 
 Representation of Fractions, 15 ; Properties of Fractions, 16 ; Multi- 
 plication of Fractions, 17 ; Division of Fractions, 19 ; Complex Frac- 
 tions, 19 ; Factoring, 20 ; Review, 23-24. 
 
 II. — Equations Pp. 25-41 
 
 Equations of One Unknown, 25-30 ; Definitions, 25, 26 ; Equivalent 
 Equations, 26 ; Linear Form, 26 ; General Solution, 27 ; Equations 
 of Two Unknowns, 31-33 ; Systems of Equations, 31 ; Simultaneous 
 Equations, 31 ; General Solution, 31 ; Formulas, 32 ; Equations of 
 Three or More Unknowns, 33-35 ; Solution, 33 ; Quadratic Equa- 
 tions, 36 ; Solution, 36 ; Review, 39-41. 
 
 III. — Kadicals Pp. 42-52 
 
 Definitions, 42 ; Addition and Subtraction, 44 ; Multiplication of Ex- 
 pressions containing Square Roots, 45 ; Division of Square Roots, 
 46 : Rationalizing the Denominator, 47 ; Radical Equations, 48 ; 
 Summary, 49; Review, 50 ; Supplementary Work, 51. 
 
 IV.— Exponents Pp. 53-73 
 
 Laws, 53 ; Fractional Exponents, 56 ; Zero and Negative Exponents, 
 63 ; Use of Zero, Negative, and Fractional Exponents, 65 ; Summary, 
 68; Review, 70; Supplementary Work, 71. 
 
 V. — Logarithms Pp. 74-89 
 
 Use of Exponents in Computation, 74 ; Definitions, 79 ; Explanation 
 of the Tables — Negative Characteristics, 80 ; Use of the Tables, 82 ; 
 Summary — Review, 88. 
 
 v 
 
vi CONTENTS 
 
 CHAP. 
 
 VI. — Imaginary and Complex Numbers . Pp. 90-99 
 
 Imaginary Numbei-s ■ — Ueal Numbers, 00 ; Complex Numbers, 91 ; 
 Addition and Subtraction — Multiplication, 92 ; Division, 03 ; Powers 
 of the Imaginary Unit, 04 ; Imaginaries as Roots of Equations, 94 ; 
 Summary, 96 ; Supplementary Work, 97. 
 
 VII. — Quadratic Equations .... Pp. 100-131 
 
 General Form, 100; General Solution — Solution by Formula, 102; 
 Literal Quadratic Equations, 103 ; Collected Methods, 104 ; Relation 
 of Roots to Coefficients, 107; Testing — Character of Roots, 108; 
 Discriminant — Factoring Quadratic Polynomials, 109 ; Graphical 
 Solution, 111 ; Higher Equations solved by the Aid of Quadratic 
 Equations, 115 ; Binomial Equations, 116 ; Summary, 117 ; Review, 
 118; Supplementary Work, 122. 
 
 VIII. — Systems of Quadratic and Higher 
 
 Equations Pp. 132-147 
 
 Simultaneous Quadratic Equations, 132 ; Simultaneous Higher Equa- 
 tions, 141 ; Summary, 142 ; Review, 143 ; Supplementary Work, 
 145. 
 
 IX. — Proportion Pp. 148-159 
 
 Definitions — Relation to the Equation, 148; Mean Proportional, 153; 
 Summary, 156 ; Review, 157 ; Supplementary Work, 159. 
 
 X. — Variation Pp. 160-169 
 
 Direct Variation — Relation to Proportion — Expressions for Direct 
 Variation, 160; Inverse Variation — Expressions for Inverse Varia- 
 tion, 161 ; Graphical Work, 163 ; Summary, 165 ; Supplementary 
 Work, 166. 
 
 XL — Series Pp. 170-191 
 
 Series — Terms, 170 ; Arithmetical Series : Common Difference, 
 171 ; Last Term — Sum of an Arithmetical Series — Formula for the 
 Sum, 172; Collected Results, 172; Geometric Series: Common 
 Ratio, 174; Last Term — Sum of a Geometric Series — Formula for 
 the Sum, 176 ; Collected Results, 176 ; Means : Arithmetical Mean 
 — Geometric Mean, 178; Other Formulas: Arithmetical Series, 
 179; Geometric Series, 182; Summary — Review, 184; Supple- 
 mentary Work, 187. 
 
 XII. — Zero. Interpretation of Results . Pp. 192-198 
 
 Zero and its Properties, 192 ; Interpretation of Results, 197. 
 XTIT. — Supplement Pp. 199-212 
 
 Geometric Problems for Algebraic Solution, 199. 
 
A SECOND COURSE IN 
 ELEMENTARY ALGEBRA 
 
 CHAPTER I 
 FUNDAMENTAL PROCESSES 
 
 1. Algebra is concerned with the study of numbers. The 
 number of objects in any set (for example, the number of 
 books on a shelf) is found by counting. Such numbers are 
 called whole numbers or integers; also, natural or absolute 
 numbers. 
 
 In arithmetic, numbers are usually represented by means of 
 the numerals, 0, 1, 2, 3 . . . 9, according to a system known as 
 the decimal notation, which we take for granted is here under- 
 stood. In algebra, numbers are also represented by letters 
 either singly or in combinations. 
 
 2. Graphical Representation. The natural integers may be 
 represented by equidistant points of a straight line , thus : 
 
 3. Addition. If two sets of objects are united into a single 
 set (for example, the books on two shelves placed on a single 
 shelf), the number of objects in the single set is called the sum 
 of the numbers of objects in the two original sets. The pro- 
 cess of finding the sum is called addition. The sign, +, be- 
 tween two number symbols indicates that the numbers are 
 to be added. In the simplest instances the sum is found by 
 
 counting. 
 
2 ELEMENTARY ALGEBRA 
 
 Thus, to find 5 + 7, we first count 5, and then count 7 more of the 
 number words next following (six, seven, eight, etc.). The number 
 word with which we end (twelve) names the sum. 
 
 4. Graphical Representation. The sum of two integers may 
 be represented graphically thus : 
 
 3 + 5 a + b 
 
 i — i — i — i — i — i — : — i — i 
 
 012345678 
 
 Theoretically, the sum of two integers can in every instance be found 
 by counting. But it is not necessary or desirable to do so when either 
 (or both) of the numbers is larger than nine. In this case, the properties 
 of the decimal notation, as learned in arithmetic, enable us to abridge the 
 process of counting, and to find the sum very easily even when the num- 
 bers are large. 
 
 5. Commutative Law of Addition. If two sets of objects are 
 to be united into a single set, the number of objects in the 
 latter is obviously the same whether the objects of the second 
 set are united with those of the first, or those of the first united 
 with those of the second. 
 
 For example, the number of books is the same whether those on the 
 first shelf be placed on the second, or those on the second be placed on 
 the first. This is true because the operation of transfer neither supplies 
 nor removes any books. 
 
 In symbols : a + b = b -f a. 
 
 This fact is called the commutative law of addition. 
 
 The letters a and b are here used to stand for integers, but later they 
 will be taken to stand for any algebraic numbers, and the law will still 
 apply. 
 
 6. Graphical Representation. The commutative law may be 
 represented graphically thus : 
 
 a + i 
 
FUNDAMENTAL PROCESSES 3 
 
 7. Addition of Two or More Integers. If more than two sets 
 of objects are united into a single set, the number of objects in 
 the resulting set is called the sum of the number of objects in 
 the original sets, and the process of finding the sum is called 
 addition. As in the case of two numbers, the sum of three or 
 more numbers may be found by counting in the simplest 
 instances, and for larger numbers, the process may be abridged 
 by use of the properties of the decimal notation. 
 
 8. The commutative law likewise applies to the sum of three 
 or more integers. That is : 
 
 The sum is the same for every order of adding the numbers. 
 
 9. Associative Law of Addition. If we have three rows of 
 books, the number of books is the same whether those in the 
 second row are first placed with the first row, and then those in 
 the third row placed with these, or those in the third row placed 
 with the second, and then all of these with the first row. 
 
 In symbols: (a + b) + c = a + (& -f c). 
 
 This fact is called the associative law of addition. 
 
 10. Graphical Representation. The associative law may be 
 represented graphically thus: 
 
 a + b 
 
 c 
 
 _^_ 
 
 a C b c "\ 
 
 V Y 
 
 a + b + c 
 
 The properties stated above are often used to abridge calculations. 
 Thus, 7 + 4 + 3 + 6, are more easily added thus : (7 + 3) + (4 + 0). In 
 actual work the change of order is made merely by the eye. 
 
 ORAL EXERCISES 
 Rearrange advantageously and add: 
 
 1. 8 + 3 + 2 + 7. 4. ±8x + 73x + 2x + 7x. 
 
 2. 91+43+9. 5. 19y + 54y + 6-y + y. 
 
 3. 87 + 26 + 13. 6. 73 b + 186 b + 14 b. 
 
4 ELEMENTARY ALGEBRA 
 
 7. 13 a + 5 a + 17 a + 5 a. 11. 279 £ + 347i + 21 £. 
 
 8. 7 a; + 12 a; + 3 a; + 18 a;. 12. 624 p-f 45p + 6p + 52>. 
 
 9. 8// + 10// + 7// + 5?/. 13. 93^ + 9 £ + 7 £ + £. 
 
 1 0. 23 a + 6 « + 2 a + 4 a, 14. 144 m + 7 /«, + 6 m + 3 m. 
 
 WRITTEN EXERCISES 
 
 Show graphically that : 
 
 1. 11 +4 + 6 = 11 + (4 + 6). 
 
 2. 8 + 5 = 5 + 8. 
 
 3. 4a + 56 = 5& + 4a. 
 
 4. 2 a + 3 a + 7 a = 2 a + (3 a + 7 a). 
 
 11. " Subtraction. It often happens that we wish to know 
 how many objects are left when some of a set are taken away, 
 or to know how much greater one number is than another. 
 The process of finding this number is called subtraction. The 
 number taken away is called the subtrahend, that from which 
 it is taken, the minuend, and the result, the difference or the 
 remainder. 
 
 12. The sign of subtraction is — . 
 
 13. Subtraction is the reverse of addition, and from every 
 sum one or more differences can at once be read. 
 
 Thus, from 5 + 7 = 12 we read at once 12 — 5 = 7, 
 
 12 - 7 = 5. 
 And from 5 + 5 = 10, we read 10 — 5 = 5. 
 
 And from a + b = c, we read c — a = b, 
 
 c — 6 = a. 
 
 Likewise, from a + b + c = d we read d — a = b + c, 
 
 d — (a + 6) = c, etc. 
 
 14. There is no commutative law of subtraction. For 7 — 4 
 is not the same as 4 — 7. In fact, the latter indicated differ- 
 ence has no meaning in arithmetic. We cannot take a larger 
 number of objects from a smaller number. 
 
FUNDAMENTAL PROCESSES 5 
 
 15. In algebra, where numbers are often represented by 
 letters, we may not know whether the minuend is larger than 
 the subtrahend or not. For example, in a — b, we do not 
 know whether a is larger than b or not. But it is desirable 
 that such expressions should have a meaning in all cases, and 
 this is accomplished by the definition and use of relative 
 numbers. 
 
 16. The First Extension of the Number System. Relative 
 Numbers. Whenever quantities may be measured in one of 
 two opposite senses such that a unit in one sense offsets a unit 
 in the other sense, it is customary to call one of the senses the 
 positive sense, and the other the negative sense, and numbers 
 measuring changes in these senses are called positive and 
 negative numbers respectively. (For examples, see First 
 Course, pp. 32-34.) 
 
 17. A number to be added is offset by an equal number to be 
 subtracted ; hence such numbers satisfy the above definition, 
 and numbers to be added are called positive, and those to be 
 subtracted are called negative. Consequently, positive and 
 negative numbers are denoted by the signs + and — 
 respectively. 
 
 Thus, + 5 means positive five, and denotes five units to be added or to 
 be taken in the positive sense. 
 
 — 5 means negative five, and denotes five units to be subtracted, or to 
 be taken in the negative sense. 
 
 18. Graphical Representation. Relative integers may be 
 represented graphically thus: 
 
 -5 -4 -3 -2 -I +1 +2 +3 +4 +5 
 
 It appears that the positive integers are represented by just the same 
 set of points as the natural or absolute integers. For this and other 
 reasons the absolute numbers are usually identified with positive numbers. 
 Although it is usually convenient to do this, we have in fact the three 
 classes of numbers : the absolute, the positive, and the negative. Thus, 
 we may consider $5 without reference to its relation to an account, or we 
 can consider it as § 5 of assets, or we may consider it as § 5 of debts. 
 
6 ELEMENTARY ALGEBRA 
 
 19. The following rules make clear in every instance 
 whether the signs -f, — denote the operations of addition or 
 subtraction, or the positive or negative character of the num- 
 bers which these signs precede : 
 
 I. If used where a sign of operation is needed, the signs +, — , 
 shall be regarded as signs of operation. 
 
 For example : 
 
 In 8 — 5, — is a sign of operation (subtraction). 
 
 In — 8 + 5, — is a sign of character, because ho sign of operation is 
 needed before the 8, but + is a sign of operation. 
 
 In the problem, " Add — 8 and + 5, 1 ' both the signs are signs of char- 
 acter, because no sign of operation is needed ; the operation has already 
 been named. 
 
 II. If it is necessary to distinguish a sign of character from a 
 sign of operation, the former is put into a parenthesis with the 
 number it affects. 
 
 Thus, — 8 + (—3), means : negative 8 plus negative 3. 
 
 III. When no sign of character is expressed, the sign plus is 
 
 understood. 
 
 Thus, 5 — 3 means : positive 5 minus positive 3. 
 Similarly, 8 a -f 9 a means : positive 8 a plus positive 9 a. 
 
 20. Absolute Value. The value of a relative number apart 
 from its sign is called its absolute value. 
 
 ORAL EXERCISES 
 
 Read the following in full, according to the agreements of 
 Sec. 19 : 
 
 9. 12_(_5). 
 
 10. -12 -(+5). 
 
 11. -7- (-9). 
 
 12. 2 a — (+3 a). 
 
 13. c + d. 
 
 14. c — d. 
 
 15. m + (—»). 
 
 16. 4ic + (— 2x). 
 
 1. 
 
 6-4. 
 
 2. 
 
 -5-8. 
 
 3. 
 
 - 8 + 20. 
 
 4. 
 
 2 a + 3 a. 
 
 5. 
 
 2b -3b. 
 
 6. 
 
 -2 a -3 a. 
 
 7. 
 
 2y+(+3y). 
 
 8. 
 
 3p-(-2p). 
 
 17. 
 
 7-9. 
 
 18. 
 
 7 + 9. 
 
 19. 
 
 -7 + (-9). 
 
 20. 
 
 2y-(-3y). 
 
 21. 
 
 -2x-(-3x). 
 
 22. 
 
 -2x + (-3x). 
 
 23. 
 
 -2 b -(-5 c). 
 
 24. 
 
 3a- (+5 y). 
 
FUNDAMENTAL PROCESSES 7 
 
 WRITTEN EXERCISES 
 Indicate, using the signs +, — : 
 
 1. The sum of positive 5 and positive '3. 
 
 2. The sum of positive a and negative b. 
 
 3 The difference of positive p and positive q. 
 
 4. The difference of negative 5 and positive 3. 
 
 ■ 5. The difference of negative x and positive y. 
 
 6. The sum of positive a and positive b. 
 
 7. The sum of negative ab and negative ab. 
 
 8. The sum of positive y and negative x. 
 
 9. The difference of positive xy and negative xy. 
 
 10. The difference of negative pq and positive mn. 
 
 21. Addition of Relative Numbers. Just as 3 lb. + 5 lb. = 8 
 lb., so 3 positive units 4- 5 positive units make 8 positive units, 
 and 3 negative units 4-5 negative units make 8 negative units. 
 
 To add units of opposite character, use is made of the de- 
 fining property of relative numbers, that a unit in one sense 
 offsets a unit in the other sense. Thus, to add 3 positive units 
 and 7 negative units we notice that the 3 positive units offset 
 3 of the negative units and the result of adding the two will 
 be 4 negative units. 
 
 That is, 
 
 (+3) + (-7) = (4-3) 4- (-3)+ (-4) =-4. 
 In general : 
 
 I. If two relative numbers have the same sign, the absolute 
 value of the sum is the sum of the absolute values of the addends, 
 and the sign of the sum is the common sign of the addends. 
 
 11. If two relative numbers have opposite signs, the absolute 
 # value of the sum is the difference of the absolute values of the 
 
 addends, and the sign of the sum is the sign of the addend having 
 the larger absolute value. 
 
8 ELEMENTARY ALGEBRA 
 
 22. More than two numbers are added by repetition of the 
 process just described. This may be done either: 
 
 (1) by adding the second number to the Jirst ; then the third 
 number to the result, and so on ; or 
 
 (2) by adding separately all the positive numbers and all the 
 negative numbers, and then adding these two results. 
 
 23. It may be verified that the Commutative and the Associa- 
 tive Laws of Addition hold also for relative integers. 
 
 24. Subtraction of Relative Numbers. Since n units of one 
 sense are offset by adding n units of the opposite sense, we 
 may subtract n units of one sense by adding n units of the 
 opposite sense. 
 
 Thus, 
 
 7-(+3) = 7 + (-3). 
 
 And, 
 
 7-(_3) = 7 + (+3). 
 
 And, 
 
 4-(+7) = 4 + (-7). 
 
 25. Accordingly subtraction may be regarded as a variety of 
 addition : To subtract a monomial, we add its opjwsite. 
 
 To subtract an algebraic expression consisting of more than 
 one term, we subtract the terms one after another. 
 
 In general, to subtract any algebraic exj)ression toe may change 
 the sign of each of its terms and add the result to the minuend. 
 
 The subtraction of a larger number from a smaller number (as in the 
 third example, Sec. 24) is made possible by the introduction of the idea 
 of relative numbers. 
 
 ORAL EXERCISES 
 
 State the sums : 
 
 1. 5 + (-3). 4. -12z + (-18z). 
 
 2. -6a+(-7a). 5. 11 x + (- 2x) + (-5x). 
 
 3. -lly + Zy. 6. -3q + 7q + (-6q). 
 
 7. How may the correctness of a result in subtraction be 
 tested ? State the differences : 
 
FUNDAMENTAL PROCESSES 9 
 
 8. 11-6. 10. -lla-(-Ga). 12. -31y-(-3y). 
 
 9. -11-6. ll. 31a- (+5 a). 13. 17 p - (- 17 p). 
 
 14. How may a parenthesis preceded by the sign' + be re- 
 moved without changing the value of the expression? One 
 preceded by the sign — ? 
 
 15. How may terms be introduced in a parenthesis preceded 
 by the sign + without changing the value of the expression ? 
 In a parenthesis preceded by the sign — ? 
 
 WRITTEN EXERCISES 
 
 Acl< 
 1. 
 
 1: 
 2a + 5 
 
 6. 
 
 c + d—5 
 
 11. 
 
 4 x — 2 z + y 
 
 
 a + 4 
 
 
 c — a" 4-5 
 
 
 2 X — ?/ + 2 
 
 2. 
 
 3a + 8 
 
 7. 
 
 x+y+2z 
 
 12. 
 
 1 + m 3 4- jr 
 
 
 a — 4 
 
 
 x — y -f- 4 z 
 
 
 1 — 7ii 3 — p 2 
 
 3. 
 
 6b + c 
 
 8. 
 
 P + (/ — 7/1 
 
 13. 
 
 ax 4- 6?/ + cz 2 
 
 
 3 b -2 c 
 
 
 ^ — <? + 2 /// 
 
 
 bx -{-ay — z 2 
 
 4. 
 
 - 3 a . + b 
 
 9. 
 
 2 a — y + z 
 
 14. 
 
 1.5 a; 4- 3.5?/ 4- 2. 
 
 
 2a-3b 
 
 
 2x + 2y — 4z 
 
 
 .5a4-6.5y— .lz. 
 
 5. 
 
 4a — 5 
 
 10. 
 
 ax-\-by + c 
 
 15. 
 
 2 a '"l""52/ — 3 2; 
 
 
 3a + 7 
 
 itract : 
 
 
 aa 4- # — c 
 
 
 i»-#2/+£« 
 
 Sut 
 
 
 
 16. 
 
 4a + 6 
 
 20. 
 
 7 x 4- 3 ,v 
 
 24. 
 
 41 x 2 - 16 f 
 
 
 2a-9 
 
 
 2 y - 4 a; 
 
 
 15 a 2 - 20 y 2 
 
 17. 
 
 8 x 4- 3 
 
 21. 
 
 4a + 3i 
 
 25. 
 
 — 11 m + AOp 
 
 
 - 3 x + 2 
 
 
 2c-5a 
 
 
 -40 m -12 p 
 
 18. 
 
 llx — Ay 
 
 22. 
 
 16 y+ 2 
 
 26. 
 
 x — 7 y -\-5 z 
 
 
 19 a 
 
 
 8 # - 10 2 
 
 
 — x + Ay — 6z 
 
 19. 
 
 12 * 
 
 23. 
 
 10 or 9 - 16 y 2 
 
 27. 
 
 p 4- ^ — m 
 
 
 - 6 « + 3 
 
 
 20 a 2 4- 4y 2 
 
 
 6^— 2 g 4- 4 ???, 
 
10 
 
 ELEMENTARY ALGEBRA 
 
 28. ax 2 + by 2 + c 
 ax 2 — by 2 — c 
 
 29. 
 
 m 2 
 — m 2 
 
 + 2p 2 
 -4p 2 
 
 -6g 2 
 + 5g 2 
 
 30. 
 
 40 x 2 
 50 a; 2 
 
 - 10 f 
 
 + 40 7/ 
 
 + z 2 
 -lz 2 
 
 31. 2.5 a + 6.3 6 -.lc 
 
 1.5 a — 3.5 b — .9 c 
 
 32. 
 
 2" ^ 
 f 35 
 
 
 33. 
 
 a 2 a; 
 a 2 a; 
 
 + & 2 2/ + A 2 
 
 + by 2 + cz 2 
 
 Remove parentheses and unite terras as much as possible : 
 
 34. 3 a-26+ (36-7 a). 
 
 35. 4m + 5— (63 — 3p). 
 
 36. (lltc + 52/)-(-3aj + 22). 
 
 37. 7 -[2 -(3 -5)]. 
 
 38. (4 a + 2 o) - [(7 a - 5 a) + (- 6 a - 17 a)]. 
 
 39. 3 + {5a;-2-(7a;-l-2 + 3a;)}. 
 
 40. ±x + \2x-[3-(7x + 5)-l+x']\. 
 
 41. a&-{2a& + c-[3c-(6-a&)]|. 
 
 42. 12 xy-\2 xy - (3 z + [4 av/ - (2 2 + a^)] } • 
 
 43. _(5 M -3^ + l)-$-2 i xy + 3.^-(a W + 2)|. 
 
 44. - {4 a&c — (2 ac + 6c) ] + \ 6 a&c + (2 ac — 6c)}. 
 
 Write the expression of Exercises 45-56, as a; minus a 
 parenthesis. Also group the terms involving x and y in each 
 expression in a parenthesis preceded by the sign — . 
 
 45. 3a + x + 2y. 
 
 46. — 5y + 7c — 8 + 05. 
 
 47. 6« 2 + 4?/ + .x--3. 
 
 48. 2 m + p + a; — 7/. 
 
 49. a + x — 6 + cy. 
 
 50. J7 + X - 7/ + &?/. 
 
 51. x — 3b+2y— c. 
 
 52. 3 b + as + ay + by. 
 
 53. &?/ + m — ?* + 3J. 
 
 54. 2p — q + x — y. 
 
 55. a//. + .(' — Z + Z?/. 
 
 56. C + fit + 9) — (a + 6) ?/. 
 
FUNDAMENTAL PROCESSES 11 
 
 26. Multiplication. To multiply two absolute integers means 
 to use the one (called the multiplicand) as addend as many times 
 as there are units in the other (called the multiplier). The 
 result is called the product. 
 
 Thus, 3 times 4 means 4 + 4 + 4. 
 
 The simpler products are obtained by actually making the additions that 
 are implied. For large numbers, as we have seen in arithmetic, the pro- 
 cess may be much abridged by use of properties of the decimal notation. 
 
 27. Commutative Law of Multiplication. The expression 3 
 times 5 means 5 +5+5, and may be indicated as follows: 
 That is, since each horizontal line 
 
 (or row) contains 5 dots, there are • • • • 
 
 all together 3 times 5 dots. But (A) • • • • • 
 each vertical line (or column) con- 
 tains 3 dots and there are 5 columns. 
 
 Hence there are 5 times 3 dots all together. But the number of 
 dots is the same whether we count them by rows or by columns, 
 hence 5 times 3 equals 3 times 5. Quite similarly, if we have 
 
 (B) 
 
 a< 
 
 t t t 
 
 a rows of dots with b dots in each row, it follows that a times 
 b equals b times a. 
 
 The result may be stated in symbols thus : 
 
 ab = ba. 
 
 This fact, called the Commutative Law of Multiplication, means 
 that the product is not altered if multiplier and multiplicand 
 are interchanged. Consequently, these names are frequently 
 replaced by the name factor applied to each of the numbers 
 multiplied. 
 
12 ELEMENTARY ALGEBRA 
 
 28. Let each dot of the block of dots of (A) above have the 
 value of 6. Since there are 3x5 dots, the value of the block 
 would be (3 x 5) x 6. 
 
 A second expression for the value of the block is obtained 
 by finding the value of the first row (viz. 5x6) and multiply- 
 ing it by the number of rows, or 3. The expression resulting 
 must be equal to that already found, or : 
 
 3x(5 x6) = (3x5)x6. 
 
 Similarly, if each dot of the block (B) above has the value 
 
 c, it follows that 
 
 a(bc) = (ab) c. 
 
 That is : The product of three absolute integers is not altered 
 if they be grouped for multiplication in any way possible without 
 changing the order. Tliis is a case of vohat is known as the 
 Associative Law of Multiplication for absolute integers. 
 
 29. By similar methods and use of these results it can be 
 proved that both the Commutative Law and the Associative 
 Law apply to all products of absolute integers. That is : 
 
 Commutative Law. The product of any number of given factors 
 is not changed, if the order of the factors be changed in any ivay. 
 
 Associative Law. The product of any number of given factors 
 in a given order is not changed if the factors be grouped in any 
 way. 
 
 The Distributive Law. From the block of dots we see that 
 
 c\ 
 
 t 
 
 • • •- 
 
 t t * + • r T T" 
 
 : ! i ! ! : 
 
 4 4 4-- — -4 4**4- 
 
 c(a + b)=ca -\-cb. 
 
 This is called the Distributive Law of Multiplication, and the 
 above proof covers the case in which a, b, and c are absolute 
 integers. 
 
FUNDAMENTAL PROCESSES 13 
 
 30. Multiplication of Relative Integers. To multiply by a 
 positive integer means to take the multiplicand as addend as 
 many times as there are units in the multiplier, and to multi- 
 ply by a negative integer means to take the multiplicand as 
 subtrahend as many times as there are units in the multiplier. 
 
 Consequently, 
 
 4-3 = 3+3 + 3 + 3 = 12. 
 4(-3)=-3 + (-3)+(-3) + (-3) = -12. 
 (_ 4)3 = _ 3 _ 3 _ 3 - 3 = - 12. 
 (-4)(-3) = -(-3)-(-3)-(-3) = 12. 
 
 So, generally, 
 
 (+a)(+6)= + a&, 
 
 (+a)(-6)=-a6, 
 
 (_a)(+&)=-a&, 
 
 (— a)(— &)= + a&. 
 
 31. In words : TJie product of two (integral) factors of like 
 signs is positive, and of two factors of unlike signs is negative ; 
 in each case the absolute value of the product is the product of the 
 absolute values of the factors. 
 
 32. We observe that the Commutative Law holds in this case 
 also. 
 
 For example : (— b) a = a ( — b). Since ba = ab, by the Commutative 
 Law for absolute integers, and by Sec. 30, ( — b)a — — ba — — ab = a(— b). 
 
 It may be shown that the Associative and the Distributive 
 Laws also hold. 
 
 ORAL EXERCISES 
 
 State the laws that are applied in the various steps of the 
 following calculations : 
 
 1. 2-8- 3-5 = 2-5. 8- 3 = 10 -24 = 240. 
 
 2. 5(17 - 2 - 6 c) =5(17 - 2) - 5(6 c) = (5 - 2) 17 - (5 - 6) c 
 
 = 10 • 17 - 30 c = 170 - 30 c. 
 
 3. 7 6(5 x + ab) = (7 6) (5 x) + (7 6) (ab) = 35 bx + 7 ab 2 . 
 
 4. (a + b) (c + d) = a(c + d) + b(c + d) =ac-\- ad + be + bd. 
 
14 ELEMENTARY ALGEBRA 
 
 State the products : 
 
 7. Sx 9. 5x 2 11. 6xyz 
 
 — 4 xy 2 x 2 —2 x£ 
 
 5. 
 
 ax 
 
 
 5b 
 
 6. 
 
 -2a 
 
 
 -3 a 2 
 
 13. 
 
 (a + b) 2 . 
 
 14. 
 
 {a - c) 2 . 
 
 15. 
 
 (x + yf. 
 
 16. 
 
 (x — a) 2 . 
 
 17. 
 
 (x-2y) 
 
 8. — a b b 10. — afy 12. — 4 aH 
 
 arc — xy 2 — 3 b 2 t 
 
 18. (x + y)(x-y). 23. {m + 2xf. 
 
 19. (2x--l) 2 . 24. (3-2d)(3+2d). 
 
 20. (2m-f«) 2 . 25. (4 - x) 2 . 
 
 21. 0-l)(a- + l). 26. -3x(x + 2). . 
 
 22. (l_y)(l + y). 27. (l-2d)(l+2d). 
 
 WRITTEN EXERCISES 
 Multiply and test (see First Course, p. "69) : 
 
 1. 
 
 2. 
 
 2a + 3 
 2a + 4 
 
 3 a - 2 & 
 2a-36 
 
 (3 x - 4 ?/) 2 . 
 (5y-7t)(Sy 
 (6 a + 13 g) 2 . 
 
 3. a? +.2 
 a-3 
 
 4. x 2 -3a- + l 
 
 a-2 
 
 (1 + 
 
 (1- 
 (&- 
 
 5. gf-3y4-4 
 
 y-2 
 
 6. J9-3J + * 3 
 p-2* 
 
 7. 
 8. 
 9. 
 
 10. 
 
 + 2 0- 11. 
 12. 
 
 .»)(2 + «)(l-.»). 
 
 3)(&4-7)(6-3). 
 
 33. Division. Division is the process of finding a number 
 called the quotient, such that when multiplied by a given num- 
 ber, called the divisor, the product is a given number, called 
 the dividend. 
 
 34. The fundamental relation of division is : 
 
 Divisor times Quotient equals Dividend. 
 
 From this it follows at once that if dividend and divisor 
 have the same sign, the quotient is positive, and if they have 
 unlike signs, the quotient is negative. 
 
 Division is usually indicated in one of the following forms: 
 f or 8 -r- 2. Both are read : " eight divided by two." 
 
FUNDAMENTAL PROCESSES 15 
 
 35. The Second Extension of the Number System; Fractions. 
 We found the natural or absolute integers by counting. We 
 defined the operation of addition for these integers and saw 
 that it was always possible. Next we defined the operation 
 of subtraction for these integers, and found that it was not 
 always possible. This led us to define relative numbers (posi- 
 tive and negative). In the system of numbers as enlarged by 
 this first extension, we saw that both addition and subtraction 
 are always possible. Then we defined multiplication for all 
 integers and saw that it was always possible. We now examine 
 the operation of division as just defined. 
 
 The operation 12 -=- 4 is possible in the system of integers 
 because there exists an integer, 3, whose product with 4 is 12. 
 But the operation 12 ~ 5 is impossible in the system of integers, 
 since there exists no integer whose product with 5 is 12. 
 This leads us to define another kind of number, the fraction. 
 This is done by dividing the unit into b equal parts, and taking 
 
 a of these parts. A symbol for the new number is -, in which 
 
 b 
 
 a and b may be any integers. This constitutes our second 
 
 extension of the number system. 
 
 CI 
 
 36. The new number, -, is called a fraction; a is called the 
 
 b 
 
 numerator and b the denominator of the fraction ; a and b together 
 are called the terms of the fraction. 
 
 37. Graphical Representation of Fractions. Fractions may be 
 represented by points of the number scale that we have already 
 had. 
 
 Thus, if the distance from zero to 1 is divided into b equal parts, and 
 then a of these parts laid off from zero, the end point represents the frac- 
 tion - . f In the figure the fraction is - . ) Fractions may also be taken 
 
 in the negative sense. Two fractions are said to be equal, or to have the 
 same value, when they are represented by the same point of the number 
 scale. 
 
16 ELEMENTARY ALGEBRA 
 
 38. If each of the b equal parts is halved, making 2 b parts, 
 and each of the a parts taken is halved, making 2 a parts, the 
 end point remains the same. 
 
 That is : 
 
 a_2o 
 
 b~2b' 
 Similarly : 
 
 a _ na 
 
 b nb 
 
 (A) In words : TJie value of a fraction is not altered if both 
 numerator and denominator is multiplied by the same integer. 
 
 (B) Reading the above equation from right to left. The 
 value of a fraction is not altered if both numerator and denomi- 
 nator be divided by a common integral factor. 
 
 Every integer may be regarded as a fraction. 
 Thus, 3 = f or § or - 1 /, etc. 
 
 39. These properties enable us to add and subtract 
 fractions. 
 
 If the given fractions have the same denominator, we say 8 fifths plus 
 3 fifths are 11 fifths (or | + f = Y) just as we say 8 ft. + 3 ft. = 11 ft. 
 
 If they have not the same denominator, they are first reduced to the 
 same denominator, and then the results added. 
 
 40. To multiply a fraction by an integer we extend the 
 definition of multiplication to this case, and we have : 
 
 3(f) = t + f + f = ¥- 
 
 c( c S) = ° + «+... to (c terms) = « + " + « " to (ca's^ca. 
 \bj b b K } b b 
 
 ( -c Hf) ==_ I ~ i '" to ^ c terms ^ 
 
 — a — a — a'-' to (c terms) _ _ca 
 ~b~ " ~b" 
 
FUNDAMENTAL PROCESSES 17 
 
 41. That is, to multiply a fraction by an integer we multiply 
 the numerator by that integer. 
 
 The fraction - is the quotient called for in the indicated division a -f- b, 
 
 because the product of the divisor b and the asserted quotient - is the 
 dividend a, as seen in 
 
 h . « _ fa* _ a . ( by property B, Sec. 38). 
 b b 
 
 In our present number system the fraction - may therefore always be 
 
 b 
 
 regarded as indicating the division of a by & (provided b is not zero). 
 
 42. The operation of division is thus seen to be possible in 
 the new number system for any integral dividend and any 
 integral divisor (except zero). 
 
 We shall see, under Division of Fractions, that if either dividend or 
 divisor or both are fractions, the operation is still possible without en- 
 larging the number system further. 
 
 43. Multiplication of Fractions. The product of two or more 
 fractions is a fraction whose numerator is the product of the given 
 numerators and whose denominator is the product of the given 
 denominators. 
 
 44. Since an integer or an integral expression may be re- 
 garded as a fraction with denominator 1, this definition applies 
 also when one or more of the factors are integral. 
 
 45. The Associative, Commutative, and Distributive Laws of 
 Multiplication may be shown to apply to fractions as well as 
 to integers. 
 
 WRITTEN EXERCISES 
 
 Perform the indicated operations, and express the result in 
 lowest terms : 
 
 ■ "+ 1 
 
 5 10 a 
 
 4<x_5a l 
 
 ' ~5 T* "" x + 1 2(3 + 1) 
 
 o 
 
 o 
 
 o 
 
 6-1 
 
 6 + 1 
 
 x-1 
 
 x + 1 
 
18 ELEMENTARY ALGEBRA 
 
 5. *L + <ty. 7. 2 3 P 
 
 2y 4x a+p ( a +P) 
 
 6. -$— + -• 8. - + — , 
 
 a — bb x* 2xr 
 
 9. J^.+ 5 
 
 ,2 
 
 1-g 2 l + g 
 
 io. %-: + '' 1 - + 
 
 (l_SB)(l-y) (y-a^-l) ">-!)(! ^-y) 
 
 12. 
 
 a 2 — (6 + c)a+&c cr—(b-\-d)a-\-bd a' 2 —(c + d)a+cd 
 2 a 1 ?/ + y 2 '_ y x 2 4- 5 x?/ 
 
 , _ 5 14 . & ac 4 
 
 13. — • 17. oc • - • ■ 
 
 7a 156 c 2 2& 3 
 
 x 3 3 xy ,. -3a 41 
 
 14. — • — — • lo. • • 
 
 y" 4Z 5 a^-9 9l 
 
 16 . *• ._•£!. ijs.. 20. ..<i-l N 
 
 3?/ 16ar 3 y 15i 3 a; \ x 
 
 21. pgr 
 
 22 
 
 Kf-*J- 
 
 a(l — a) 1 -f-a 
 
 l+2a + a 2 ' l-2a + a 2 ' 
 
 1-a 2 1 - & 2 /a &_\ 
 
 1 + 2 6 + 6 2 ' 6 2 - 2 a& + a 2 \JL - a 1 - &/ 
 t + 1 * _ 1 4« 2 \^ + 2i + l 
 
 23. 
 
 24 
 
 '*-l ? + l 1-^V 4< 
 
 tf + £-l_£ + lY!!-.£ 
 
 ^r v 2 p v J\'p v 
 
 46. Reciprocals. If the product of two numbers is 1, each 
 is called the reciprocal of the other. 
 
 « ._ b _. a fe «& _ j 
 
 Thus, the reciprocal of - is - , since — • - = — 
 
 b a b a ab 
 
FUNDAMENTAL PROCESSES 19 
 
 47. Division of Fractions. To divide - by - means to find 
 
 b a 
 
 a number q such that - ■ q = — • 
 
 a b 
 
 (1 c 
 
 Multiplying both members by -, the reciprocal of -, we 
 
 , , . cl a ad 
 
 obtain Q = - ' v = v~ ' 
 
 c b be 
 
 That is : To obtain the quotient, multiply the dividend by the 
 reciprocal of the divisor. 
 
 We have thus seen that in the number system as now extended the 
 four fundamental operations are possible when any or all of the numbers 
 involved are fractions. (Division by zero is always excepted.) 
 
 48. Complex Fractions. If one or both of the terms of a frac- 
 tion are themselves fractions, the given fraction is called a 
 complex fraction. Since a fraction indicates division, a com- 
 plex fraction may be simplified by performing the indicated 
 division of its numerator by its denominator. 
 
 WRITTEN EXERCISES 
 Divide, and express results in lowest terms : 
 
 5. ^2c 2 . 9. £+f£. 
 
 46 2 8 
 
 7h 2 m v° 
 
 3x n -6 . -18 
 
 1. 
 
 2 " 6 
 
 
 2. 
 
 ab 
 
 Ab 2 
 
 18 c ' 
 
 Sac 
 
 3. 
 
 ax 2 
 2tf ' 
 
 5x 
 
 bx 3 ^ 
 ay 2 
 
 4. 
 
 3tf 
 
 2x 
 
 
 4t/ + l 2x— 1 X s x 7 
 
 x 5 a 
 
 8 - r 12 - »y 
 
 9 m 3 x 2 6 
 
20 ELEMENTARY ALGEBRA 
 
 4rc 1 f 
 
 13. Tjf 14. f 15. l" 
 
 16 . f^. a V/i_r 
 «+j^(i_™ 
 
 18. 
 
 1 1 
 
 cr/ \tr a 3 
 
 Perform the operations indicated : 
 
 19. *_ J_. 20. t+J— 21. f£=2+lV/ r i---2£-Y 
 
 1+1 1+1 \P+<! J \ P+9J 
 
 x t 
 
 49. Factoring. The process of finding two or more factors 
 whose product is a given algebraic expression, is called factoring 
 the given expression. 
 
 In division one factor (the divisor) is given and another factor (the 
 quotient) is to he found such that the product of the two factors is the 
 given expression (the dividend). Division is thus really a variety of 
 factoring, although the name "factoring" is not usually applied to it. 
 As a rule, when an expression is to be factored, none of the factors is 
 specified in advance, and any set of factors is acceptable on the sole con- 
 dition that their product is the given expression. 
 
 50. Several type products are of special importance. These 
 have been treated in the First Course (pp. 76-95, 175-188), 
 and are collected here for reference. 
 
 I. xy + xz = x(y + z). 
 
 II. x 2 ±2xy + f = (x±$f: >_+ ; 
 
 III. x>-tf=(x + y)(x-y). 
 
 IV. x 2 + (a + b)x + ab = (x + a) (x + &). 
 V. X s + 3 x-y + 3 xy°- + f = (x + y) 3 . 
 
 VI. or 5 + ?/ 3 = (a + y) (x 2 — xy + y 2 ). 
 
 VII. x>-y* = tx-y)tx 2 + xy + tf). 
 
FUNDAMENTAL PROCESSES 21 
 
 In all of these formulas the letters may, of course, represent either 
 positive or negative numbers. 
 
 Detailed treatment of these types is given in the First Course, together 
 with exercises under each type. The following set of miscellaneous exer- 
 cises covers all the types. 
 
 ORAL EXERCISES 
 Factor : 
 
 1. ax + ay. 26. 1—2 (x + ?/) + (x + y) 2 . 
 
 2. ax + a. 27. 49 + 14 x + x 2 . 
 
 3. ax + a 2 . 28. 9 - 12 y 2 + 4 y\ 
 
 4. abx — ay. 29. 16 - 40 z + 25 z 2 . 
 
 5. x 2 + ax. 30. a 2 b 2 + 2 abed + c 2 d 2 . 
 
 6. x^ + ax 2 . 31. \ + t + t 2 . 
 
 7. a 2 J? — ax. 32. ?/ + .4 / + .04. 
 
 8. mx + my + m- 33. m 2 a? 2 + 4 mx + 4. 
 
 9. jxv 2 + jrx -{- }ig. 
 
 10. a 8 — ar + x. 
 
 11. x 2 y 2 + xy + y. 
 
 12. (a-f-6)ic— (a + &)?/. 
 
 13. s 2 £ + ^ 2 + s 2 * 2 . 
 
 14. ±gt 2 + gt. 
 
 15. a&c — acd + bed. 
 
 16. a 2 x 2 -b 2 y 2 . 
 
 17. 4a: 2 -9z 2 . 
 
 18. atbW-c 2 . 
 
 19. l-p 2 q 2 y 2 . 
 
 20. m 2 p 2 -sH 2 . 
 
 21. (« + &) 2 - (c + tf) 2 . 
 
 22. 1 - (m + p)\ 
 
 23. a 2 + 2 abc + 6 2 c 2 . 
 
 24. a 4 .*; 4 + 2 a 2 <B? + 1. 
 
 25. a 2 + 2 a (6 + c) + (6 + c) 2 . 
 
 34. 
 
 16 2 
 
 35. 
 
 x 2 — 5 x + 6. 
 
 36. 
 
 ar + 7a; + 12. 
 
 37. 
 
 :c 2 - x - 12. 
 
 38. 
 
 a 2 + x — 6. 
 
 39. 
 
 « 2 - 12 x + 35. 
 
 40. 
 
 s 2 _ 2 x _ 35. 
 
 41. 
 
 12 a 2 + 7 a? + 1. 
 
 42. 
 
 15y 2 -2x-l. 
 
 43. 
 
 6 a,- 2 -12a; + 6. 
 
 44. 
 
 3?/ 2 + 8?/ + 5. 
 
 45. 
 
 a 3 — & 3 . 
 
 46. 
 
 a 3 + b\ 
 
 47. 
 
 8 - a 3 6 3 . 
 
 48. 
 
 64 — a; 3 ?/ 3 . 
 
 49. 
 
 8 a? - 12 a; 2 + 6 x - 1. 
 
 50. 
 
 27y s + 27tf + 9y + l 
 
22 ELEMENTARY ALGEBRA 
 
 WRITTEN EXERCISES 
 Factor : 
 
 1. x 2 — 49. 26. my + 4 s 4 . 
 
 2. t 2 -7t + 6. 27-- — + — -1 
 
 3. of-3tf- + y. " 8~ 4 2 
 
 4. 4v 2 -9?* 2 . 28. .125 r 3 + .75 z 2 + 1.5 a,- + 1. 
 
 5. 1 + 10 a + 25 a 2 . 29. (2 m + If -(3 m - l) 2 . 
 
 6. a 3 + 3a 2 & + 3«5 2 +6 3 . 30. 8a 3 + 12a 2 £ + 6a£ 2 + £ 3 . 
 
 7. (x + l) 2 +2(a; + l) + l. 31. ^ - 16. 
 
 8 1_? + I. 32 " f '7/ 2 -Hoz/ + 24. 
 
 a a 2 33. 4 /r - 12 lit + 9£ 2 . 
 
 9. ^-2 a -15. 34. 1-m 3 . 
 
 10. a 2 & 2 -l. 35. (2a + 6) 2 -(3a-2&) 2 . 
 
 11. x 4 + 6x 2 y + 9y 2 . 36 . 8 a 3 -32 a 5 . 
 
 12. v s -t\ 37 10 m 2 .r - 60 mx + 90 x. 
 
 13. 1 — »y. 
 
 14. a 4_4 a ^2. 
 
 38. ^ 3 + 27. 
 
 39. a 4 + 3 a 2 -28. 
 
 15. (x + y) 4 — 1. 
 
 16. „*-* 40 - 8» s -60^ + 150W 
 
 17. » 4 — 2«y + ?/ 4 . 
 
 18. (m+p) 3 — 1. 
 
 19. (a; + 7 / ) 2 -( i > + ^ 
 
 20. (a + ^/' + iB 3 . 
 
 -125w 3 . 
 
 41. .s 3 -64. 
 
 42. 56 a 2 - 68 a + 20. 
 
 43. 24 a^ + av- 10. 
 
 2i. ajs + fa^ + fa + f 44 - 15 ^ - 34 as- 16. 
 
 22. a 3 -lia 2 + fa-i 45 - aV-7cK! + 12. 
 
 23. cc 2 + (a + &)x- + a6. 46. y 4 - 13 ?/ 2 + 42. 
 
 24. y 2 + {ac-bd)y-abcd. 47. (a + 1) 2 - (a - l) 2 . 
 
 25. acx 2 + (c6 + ad)x + 6(i. 48. a 2 -^-}. 
 
 Calculate : 
 
 49. 27 2 -25 2 . 50. 387 2 -377 2 . 51. 26 3 -25 3 . 
 
 Note. Chapters I and II ai'e themselves in the nature of summaries ; 
 
 consequently, no summaries of these chapters are given here. 
 
^\DAMENTAL PROCESSES 23 
 
 REVIEW 
 
 WRITTEN EXERCISES 
 
 Perform the indicated operations, expressing fractional re- 
 sults in lowest terms: 
 
 1. 2a -36 4- 4 a + 11 c — 2d + 6-8a-9& + 3 c — 4 — 5 d 
 
 4-2a-&. 
 
 2. afy + 3 a:?/ 2 + 4 a,- 2 - 2a# - 3 x 2 y + 2 x-y 2 -4xy + 8 xy 2 
 
 — 7y 2 — 3 x 2 y 2 . 
 
 3. 5 x - ly — (3a; + 4y- 2) + 3 +(8oj - 7)- {2x — Sy 
 
 + 13)4-8(2 a;-l). 
 
 4. 4<c-{2c-(3a;4-2# + 5c)}. 
 
 5. 5ac-[4& + 2c[3a-5&-(6a-2&- 7c + 3)] } . 
 If Z = 4a; + 2y-3, M=x-9y + l, R = y-5x,find: 
 
 6. Z + Jf+#. 8. LM-3E. 
 
 7. 3£-2Jf+#. 9. V--M 2 . 
 
 10. (Z + 7lf) 2 . 
 
 11. 3L 2 -5ME. 
 
 If X = 7a4-2&, F=2a-96 + 3, Z = 4 & -7a-3, find: 
 12. X-(3F-2Z). 13. 2F-[Z+3(4X-8F)]. 
 
 i4. z 2 -i6F 2 . 15. r 2 + 4 rz. i6. xra. 
 
 , _ a -f- 5 a 2 + 75 , a — 5 
 
 17. -\ • 
 
 a — 5 ctr — 25 a + 5 
 
 6 m 4- 4 /? _ 3 w 2 + 4 jj 2 ra — 2p 
 157» 15 mp 
 
 9 -f- a a — 3 
 
 18. 
 
 19. 1 
 21. 
 
 1 + a 2 a + 1 
 2 . 1 
 
 5_p 
 20. 
 
 H) (-+ * 
 
 +w 
 
 p + 9t + 20 t 2 + 7t + 12 «■ + 11 1 + 28 
 
 1 /Ca 8 + 9a : 
 
 22 . Vaf_3a ; + 14 
 Vc 3 c 2 c 
 
 23. 
 
 3a 2 V 4a + 2 
 
 24. (& + 3) 3 -(&-2) 3 . 
 
 25. (x n + 3a;"- 1 + 5a; n - 2 )(ar 2 + 4a;). 
 10 a,- 3 5 x 3 5 aA 15 x 
 
 26. 
 
 3 ?n 2 9 m 2 
 
 w 
 
 war 
 
24 ELEMENTARY ALGEBRA 
 
 27. 
 28. 
 
 33. 
 
 34. 
 
 35 + 12 
 
 5y 2 , . 
 
 '£ + ^ 2 
 » + 2 
 
 . 56 + 1 
 
 xj 
 
 5g + g 2 
 
 15 + 8 
 1 3 
 
 " 2± + llg + g* 
 
 «« 2 ™ g— 1 o+l 
 
 29 - o b' 30 - ^m — ^-r' 
 
 2 3 g+1 g— 1 
 
 a 8 a q — lq + 1 
 
 Q1 /3 a llo\/3a 6\ /3 a 2 6\ 2 
 
 32. 
 
 6 3 a J \ b 3 a) \b 3 a 
 2 13 
 
 <e 2 + 4» + 3 a; 2 + 3z + 2 a 2 + 5x- + 6 
 
 or?/ 2 a* 3 
 a 3 + 1 ' a + 1 
 
 4a + 7 . 16a 2 + 56a + 49 
 a 2 -16x + 64 ' a 2 -64 
 
 35. -i^+ § + ^- 
 
CHAPTER II 
 
 EQUATIONS 
 
 EQUATIONS WITH ONE UNKNOWN 
 
 51. Two algebraic expressions are equal when they represent 
 the same number. 
 
 52. If two numbers are equal, the numbers are equal which 
 result from : 
 
 1. Adding the same number to each. 
 
 2. Multiplying each by the same number. 
 
 Subtraction and division are here included as varieties of addition and 
 multiplication. 
 
 53. The equality of two expressions is indicated by the 
 
 symbol, =, called "the sign of equality." 
 
 54. Two equal expressions connected by the sign of equality 
 form an equation. 
 
 55. Such values of the letters as make two expressions equal 
 are said to satisfy the equation between these expressions. 
 
 56. Equations that are satisfied by any set of values what- 
 soever for the letters involved are called identities. 
 
 57. Equations that are satisfied by particular values only 
 are called conditional equations, or, when there is no danger of 
 confusion, simply equations. 
 
 58. The numbers that satisfy an equation are called the 
 roots of the equation. 
 
 3 25 
 
26 ELEMENTARY ALGEBRA 
 
 59. To solve an equation is to find its roots. 
 
 60. The letters whose values are regarded as unknown are 
 called the unknowns. 
 
 61. The degree of an equation is stated with respect to its 
 unknowns. It is the highest degree to which the unknowns 
 occur in any terra in the equation. Unless otherwise stated, 
 all the unknowns are considered. 
 
 62. An equation of the first degree is called a linear equation. 
 
 63. An equation of the second degree is called a quadratic 
 equation. 
 
 64. An equation of the third or higher degree is called a 
 higher equation. 
 
 65. In order to state the degree of an equation its terms 
 must be united as much as possible. 
 
 66. Terms not involving the unknowns are called absolute 
 terms. 
 
 67. Equivalent Equations'. If two equations have the same 
 roots, the equations are said to be equivalent. If two equa- 
 tions have together the same roots as a third equation, the two 
 equations together are said to be equivalent to the third. 
 
 68. The Linear Form, ax + b. Every polynomial of the first 
 degree can be put into the form ax + 1>. That is, by rearrang- 
 ing the terms suitably, it can be written as the product of x by 
 a number not involving x, plus an absolute term. Hence, the 
 form ax + 6 is called a general form for all polynomials of the 
 fir.st degree in x. 
 
 69. Every equation of the first degree in one unknown can 
 
 be put into the form : 
 
 ax + 6 = 0. 
 
 Consequently this is called a general equation of the first degree 
 in one unknown. 
 
EQUATIONS 27 
 
 i 
 
 70. General Solution. From the equation ax -\-b = 0, (1) 
 
 we have ax = — b, (2) 
 
 and hence, x = (3) 
 
 a 
 
 71. is the general form of the root of the equation of 
 
 a ' ■ 
 
 the first degree. There is always one root, and only one. 
 
 The advantage of a general solution like this is that it leads to a for- 
 mula which is applicable to all equations of the given form. 
 
 In words : When the equation has been put into the form 
 ax -\-b = the root is the negative of the absolute term divided by 
 the coefficient of x. 
 
 Test. The correctness of a root is tested by substituting it 
 in the original equation. 
 
 If substituted in any later equation, the work leading to the equation 
 is, of course, not covered by the test. 
 
 Results for problems expressed in words should be tested 
 by substitution in the conditions of the problem. 
 
 If tested by substitution in the equation only, the correctness of the 
 solution is tested, but the setting up of the equation is not tested. Nega- 
 tive results that may occur in such problems are always correct as solu- 
 tions of the equations, but they are admissible as results in the concrete 
 problem only when the unknown quantity is such that a unit of the un- 
 known quantity is offset by a unit of its opposite. 
 
 For example, if the unknown measures distance forward, a negative 
 result means that a corresponding distance backward satisfies the con- 
 ditions of the problem. But, if the unknown is a number of men, a 
 negative result is inadmissible, since no opposite interpretation is possible. 
 
 ORAL EXERCISES 
 Solve for x : 
 
 1. 3x = 15. 4. .r--6 = 10 
 
 2. 2 a; = 11. 5. .r + 6 = 12. 
 
 3. 4Jx = 9. 6. 2 £ + 1 = 13. 
 
28 ELEMENTARY ALGEBRA 
 
 Solve for t : 
 
 7. 6* = 36. 10. 3t-8 = 22. 
 
 8. t — 5 = 20. 11. at = ab. 
 
 9. 2 t + 5 = 25. 12. a« + 6 = c. 
 
 Solve for y : 
 
 13. 3y-l = 2. 16. by = bc. 
 
 14. 2y — l = 7. 17. % = 6 + c. 
 
 15. 5^ + 5 = 35. 18. ay-b = c. 
 
 Solve for a : 
 
 19. 6 a = 18. 22. 3a + l = 13. 
 
 20. 2|a = 10. 23. 2 a- 5 = 15. 
 
 21. a + l=13. 24. 5 a — 6 = <?. 
 
 WRITTEN EXERCISES 
 
 Solve for x : 
 
 1. x- 75 = 136. 11. 325 a; + 60 = 400. 
 
 2. 2 a? -15 = 45. 12. 125 a; + 5 = 10. 
 
 3. 3.r+6 = 48. 13. 8 a? + 625 = 105. 
 
 4. 5 a; — 10 = 55. 14. ax + 6.c = c. 
 
 5. 1.5 a; — 5 = 70. 15. abx + cwc = a&. 
 
 6. 3.5 x — 5 = 100. 16. c;r + fte = c + d. 
 
 7. 2.1 as — 41 = 400. 17. mx + px = p + q. 
 
 8. 1.3 as + .1 = 1.79. 18. lx + mx = l — m. 
 
 9. .25 a; +.50 =3.25. 19. ax + bx = 2(a + b). 
 10. .11 x + .11 = 1.32. 20. 2cx + dx = l. 
 
 Solve and test : 
 
 21. 2(*-l)+3(2a; + 5)=0. 
 
EQUATIONS 29 
 
 22 . 4fe±a + 8_ 8 . + t 24 . »1±1_1=|1. 
 
 5 4 2/ — 2 4 — 4 ?/ 
 
 23. ^l + x ^l + ^zl=x. 25. l+3p sa 17-5p > 
 
 2 3 4 P P 
 
 Solve for c : 
 
 26. 5c + 3a = a C . 2g v==ct + 9*. 
 
 27. 
 
 30. 
 
 4 c - 1 2 c + 5 
 
 L' 
 
 3 m 6 m 29. (a + c)(a — c) =— (c + a) 2 - 
 
 3 C c -2 = (c-2)(2c-7) . ^ =z 
 
 c-3 c-4 c*-7c + 12 ' 1 + ct 
 
 32. An inheritance of $ 2000 is to be divided between two 
 heirs, A and B, so that B receives $ 100 less than twice what 
 A receives. How much does each receive ? 
 
 33. To build a certain staircase 20 steps of a given height 
 are required. If the steps are made 2 inches higher, 16 steps 
 are required. Find the height of the staircase. 
 
 34. In a certain hotel the large dining room seats three 
 times as many persons as the small dining room. When the 
 large dining room is f full and the small dining room ^ full, 
 there are 100 persons in both together. How many does each 
 room seat? 
 
 35. Tickets of admission to a certain lecture are sold at two 
 prices, one 25 cents more than the other. When 100 tickets 
 at the lower price and 60 at the higher price are sold, the total 
 receipts are $ 95. Find the two prices. 
 
 36. Originally, \^- of the area of Alabama was forest land. 
 One third of this land has been cleared, and now 20 million 
 acres are forest land. Find the area of Alabama in million 
 acres. 
 
 37. In a recent year the railroads of the United States 
 owned 70,000 cattle cars. Some of these were, single-decked, 
 and others double-decked. There were 44,000 more of the 
 former than of the latter. Find how many there were of each. 
 
30 ELEMENTARY ALGEBRA 
 
 38. The average number of sheep carried per deck is 45 
 larger thah the average number of calves. If a double-decked 
 car has the average number of calves on the lower deck and of 
 sheep on the upper deck, it contains 195 animals. Find the 
 number of sheep and of calves. 
 
 39. The average number of inhabitants per square mile for 
 Indiana is \ of that for Iowa, and that for Ohio is 32 greater 
 than that for 'Indiana, and 62 greater than that for Iowa. 
 Find the number for each state. 
 
 40. Lead weighs |4 times as much as an equal volume of 
 aluminium. A certain statuette of aluminium stands on a base 
 of lead. The volume of the base is twice that of the statuette, 
 and the whole weighs 282 oz. Find the weight of the statuette 
 and of the base. 
 
 41. A man inherits $10,000. He invests some of it in 
 bonds bearing 3^ °J interest, the rest in mortgages bear- 
 ing 51 % interest per annum. His entire annual income 
 from these investments is $ 510. Find the amount of each 
 investment. 
 
 42. A pile of boards consists of inch boards and half-inch 
 boards. There are 80 boards and the pile is 58 in. high. 
 How many boards of each thickness are there ? 
 
 43. How much water must be added to 30 oz. of a G °J solu- 
 tion of borax to make a 4 % solution ? 
 
 44. How much acid must be added to 10 quarts of a 2 °J 
 solution to make a 5% solution ? 
 
 45. A hardware dealer sold a furnace for $180 at a gain of 
 5 %. What did the furnace cost him. 
 
 46. A merchant sold a damaged carpet for $42.50 at a loss 
 of 15 °Jo- What did the carpet cost him ? 
 
 47. A collector remitted $475 after deducting a fee of 5%. 
 How many dollars did he collect ? 
 
 48. The amount of a certain principal at 4 % simple interest 
 for 1 year was $410. AVhat was the principal? 
 
EQUATIONS 31 
 
 EQUATIONS WITH TWO UNKNOWNS 
 
 72. Systems of Equations. Two or more equations con- 
 sidered together are called a system of equations. 
 
 73. Simultaneous Equations. Two or more equations are said 
 to be simultaneous when all of them are satisfied 'by the same 
 values of the unknowns. 
 
 74. All systems. of two independent simultaneous equations 
 of the first degree in two unknowns can be solved by the 
 method of addition and subtraction, which consists in multiplying 
 one or both of the given equations by such numbers that the 
 coefficients of one of the unknowns become equal. Then by 
 subtraction this unknown is eliminated, and the solution is 
 reduced to that of a single equation. 
 
 If the coefficients of one unknown are made numerically 
 equal, but have opposite signs, the equations should be added. 
 
 75. Occasionally the method of substitution is useful. • This 
 consists in expressing one unknown in terms of the other by 
 means of one equation and substituting this value in the other 
 equation, thus eliminating one of the unknowns. 
 
 This may be the shorter method when an unknown in either 
 equation has the coefficient 0, + 1, or — 1. 
 
 Examples of the use of these methods may be found in the 
 First Course, pp. 117, 119. 
 
 General Solution. A general form for two equations of tlie 
 first degree is 
 
 ax -\-by = e, (1) 
 
 cx + dy=f. (2) 
 
 From these it is possible (without knowing the values of a, 
 b, c, d, e, /) to find a general form for the solution, namely : 
 
 x= de -V 1 (3) 
 
 ad — be 
 
 J ad -be w 
 
32 
 
 ELEMENTARY ALGEBRA 
 
 These results are the formulas for the roots of any system 
 of two independent linear simultaneous equations with two 
 unknowns. 
 
 76. The application of these formulas is made easier by 
 noticing how they are formed from the known numbers in the 
 equations. 
 
 1 . The denominator is the same in each result and is made 
 up from the coefficients as follows : 
 
 Coefficients of x 
 a 
 
 Coefficients of y 
 b 
 
 2. The numerator of the value of x is made up thus 
 
 Absolute Terms 
 e 
 
 Coefficients of y 
 b 
 
 3. The numerator of the value of y : 
 
 Coefficients of x 
 a 
 
 Absolute Terms 
 e 
 
 Examples of the use of these formulas may be found in the 
 First Course, pp. 223-224. 
 
 The graphical representation of all the solutions of one equation in 
 two unknowns and the graphical solution of a system of two equations 
 in two unknowns may be reviewed at this point if desired. (See First 
 Course, pp. 170-173, 207-213, 232-234.) 
 

 
 EQUATIONS 
 
 „ , WRITTEN 
 Solve : 
 
 EXERCISES 
 
 1. 
 
 x + y = 5, 
 
 a; - ?/ = 3. 
 
 
 11. 
 
 4 x — 3 y = 3, 
 3 a; - 4 y = - 3. 
 
 ^2. 
 
 a; +■ y = 5, 
 a; - y = J . 
 
 
 12. 
 
 4:X + 2y = l, 
 3x-2y=*. 
 
 3. 
 
 2x- + y = 3, 
 a; + y = 2. 
 
 
 13. 
 
 12 a; - 11 y = 87 
 
 4 a; + 2 ?/ = 46 
 
 ^4. 
 
 a;-y = l. 
 
 
 14. 
 
 7 a; - 2 ?/ = 3, 
 7a;-4?/ = — 1. 
 
 5. 
 
 2x + 2y = 8, 
 
 2x-y = 2. 
 
 
 15. 
 
 9 cc — 3y = — 6, 
 
 8 a; - 2 y = - 6. 
 
 6. 
 
 3x — y = — 5, 
 2x-y = -3. 
 
 
 16. 
 
 oa; + y = l, 
 
 6a; + y = 2. 
 
 —■7. 
 
 4x-3y = 7. 
 3x-4y = 7. 
 
 
 17. 
 
 ax -\-by = c, 
 px + qy = tf. 
 
 8. 
 
 5x + y = 9, 
 3 x -f- y — 5. 
 
 
 18. 
 
 a; — m?/ = a, 
 a; + py = b. 
 
 ^,9. 
 
 4 a; + 5 y = 22, 
 3 x + 2 y = 13. 
 
 
 19. 
 
 ax — by = e, 
 ex — dy = f. 
 
 10. 
 
 a; - 5 y = - 22, 
 
 5# — y = 10. 
 
 
 20. 
 
 ax — y — b, 
 cx-\-y = d. 
 
 33 
 
 EQUATIONS WITH THREE OR MORE UNKNOWNS 
 
 77. The definitions and methods for the solution of two 
 equations with two unknowns may be applied equally well to 
 a greater number of equations and unknowns. 
 
 To solve three linear equations with three unknowns, elimi- 
 nate one unknown from any pair of the equations and the 
 same unknown from any other pair; two equations are thus 
 formed which involve only two unknowns and which may be 
 solved by methods previously given. 
 
 Four or more equations with four or more unknowns may 
 be solved similarly. 
 
34 ELEMENTARY ALGEBRA 
 
 WRITTEN EXERCISES 
 Solve and test : 
 
 1. 2x + 3y = 19, 7. 5x = 3y, 
 3x — 4:y = 3. 2x + Sy = ±. 
 
 2. 5x-2y = 46, 8 - ax-by = c,' 
 
 x + y = 12. cx + ay = b. 
 
 3. 8x + 7y = ll, 9 - 4a 2 .« + 5m/ = 3 5 
 2x-3y = 17. 6ax + 7y = 2. 
 
 10. ax + by = c, 
 
 4. 4 a,* — by = l, 2 i 1,2 2 
 
 ^ era: + 6-?/ = c\ 
 
 3 x - 5 y = 2. 
 
 5. 7 a; 4- 9?/ 4- 1 = 0, 
 
 11. a? 4- &?/=!, 
 
 3^ + 4^4 = 0. a V ~ ' 
 
 6. 17 a; 4-13 j/- 5 = 0, 12. fa + |?/ = i, 
 
 3aj + 6y + 3 = 0. f* + T 7 62/ = i- 
 
 13. 2(2x + 3y)- 5 ^ + 3) -^ = 9, a? + y = l. 
 
 8 4 
 
 14. 4?/ + 3(j/-a;-2)4-20a- = 0, 
 3(,_5)-2(. C -|) + 2 J = 0. 
 
 15. 4 a- 2 i/4-2 = 3, 18. 3x -2 2 + 5 = 0, 
 a 4- 3 ?/ 4- 2 z = 13, 2x4-3?/- 21=0, 
 -8x + 12y + z = 21. 4 1/ + 7 2 - 69 = 0. 
 
 16. 
 
 5 a; 4- 4 y + 2 2 = 17, 19- » + y==-i 
 
 
 3^-22/4-52 = 2, a 
 2a;-2/4-32 = 2. 2/ + z = i> 
 
 
 
 
 17. 
 
 x — y — z = a, 1 
 3 t/ — a; — z = 2 a, » + 2 = - • 
 72 — y — x = 4 a. 
 
 
 20. 
 
 • + y + z =2c, 
 
 a + b b — c a + c 
 
 
 x y z -*a 
 
 
 a — b 6 — c a — c 
 
 
 * , V 2 o o 
 
 2 = 2 a — 2 c. 
 
 a — 6 c — ?> « + o 
 
EQUATIONS 35 
 
 21. x — y — z — 2v:= — 12, 
 3x~y-2z + 8w = 4Q, 
 
 4 x — 4 ?/ + 7 z — 5 w = 52, 
 3 a: — y + 2 z + io = 44. 
 
 22. A merchant bought a certain number of platters for 
 $ 366. Three were broken during shipment. He sold \ of the 
 remainder at a profit of 25%, for $75. Find the number of 
 platters bought and the price per platter. 
 
 23. A certain hall contains both gas jets and electric lights. 
 When 60 gas jets and 80 electric lights are used, the cost for 
 an evening is $4. If 90 gas jets and 60 electric lights are 
 used, the cost is $ 4.05. Find the cost per gas jet and electric 
 light. 
 
 24. A tailor paid $12 for 4 yd. of cloth and 8 yd. of lining. 
 At another time he paid $21 for 6 yd. of the cloth and 16 yd. 
 of the lining. Find the price of each per yard. 
 
 25. A certain train runs 25 mi. per hour on the level, 15 mi. 
 per hour on up grade, and 30 mi. per hour on down grade. It 
 goes from A to B, 200 mi., in 8 hr. 48 min., and from B to A 
 in 9 hr. 12 min. How many miles are level, up grade, and 
 down grade respectively between A and B ? 
 
 26. Two wheelmen are 328 ft. apart and ride toward each 
 other. If A starts 3 seconds before B, they meet in 14 seconds 
 after A starts ; or if B starts 2 seconds before A, they meet in 
 14 seconds after B starts. Find the rate of each. 
 
 27. A man had a portion of his capital invested in stocks 
 paying 6 c / dividends, the remainder in mortgages paying 5 %. 
 His annual income was $ 700. The next year the dividend on 
 the stock was reduced to 5 %, but by reinvestment he replaced 
 his old mortgages by new ones paying 51%. His income for 
 this year was $ 690. How much had he invested in stocks ; 
 also in mortgages ? 
 
 28. The sum of the digits in a certain number of two figures 
 is 13, and if the result of multiplying the tens' digit by \\ 
 is added to the number itself, there results a number with the 
 same digits in reverse order. Find the number. 
 
36 ELEMENTARY ALGEBRA 
 
 QUADRATIC EQUATIONS 
 
 78. Quadratic Equations. Equations of the second degree 
 are called quadratic equations. 
 
 A general form for quadratic equations in one unknown is 
 
 ax 2 + bx + c = 0, 
 
 in which a, b, c represent any known numbers, except that a 
 may not be zero. 
 
 79. Solution of Quadratic Equations. (1) The incomplete 
 quadratic equation x 2 = a is solved by extracting the square 
 
 root of both members. The roots are : x = ± Va. 
 
 (2) The incomplete quadratic equation ax 2 + bx = is solved 
 
 by factoring. The roots are x = and x = 
 
 a 
 
 (3) Complete quadratic equations are solved by completing 
 the square. 
 
 The process consists of two main parts : 
 
 (a) Making the left member a square while the right member 
 does not contain the unknown. 
 
 This is called completing the square. 
 
 It is based upon the relation (x + a ) 2 = x 2 + 2 ax + a 2 , in which it ap- 
 pears that the last term, a 2 , is the square of one half of the coefficient of c. 
 
 (&) Extracting the square roots of both members and soloing the 
 resulting linear equations. 
 
 Square roots which cannot be found exactly should be indi- 
 cated. 
 
 
 EXAMPLE 
 
 
 Solve : 
 
 ^ _ 8 x + 9 = 0. 
 
 (0 
 
 Transposing, 
 
 a;2_8a;=-0. 
 
 (2) 
 
 Completing the square, 
 
 x 2 -8x+ 16 =-9 + 10. 
 
 (3) 
 
 Rearranging, 
 
 (x-4) 2 = l. 
 
 (4) 
 
 Extracting the square root, 
 
 x — 4 = ± V7. 
 
 (5) 
 
 Solving (/;) for x, 
 
 x — 4 ± V7. 
 
 (6') 
 
EQUATIONS 37 
 
 (4) If any quadratic equation has zero for the right member, 
 and if the polynomial constituting the left member can be fac- 
 tored, the quadratic is equivalent to two linear equations whose 
 roots can readily be found. (See First Course, pp. 186-187.) 
 
 ORAL EXERCISES 
 Solve : 
 
 1. « 2 = 16. 13. x 2 + 2 a; + 1 = 0. 
 
 2. r = 64. 14. y 2 -2y + l=0. 
 
 3. z 2 = 8. 15. z 2 - 3 2 + 2 = 0. 
 
 4. x 2 = a 2 . 16. 3x 2 = 6x. 
 
 5. if = a 2 b 2 . 17. (x- l)(x - 2) = 0. 
 
 6. 2 a? = 8. 18. x 2 - 5 £ + 6 = 0. 
 
 7. 3 f = 27. 19. (x - 7)(as + 1) = 0. 
 
 8. D2 2 = 125. 20. x 2 + a; + \ = 0. 
 
 9. 3* 2 = 75. 21. s 4 -16 = 0. 
 
 10. x(x-l) = 0. 22. t 2 -t + \ = 0. 
 
 11. a 2 + a; = 0. 23. £ _p 2 + j» + 1 = 0. 
 
 12. ?/ 2 -4 = 0. 24. z(ar-2z-3) = 0. 
 
 WRITTEN EXERCISES 
 Solve: 
 
 1. 3 or = 18. 3. x 2 -5a; + 6 = 0. 5. t 2 -2t-6 = Q. 
 
 2. x 2 — 5x = Q. 4. ar 9 +4a;-3 = 0. 6. Sjpr=zop. 
 
 7. ar 9 + ll.i' + 24 = 0. s = 9 .2x ±35 
 
 • 4 »- 8" 5 
 
 8. 15x 2 -13x + 5 = 0. 
 
 11. -J_+^L_ ±- = o. 
 
 9. 15/ + 134?/ + 288=0. y-3 1-y y-2 
 
 12. 7(7 - z)(z - 6) + 3(5 - 2) (2 - 2) - 40 = 0. 
 
 13. 
 
 6 — 2 mj 5 + mj w — 5 
 
 ^+ ; 
 
 io — 2 3 + (o 2 — to 
 
38 ELEMENTARY ALGEBRA 
 
 14. Find two numbers whose sum is 10, and the sum of 
 whose squares is 68. 
 
 Suggestion. Let x represent one number and 10 — x the other. 
 
 15. A room is 3 yd. longer than it is wide ; at $ 1.75 per 
 square yard, carpet for the room costs $49. Find the 
 dimensions of the room. 
 
 16. A man bought for $300 a certain number of oriental 
 rugs at the same price each. If he had bought rugs each 
 costing $40 more, he would have obtained 2 fewer rugs. How 
 many rugs did he buy ? 
 
 17. A dealer bought a number of similar tables for $153. 
 He sold all but 7 of them at an advance of $ 1 each on their 
 cost, thus receiving $ 100. How many tables did he buy ? 
 
 18. A man invested $6000 at a certain rate of simple 
 interest during 4 years. At the end of that time he reinvested 
 the capital and the interest received during the 4 years, at the 
 rate of interest 1 °J lower than at first. His annual income 
 from the second investment was $ 372. What was the original 
 rate of interest? 
 
 19. A rectangle of area 84 sq. in. is 5 in. longer than it is 
 wide. Find its dimensions. 
 
 20. A certain number of men hire an automobile for $156. 
 Before they start, two others join them, sharing equally in the 
 expense. The amount to be paid by each of the original hirers 
 is thus reduced by $13. How many men were there at first'.' 
 
 21. A man rows down a stream a distance of 21 mi. and 
 then rows back. The stream flows at 3 mi. per hour and the 
 man made the round trip in 13i hours. What was his rate of 
 rowing in still water ? 
 
 22. The product of a number and the same number increased 
 by 40 is 11,700 ; what is the number ? 
 
 23. If each side of a certain square is increased by 5 the area 
 becomes 64 ; what is the length of a side ? 
 
 24. Find two numbers whose sum is 16 and the difference 
 of whose squares is 32. 
 
EQUATIONS 39 
 
 REVIEW 
 
 WRITTEN EXERCISES 
 
 Solve : 
 
 1 a + 3_z-l 4. 2 x + 3(4 x -1) = 5(2 x +7). 
 
 aj — 1 a; + 3 
 
 2. ar -14 a; + 33 = 0. 2a; 
 
 ~ = 6. 
 
 3. a + -=c. 6. 14aj-5(2a;+4) = 3(aj+l). 
 
 x 
 
 1.. (5x- 2)(6 a; + l)-(10a; + 3)(3 a + 10) = 0. 
 8. (2 x - 3)(x + 1) - (3 x - 7)(x- - 4) = 36. 
 
 - 1 +_^ = -^_. 12. *=7, 
 
 a; + 2 x-2 2x — S # 
 
 2a?-10y = 3y + 2. 
 
 10. 2av+5y = 6, 
 
 4x' + ll?/ = 3. 13. 
 
 7y-6 _l 
 3a + 4 2' 
 
 11. 5 a; — 3y = l, 4y — 5 _1 
 
 13a;-8y = 9. 2a; + l _ 3' 
 
 ,. 3x- + 2i/ + 5 a?-3y-13 , y-3a; + 3 Q 
 14 ' 2 + 5 + '" 6 =S ' 
 
 2 x — 4 ?/ + 6 _ _r> 
 ?/ — 5 a; + 11 " 
 
 15. 6z = 43-5y, 16. ?~ 2 =-, 
 
 3 Z = 37-4a?, * + ? 5 
 
 . 4y = 55-5a;. ' *=* = * 
 
 ?/ + z 3 
 
 17. 3a; + 4y + 2s = -4, »-y_l 
 2 a? -5 y- z = 9, ^T z ~1' 
 -4a; + 2y + 3« = -23. 
 
 18. («-2)(a; + 3) = (a;-l)(z-l), 
 
 ( z + 8)(y-2)-(y + 2)(z.+ 2) = 0, 
 y(3-2aj) + (2y-3)(l+aj) = 0. 
 
 19. 
 
 3 x 2 + 14 , 1 + 3 a; 3a;_ n 
 
 2 a; + 10 7(4 -aj) 5 + a; 7 
 
40 ELEMENTARY ALGEBRA 
 
 20. In a certain election 36,785 votes were cast for the 
 three candidates A, B, C. B received 812 votes more than 
 twice as many as A ; and C had a majority of one vote over 
 A and B together. How many votes did each receive ? 
 
 21. In a certain election there were two candidates, A and 
 B. A received 10 votes more than half of all the votes cast. 
 B received 4 votes more than one third of the number received 
 by A. How many votes did each receive ? 
 
 22. A group of friends went to dine at a certain restaurant. 
 The head waiter found that if he were to place five persons 
 at each table available, four would have no seats, but by plac- 
 ing six at each table, only three persons remained for the last 
 table. How many guests were there, and how many tables 
 were available ? 
 
 23. A flower bed of uniform width, is to be laid out around 
 a rectangular house 20 ft. wide and 36 ft. long. What must 
 be the width of the bed, in order that its area may be one 
 third of that of the ground on which the house stands? 
 
 24. Wood's metal, which melts in boiling water, is made 
 up of one half (by weight) of bismuth, a certain amount of 
 lead, half that much zinc, and half as much cadmium as zinc. 
 How much of each in 100 lb. of Wood's metal ? 
 
 25. If in the preceding exercise three fourths as much cad- 
 mium as zinc is used, a metal is formed that melts at a still 
 lower temperature. How many pounds of each constituent 
 metal are there in 100 lb. of this metal ? 
 
 26. A cask contains 10 gal. of alcohol. A certain number 
 of quarts are drawn out ; the cask is then filled up with water 
 and the contents thoroughly mixed. Later twice as many 
 quarts are drawn out as the previous time and the cask filled 
 up with water. There now remains only 4.8 gal. of alcohol 
 in the mixture. How many gallons were drawn out the first 
 time? 
 
 27. A certain fraction not in its lowest terms has the value 
 \. If the numerator is diminished by 2, and the denominator 
 increased by 2, the value of the resulting fraction is the same 
 
EQUATIONS 41 
 
 as that which results when the numerator of the given frac- 
 tion is doubled and the denominator multiplied by 5. Find 
 the fraction. 
 
 28. A's investments amount to $4000 less than B's, and 
 C's amount to $6000 more than B's. A's average rate of 
 annual income from his investments is one half of 1 % more 
 than B's, and C's is one half of 1 % less than B's, and A's 
 annual income is $80 less than B's, and C's annual income 
 is $120 more than B's. Find the amount of each man's 
 investments, and each man's average rate, of income. 
 
CHAPTER III 
 
 RADICALS 
 
 DEFINITIONS AND PROPERTIES 
 
 80. Rational Numbers. Integers and other numbers expres- 
 sible as, the quotient of two integers are called rational numbers. 
 
 Thus, 5 and .2, which is expressible as j%, are rational numbers. 
 
 81. Irrational Numbers. Any number not rational is called 
 an irrational number. 
 
 Thus, \/2, V3, VlO, — , 1+V3, V2— \/3, 7r, are irrational numbers. 
 V5 
 
 82. An indicated root of any number is called a radical. 
 
 Thus, V5, y/8, \-r~i ^ a + x2 ' are radicals. 
 
 In the present chapter all roots that cannot be exactly extracted by 
 inspection ars indicated. Methods for finding approximate numerical 
 values of certain roots are given later. 
 
 83. Surd. An irrational number that is an indicated root of 
 a rational number is sometimes called a surd. 
 
 Thus, V2, Vo, \/T, are surds. 
 
 84. An expression involving one or more radicals is called 
 a radical expression. 
 
 Thus, 5 + 2V3, -^- — 1, + V _ a , are radical expressions. 
 y/x . 2 - V3 b 
 
 85. Some Properties of Radicals. A few important proper- 
 ties of radicals are given here. The fuller treatment is con- 
 tained in the chapter on Exponents. 
 
 42 
 
RADICALS 43 
 
 86. I. Vo • V& = Va6. 
 
 For example, y/2 • V3 = V6. 
 
 That this is true may be seen by squaring both members. 
 
 Thus, ( V2 • V3) ( V2 ■ V3) = V6 ■ a/6, 
 
 or, V2 • a/2 • a/3 • a/3 = a/6 • a/6, 
 
 or, 2-3 = 6, which is known to be true. 
 
 In the same way, it may be seen that for every a and ft, Va • Vb = Vab. 
 
 In words : 
 
 The product of tivo square roots is the square root of the 
 product of the numbers. 
 
 WRITTEN EXERCISES 
 Show by squaring that : 
 
 1. V3 • V5 = a/15. 5. V2a • V3& = V67T&. 
 
 2. a/4 • a/7 = V28. 6. Va? • V5y = a/5 afy. 
 
 3. a/3 • \/7 = V2T. 7. V2 ■ V3 • a/5 = a/30. 
 
 4. VB • a/IT= a/55. 8. Vet • V& • Vc = V«6c. 
 
 87. II. V^6= Va 2 Vb = ay/b. 
 
 In words : 
 
 Square factors may be taken from under the radical sign. 
 
 Thus, a/18 = V9 • a/2 = VP . V2 = 3 V2. 
 
 WRITTEN EXERCISES 
 
 Take all square factors from under the radical sign : 
 
 1. V20. ■ 5. a/45. 9. a/12. 13. V8~a 2 . 
 
 2. a/27. 6. V75. 10. a/40. 14. Vrf. 
 
 3. a/50. 7. a/24. 11. a/500. 15. V48^y. 
 
 4. a/48. 8. a/32. 12. a/128. 16. V46aV. 
 
44 ELEMENTARY ALGEBRA 
 
 88. III. a Vb = Va 2 ■ V& = Va 2 6. 
 
 In words : 
 
 Any factor outside the radical sign may be placed under the 
 radical sign provided the factor is squared. 
 
 For example : 
 
 3 V2 = V9 • V2 = Vl8. 
 
 WRITTEN EXERCISES 
 Place under one radical sign : 
 
 1. 6V2. 5. 3-V7-2. 9. 4V2-3. 13. tVg. 
 
 2. 5V3. 6. 5.V3-V2. 10. &V2. 14. rVirr. 
 
 3. 2-V3. 7. 2.V3-VH. 11. 2xVS^c. 15. fVl8ay. 
 
 4. 5-V7. 8. 5-V3-V7. 12. a6V6c. 16. aV&-a. 
 
 PROCESSES 
 
 89. Preparatory. 
 
 Read and supply the blanks : 
 
 1. 3a + 2a=( -)a. Similarly, 3V2 + 2V2 = ( )V'2. 
 
 2. 5a-3a=( )a. Similarly, 5V2-3V2 = ( )V2. 
 
 3. 7V3 + 3V3 = ( )V3. 4. 8V5-6V5=( )V5. 
 5. V75- V12 = 5V3-2V3 = ( )V3. 
 
 90. Addition and Subtraction of Radical Expressions. Radi- 
 cals can be united by addition or subtraction only when the 
 same root is indicated and the expressions under the radical 
 sign are the same in each. 
 
 When the expression cannot be put into this form the sum 
 or the difference can only be indicated. 
 
RADICALS 45 
 
 91. To add or subtract radical expressions having the same 
 radical part, add or subtract the coefficients of their radical parts. 
 
 For example : 
 
 2 V3 + 3 V3 = 5 V3. 
 
 2 Vl2 + V300 = 4 VS + 10 V3 = 14 V3. 
 
 Add y/2, - V8, \Vl6, \V^54 : 
 
 -V8=-2V2; ¥l6 = 2</2; v^54 = - 3 y/2. 
 
 .'. V2 - V8 + \Vl6 + V^54 = v / 2-2v / 2 + 2\/2-3v / 2 
 
 = _ V2-^. 
 
 If the numerical value of the sum or the difference is needed, it can be 
 found approximately by methods given later. 
 
 WRITTEN EXERCISES 
 Find the sum : 
 
 1. V2, V8, VI8. 8. V6, V24, V63. 
 
 2. V75, - V12, - V3. 9. V108, - Vl2, V48. 
 
 3. V8, V5, - V18. 10. Vt5, V48, - V27. 
 
 4. V128, -2V50, V72. 11. V80, V20, - \/45. 
 
 5. -^40, _ J/320, ^135. 12. VS, - V99, Vl21. 
 
 6. 8 V48, - 1 Vl2, 4 V27. 13. 5 V24, - V54, 3 V96. 
 
 7. V72, -3V9, 6V243. 14. V27x 4 , -V64x 4 , VlGoA 
 
 92. Multiplication of Radical Expressions containing Square 
 Roots. In multiplying expressions containing indicated square 
 roots, make use of the relation Va • V& = V«&. 
 
 EXAMPLES 
 
 1. 3-4V5 2. 2- V3 
 
 6 + 2V3 5 + 2V3 
 
 18-24V5 10-5V3 
 6V3-8VT5. 4V3-6 
 
 18-24v / 5 + 6v / 3-8Vl5. 10- V3 - 6 = 4 - a/3. 
 
4G ELEMENTARY ALGEBRA 
 
 , , . , WRITTEN EXERCISES 
 
 Multiply : 
 
 1. 2 + V5by2-V5. 7. 4 + V5byVlO. 
 
 2. 1+V3by2+V5. 8. 3-Vl5by2+VK 
 
 3. 2+V3by2 + V3. 9. 1+V2byl-VB. 
 
 4. V2 + V3 by 1 - a/3. 10. 2 V3 - 3 v'5 by V3 - VS. 
 
 5. V3-V5by V3 + V5. 11. V14 + V7 by V8 - V21. 
 
 6 . Vf> - V6 by V5 - V6. 12. V5 - a/48 by a/5 + Vl2. 
 
 93. Division of Square Roots. The quotient of the square 
 roots of two numbers is the square root of the quotient of the 
 
 Va _ ja 
 numbers. In symbols, —= — \ , " 
 
 V5 . |5 
 
 Thus, ^ = a , - , because, multiplying each member by itself, 
 V6 ^ 
 
 V6 V6 
 
 V0-V6 >6 ^ 
 
 or 
 
 
 or - = - , which is evidently true. 
 
 6 
 
 ORAL EXERCISES 
 
 Read each of the following as a fraction under one radical 
 sign: 
 
 1. VS. 3. ^*. 5. -!_ 7. -^t 
 
 Vo V7 V5 VIS 
 
 2 vi. 4 . j_ = yr. 6 . vs. 8 yjo. 
 
 V7 V2 V2 V8 V20 
 
RADICALS 47 
 
 94. Rationalizing the Denominator. Multiplying both numer- 
 ator and denominator of a fraction by an expression that will 
 make the denominator rational is called rationalizing the de- 
 nominator. 
 
 Thus, multiplying both numerator and denominator of by V'2, we 
 
 obtain "^ 
 
 V3 _ V2 • V3 _ V6 
 
 V2 V2 • V2 2 
 
 WRITTEN EXERCISES 
 Rationalize the denominator of: 
 
 1. 1 . 
 
 V2 
 
 4. 2 - 
 V3 
 
 7 V5 
 V7 
 
 10. 10 
 
 V5 
 
 2. JL. 
 
 V3 
 
 5. 3 - 
 
 V7 
 
 8. A. 
 V3 
 
 ii. A 
 V3 
 
 3. J-. 
 
 o. • 
 
 V3 
 
 9. A 
 
 V3 
 
 12. A 
 
 V2 
 
 95. Rationalizing Factors. When the denominator is of the 
 form Va + V& or a + V&, the rationalizing factor is the same 
 binomial with the connecting sign changed, often called the 
 conjugate binomial. 
 
 It is not necessary in elementary algebra to take up the rationalizing 
 of more complicated denominators. 
 
 EXAMPLES 
 
 3 
 
 1. Rationalize the denominator in 
 
 2-V5 
 The conjugate of 2 — Vh is 2 + Vb. 
 
 Then, 3 •=■ 3(2+V5) J + 3V5 = _ (fl + 3^ 
 
 2 - Vo (2 - V5) (2 + V5) ^ - 5 
 
48 ELEMENTARY ALGEBRA 
 
 2 + V3 
 
 2. Rationalize the denominator in 
 
 V3 + V5 
 The conjugate of V3 + Vb is V3 — VE. 
 
 Then 2+V3 (2 + \/3)(V3 -VI) _ 2>/3 + 3 - 2V5 - VT5 
 
 V3+V5 (•v / 3 + v / 5)(V3-V'5) 3 ~ 5 
 
 3 + 2V3-2V/5-V15 
 
 ~~ 2 
 
 WRITTEN EXERCISES 
 
 Rationalize the denominators : 
 
 ! 2 + VF . 5 5 9 3+V5 
 
 3+V3* V3+V7 3-V5 
 
 3V3 + 2V2 . 6 2-Vg 1Q 8-5V2 
 
 V3-V2 3-V5 3-2V2 
 
 1 
 
 3 n 2 + 4V7 
 
 1_V2 V5 + V2 2V7-1 
 
 3 g V5 + 2V2 , 12 2VT5-6 
 
 2-V5 " 4-2V2. V5+2V2 
 
 RADICAL EQUATIONS 
 
 96. To solve eqxiations in which only a single square root 
 occurs, transpose so that the square root constitutes one member. 
 Square both members and solve the resulting equation. 
 
 EXAMPLE 
 
 Solve: 2x-3=Vx 2 + 6x-6. (2) 
 
 Squaring both members, 4 X 2 — 12 X + 9 = X 2 + 6 X — 6. (£) 
 
 Collecting terms, 3 X 2 — 18 X + 15 = 0. (5) 
 
 Dividing both members by 3, X 2 — 6 X + 5 = 0. (4) 
 
 Solving (4), x = 5orl. (5) 
 
 Test. On trial, it appears that 5 satisfies the given equation, tak ing th e 
 radical as positive, while 1 satisfies the equation 2 x— 3 = — Vx 2 + 6 x — 0. 
 
RADICALS 49 
 
 1. It must be remembered that the equation resulting from squaring 
 will usually not be equivalent to the given equation (First Course, Sec. 
 232, p. 186). It may have additional roots, and trial must determine 
 which of the roots found satisfy the given equation. 
 
 2. In order that the given problem may be definite, the radical must 
 be taken with a given sign. If every possible square root is meant, two 
 different equations are really given for solution. Thus, unless restricted, 
 
 2 x = V4 — 6 x is a compact way of uniting the two different equations, 
 
 2 x = + V4 — 6 x, and 2 x = — V4 — 6 x. If solved as indicated above, 
 it appears that the first is satisfied when x = \, the second when x = — 2. 
 
 3. In the exercises of the following set the radical sign is to be un- 
 derstood to mean the positive square root. 
 
 WRITTEN EXERCISES 
 
 - Solve: 
 
 1. x = Vl0x + 7. 7. 30 = x - 29 Vx. 
 
 2. x = Vb + x - bx. 8. x = 2 + V'S-llx. 
 
 3. 3 x - 7 Vx = - 2. 9. x- VaT+2 = 3. 
 
 4. Vx + 5 - x = - 1. 10. 8 x + 1 = VxT3«. 
 
 Vr-1 
 
 a; 
 
 3 16 
 
 11. V100 - x 2 = 10 
 
 x. 
 
 6. x -(- 5 V37 - x = 43. 12. x + V2 s - x 2 = 6. 
 
 SUMMARY 
 I. Definitions. 
 
 1. Rational numbers are integers and other numbers ex- 
 pressible as the quotient of two integers. Sec. 80. 
 
 2. An irrational number is any number not rational. 
 
 Sec. 81. 
 
 3. A radical is an indicated root of a number. Sec. 82. 
 
 4. A surd is an irrational number that is an indicated root 
 of a rational number. Sec. 83. 
 
 5. A radical expression is an expression involving one or 
 more radicals. . Sec. 84. 
 
50 ELEMENTARY ALGEBRA 
 
 II. Properties and Operations. 
 
 1. The product of two square roots is the square root of the 
 product of the numbers. Sec. 86. 
 
 2. Square factors may be taken from under the radical sign. 
 
 Sec. 87. 
 
 3. Any factor outside the radical sign may be placed under* 
 the radical sign provided the factor is squared. Sec. 88. 
 
 4. Radical expressions whose radical parts are the same may 
 be added or subtracted by adding or subtracting the coefficients 
 of their radical parts. Sec. 91. 
 
 5. The quotient of two square roots is the square root of the 
 quotient of the numbers. Sec. 93. 
 
 6. Multiplying both numerator and denominator of a frac- 
 tion by a factor that will make the denominator rational is 
 called rationalizing the denominator. Sec. 94. 
 
 7. Radical equations involving only a single square root are 
 solved by transposing so that the square root constitutes one 
 member, squaring and solving the resulting equation. Sec. 96. 
 
 
 
 REVIEW 
 
 
 Simplify : 
 
 WRITTEN EXERCISES 
 
 1. 
 
 V5 
 V60 
 
 4. — - • 
 
 V40 
 
 2. 
 
 v 8 v 8" 
 
 5. V50+V128. 
 
 3. 
 
 V6 • V125. 
 
 6. VS-V5. 
 
 7> 3V3 + 2V2 
 V3-V2 
 
 8. V6h-V2. 
 
 9. (2 + V3) 2 . 
 
 10. (5 + V7)(5-V7). 11. (2 V3 + 3V5)WI5. 
 
 12. (V6 + V15)(V8-V20). 
 Solve : 
 
 13. Vx+5==x — 7. 
 
 16. x + ^Jx + 5 — 2x — 1. 
 
 14. V 2.r + 7 = |. 17. .-2 + V2~; = 0. 
 
 , e a-10 . x-1 , V2.X--1 .. 
 
 15. —• =U 
 
 2 3 2 18. x+Vd-x-=A. 
 
RADICALS 51 
 
 SUPPLEMENTARY WORK 
 
 ADDITIONAL EXERCISES 
 
 Express with rational denominators, and with at most one 
 radical sign in the dividend : 
 
 1. 
 
 •V12-5- V3. 
 
 2. 
 
 V7- vn. 
 
 3. 
 
 2V24-5-2V6. 
 
 4. 
 
 2 -=- 3 V5. 
 
 X, 
 
 4 
 
 
 V5-1 
 
 6. 
 
 l-:-(V2-10).' 
 
 7. 
 
 V2 - (V2 - V3). 
 
 8. 
 
 (2V6 + 5V12)- V6. 
 
 9. 
 
 (5 V18 - 8 V50) -5- 2 V2 
 
 10. 
 
 11. 
 
 12. 
 
 13. 
 
 14. 
 
 8-5V2 
 3-2V2* 
 
 3 + V5 
 3- V5 
 
 V3 _ 
 
 V5-V3 
 
 2.+ 4V7 
 2V7-1 
 
 4V7+3V2 
 5V2 + 2V7* 
 
 Square Root of Binomials of the Form a +- V6 
 
 Binomials of the form a + V& can often be put into the form 
 x + y + 2 %/.»?/, or ( V.c + Vj/) 2 , and hence the square root, 
 V*c + Vy, of the binomial can be written at once. 
 
 EXAMPLES 
 
 1. Find the square root of 4 + 2V3. 
 
 4 + 2VS = 3 + 1 + 2V3TT. 
 Hence, a; + ?/ =.3 + 1 and a;?/ = 3 -'1, from which x = 3 and y = 1. 
 
 .-. V4 +2V3= ± (V3 + VI) = ± (V3 + 1). 
 
 The coefficient of the radical must be made 2 in order to apply the 
 formula x + y +'2\'xy. 
 
52 ELEMENTARY ALGEBRA 
 
 2. Find the square root of 3 — V8. 
 
 3 -V8 = 3 - V4T2 = 3 - 2 V2. • 
 . ■. x + y = 3, and xy = 2. ,\ a; = 2, y = 1 by inspection. 
 
 .-. V3-V8= ±(V2-Vl) = ±(\/2 - 1). 
 
 3. Find the square root of 7 + 4 V3. 
 
 7 +4V3 = 7 + 2vTT3; x + y = 7, xy = 12; .-. x = 4, ?/ = 3. 
 
 .-. V7 + 4V3 = ± (VT + V3) = ± (2 + V3). 
 
 The square root as a whole may be taken positively or negatively, as in 
 the case of rational roots. 
 
 The solution of these problems depends upon finding two numbers 
 whose sum and product are given. This can sometimes be done by 
 inspection, but the general problem is one of simultaneous equations. 
 See Sec. 168, p. 132. 
 
 WRITTEN EXERCISES 
 Find the square root of : 
 
 1. 11 + 6V2. 4. 41 - 24V2. 7. 17 + 12V2. 
 
 2. 8-2V15. 5. 21 -V5. 8. IV5 + 3 1 , 
 
 3. 49 - 12 V10. 6. 2| - |V3. 9. 56 - 24 VE. 
 
CHAPTER IV 
 
 EXPONENTS 
 
 LAWS OF EXPONENTS 
 
 97. Preparatory: 
 
 1. What is the meaning of a 2 ? Of « 4 ? Of a 7 ? Of a 11 ? 
 Of a" ? 
 
 2. What is the meaning of Vo 2 ? Of tya??. Of K/a™? 
 Of a/^'? 
 
 3. « 2 . « 3 = ? a 3 • a 2 = ? a 5 • a 3 = ? a 8 • a 5 = ? 
 
 4. a 3 -j-a 2 =? a 4 -s-a 2 =? a s -»-a 2 = ? b w ---b i =? 
 
 5. (a 2 ) 2 = ? (a 2 ) 3 =? (c 5 ) 2 =? (V°) 3 =? 
 
 98. We shall soon define negative and fractional exponents, 
 but until this is done literal exponents are to be understood to 
 represent positive integers. 
 
 99. Law of Exponents in Multiplication. 
 I. a m - a r = a m+r . 
 For a m = a ■ a ■ a • • • to m factors, 
 
 and a r = a ■ a ■ a ■■■ to r factors. 
 
 .-. a m • a r — (a • a ■ a • •• to m factors)(a .a • a ••• to r factors) 
 = a ■ a ■ a ■ a • • • to m + r factors 
 = a m+r , by the definition of exponent. . 
 Similarly, a m • a r ■ aP •■■ - a m+r+ P •■-. 
 
 Multiply: 0RAL EXERCISES 
 
 1. a 2 -a\ 4. m l -m 5 . 7. (-l) 3 -(-l) 5 . 10. 2 3 • 2 3 • 2 2 . 
 
 2. a 3 -a\ 5. x>x. 8. 6 2 • 6 2 . 11. 7 • 7 2 • 7 ■". 
 
 3. a 5 -a 7 . 6. 2 3 -2 4 . 9. 5 • 5 • 5 2 . 12. 3-3 5 -3 2 . 
 13. (_l) 2 .(-l) 3 .(-l)l 14. (_«) 2 . (-«)<•(-«). 
 
54 ELEMENTARY ALGEBRA 
 
 100. Law of Exponents in Division. 
 
 II. — = a m - r , if m > r. 
 
 a r 
 
 For a m = a • a • a •■• in factors, 
 
 and a T = a ■ a ■ a ■■■ r factors. 
 
 a m _ a • a ••• m factors 
 a r a ■ a ■■• r factors 
 
 = a ■ a ■ a ••• in — r factors, canceling the r factors from both 
 terms 
 
 = a m ~ r , by definition of exponent. 
 Divide: 0RAL EXERCISES 
 
 1. 
 
 a 4 
 
 a 2 ' 
 
 5. 
 
 a 8 
 
 'a 5 
 
 9. 
 
 6 3 
 6 2 ' 
 
 13. 
 
 x*y m 
 
 X 
 
 2. 
 
 a 5 
 a 3 ' 
 
 6. 
 
 (-a) 5 , 
 (-a) 
 
 10. 
 
 5 4 
 5 3 ' 
 
 14. 
 
 V 
 
 3. 
 
 a 7 
 a 5 ' 
 
 7. 
 
 (-1) 5 . 
 (-1) 3 
 
 11. 
 
 4r 
 
 15. 
 
 . t 
 
 4. 
 
 2 4 
 2 2 ' 
 
 8. 
 
 (a6) 2 
 (aft) 
 
 12. 
 
 X 
 
 16. 
 
 TT'fi 
 
 r 
 
 101. Laws of Exponents for Powers. 
 
 III. (a m y = a mr . 
 
 For (a m ) r = a m • a m • a m ■•• to r factors 
 
 = (a • a •■■ to m factors)(a • a ••• to m factors) to 
 r such parentheses 
 
 = a • a- a ■■■ to mr factors 
 
 = a""', by definition of exponent. 
 
 ORAL EXERCISES 
 
 Apply this law to : 
 
 1. (4 3 ) 2 . 4. (« 3 ) 2 . 7. O 2 ) 5 . 10. [(ft) 4 ] 5 . 
 
 2. (3 2 ) 5 . 5. (a 2 ) 3 . 8. (a 3 ) 8 . 11. [(-a) 5 ] 2 . 
 
 3. (-/. 6. (a 5 ) 2 . 9. (</'/. 12. [(-b/j ; . 
 
EXPONENTS 55 
 
 IV. (ab)» = a"b". 
 
 For (ab) n = (nb) • (ah) ••• to n factors 
 
 = (a • a ••• to n factors)(6 • b ••• to n factors) 
 = a»l>'\ by definition of exponent. 
 
 Similarly, (abc •••)" = a"b n c i •••. 
 
 ORAL EXERCISES 
 
 Apply this law to : 
 
 1. (8 • 3) 2 . 4. (ab) 5 . 7. {mn)p. 10. (afy) 3 . 
 
 2. (4 • 5) 2 . 5. (cdf. 8. (.>•>,)*. 11. (Vy 2 ) 5 . 
 
 3. (2-5) 3 . 6. (abcy. 9. (a&f. 12. (aa 8 ) 6 . 
 
 V. 
 
 ?Y — — 
 
 6/ "6" 
 
 (f)"= 
 
 For = - • - • • • to n factors 
 
 b b 
 
 a • a ■■■ to n factors 
 b • b ••• to n factors 
 
 = — , by definition of exponent. 
 b n 
 
 ORAL EXERCISES 
 
 Apply this law to : 
 
 i- ay- . ftY. 12 . 
 
 (f- 
 
 2- (i) 3 - ' 
 
 3- (I) 5 - 9. f C X 13. ^Y 
 
 *• (t) 4 ' 
 
 6 
 
 5- ("f) 2 - in W. 14. U 
 
 6- (-f) 
 
 3\3 
 
 y 
 
 5d 
 
 7 
 
 11. W 15. 
 
 n 
 
5G ELEMENTARY ALGEBRA 
 
 102. Collected Laws of Exponents. 
 
 I. a m • a r = a m+r . Sec. 99. 
 
 II. « m -v- a r = a m - r . (m > r.)' Sec. 100. 
 
 III. (a m ) r = a mr . Sec. 101. 
 
 IV. (ab) n = a"b n . Sec. 101. 
 
 V f«V = -- Sec. 101. 
 
 \jbj b n 
 
 FRACTIONAL EXPONENTS 
 
 103. Hitherto we have spoken only of positive integers as 
 exponents, the exponent meaning the number of times the 
 base is used as a factor. This meaning does not apply to 
 fractional and negative exponents, because it does not mean 
 anything to speak of using a as a factor f of a time, or — 6 
 times. But it is possible to find meanings for fractional and 
 negative exponents such that they will conform to the laws of 
 integral exponents. 
 
 104. Preparatory. 
 
 Find the meaning of a 2 . 
 
 i i i + i 
 Assuming that Law I applies, a? • a % = a? * = a, 
 
 or, (a*) 2 = a. 
 
 That is, a? is one of the two equal factors of a, 
 
 or, a 2 = va. 
 
 Thus, the fractional exponent \ means square root. 
 
 ill i+i+i 
 Similarly, a* • a* • a 3 = a 3 3 3 = a, 
 
 i 
 or, (a 3 ) 3 = a. 
 
 That is, a 3 is one of the three equal factors of a, 
 
 or, a*= Va. 
 
 Thus, the fractional exponent £ means cube root. 
 
EXPONENTS 57 
 
 WRITTEN EXERCISES 
 Find similarly the meaning of : 
 
 1. dfl, 3. X*. 
 
 
 
 5. ri*. 7. 
 
 i 
 vnJ. 
 
 2. </■■. 4. &A- 
 
 
 
 6. c*. 8. 
 
 a?\ 
 
 1 
 105. The meaning of a» 
 
 is 
 
 found as follows : 
 
 
 Assuming that Law I applies 
 
 » 
 
 
 
 
 i i i 
 
 a n • a 11 • a" ••• to n 
 
 factors 
 
 - -1 1 1- - • • to n terms. 
 
 — qU n n 
 
 1 
 n • — 
 
 = a » 
 
 
 
 
 
 
 »1 
 
 = a 1 — a. 
 
 l i 
 
 That is, a n is one of n equal factors of a, or a" = Va. 
 
 106. Preparatory. 
 
 2 
 
 Find the meaning of a?. 
 
 2 2 2 fi 2 
 
 Assuming that Law I applies, a 5 • cfi • a* = a? = a 2 , or (a?) 3 = a 2 . 
 That is, a? is one of the three equal factors of a 2 , or a T = Va 2 . 
 
 p 
 
 107. The meaning of a q is found as follows : 
 
 Assuming that Law I applies, 
 
 '-' 2 P M P + E + P... to « terms 
 
 a « . a? . a« ••• to g factors = a « « « 
 
 p 
 
 9 ■ - 
 
 — a « = a?. 
 
 That is, a* is one of the q equal factors of up, 
 
 p 
 
 or, ai=VaP. 
 
 P -. V 
 
 Similarly, • ai =a" ' P = (Va)P. 
 
 In words : 
 
 a with the exponent -- denotes the qth root of the pi/i power of 
 a, or the \>th power of the ([th root of&. 
 
58 ELEMENTARY ALGEBRA 
 
 This definition applies when p and q are positive integers. The mean- 
 ing of negative fractional exponents is found in Sec. 116, p. 64. 
 
 ORAL EXERCISES 
 
 1. State the meaning of ; 
 
 1. 5 3 . 3. 6*. 
 
 5. 4i 
 
 7. 8*. 
 
 2. a*. 4. ai 
 
 6. 6*. 
 
 8. c\ 
 
 Find the value of : 
 
 
 
 9. 8s. io. 16*. 
 
 11. 25* 
 
 12. 32*. 
 
 WRITTEN EXERCISES 
 Express with fractional exponents : 
 
 1. </a*. 
 
 2. Va 2 . 
 
 3 / * 
 
 3. vraw. 12 
 
 4. VaV&. 
 
 5. Va&. 
 
 s/32 « 5 6 8 
 
 13. ^33. 
 
 6. -Vatfxy. e/64* 5 
 
 14. 
 
 3/o; 4 w 2 
 
 >■ # 
 
 .'/ 
 
 15. Va + &. 
 
 \-— • 16. Va 3 +6 3 . 
 
 9. Vl6 ^. 17. VP • Va. 
 
 19. 
 
 Va. 
 
 20. 
 
 V(«-6) 2 . 
 
 21. 
 
 Va-Vo. 
 
 22. 
 
 Va™ 
 
 23. 
 
 VcT». 
 
 24. 
 
 W\ 
 
 25. 
 
 Va m ". 
 
 26. 
 
 2 "Va 2 ™. 
 
 27. 
 
 "Va»6' w . 
 
 28. 
 
 Va • Va. 
 
 29. 
 
 Vaoc. 
 
 30. 
 
 Va' • Va7 
 
 10. VmWVp. 18. V-a-V-6. 
 
 108. The definition of positive fractional exponents has been 
 found as a consequence of the assumption that Law I applies 
 to them. It can be shown that the other laws of Sec. 102, 
 p. 56, also apply to this class of exponents, as thus defined, 
 and we shall so apply them, although the proof is omitted here. 
 
EXPONENTS 59 
 
 109., According to Law I (Sec. 99), when the bases are the 
 same, the exponent of the product is found by adding the exponents. 
 
 1 3 1.8 5 
 
 For example : a 2 ■ a 4 " = a- * = a 4 ". 
 
 A general formula for this statement is, 
 
 m p mq np mq+np 
 
 a" • a q = a" q nq — a m • 
 The number a, or the base, must be the same in all factors. When it 
 
 1 3 
 
 is not, as in a 2 • 6 ¥ , the product cannot be found by adding the exponents. 
 
 WRITTEN EXERCISES 
 
 
 
 Find the products : 
 
 
 
 
 1. a$ • ai 6. 
 
 a' 3 • a* 
 
 11. 
 
 i i 
 
 x a • x b . 
 
 2. 4 2 -4l 7. 
 
 3 3 
 
 m* • m 3 . 
 
 12. 
 
 m p 
 
 x n • x n . 
 
 3. 7*. 7*. 8 - 
 
 a^ • a?. 
 
 
 
 4. a 2 " • ai 9. 
 
 i i 
 
 13. 
 
 2 1 
 pi . pZ . p. 
 
 5. &* . &i. io. 
 
 1 1 
 
 a" • a" 1 . 
 
 14. 
 
 4 3 1 
 
 a 5 • a 4 • a 2 ". 
 
 110. According to Law II (Sec. 102), when the bases are the 
 same, the exponent of the quotient is found by taking the difference 
 between the exponents. 
 
 1 1 2_1 1 
 
 For example : a 3 -s- a 2 = a 3 . 2 = a*. 
 
 a h- a* = a * = a*. 
 A general formula for this statement is, 
 
 m p m p mq np 
 
 a n ^-a q = a" 1 =a ng . 
 
 - is here supposed to be greater than -, but this restriction will 
 n q 
 
 be removed later. 
 
GO ELEMENTARY ALGEBRA 
 
 WRITTEN EXERCISES 
 Find the quotients : 
 
 1. 
 
 3 1 
 
 7. 
 
 3 2 
 
 12. 
 
 
 2. 
 
 a -h a T2 . 
 i 
 
 8. 
 
 2 1 
 
 x 5 -:- a; 7 . 
 
 13. 
 
 i i 
 
 m a -7- m b . 
 
 3. 
 
 a -i- a n . 
 p 
 
 9. 
 
 a&-j-(a£) 7 . 
 
 14. 
 
 „2 1 
 
 6* -=-6*. 
 
 4. 
 
 5. 
 
 a -i- a 7 . 
 
 4 3 
 
 a 5 -s- a*. 
 
 10. 
 
 W " W 
 
 15. 
 
 3 1 
 
 m 8 -7- m' 
 
 6. 
 
 1 2 
 
 X B -i-X s . 
 
 11. 
 
 •> i 
 5 3 -=-5 2 . 
 
 16. 
 
 4 3 
 
 J9 ? H-p 4 . 
 
 111. According to Law III (Sec. 102), when an exponent is 
 applied to a base having an exponent, the product of the exponents 
 is the exponent of the result. 
 
 i 2.1 
 
 For example : (a 2 )' 2 = a 2 — a x = a. 
 
 1 2.1 2 1_ 
 
 («2)4 = a ^ = a l =a^. 
 
 12 1.2 _2 
 
 l3\~5 — n.S 5 — ,7 13. 
 
 (a 3 ) 5 = a 5 
 
 A general formula for this statement is, 
 
 m V- m p mp 
 
 (a») q =a n '* = «"*. 
 
 ~. ,.„ ORAL EXERCISES 
 
 Simphty : 
 
 1. (2*)*. 5. (b^)K 9. (at*)*. 13. (a*)* 
 
 2. (3*) 2 . 6. (a*)i 10. 0/ 9 )*. 14. («*) 4 . 
 
 3. (3*) 3 . 7. (a^)i 11. (6*)*. 15. (10*)*. 
 
 4. (5*) 2 ; 8. (cfyk 12. (3*)i 16. [( a+ &)£]& 
 17. [(a-byy. 18. [K-6; ! j'. 19. [(.e"->/>> 
 
EXPONENTS 61 
 
 112. According to Law IV (Sec. 102), an exponent affecting 
 
 a product is applied to each factor, and according to Law V 
 
 (Sec. 102), an exponent affecting a fraction is applied to both 
 
 numerator and denominator. 
 
 i ii 
 For example : (ab) 2 — a 2 b^. 
 
 (abh 2 ) 2 ' = ahh. 
 
 (8 x 6 y 2 z) 3 - 8*x*ySz* = 2 x 2 y^z^. 
 
 m p mp mp p 
 
 (a m b n c)i = a i ■ b'"> • c«. 
 
 (y s ) f 2/ f 
 
 A general formula for this statement is 
 
 mr ' pr 
 
 5 
 
 n 
 
 mr ' vr ~ 
 
 a" 
 
 — = a ns h- b qs , or 
 
 p J > in_ 
 
 b q " 
 
 WRITTEN EXERCISES 
 
 Simplify : 
 
 1. 
 
 (a?b*)k 
 
 2. 
 
 (arty. 
 
 
 1 
 
 3. 
 
 (a m b n ) p . 
 
 4. 
 
 (x^y^y. 
 
 5. 
 
 (a 3 6-f) 2 
 
 7. ($Qx«y)i ii. (tt\. 
 
 {32 a 5 b w y 
 
 [~~^~J ' 12 - ( 16 A 8 )*- 
 
 /G4^ 12 \* 13 - (« 6 ^ 915 ) 3 - 
 
 y 
 
 14. (27a}Wy. 
 
 i 
 
 f9 aW 
 
 6. (a* • 6*) 8 . 10 " V « 7 ' IB- (mW) M - 
 
62 ELEMENTARY ALGEBRA 
 
 113 When the bases are different and the fractional expo- 
 nents are different, the exponents must have a common denomi- 
 nator, before any simplification by multiplication or division 
 is possible. 
 
 For example : a~ 2 b 3 - a ? 6 5 = («^» 2 )*. 
 
 A general formula for this statement is, 
 
 m p mq np 1 
 
 a "&« = aF 9 • b" 9 = (a mq . b"")" 9 . 
 
 This is called simplifying by reducing exponents to the same 
 order. 
 
 WRITTEN EXERCISES 
 
 Simplify by reducing the exponents to the same order : 
 
 1. 
 
 1 ,i 
 a 2 • b 3 . 
 
 5. 
 
 1 ,3 
 
 a 3 • &*. 
 
 9. 
 
 lii 
 
 2. 
 
 2 . 3 
 
 a 3 • &*. 
 
 6. 
 
 &* • 6*. 
 
 10. 
 
 2 3 1 
 
 m 3 • n 5 -i-p*, 
 
 3. 
 
 i , i 
 ■v* • b 3 . 
 
 7. 
 
 5* • &*. 
 
 11. 
 
 i ii 
 
 W n ■ Q m • It. 
 
 4. 
 
 i , i 
 
 8. 
 
 i , i 
 a? -=- &*.■ 
 
 12. 
 
 p m 1 
 
 x 9 • y n ■ z n . 
 
 114. It is usually preferable to indicate roots by fractional 
 
 exponents instead of by radical signs, since operations are thus 
 
 more easily seen. 
 
 COMPARISON 
 By Radicals By Exponents 
 
 1. Vffl y/a = VopVa? = y/a 3 a' z — VaK a^a* = a*a* = a* + 6 = s . 
 
 WRITTEN EXERCISES 
 
 Simplify by use of fractional exponents as in the examples 
 above : 
 
 l. 22 • 5*. 3. VSla^. 5. 3* .2.7*. 
 
 2. Vy 4 ^-. 4. V5 • VT5. 6. 3 • 5* ■ 2V3. 
 
EXPONENTS 63 
 
 7. V8-3V2. 21. (16 x 4 jj)K 34 /49a 4 ^ 
 
 ' V04 6 6 
 
 8. 2V3 -3 • V10. 22 V32a a & 10 . 
 
 9. 2a/3-3v/2. oq , ,,,. 4 2 
 
 23. Vob 7/r/r. 
 
 35. V a 8 V25&. 
 
 36. ^x 2 y^/2oyz. 
 
 10. 52 • 3 • 5i nA ,— — — . 
 
 24. Vb4 mV, 6 
 
 11. V7 • 11* ' . 37 - Va ^- 
 
 25. (49a 4 ^. 
 
 12. -8V2-12V3. 38. V^Irf. 
 
 26. a/2o¥ 2 . _ 
 
 13. -VI2-2V3. 39. V</8^. 
 
 27. V32 • V2. 
 
 14. aVb-bVa. 40. V^/sTa 4 . 
 
 28. (a 2 bVxJ 2 f. 
 
 15. (V2-V3)2V3. 41. V^^9¥ 2 . 
 
 *,* „ ^ 29 - ( m3 ^ 7 ) 4 ' 
 
 16. 3^(6^ - 2 • 5*). 42. V^^/27 a 3 . 
 
 3A - 30. V6a& • V2a. 
 
 17. (a-Vxy. 43 _ a (a 2 6)^.6(a^)i 
 
 ,„ /-en 31. [2a(4a 2 )*] 2 . t „ 
 
 18. Va 6 6 8 . L J 44. 3V9 a- 3*. 
 
 19. v8a°^ fa . 45. 5 v a¥ • 2 Vacc. 
 
 20. (27 ay 2 )*. 33. Vm 2 ^/ m ! 46 . 3 V8 • 2 ^6 • 3 </54. 
 
 MEANING OF ZERO AND NEGATIVE EXPONENTS 
 
 115. The meaning of a n may be found as follows: 
 
 Assuming that Law I holds for a , a 5 • a = a 5+0 
 
 = a 5 . 
 
 a 5 
 Dividing by a 5 , a = — = 1. 
 
 « s 
 That is: 
 Any number (not zero) with the exponent zero equals 1. 
 
 Thus, 5" = 1, 10° = 1, (Y)° = 1, 
 
5. 
 
 (-3)°. 
 
 10. 
 
 \ ■ a . 
 
 15. 
 
 G)°- 
 
 6. 
 
 (*)•■ 
 
 11. 
 
 a 5 a°. 
 
 16. 
 
 .9 -s- 100 n . 
 
 7. 
 
 (-^r)°- 
 
 12. 
 
 a m ■ b°. 
 
 17. 
 
 9i • 5 n . 
 
 8. 
 
 3 2 • 3 n . 
 
 13. 
 
 a 5 -=- a . 
 
 18. 
 
 OO 9J0 
 
 9. 
 
 3 2 • 5°. 
 
 14. 
 
 4 3 --4°. 
 
 19. 
 
 (2 3 • 3 2 )°. 
 
 G4 ELEMENTARY ALGEBRA 
 
 „...,- ORAL EXERCISES 
 
 State the value oi : 
 
 1. (a b)°. 
 
 2. .9°. 
 
 3. 100 n . 
 
 be 
 
 116. The meaning of the negative exponent may be found 
 as follows : 
 
 Assuming that Law I holds for negative exponents, 
 
 5-3 . 5+3 --5-3+3- 50-1. 
 
 That is, 5~ 3 is a multiplier such that its product with 5+ 3 is 1. But if 
 the product of two numbers is 1, one is the reciprocal of the other. 
 
 Therefore, 5 -8 is the reciprocal of 6 8 which is — . 
 
 6 
 Expressed in general terms : 
 
 a~ n ■ a n - «-"+"■ 
 
 = «° = 1. 
 
 In words : 
 
 a~ n means —, for all values of n, positive or negative, inte- 
 a n 
 gral or fractional. 
 
 ORAL EXERCISES 
 Find similarly the meaning of : 
 1. 4" 3 . 3. (f)- 2 . 5. a- 3 . 7. a)~l 
 
 3 
 
 2. 2-\ 4. (-5)" 6 . 6. a"*. 8. (-f) • 
 State the value of : 
 
 9. 4"~*. 13. 16 _ L 17. 32~*. 21. or 3 . 
 
 10. 8"*. 14. 16~*. 18. 2T*. 22. (a )" 2 . 
 
 -._?. 
 
 .3 
 
 11. 8~*. 15. .125"*. 19. .36 * 23. a •. 
 
 12. (.2°)^ 16. .125"i 20. 64~l 24. a"5. 
 
EXPONENTS 65 
 
 WRITTEN EXERCISES 
 
 Perform the operations indicated : 
 
 1. (4 2 -5 4 )~i . 4. 2"*-2i 7 
 
 „_1 „3 
 
 2. 2 14 -2- 6 . 5. 2T*-2T*. 
 
 5° 
 8. 10- 5 • 10 2 • 10 n . 
 
 3. 2*.2~* 6. 3«.3"2.4°. 9. 10 r -10- 7 . 
 
 10. 27"*- 9*. 16. (a- 4 ) 2 . 17. a°-aK 
 
 11. -^18^.^18^. is. i!( ra-^H)). 
 
 12. -v'io^ 4 • ^lol a / , , 
 
 2 19. a 2 -a 2 (or a 2 2 ). 
 
 13. (10- 4 • io- 3 y. 1 
 
 20. ^(or«H4)). 
 a -2 
 
 15. 10 2 • 10* • 10* • (A) . 21. (<T*)~* 22. (x-*) s 
 
 14. 10'* ■ 10= 
 
 USE OF ZERO, NEGATIVE, AND FRACTIONAL EXPONENTS 
 
 117. We have defined zero and negative exponents so that 
 Law I holds for them. It can be shown that the other four 
 laws hold for these exponents as defined, but the laws will be 
 applied here without proof. 
 
 118. The relation a~ n • a n = a = 1 can be m used to change the 
 form of expressions. 
 
 I. To free an expression from a negative exponent, multiply 
 both numerator and denominator by a factor that will so combine 
 with the factor having the negative exp>onent as to produce unity 
 in accordance with the relation just mentioned. If more than one 
 negative exponent is involved, apply the process for each. 
 
 73 . 7- 3 • 2* 2* 
 
 For example : 7~ 3 • 2 4 = ■ 
 
 i- 1 
 
 h5 . «3 7-,5~3 
 
 = b 5 a 3 . 
 
 73 73 
 
 X s b 6 ■ a* b 5 a z 
 
 b~ 5 b 5 • 6- 5 1 
 
 Tkr 2 _ fir 2 • ,r~ 2 _ fi 
 r 5 ~ fix 2 ■ r 5 ~ x 2 ' 
 
06 ELEMENTARY ALGEBRA 
 
 
 
 
 WRITTEN EXERCISES 
 
 
 
 Free fr< 
 
 Dm 
 
 negative exponents : 
 
 
 
 
 
 o_2 
 1. 5-. 
 
 2 - 3 
 
 
 4. 
 
 a~ l b- 2 
 x~ A 
 
 7. 
 
 2cCK 
 
 10. 
 
 ax~* 
 
 O" 5 
 
 2. — • 
 
 Zr 3 
 
 
 5. 
 
 a~ x 
 
 8. 
 
 ab 
 cr*' 
 
 11. 
 
 By 3 
 
 Q_2 
 
 3. 3 ' 
 
 5" 
 
 4" 
 
 -3 
 
 i 
 
 6. 
 
 a- 3 
 b s ' 
 
 9. 
 
 a-'b- 2 . 
 
 12. 
 
 1 
 
 a~ 3 x~ 3 
 
 II. To free an expression from a fractional form, multiply 
 both numerator and denominator by a factor that, in combination 
 with the denominator, will produce unity. If more than one such 
 form is involved, apply the process for each. 
 
 For example : 
 
 b* b-W 
 
 Is JL a 3 4 5 ■ 2 3r 2 y 8 a 8 
 
 a& °' 4-5 xV 3 45.4- 5 x-hfzhj-* 
 
 — 45 . 2 + x ~~y a ' 
 
 (x-' 2 x--)(2/V 3 ) 
 
 = 2.4 5 + a 3 x- 2 «/ 3 . 
 
 WRITTEN EXERCISES 
 Free from fractional forms : 
 
 1. 
 
 3 
 
 5 2 ' 
 
 3. 
 
 a 
 
 bx 3 
 
 5. i + »: 
 
 5 2 2 3 
 
 n 
 
 ^ -1 a; -3 
 
 2. 
 
 1 
 
 a 2 6 3 ' 
 
 4. 
 
 a~ 2 
 
 r 3 ' 
 
 6. * + £• 
 
 y 3 X s 
 
 8. 2+*l 
 
 6 6" 1 
 
EXPONENTS 67 
 
 III. To transfer any specified factor from the numerator into 
 the denominator, or vice versa, multiply the numerator and de- 
 nominator by a factor that, in combination with the factor to be 
 transferred, will produce unity. 
 
 EXAMPLES 
 
 1. Transferring the factors of the denominator to the numerator : 
 
 x*y~ s x~ i y i x i y~ 3 
 
 2. Transferring the literal factors of the numerator to the denomi- 
 nator : 
 
 5 am _ 5 g-HrWb - _ 5 
 4 a 4 6- 3 4 a- :i b-Wb- s 4 ab~ 5 ' 
 
 WRITTEN EXERCISES 
 In the following expressions : 
 
 (a) Transfer all literal factors to the numerator. 
 
 (b) Transfer all literal factors to the denominator. 
 
 1. 
 
 a 2 x? 
 
 4. 
 
 I, 2 
 
 '5 a *b* 
 8ah~i 
 
 7. 
 
 r -I 
 5a 3 
 
 abx 
 
 2. 
 
 6 ay 2 z~ 5 
 11 ct 3 x~ 3 y* 
 
 5. 
 
 a s b-°c* 
 ah~ 5 c~* 
 
 8. 
 
 b x -\?/- 
 5pq 
 
 3 
 
 4 a W 
 
 17a 3 6 5 c- 3 ' 
 
 6. 
 
 7(cr 7 b- 5 )\ 
 6 a~M~i 
 
 9. 
 
 3 
 
 a* 
 
 119. The laws of exponents enable us to perform operations 
 with polynomials containing fractional and negative ex- 
 ponents. 
 
 Thus : (a* + 6^)2 _ ( ffi !)2 + 2 ah^ + (ft*) 2 
 
 = J + 2 a*6* + b. 
 
 OF THE 
 
08 
 
 ELEMENTARY ALGEBRA 
 
 WRITTEN EXERCISES 
 Perform the indicated operations : 
 
 1. 
 
 ;o* - ity. 
 
 
 
 8. 
 
 a 5 — x* 
 a^ — x 2 
 
 
 2. 
 
 :4 oW + 2 3 ) 2 . 
 
 
 
 9. 
 
 (x n -y n f. 
 
 
 3. 1 
 
 >-* + &*)*. 
 
 
 
 10. 
 
 (a 2n b 3r - If. 
 
 
 4. ( 
 
 a" - 3 b'f. 
 
 * 4 - y 3 )0 
 
 * + y 
 
 ■ f ). 
 
 11. 
 
 a p • b q 
 
 
 5. 1 
 
 a v+\ . ^-i 
 
 
 6. 
 
 >-" + l)(ar» 
 
 -i). 
 
 
 12. 
 
 (a n + t n ) 2 . 
 
 
 7. 1 
 
 [SB* + 3)(a£ + 
 
 5). 
 
 
 13. 
 
 (u 3 & 2 + o: 2 )(a 3 6 2 - 
 
 -a; 2 ). 
 
 Express as a product of two factors 
 
 14. x 6 
 
 m 
 
 -4 
 
 15. a 8 -2o 4 a;' + a;l 
 
 16. a 2n + 2a n b n + b 2n . 
 
 17. sb*» — 4 x 2 "?/ 2 " + 4 ?/*". 
 
 18. y~ 7 — x~ 10 . 
 
 19. of — 4. 
 
 20. l + 8x~% + 16x- 5 . 
 
 21. x 12 + 6 a?y~* -f 9 w" 7 - 
 
 SUMMARY 
 
 I. Definitions. 
 
 Meaning of the Fractional Exponents, a" denotes the qth. 
 root of the ^>th power, or the pth power of the gth root of a. 
 
 Sec. 107. 
 
 Meaning of the Exponent Zero. Any number (not 0) with 
 the exponent zero equals 1. Sec. 115. 
 
 Meaning of Negative Exponents. a~" means — for all values 
 
 a" 
 
 of n, positive or negative, integral or fractional. 
 
 Sec. 116 
 
EXPONENTS 69 
 
 U Laws of Exponents. 
 
 1. a m -a r = a m+r . Sec. 99. 
 
 2. a m -=- « r = a m " r . Sec. 100. 
 
 3. (a m ) r = a mr . Sec. 101. 
 
 4. (aft)" = a n 6 n . Sec. 101. 
 
 5. f«Y = — • Sec. 101. 
 
 These laws apply for all values of the exponents, m, n, r, 
 positive, negative, integral, or fractional. 
 
 III. Processes with Exponents. 
 
 1. When the bases of the factors are the same, the exponent 
 of the product is found by adding the given exponents (Law I). 
 
 Sec. 109. 
 
 2. When the bases of the expressions are the same, the 
 exponent of the quotient is found by subtracting the exponent 
 of the divisor from the exponent of the dividend (Law II). 
 
 Sec. 110. 
 
 3. When an exponent is applied to a number having an ex- 
 ponent, the product of the exponents is taken as the exponent 
 of the result (Law III). Sec. 111. 
 
 4. An exponent affecting a product is applied to each factor 
 (Law IV). Sec. 112. 
 
 5. An exponent affecting a fraction is applied to both nu- 
 merator and denominator (Law V). Sec. 112. 
 
 6. When the bases are different and the fractional exponents 
 are different, the exponents must have, or be made to have, a 
 common denominator, before any simplification by multiplica- 
 tion or division is possible. . Sec. 113. 
 
 7. To free an expression from a negative exponent, multiply 
 both numerator and denominator by a factor that will so com- 
 bine with the factor having the negative exponent as to produce 
 unity, in accordance with the relation a - " • a" = a n . Sec. 118. 
 
70 ELEMENTARY ALGEBRA 
 
 8. To free an expression from the fractional form, multiply 
 numerator and denominator by a factor that, in combination 
 with the denominator, will produce unity. If more than one 
 such form is involved, apply the process for each. Sec. 118. 
 
 9. To transfer any specified factor from the numerator into 
 the denominator, or vice versa, multiply the numerator and 
 denominator by a factor that, in combination with the factor 
 to be transferred, will produce unity. Sec. 118. 
 
 REVIEW 
 
 WRITTEN EXERCISES 
 
 Express with positive exponents : 
 
 1. m~*n s . 2. ±x~%y- l z. 3. 3 a~ 5 b s . 4. 17 aT%- 7 z~*. 
 
 Transfer all literal factors from the denominator to the 
 numerator : 
 
 5. *t. 6. -J*-. 7. 1 . 8. 5 
 
 x~i' arh- 3 6x~ 2 y* a V * 
 
 Multiply : 
 
 9. (2+V^+T) 2 . 12. p.p-K 
 
 10. V5 . V&. 13. (or 1 - b- l )(aT? - ft - *). 
 
 11. 5^/mr 3 . 2m- 1 . 14. (a- 2 - 1)(J -pi). 
 Remove the parentheses : 
 
 <n-y 21. 
 
 \ _p 
 
 16. (i/a^) 3 . pnu X 9 
 
 in \ rfi ) 10n / P9 \ x2y 
 
 19 - ^ I • 22. (»*)*. 
 
 Vlf) 20. [x^J* 1 *. 23. [s~ l ) 
 
EXPONENTS 71 
 
 SUPPLEMENTARY WORK 
 
 ADDITIONAL EXERCISES 
 Perform the operations indicated: 
 
 1 
 
 2. 
 3. 
 
 4. 
 5. 
 6. 
 
 7. 
 
 ' Xo. zn^ • — ' 
 
 16 a* tor 4 )*. 
 5aafy*)*. 
 
 17. a; -3 • x* • .t 2 • Vic. 
 18. 
 
 Vx cc 3 Vas 2 . 
 
 
 .1.2 
 
 -2B0a^-V»)* 20 - [(«"¥]" 
 
 -243 o-V 1 ^)"*. 21 ' l V «ft-* V ^{ 4 . 
 
 '"^i^T*. 22 - K^ 2 )*!"" 1 - 
 
 i i 
 
 8 
 
 9. Vz-V'V* 23 - ^« 2V ^ 
 
 io. v^^. 24 - t>rT°-[(-o c p- 
 
 n q H* , o K* 2 5- -a 3 &-V 3 cZ 5 --a- 2 6 5 c 4 d- 5 . 
 
 11. 3a 2 6 3 c _1 • 2a 4 6 5 c. 
 
 , 3 , 26. a 2 x — 3 w 6 • oW. 
 
 12. 10 a&~M -f- 2 o-W. 
 
 no m *-2„ 2 *. 3 27 - 3 *~*5 ** ' 10 **• 
 
 13. a m & 2n c -2 -4- ab n z~ 3 . 
 
 14. (3 a" 2 -5- 6- 2 )- 5 . 28. V a-lv/a3 ^«^- 
 
 15. (ar-ffH-a^a*. 29. Ix"^^- 2 )"^^ 1 )/)^. 
 
 Simplify : 
 
 30 2 * + a! 
 
 "^2 31. ^af- 1 -*-;*;"* 1 ] 2 ". 
 
 / cc 4- 2 w 
 **~\J>x + iy) 32 - [(a^) P " ? (« 93 ) ? ] p: 
 
 3s 
 
72 ELEMENTARY ALGEBRA 
 
 (•) ^I + VIo + VI-^- • 
 
 34. lV45 + 4Vf- V12K 
 
 35. Express the product Va 2 Va 3 " as a single radical. 
 
 36. Divide 2x*y- s -5xiy- 2 + 7x*y- l -5x% + 2x i y by 
 
 Find equivalent expressions with rational divisors : 
 
 37. 3V2^2V36. 48 (V5-2)(3+V5) , 
 
 5-Vo 
 
 p- Vg 
 
 38. &V(?-^Va6. 
 
 39. V40 ic 3 ?/ -5- a; V5 ?/. 49 - 
 
 p+ Vg. 
 40. .-^y + sf^ z+V^l 
 
 50. 
 
 41. 2^2a 2 -\/4a. ' z-V^T 
 
 42. _|Vf^ T VV3. 51 V5 
 
 Va — V& 
 
 43. 6A/54.r 2 -2A/'2x 2 . 
 
 44. a 2 ^48^-r-2a&^3aP: 52. Vs-Vs + y , 
 
 Va; + Va; + 2/ 
 
 45. 4 OOJ -5- V('.*'. 
 
 / „ Va 2 4-/> 2 + & 
 
 46. V« + ft + YJLzJ. M " Va^+F-& 
 Va+6— Va— & 
 
 47. 
 
 1 ** 3Va-3 + V.r-f 3 
 
 54. 
 
 a _ Va 2 - x 2 3 Va; — 3 — Va; + 3 
 
EXPONENTS 73 
 
 V^+T+V^T J3 + V5-V5-V5 
 . fifi' j • 
 
 55. ' — • 56 
 
 1 
 
 57. 
 
 2 + Vo - V2 
 
 Suggestion. Rationalize the denominator in two steps, using as first 
 
 factor ,- _ 
 
 2_(\/5- V2). 
 
 Solve : - 
 
 58. 3*+ Va*-2a; + 5 = l. 59 - 5 = a 2 -± 2 ab + b % ' 
 
 ^_ _i_ : / ,,. . ,„ . ... .. . , , 2a& 
 
 + - 
 
 a + a+6 
 
 61. Vz(2a-&+ Va) = 3a 2 — a&. 
 
CHAPTER V 
 
 LOGARITHMS 
 
 MEANING AND USE OF LOGARITHMS 
 
 120. Use of Exponents in Computation. By applying the 
 laws of exponents certain mathematical operations may be 
 performed by means of simpler ones. The following table of 
 powers of 2 may be used in illustrating some of these simpli- 
 fications : 
 
 1 =2° 
 
 32 = 25 
 
 1024 = 2io 
 
 32768 = 215 
 
 2 = 2i 
 
 64 = 26 
 
 2048 = 2U 
 
 65536 = 2!6 
 
 4 = 22 
 
 128 = 2 7 
 
 4096 = 2i2 
 
 131072 = 2" 
 
 8 = 2 3 
 
 256 = 28 
 
 8192 = 213 
 
 262144 = 2" 
 
 16 = 2* 
 
 512 = 29 
 
 16384 = 2" 
 
 524288 = 219 
 
 121. Application of Law I, Sec. 99, p. 53. 
 
 EXAMPLES 
 
 1. Find: 8-32. 
 
 From the table, 8 = 2 3 , (1) 
 
 and 32 = 25. (g) 
 
 Then, 8 • 32 = 23 • 2 5 = 2 8 , (J) 
 
 and, according to the table, 2 8 = 256. (^) 
 
 2. Find: 2048-64. 
 
 From the table, 2048 = 2 11 , (i) 
 
 and 64 = 2 G . (2) 
 
 Then, 2048 • 64 = 2ii . 2 6 _ 2", (3) 
 
 and, according to the table, 2 17 = 131072. (4) 
 
 Thus the process is simply one of inspection. In the above example 
 we merely added 11 and 6 and looked in the table for the number 
 opposite to 2 17 . 
 
 74 
 
LOGARITHMS 75 
 
 ORAL EXERCISES 
 State the following products by reference to the table : 
 
 1. 16 • 256. 5. 32 • 32. 9. 128 • 512. 
 
 2. 32-128. 6. 64-64.. 10. 128-1024. 
 
 3. 64-512. 7. 32-2048. 11. 8-16384. 
 
 4. 8 - 2048. 8. 16 • 4096. 12. 32 - 4096. 
 
 122. Application of Law II. Sec. 100, p. 54. 
 
 EXAMPLES 
 
 (3) 
 (4) 
 
 1. Find:. 256 . 
 32 
 
 
 
 From the table, 
 
 
 256 = 28, 
 
 and 
 
 
 32 = 2 5 . 
 
 Hence, 
 
 256 
 32 
 
 98 
 
 = — = 2 8 -5=2 8 , 
 
 2 5 ' 
 
 and, according 1 
 
 to the table, 
 
 2 3 =8. 
 
 2. Find: 65536 . 
 2048 
 
 
 
 As above, 
 
 65536 
 2048 
 
 016 
 
 = — = 2 5 = 32. 
 
 2ii 
 
 ORAL EXERCISES 
 By use of the table determine the value of the following : 
 , 1024 „ 32768 32 • 2048 
 
 1. • 6. • 5. • 
 
 128 1024 512 
 
 2 8192 64 - 512 128- 131072 
 
 64 ' ' 16-128' ' 64-8-8192 ' 
 
 123. Application of Law III, Sec. 101, p. 54. 
 
 EXAMPLES 
 
 1. Find: 16 3 . 
 
 By the table, 16 = 2 4 . (1) 
 
 Hence, 16 3 = (2 4 ) 3 = 212. (£) 
 
 and, according to the table, 21' 2 = 4096. (5) 
 
7. V8192. 
 
 10. 
 
 11. 
 12. 
 
 512 1 . 
 
 8. ^4096. 
 
 a/32768. 
 
 9. ^65536. 
 
 a/1024. 
 
 76 ELEMENTARY ALGEBRA 
 
 2. Find: a/1024. 
 
 As above, 1024 = (1024)* = (2 1 *)* _ 25 = 32, according to the table. 
 
 3. Find : a/32768. 
 
 As above, \^2768 = (215) £ _ 2 3 = 8. 
 
 ORAL EXERCISES 
 By use of the table find the value of : 
 
 1. 32 3 . 4. 64 2 . 
 
 2. 3 5 . 5. 256 2 
 
 3. 32 5 . 6. 16 4 . 
 
 124. The examples and exercises above show that the laws 
 of exponents furnish a powerful and remarkably easy way of 
 making certain computations. 
 
 In the above illustrations we have used a table based on the number 2, 
 and have limited the table to integral exponents ; but for practical pur- 
 poses a table based on 10 is used and is made to include fractional 
 exponents. 
 
 For example : 
 
 1. It is known that approximately, 
 
 2 = lO 1 ^ or 10- 3 (more accurately 10- 301 ). 
 From this we can express 20 as a power of 10, for 
 
 20 = 10 • 2 = 10 1 • 10- 301 = 10 1 - 301 . 
 Similarly, 200 = 10 • 20 = 10 1 • lO 1 *" = 10 2 - 301 , 
 
 and 2000 = 10 • 200 = 10 1 • lO*** = 10 3 - 301 . 
 
 2. It is known that approximately 763 = 10 2 -88. 
 Then 
 and 
 
 Similarly, 
 and 
 
 7630 = 
 
 = 10- 
 
 763= 10i -10 2 -88 =10 3 -88, 
 
 6300 = 
 
 = 100 
 
 . 763 = 10 2 • 10 2 -88 = 10^-88. 
 
 76.3 = 
 
 _ 763 
 10 
 
 1<V 88 
 _ * u _ JQ2.88-1 — 101-88 
 
 101 
 
 7.63 = 
 
 763 
 "100" 
 
 1fl2 88 
 _ 1" _ 1Q2.88-2— 100-88 
 
 10 2 
 
LOGARITHMS J 77 
 
 WRITTEN EXERCISES 
 
 Given 48 = 10 168 ; express as a power of 10 : 
 1. 480. 2. 4800. 3. 48,000. 4. 4.8. 
 
 Given 649 = 10 2 - 81 ; express as a power of 10 : 
 
 5. 6490. 7. 649,000. 9. 6.49. 
 
 6. 64,900. 8. 64.9. 10. 649,000,000. 
 
 Given 300 = 10 247 ; express as a power of 10 : 
 
 11. 3. 13. 3000. 15. 300,000. 
 
 12. 30. 14. 30,000. 16. 3,000,000. 
 
 125. The use of the base 10 has several advantages. 
 
 I. The exponents of numbers not in the table can readily be 
 found by means of the table. 
 
 To make this clear, let us suppose that a certain table expresses all 
 integers from 100 to 999 as powers of 10 ; then 80, although not in this 
 table, can be expressed as a power of 10 by reference to the table. 
 
 For, 30 = ^- , and since 300 is in the supposed table we may find 
 10 
 
 1 02-47 
 
 by reference to the table that 300 = 10 2 -* 7 , and hence, 30 = ^— = 10 1 - 47 . 
 J 10 1 
 
 Similarly, 3.76 is not in the supposed table, but 376 is and 
 
 3.76 = -— = -— • Therefore it is necessary only to subtract 2 from 
 100 10 2 
 
 the power of 10 found for 376 in order to find the power of 10 equal to 
 
 3.76. 
 
 Similarly, 4680 is not in the table, but 468 is and 4680 = 408 ■ 10 1 . 
 Therefore, it is necessary only to add 1 to the power of 10 found for 468 
 in order to find the power of 10 equal to 4680. 
 
 Such a table would not enable us to express in powers of 10 numbers 
 like 4683, 46.83, and 356,900, but only numbers of 3 or fewer digits, which 
 may be followed by any number of zeros. 
 
 Similar conditions would apply to a table of powers for numbers 
 from 1000 to 9999, from 10,000 to 99,999, and so on. 
 
 II. The integral part of the exponent can be written with- 
 out reference to a table. 
 
78 * ELEMENTARY ALGEBRA 
 
 For example : 
 
 1. 879 is greater than 100, which is the second power of 10, and less 
 than 1000, or the third power of 10. That is, 879 is greater than 10' 2 but 
 less than 10 3 . Therefore the exponent of the power of 10 which equals 
 879 is 2. + a decimal. 
 
 2. Similarly, 87.9 lies between 10 and 100, or between 10 1 and 10' 3 , 
 hence the exponent of the power of 10 that is equal to 87.9 is 1. + a 
 decimal. 
 
 ORAL EXERCISES 
 
 State the integral part of the exponent of the power of 10 
 equal to each of the following : 
 
 l. 35. 4. 25. 7. 25.5. 
 
 2 350. 5. 2500. 8. 3G5.5. 
 
 3. 36.5. 6. 36,500. 9. 17.65. 
 
 III. If two numbers have the same sequence of digits but 
 differ in the position of the decimal point, the exponents of 
 the powers of 10 which they equal have the same decimal 
 part. 
 
 For example : 
 
 Given that 274.3 = 10 2 - 43 , 
 
 we have 27.43 = ^M = ™^ = 10 1 « 
 
 10 10i 
 
 also 2743 = 10 • 274.3 = 10 1 • 10 2 - 43 = 10 3 - 43 , 
 
 also 274,300 = 1000 • 274.3 = 10 3 • 10 2 * 3 = 10 5 - 43 . 
 
 In each instance the decimal part of the exponent is the same. It is 
 evident that this will be the case in all similar instances, for shifting the 
 decimal point is equivalent to multiplying or dividing repeatedly by 10, 
 which is equivalent to changing the integral part of the exponent by 
 adding or subtracting an integer. 
 
 ORAL EXERCISES 
 
 Given 647 = 10 281 , state the decimal part of the exponent 
 of the power of 10 that equals : 
 
 1. 64.7. 3. 6470. 5. 647,000. 
 
 2. 6.47. 4. 64,700. 6. 6,470,000. 
 
LOGARITHMS 79 
 
 Given 568.1 = 10 2 - 75 , state the decimal part of the exponent 
 of the power of 10 that equals : 
 
 7. 56.81. 9. 5681. 11. 568,100. 
 
 8. 5.681. 10. 56,810. 12. 56,810,000. 
 
 126. Logarithms. Exponents when used in this way for 
 computation are called logarithms, abbreviated log. 
 
 127. The number to which the exponents are applied is 
 called the base. 
 
 For the purposes of computation the base used is 10. 
 
 According to the above definition the equation 30 = 10 lAS may be 
 written log 30 = 1.48, which is read " The logarithm of 30 is 1.48." These 
 equations mean the same thing ; namely, that 1.48 is (approximately) 
 the power of 10 that equals 30. 
 
 WRITTEN EXERCISES 
 Write the following in the notation of logarithms : 
 
 1. 700 = 10 284 . 3. 6 = 10 077 . 5. 361 = 10 255 . 
 
 2. 75 = 10 187 . 4. 50 = 10 169 . 6. 45 = 10 165 . 
 
 Write the following as powers of 10 : 
 
 7. log 20 = 1.3. 9. log 3 = 0.47. 11. log 111 = 2.04. 
 
 8. log 500 = 2.70. 10. log 7 = 0.84. 12. log 21 = 1.32. 
 
 128. Characteristic and Mantissa. The integral part of a 
 logarithm is called its characteristic, and the decimal part its 
 mantissa. 
 
 ORAL EXERCISES 
 
 1-12. State the characteristic and the mantissa in each of 
 the logarithms in Exercises 1-12 above. 
 
 129. According to Sec. 125, II (p. 77), the characteristics 
 of logarithms can be determined by inspection, consequently 
 tables of logarithms furnish only the mantissas. 
 
80 ELEMENTARY ALGEBRA 
 
 EXPLANATION OF THE TABLES 
 
 130. The use of the tables, pp. 84 and 85, is best seen from 
 an example. 
 
 Find the logarithm of 365. 
 
 The first column in' the table, p. 84, contains the first two figures of 
 the numbers whose mantissas are given in the table, and the top row con- 
 tains the third figure. 
 
 Hence, find 36 in the left-hand column, p. 84, and 5 at the top. 
 
 In the column under 5 and opposite to 36 we find 5623, the required 
 mantissa. 
 
 Since 365 is greater than 100 (or 10 2 ) but less than 1000 (or 10 3 ), the 
 characteristic of the logarithm is 2. 
 
 Therefore, log 365 is 2.5623. 
 
 WRITTEN EXERCISES 
 By use of the table find the logarithms of : 
 
 1. 25. 5. 99. 9. 9.9. 13. 1000. 
 
 2. 36. 6. 86. 10. 8.6. 14. 5000. 
 
 3. 50. 7. 999. 11. 33,000. 15. 505. 
 
 4. 75. 8. 800. 12. 99,900. 16. 5.05. 
 
 131. Negative Characteristics. An example will serve to 
 show how negative characteristics arise : 
 
 From log 346 = 2.5391, we find, 
 
 log 34.6 = log — = log 346 - log 10 = 2.5391 - 1 = 1.5391. 
 
 log 3.46 = log — = 1.5391 - 1 = 0.5391. 
 
 log .346 = log — = 0.5391 - 1. 
 & & 10 
 
 In the last line we have a positive decimal less 1, and the result is a 
 negative decimal ; viz. — .4609. But to avoid this change of mantissa, it 
 is customary not to carry out the subtraction, but simply to indicate it. 
 It might be written — 1 + .5391, but it is customarily abridged into 
 1.5391. The mantissa is kept positive in all logarithms. The loga- 
 rithm 1.5391 says that the corresponding number is greater than 10 - 1 
 (or Jj), but less than 10° or 1. 
 
LOGARITHMS 81 
 
 We now write log .346 = 1.5391 
 
 » 
 
 Similarly, log .0346 = log — = 1.5391 - 1 = 2.5391. 
 
 Thus we see that the mantissa remains the same, no matter how the 
 position of the decimal point is changed. The mantissa is determined 
 solely by the sequence of digits constituting the number. 
 
 The characteristic is determined solely by the position of the decimal 
 point. Xegative characteristics, like positive ones, are determined by 
 inspection. 
 
 EXAMPLES 
 
 1. What is the characteristic of log .243 ? 
 
 .243 is more than .1 or 10- 1 , but less than 1 or 10°. Hence, 
 
 .243 = 10 _1 + a decimal. 
 
 The characteristic is 1. 
 
 2. Find the characteristic of log .0593. 
 
 .0593 is more than .01 or 10~ 2 , but less than .1 or 10 _1 . Hence, 
 
 .0593 = 10-2+ a decimal. 
 
 The characteristic is — 2, or 2. 
 
 3. Similarly, since .00093 is greater than .0001 or 10 ~ 4 , but 
 less than .001 or 10~ s , 
 
 log .00093 = 4+ a decimal. 
 
 132. The characteristic having been determined, the mantissa 
 
 is found from the table in the usual way. 
 
 For example : 
 
 log .243 = 1.3856, 
 
 log .0593 = 2.7731, 
 
 log .00093 = 4.9685. 
 
 WRITTEN EXERCISES 
 
 Find the logarithms of : 
 
 1. .35. 3. .105. 5. .0023. 7. .00342. 
 
 2. ,j.634. 4. .027. 6. .0123. 8. .0004. 
 
 133. In finding the number corresponding to a logarithm 
 with negative characteristic, the same method is followed as 
 when the characteristic is positive. The mantissa determines 
 the sequence of digits constituting the number ; the character- 
 istic fixes the position of the decimal point. 
 
32 
 
 ELEMEXTARY ALGEBRA 
 
 For example : 
 
 If log n — 2.595o. the digits of n are 394. The characteristic 2 says 
 that n is greater than 10-- out less than 10 -1 (or .1). Hence. 
 
 the decimal point must be so placed that n has no tenths but some 
 hundredths. Therefore n = .0394. 
 
 WRITTEN EXERCISES 
 By use of the tables find the numbers whose logarithms are 
 
 1. 1.6232. 
 
 2. 1.4914. 
 
 3. 2.4281. 
 
 4. 1.9196. 
 
 5. 0.9196. 
 
 6. 3.4281. 
 
 Find n if : 
 
 19. log n = 1.9289. 
 
 20. log n= 0.9289. 
 
 7. 2.0792 
 
 8. -.7076 
 
 9. 4.9196 
 
 10. 3.2201 
 
 11. 0.2201 
 
 12. 1.2201 
 
 13. 3.0969. 
 
 14. 1.6972. 
 
 15. 3.9284. 
 
 16. 2.9284. 
 
 17. 1.9284. 
 
 18. 5.7832. 
 
 21. log (n — 1) = 3.9294. 
 
 22. log (| n) = 1.6128. 
 
 I. 10 s * • 10 r = 10"^ r . 
 
 USE OF THE TABLES FOR COMPUTATION 
 
 134. For use in computation by logarithms the laws of 
 exponents may be expressed thus : 
 
 Tfte logarithm of a product is the 
 
 sum of the logarithms of the factors. 
 
 Thie logarithm of a quotient is the 
 logarithm of the dividend minus the 
 logarithm of the divisor. 
 
 The logarithm of a number with an 
 exponent is the product of the exponent 
 and the logarithm of the number. 
 
 II. 
 
 10"* 
 
 — =1' . 
 
 irr 
 
 III. (10-)' = 10" 
 
 Since r may be positive or negative, integral or fractional, Law III 
 provides not only for raising to integral powers, but also for finding recip- 
 rocals of such powers, and for extracting roots. 
 
LOGARITHMS 83 
 
 EXAMPLES 
 
 1. Multiply 21 by 37. 
 
 1. log 21 = 1.3222, table, p. 84. 
 
 2. log 37 = 1.5682, table, p 84. 
 
 3. Adding, log 21 + log 37, or log (21 x 37)= 2.8904 (Sec. 134). 
 
 4. .•. 21 x 37 = 777 from table, p. 85. 
 
 2. Divide 814 by 37. 
 
 1. log 814 = 2.9106, table, p. 85. 
 
 2. log 37 = 1.5682, table, p. 84. 
 
 3. .-. log 814 - log 37 = 2.9106 - 1.5682 = 1.3424. 
 
 4. .-. 814 -h 37 = 22, table, p. 84. 
 
 3. Extract the cube root of 729. 
 
 1. ^729 = (729)i 
 
 2. log 7293 _ i log 729 (Sec. 134). 
 
 3. log 729 = 2.8627, table, p. 85. 
 
 4. i of 2.8627 = 0.9542. 
 
 5. .-. (729)3 or .3/729 _ g ? table, p. 85. 
 
 Compute by use of logarithms: 
 
 1. 8 x 15. 6. 5 4 . 
 
 2. 41 x 23. 7. 31 2 . 
 
 3. 37 x 17. 8. V414. 
 
 4. 12x17. 9. V343. 
 
 5. 893-5-19. 10. 940 h- 47. 15. V216 
 
 Since the logarithm is approximate, the result in general is approxi- 
 mate. Thus, log \/(256) = 0.60205, which is not the logarithm of 4, but 
 is sufficiently near to be recognized in the table. 
 
 11. 
 
 19 2 . 
 
 12. 
 
 7 3 . 
 
 13. 
 
 4 5 . 
 
 14. 
 
 V196. 
 
 1 c 
 
 3 /„ . n 
 
84 
 
 ELEMENTARY ALGEBRA 
 
 N 
 10 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 0000 
 
 0043 
 
 0086 
 
 0128 
 
 0170 
 
 0212 
 
 0253 
 
 0294 
 
 0334 
 
 0374 
 
 11 
 
 0414 
 
 0453 
 
 0492 
 
 0531 
 
 0569 
 
 0607 
 
 0645 
 
 0682 
 
 0719 
 
 0755 
 
 12 
 
 0792 
 
 0828 
 
 0864 
 
 0899 
 
 0934 
 
 0969 
 
 1004 
 
 1038 
 
 1072 
 
 1106 
 
 13 
 
 1139 
 
 1173 
 
 1206 
 
 1239 
 
 1271 
 
 1303 
 
 1335 
 
 1367 
 
 1399 
 
 1430 
 
 14 
 
 1461 
 
 1492 
 
 1523 
 
 1553 
 
 1584 
 
 1614 
 
 1644 
 
 1673 
 
 1703 
 
 1732 
 
 15 
 
 17G1 
 
 1790 
 
 1818 
 
 1847 
 
 1875 
 
 1903 
 
 1931 
 
 1959 
 
 1987 
 
 2014 
 
 16 
 
 2041 
 
 2068 
 
 2095 
 
 2122 
 
 2148 
 
 2175 
 
 2201 
 
 2227 
 
 2253 
 
 2279 
 
 17 
 
 2304 
 
 2330 
 
 2355 
 
 2380 
 
 2405 
 
 2430 
 
 2455 
 
 2480 
 
 2504 
 
 2529 
 
 18 
 
 2553 
 
 2577 
 
 2601 
 
 2625 
 
 2(i48 
 
 2672 
 
 2695 
 
 2718 
 
 2742 
 
 2765 
 
 19 
 
 2788 
 
 2810 
 
 2833 
 
 2856 
 
 2878 
 
 2900 
 
 2923 
 
 2945 
 
 2967 
 
 2989 
 
 20 
 
 3010 
 
 3032 
 
 3054 
 
 3075 
 
 3096 
 
 3118 
 
 3139 
 
 3160 
 
 3181 
 
 3201 
 
 21 
 
 3222 
 
 3243 
 
 3263 
 
 3284 
 
 3304 
 
 3324 
 
 3345 
 
 3365 
 
 3385 
 
 3404 
 
 22 
 
 3424 
 
 3444 
 
 3464 
 
 3483 
 
 3502 
 
 3522 
 
 3541 
 
 3560 
 
 3579 
 
 3598 
 
 23 
 
 3617 
 
 3636 
 
 3655 
 
 3674 
 
 3692 
 
 3711 
 
 3729 
 
 3747 
 
 3766 
 
 3784 
 
 24 
 
 3802 
 
 3820 
 
 3838 
 
 3856 
 
 3874 
 
 3892 
 
 3909 
 
 3927 
 
 3945 
 
 3962 
 
 25 
 
 3979 
 
 3997 
 
 4014 
 
 4031 
 
 4048 
 
 4065 
 
 4082 
 
 4099 
 
 4116 
 
 4133 
 
 26 
 
 4150 
 
 4166 
 
 4183 
 
 4200 
 
 4216 
 
 4232 
 
 4249 
 
 4265 
 
 4281 
 
 4298 
 
 27 
 
 4314 
 
 4330 
 
 4346 
 
 4362 
 
 4378 
 
 4393 
 
 4409 
 
 4425 
 
 4440 
 
 4456 
 
 28 
 
 4472 
 
 4487 
 
 4502 
 
 4518 
 
 4533 
 
 4548 
 
 4564 
 
 4579 
 
 4594 
 
 4609 
 
 29 
 
 4624 
 
 4639 
 
 4654 
 
 4669 
 
 4683 
 
 4698 
 
 4713 
 
 4728 
 
 4742 
 
 4757 
 
 30 
 
 4771 
 
 4786 
 
 4800 
 
 4814 
 
 4829 
 
 4843 
 
 4857 
 
 4871 
 
 4886 
 
 4900 
 
 31 
 
 4914 
 
 4928 
 
 4942 
 
 4955 
 
 4969 
 
 4983 
 
 4997 
 
 5011 
 
 5024 
 
 5038 
 
 32 
 
 5051 
 
 5065 
 
 5079 
 
 5092 
 
 5105 
 
 5119 
 
 5132 
 
 5145- 
 
 5159 
 
 5172 
 
 33 
 
 5185 
 
 5198 
 
 5211 
 
 5224 
 
 5237 
 
 5250 
 
 5263 
 
 5276 
 
 5289 
 
 5302 
 
 34 
 
 5315 
 
 5328 
 
 5340 
 
 5353 
 
 5366 
 
 5378 
 
 5391 
 
 5403 
 
 5416 
 
 5428 
 
 35 
 
 5441 
 
 5453 
 
 5465 
 
 5478 
 
 5490 
 
 5502 
 
 5514 
 
 5527 
 
 5539 
 
 5551 
 
 36 
 
 5563 
 
 5575 
 
 5587 
 
 5599 
 
 5611 
 
 5623 
 
 5635 
 
 5647 
 
 5658 
 
 5670 
 
 37 
 
 5682 
 
 5694 
 
 5705 
 
 5717 
 
 571".) 
 
 5740 
 
 5752 
 
 5763 
 
 5775 
 
 5786 
 
 38 
 
 5798 
 
 5809 
 
 5821 
 
 5832 
 
 5843 
 
 5855 
 
 -,S66 
 
 5877 
 
 5888 
 
 5899 
 
 39 
 
 5911 
 
 5922 
 
 5933 
 
 5944 
 
 5955 
 
 5966 
 
 5977 
 
 5988 
 
 5999 
 
 6010 
 
 40 
 
 6021 
 
 6031 
 
 6042 
 
 6053 
 
 6064 
 
 6075 
 
 6085 
 
 6096 
 
 6107 
 
 6117 
 
 41 
 
 6128 
 
 613S 
 
 DIVJ 
 
 6160 
 
 6170 
 
 6180 
 
 6191 
 
 6201 
 
 6212 
 
 6222 
 
 42 
 
 6232 
 
 6243 
 
 6253 
 
 6263 
 
 6274 
 
 6284 
 
 6294 
 
 6304 
 
 6314 
 
 6: ;■_'.", 
 
 43 
 
 6335 
 
 6345 
 
 6355 
 
 6365 
 
 6375 
 
 6385 
 
 6395 
 
 6405 
 
 6415 
 
 6425 
 
 44 
 
 6435 
 
 6444 
 
 6454 
 
 6464 
 
 6474 
 
 6484 
 
 6493 
 
 6503 
 
 6513 
 
 6522 
 
 45 
 
 6532 
 
 6542 
 
 6551 
 
 6561 
 
 6571 
 
 6580 
 
 6590 
 
 6599 
 
 6609 
 
 6618 
 
 46 
 
 6628 
 
 6637 
 
 6646 
 
 6056 
 
 666,5 
 
 6675 
 
 6684 
 
 6693 
 
 6702 
 
 6712 
 
 47 
 
 6721 
 
 6730 
 
 6739 
 
 6749 
 
 6758 
 
 6767 
 
 6776 
 
 6785 
 
 6794 
 
 6803 
 
 48 
 
 6812 
 
 6821 
 
 6830 
 
 6S39 
 
 6848 
 
 6.S57 
 
 6S66 
 
 6875 
 
 6SS4 
 
 6893 
 
 49 
 
 6902 
 
 6911 
 
 6920 
 
 6928 
 
 6937 
 
 6946 
 
 6955 
 
 6964 
 
 6972 
 
 6981 
 
 50 
 
 6990 
 
 6998 
 
 7007 
 
 7016 
 
 7021 
 
 7033 
 
 7042 
 
 7050 
 
 7059 
 
 7067 
 
 51 
 
 7076 
 
 7084 
 
 7093 
 
 7101 
 
 7110 
 
 7118 
 
 7126 
 
 71 35 
 
 7143 
 
 7152 
 
 52 
 
 7160 
 
 7168 
 
 7177 
 
 71S5 
 
 7193 
 
 7202 
 
 7210 
 
 7218 
 
 722(5 
 
 7235 
 
 53 
 
 7243 
 
 7251 
 
 72:,! I 
 
 7267 
 
 7275 
 
 7284 
 
 7292 
 
 7300 
 
 7308 
 
 7316 
 
 54 
 
 7324 
 
 7332 
 
 7340 
 
 7348 
 
 7356 
 
 7364 
 
 7372 
 
 7380 
 
 7388 
 
 7396 
 
 N 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
LOGARITHMS 
 
 85 
 
 N 
 55 
 
 
 
 1 
 
 2 
 
 3 
 
 1 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 7404 
 
 7412 
 
 7419 
 
 7427 
 
 7435 
 
 7443 
 
 7451 
 
 7459 
 
 7466 
 
 7474 
 
 56 
 
 7482 
 
 7490 
 
 7497 
 
 7505 
 
 7513 
 
 7520 
 
 7528 
 
 7536 
 
 7543 
 
 7551 
 
 57 
 
 7559 
 
 7566 
 
 7574 
 
 7582 
 
 7589 
 
 7597 
 
 7604 
 
 7612 
 
 7619 
 
 7627 
 
 58 
 
 7634 
 
 7642 
 
 7649 
 
 7657 
 
 7664 
 
 7672' 
 
 7679 
 
 7686 
 
 7694 
 
 7701 
 
 59 
 
 7709 
 
 7716 
 
 7723 
 
 7731 
 
 7738 
 
 7745 
 
 7752 
 
 7760 
 
 7767 
 
 7774 
 
 60 
 
 7782 
 
 7789 
 
 7796 
 
 7803 
 
 7810 
 
 7818 
 
 7825 
 
 7832 
 
 7839 
 
 7846 
 
 61 
 
 7853 
 
 7860 
 
 7868 
 
 7875 
 
 7-S.S2 
 
 7889 
 
 7896 
 
 7903 
 
 7910 
 
 7917 
 
 62 
 
 7924 
 
 7931 
 
 793S 
 
 7945 
 
 7952 
 
 7959 
 
 7966 
 
 7973 
 
 7980 
 
 7987 
 
 63 
 
 7993 
 
 8000 
 
 8007 
 
 8014 
 
 8021 
 
 8028 
 
 8035 
 
 8041 
 
 8048 
 
 8055 
 
 64 
 
 8062 
 
 8069 
 
 8075 
 
 8082 
 
 8089 
 
 8096 
 
 8102 
 
 8109 
 
 8116 
 
 8122 
 
 65 
 
 8129 
 
 8136 
 
 8142 
 
 8149 
 
 8156 
 
 8162 
 
 8169 
 
 8176 
 
 8182 
 
 8189 
 
 66 
 
 8195 
 
 8202 
 
 8209 
 
 8215 
 
 8222 
 
 8228 
 
 8235 
 
 8241 
 
 8248 
 
 8254 
 
 67 
 
 8261 
 
 8267 
 
 8274 
 
 8280 
 
 8287 
 
 8293 
 
 8299 
 
 8306 
 
 8312 
 
 8319 
 
 68 
 
 8325 
 
 8331 
 
 8338 
 
 8344 
 
 8351 
 
 8357 
 
 8363 
 
 8370 
 
 8376 
 
 8382 
 
 69 
 
 8388 
 
 8395 
 
 8401 
 
 8407 
 
 8414 
 
 8420 
 
 8426 
 
 8432 
 
 8439 
 
 8445 
 
 70 
 
 8451 
 
 8457 
 
 8463 
 
 8470 
 
 8476 
 
 8482 
 
 8488 
 
 8494 
 
 8500 
 
 8506 
 
 71 
 
 8513 
 
 8519 
 
 8525 
 
 8531 
 
 8537 
 
 8543 
 
 8549 
 
 8555 
 
 8561 
 
 8567 
 
 72 
 
 8573 
 
 8579 
 
 S5S5 
 
 8591 
 
 8597 
 
 8603 
 
 8609 
 
 8615 
 
 8621 
 
 8627 
 
 73 
 
 8633 
 
 8639 
 
 8645 
 
 8651 
 
 8657 
 
 8663 
 
 8669 
 
 8675 
 
 8681 
 
 8686 
 
 74 
 
 8692 
 
 8698 
 
 8704 
 
 8710 
 
 8716 
 
 8722 
 
 8727 
 
 8733 
 
 8739 
 
 8745 
 
 75 
 
 8751 
 
 8756 
 
 8762 
 
 8768 
 
 8774 
 
 8779 
 
 8785 
 
 8791 
 
 8797 
 
 8802 
 
 76 
 
 8808 
 
 8814 
 
 8820 
 
 8825 
 
 8831 
 
 8837 
 
 8842 
 
 8848 
 
 8854 
 
 8859 
 
 77 
 
 ssi;.-. 
 
 8871 
 
 8876 
 
 8882 
 
 8887 
 
 8893 
 
 8899 
 
 8904 
 
 8910 
 
 8915 
 
 78 
 
 8921 
 
 8927 
 
 8932 
 
 8938 
 
 8943 
 
 8949 
 
 8954 
 
 8960 
 
 8965 
 
 8971 
 
 79 
 
 8976 
 
 8982 
 
 8987 
 
 8993 
 
 8998 
 
 9004 
 
 9009 
 
 9015 
 
 9020 
 
 9025 
 
 80 
 
 9031 
 
 9036 
 
 9042 
 
 9047 
 
 9053 
 
 9058 
 
 9063 
 
 9069 
 
 9074 
 
 9079 
 
 81 
 
 9085 
 
 9090 
 
 9096 
 
 9101 
 
 9106 
 
 9112 
 
 9117 
 
 9122 
 
 9128 
 
 9133 
 
 82 
 
 9138 
 
 9143 
 
 9149 
 
 9154 
 
 9159 
 
 9165 
 
 9170 
 
 9175 
 
 9180 
 
 9186 
 
 83 
 
 9191 
 
 9196 
 
 9201 
 
 9206 
 
 9212 
 
 9217 
 
 9222 
 
 9227 
 
 9232 
 
 9238 
 
 84 
 
 9243 
 
 9248 
 
 9253 
 
 9258 
 
 9263 
 
 9269 
 
 9274 
 
 9279 
 
 9284 
 
 9289 
 
 85 
 
 9294 
 
 9299 
 
 9304 
 
 9309 
 
 9315 
 
 9320 
 
 9325 
 
 9330 
 
 9335 
 
 9340 
 
 86 
 
 9345 
 
 9350 
 
 9355 
 
 9:360 
 
 9365 
 
 9370 
 
 9375 
 
 9380 
 
 9385 
 
 9390 
 
 87 
 
 9395 
 
 9400 
 
 9405 
 
 9410 
 
 9415 
 
 9420 
 
 9425 
 
 9430 
 
 9435 
 
 9440 
 
 88 
 
 9445 
 
 9450 
 
 9455 
 
 9460 
 
 9465 
 
 9469 
 
 9474 
 
 9479 
 
 9484 
 
 9489 
 
 89 
 
 9494 
 
 9499 
 
 9504 
 
 9509 
 
 9513 
 
 9518 
 
 9523 
 
 9528 
 
 9533 
 
 9538 
 
 90 
 
 9542 
 
 9547 
 
 9552 
 
 9557 
 
 9562 
 
 9566 
 
 9571 
 
 9576 
 
 9581 
 
 9586 
 
 91 
 
 9590 
 
 9595 
 
 9600 
 
 9605 
 
 9609 
 
 9614 
 
 9619 
 
 9624 
 
 9628 
 
 9633 
 
 92 
 
 9638 
 
 9643 
 
 9647 
 
 9652 
 
 9657 
 
 9661 
 
 9666 
 
 9671 
 
 9675 
 
 9680 
 
 93 
 
 9685 
 
 9689 
 
 !Ki94 
 
 9699 
 
 9703 
 
 970S 
 
 9713 
 
 9717 
 
 9722 
 
 9727 
 
 94 
 
 9731 
 
 9736 
 
 9741 
 
 9745 
 
 9750 
 
 9754 
 
 9759 
 
 9763 
 
 9768 
 
 9773 
 
 95 
 
 9777 
 
 97S2 
 
 9786 
 
 9791 
 
 9795 
 
 9800 
 
 9805 
 
 9809 
 
 9814 
 
 9818 
 
 96 
 
 9823 
 
 9827 
 
 9832 
 
 9836 
 
 9841 
 
 9S45 
 
 9850 
 
 9854 
 
 9859 
 
 9863 
 
 97 
 
 9S68 
 
 9872 
 
 9877 
 
 9881 
 
 9886 
 
 9890. 
 
 9894 
 
 9899 
 
 9903 
 
 9908 
 
 98 
 
 9912 
 
 9917 
 
 9921 
 
 9926 
 
 9930 
 
 9934 
 
 9939 
 
 994:'. 
 
 9948 
 
 9952 
 
 99 
 
 N 
 
 9956 
 
 9961 
 
 9965 
 
 9969 
 
 9974 
 
 9978 
 
 99S3 
 
 9987 
 
 9991 
 
 9996 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
86 ELEMENTARY ALGEBRA 
 
 135. How to calculate the logarithm of a number not found 
 in the table is best seen from an example. 
 
 EXAMPLE 
 Find the logarithm of 257.3 (see 16th line of table, p. 84) : 
 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 25 
 
 3979 
 
 3997 
 
 4014 
 
 4031 
 
 4048 
 
 4i Mi:. 
 
 4082 
 
 4099 
 
 4116 
 
 4133 
 
 The numbers nearest to 257.3 whose logarithms are given in the table 
 are 257 and 258. We have 
 
 log 258 = 2.4116 
 
 log 257 = 2.4099 
 
 Difference, .0017 
 
 That is, an increase of 1 in the number causes an increase of* .0017 in the 
 logarithm. Assuming that an increase of .3 in the number would cause 
 an increase of .3 of .0017, or .0005, in the logarithm, then 
 
 log 257.3 = 2.4099 + .0005 = 2.4104. 
 
 Notes : 1. The product of .3 and .0017 is .00051, but we take only four 
 places, because the mantissas as given in the table are expressed to four 
 places only. If the digit in the fifth decimal i>lace of the correction is 
 more than 5, we replace it by a unit in the fourth place. 
 
 2. The difference between two succeeding mantissas of the table 
 (called the tabular difference) can be seen by inspection. 
 
 3. What is written in finding the logarithm of 257.3 should be at most 
 the following : 
 
 log 257 = 2.4099 tabular difference 17 
 
 correction for .3 = 5 .3 
 
 log 257.3 = 2.4104 IT 
 
 4. The corrections are made on the assumption that the change in the 
 logarithm is proportional to the change in the number. This is suffi- 
 ciently accurate when used within the narrow limits here prescribed. 
 
 WRITTEN EXERCISES 
 Find the logarithm of : 
 
 1. 1235. 5. 1425. 9. 3.142. 13. .4071. 17. .3002. 
 
 2. 23.5. 6. 1837. 10. 1.414. 14. 85.51. 18: 9009. 
 
 3. 2.36. 7. 6720. 11. 1.732. 15. .OUT.. 19. 12.02. 
 
 4. .0237. 8. 67.25. 12. .6220. 16. .ll'67. 20. 5.008, 
 
LOGARITHMS 
 
 81 
 
 136. To calculate the number whose logarithm is given 
 apply the table as follows: 
 
 (1) If the given logarithm is in the table, the number can 
 be seen at once. 
 
 (2) If the given logarithm is not in the table, the number 
 corresponding to the nearest logarithm of the table may be 
 taken. 
 
 A somewhat closer approximation may be found by using 
 the method of the following example : 
 
 EXAMPLE 
 
 Find the number whose logarithm is 1.4271. The mantissas 
 nearest to this are found in the 17th line of table, p. 84. 
 
 26 
 
 
 
 4150 410G 4183 4200 4210 
 
 8 
 
 4232 4249 4265 4281 4298 
 
 The process is the reverse of that of finding the logarithm. 
 
 The next smaller mantissa in the table is 4205, corresponding to the 
 number 267. The difference between this mantissa and the given man- 
 tissa is 6. The tabular difference between 4205 and the next larger man- 
 tissa is 10. An increase of 10 in the logarithm corresponds to an increase 
 of 1 in the number. Hence, an increase of in the logarithm corresponds 
 to an increase of T R 5 of 1, or .4, in the number. This means .4 of one unit 
 in the number 207. What its place value is in the final result depends 
 upon the characteristic. The digits of the result are 2674. 
 
 The characteristic 1 shows that the desired number is greater than the 
 first power of 10, but less than the second power of 10 or 100. Hence, 
 the decimal point must be placed between and 7, and the final result is 
 26.74. 
 
 Notes: 1. For reasons similar to those of Note 1, p. 86, the correction 
 should be carried to one place only. 
 
 2. At most the following should be written : 
 
 logn = 1.4271 tab. diff. 16 
 
 mantissa for 207 = .4265 T 6 e = A 
 
 diff. 
 
 Therefore, 
 
 
 n = 26.74. 
 
88 ELEMENTARY ALGEBRA 
 
 WRITTEN EXERCISES 
 Find the number whose logarithm is : 
 
 1. 0.7305. 4. 2.9023. 7. 1.1962. 10. 3.9485. 
 
 2. 0.5029. 5. 3.1467. 8. 2.0342. 11. 4.6987. 
 
 3. 1.4682. 6. 3.6020. 9. 3.3920. 12. 2.6376. 
 
 SUMMARY 
 
 I. Definitions. 
 
 1. Exponents indicating the powers of a base and used for 
 the purposes of calculation are called logarithms. Sec. 126. 
 
 2. The integral part of a logarithm is called its characteristic, 
 and the decimal part its mantissa. Sec. 128. 
 
 II. Laws. 
 
 1. 10 m • 10 r = 10 m+r . The logarithm of a product is the sum 
 of the logarithms of the factors. 
 
 10 m 
 
 2. — = 10 m_r . The logarithm of a quotient is the logarithm 
 
 of the dividend minus the logarithm of the divisor. 
 
 3. (10 m ) r = 10 mr . The logarithm of a number with an ex- 
 ponent is the product of the exponent and the logarithm of 
 the number. Sec. 134. 
 
 REVIEW 
 
 WRITTEN EXERCISES 
 
 Find exactly or approximately by use of logarithms the 
 value of: 
 
 1. V2. 4. V7. 7. V756. 10. (1.03) 7 . 
 
 2. V5. 5. V92U 8. ^812. 11. (1.04) 10 . 
 
 3. i/d. 6. a/656. 9. (1.5) 4 . 12. (1.06) 9 . 
 
 13. 
 14. 
 
 M?§). 15 . V (624)(598)(178). 
 
 779 
 
 (732X774), 16 J (651)(654)(558) 
 
 (731) (671) ' * 763 
 
LOGARITHMS 89 
 
 17. It is known that the volume of a sphere is § Ttr 3 , r being 
 the length of the radius. Using 3.14 as the approximate 
 value of ir, find by logarithms the volume of a sphere of radius 
 7.3 in. 
 
 18. Find, as above, the volume of a sphere whose radius is 
 36.4 ft. 
 
 Calculate by logarithms : 
 
 (132)(1837) 6 . ^^578. 
 
 167 
 
 (2076) (379) 
 173 
 
 (15.61) 2 
 (700)3 
 
 (3059)(349) /(7688)(7719) 
 
 (19)(23)(2443) * \ (248)(249) " 
 
 J(294)(1842) > 9 (217.6)(.00681). 
 
 307 
 
 3X675) 1Q [V278.2 (2.578)1* , 
 
 113 V.00231 • V76.19 
 
 11. Given a = 0.4916, c = 0.7544, and b = c 2 - a 2 . Find &. 
 Suggestion. 6 = (c — a)(c + a). 
 
 12. It is known that in steam engines, the piston head's 
 average velocity (c) per second is approximately given by the 
 formula : 
 
 31 
 
 = «>w& 
 
 15' 
 
 where s denotes the distance over which the piston moves 
 (expressed in the same unit as c), and p the number of pounds 
 pressure in the cylinder. 
 
 (1) Find c, if s = 32.5 in., p = 110 lb. 
 
 (2) Findp, if c = 15 ft., s = 2.6 ft. 
 
CHAPTER VI 
 IMAGINARY AND COMPLEX NUMBERS 
 
 137. Imaginary Numbers. The numbers denned in what pre- 
 cedes have all had positive squares. Consequently, among 
 them the equation ar = — 3, which asks, " What is the number 
 whose square is — 3 ? " has no solution. 
 
 A solution is provided by defining a new number, V— 3, as 
 a number whose square is —3. Similarly we define V — a, 
 where a denotes a positive number, as a number whose square 
 is —a. 
 
 The square roots of negative numbers are called imaginary 
 numbers. 
 
 138. If a is positive, V — a may be expressed Va V — I. 
 Similarly, V— 5 = Vo (— 1) = Vo V— 1. 
 
 V-49 = V49(- 1) = 7 V^T. 
 
 139. Real Numbers. In distinction from imaginary num- 
 bers, the numbers hitherto studied are called real numbers. 
 
 WRITTEN EXERCISES 
 Express as in Sec. 138 : 
 
 1. V^9. 4. V^IOO. 7. V^18. 10. a/^12. 
 
 2. V-16. 5. -V^64. 8. -V-32. 11. V-50. 
 
 3. V-25. 6. V^8. 9. -V— 7. 12. —V- 
 
 i •>. 
 
 140. The positive square root of — 1 is frequently denoted 
 by the symbol i ; that is, V— 1 = i. 
 
IMAGINARY AND COMPLEX NUMBERS 91 
 
 Using this we write : 
 
 V^5 = V5 • i. 
 
 V^-49 = ±li. 
 
 V- 75a 2 6 = V3 . 25 a 2 6' .-1=5 aV3& • i. 
 
 Note. Throughout this chapter the radical sign is taken to mean the 
 positive root only. 
 
 WRITTEN EXERCISES 
 
 Rewrite the following, using the symbol i as in Sec. 140 : 
 
 9. 12 -V-~9. 
 
 1. 
 
 2 + V-4. 
 
 5. 25-V-25. 
 
 2. 
 
 3-V-9. 
 
 6. 5— V— 3. 
 
 3. 
 
 4+V-4. 
 
 7. 3 + V-6. 
 
 4. 
 
 5_V-16. 
 
 8. 7 + V-12. 
 
 
 13. -V-6 2 c. 
 
 15. 
 
 10. 2V-100. 
 
 11. 4 V— (a + 6). 
 
 12. Va + V— 6 2 c 2 . 
 
 .'• 
 
 + #— V— xy* 
 
 14. a + V- (a* + x 2 ). 16. p 2 + V- (p -h q) 3 . 
 
 141. Complex Numbers. A binomial one of whose terms is 
 real and the other imaginary is called a complex number. 
 
 The general form of a complex number is a + bi, where a 
 and b may be any real numbers. 
 
 Note. Complex numbers are also simply called imaginary, any ex- 
 pression which involves i being called imaginary. Single terms in which 
 i is a factor (those which we have called imaginary above) are often 
 called pure imaginaries, while the others are called complex imaginaries. 
 Thus, V— 2, 3V— a, 5 i are pure imaginaries and 1 — V— 3, a — V— b 
 are complex imaginaries. 
 
 ORAL EXERCISES 
 
 1. Name the real term and the imaginary term in each 
 exercise of the last set. 
 
 2. Name the values of a and b in each exercise. 
 
 142 Processes with Imaginary and Complex Numbers. After 
 introducing the symbol i for the imaginary unit V— 1, the 
 operations with imaginary and complex numbers are performed 
 like the operations with real numbers. 
 
92 ELEMENTARY ALGEBRA 
 
 I. Addition and Subtraction. 
 
 EXAMPLE 
 
 AddV-9, -V^25, V^3. 
 
 V^9 = 3 i. 
 
 ■ ( 
 
 \ 
 
 - V- 25 = - 5 i. 
 \/="3 = - V3.i. 
 .-. the sum is (3 - 5 - \/S)i = - (2 + V3)i. 
 
 WRITTEN EXERCISES 
 Add: 
 
 1. 2 i, 3 i, -i. 6. 3 + 4 i, 2 - 3 i, 5 + 5 i. 
 
 2. Vl6i, -2i. 7. V-9z 2 , -V-8« 2 . 
 
 3. V=16, -2V :r l. 8. V-(a + &) 2 ,-V(6+c) 2 . 
 
 4. V^4, V^, a/^1. 9. 2V-32a 3 , 3V^8a 3 , &V2L 
 
 II. Multiplication. 
 
 To multiply complex numbers we apply the fact that V— 1 
 . V^L = — 1, or ? 2 = — 1, since the square of the square root 
 of a number is the number itself. 
 
 EXAMPLES 
 
 Multiply : 
 
 1. V^l6 by V :r 9. 
 
 V^16 = 4V^l = 4i\ 
 
 ■v/^9 =3V- 1 =Si. 
 
 .: the product is 12(\/^T)' 2 = (12) (- 1) = - 12. 
 This may be written (4 i)(3 i) = 12 1 2 = - 12. 
 
 2 . 3 - V^ by 2 - V^ 3- a + bibya-bi. 
 
 S-VSi a + bi 
 
 2 - V5 i a ~ M 
 
 6-2V3i a2 + aM 
 
 - 3V5 i + a/151 2 -abi- b 2 i? 
 
 6-(2\/3 + 3V5)f- \/l5. " 2 + 6 * 
 
IMAGINARY AND COMPLEX NUMBERS 93 
 
 143. a + bi and a — bi are called conjugate complex numbers. 
 
 WRITTEN EXERCISES 
 Multiply : 
 
 1. 5-3tby5 + 3i. 7. 4 + ? - by5 — i. 
 
 2. 3 + V^3 by 2 + V^5. 8. a + xi by a — xi. 
 
 3. 5 _ 2 V^l by 3 + 2 V^l: 9. a 2 + bH by a 2 - b 2 L 
 
 4. 5 + V^3 by 5 — V 3 ^. 10. Vr + 3iby Vr — 3i. 
 
 5. 3- V^by 3 + 2V := ~2. 11. V^o by V^ by V^5. 
 
 6. 1 — V^7 by 2 + 3V^7. 12. V— a by V— & by — ci. 
 
 III. Division. 
 
 Fractions (that is, indicated quotients) may be simplified by 
 rationalizing the denominator (Sec. 94, p. 47). 
 
 For example : 
 
 V3~7 = V^T V=T5 _ V7 • V5 (~ I) 2 = V35 . 
 
 V^~5 V^5 V^5 - 5 5 
 
 2+ V3"3 _ (2 + V^S) (3 + V~^5) _ 6 + 3\/^3"+2\/^"5-vT5 
 
 3+ V^5~(3_ V^5)(3 + V^5)~ 9 ~ (- 5 ) 
 
 = ^ (6 + 3 V^3 + 2 V^5"- VlT) . 
 
 g + yi _ (x + yQ 2 _ a; 2 + 2 syi — y 2 
 x — yi (x- yi) (as + yi) ~ x 2 + y 2 
 
 WRITTEN EXERCISES 
 Write in fractional form and rationalize the denominators : 
 
 7. a-r- (a — &i). 
 
 8. (a + bi) -s- (a — 6t). 
 
 9. (3 + 6»)-i-(5+4t). 
 10. (V3-9i)-s-(V2-9i). 
 
 11. (x-V^Ty+ix+V^Ty 
 
 13. (V^-2H-V-5)-(V^ r 5"-V-2). 
 
 1. 
 
 V-6.-S-V2. 
 
 2. 
 
 1 -f- (a + aa"). 
 
 3. 
 
 V-3-5- V-5. 
 
 4. 
 
 Vra-r V — a. 
 
 5. 
 
 l + (2-V-3). 
 
 6. 
 
 4V-l^--2V-4 
 
04 ELEMENTARY ALGEBRA 
 
 144. Powers of the Imaginary Unit. Beginning with r = — 1 
 and multiplying successively by i we find : 
 
 p = f 2 .»•=•_£. f = i 6 • i = — 1 ■ i = — i. 
 
 i* = i?. r°=-l(-l) = +l. i 8 =i*.i 4 =(+l) 2 =+l. 
 
 i 5 = i 4 • j = i. ?' 9 = i' 8 • i = «'. 
 
 145. By means of the values of P, i 3 , i*, any power of i can 
 be shown to be either ± i or ±1. 
 
 For example : * 63 = j' 60 ■ P = 0' 4 ) 15 • *' 3 = l 15 • *' 3 = «' 3 = - *'• 
 
 WRITTEN EXERCISES 
 Simplify similarly : 
 
 1. /'•'. 4. i 16 . 7. i :A . 10. ;' 143 . 
 
 2. i 10 . 5. r 1 . 8. l 56 . 11. i m . 
 
 3. P. 6. J -27 . 9. i 198 . 12. i 3001 . 
 
 Perform the operations indicated: 
 
 13 . (1 + , )2 is. (l-O 3 ^. it. (1 + 0-* 
 
 16 . (=1±**\\ is, ^ 
 
 14. a-/) 3 . v 2 
 
 19. (l + o -(i-0 2 - 20 - (i + o'-a-o 2 - 
 
 IMAGINARIES AS ROOTS OF EQUATIONS 
 
 146. Complex numbers often occur as roots of quadratic 
 equations. 
 
 EXAMPLE 
 
 Solve : x* + x + 1 = 0. (1) 
 
 x* + x = - 1. (2) 
 
 Completing the square, X 2 + X + ^ = \ — 1. (5) 
 
 .■.3! + l=±V=l'. (*) 
 
 Test: (- \ ± jV3 -i) 2 + (- i ± 'W5-0 + 1 =0. 
 
IMAGINARY AND COMPLEX NUMBERS 95 
 
 WRITTEN EXERCISES 
 
 Solve and test, expressing the imaginary roots in the form 
 a + hi : 
 
 1. a- 2 + 5 = 0. 16. 12£ 2 + 24 = 0. 
 
 2. x* + 2x + 2 = 0. 17. 6w 2 +30 = 0. 
 
 3. x* — x + l=Q. 18. 8t 2 + t + 6 = 0. 
 
 4. 8* + x + 5 = 0. 19. 7x 2 + x + 5 = 0. 
 
 5. x 2 + 2x + 37 = 0. 20. 6or + 3a; + l=0. 
 
 6. x 2 -Sx + 25 = 0. 21. 4x 2 + 4a; + 3 = 0. 
 
 7. x 2 - 6 a; + 10 = 0. 22. 12 a 2 + a; + 1 = 0. 
 
 8. m 2 + 4m + 85 = 0. 23. 8 v 2 + 3 v + 6 = 0. 
 
 9. or 2 + 10 a + 41=0. 24. w 2 + 5^o + 6 = 0. 
 
 10. x 2 + 30 .c + 234 = 0. 25. 9z 2 + 2z + 5 = 0. 
 
 11. y 2 _4y + 53 = 0. 26. 7 x 2 - 3 a + 4=0. 
 
 12. z 2 -6z + 90 = 0. 27. 15z 2 + 5z-l=0. 
 
 13. p 2 + 20p + 104 = 0. 28. 16ar-8a; + l = 0. 
 
 14. 2. r 2 + 4^ + 3 = 0. 29. 10a? -2 a + 3 = 0. 
 
 15. 3ar + 2x + l = 0. 30. 7f-t + 1 = 0. 
 
 147. The occurrence of imaginary roots in solving equations 
 derived from problems often indicates the impossibility of the 
 given conditions. 
 
 EXAMPLE 
 
 A rectangular room is twice as long as it is wide; if its 
 length is increased by 20 ft. and its width diminished by 2 ft., 
 its area is doubled. Find its dimensions. 
 
 Solution. 1. Let x = the width of the room, and 2x its length. 
 
 2. Then (2 x + 20) (»- 2) =2 • 2x ■ x, or*- - 8x + 20 = 0. 
 
 3. Solving (2), x = 4 ± 2 i. 
 
 The fact that the results are complex numbers shows that no actual 
 room, can satisfy the conditions of the problem. 
 
96 ELEMENTARY ALGEBRA 
 
 WRITTEN EXERCISES 
 
 Solve and determine whether or not the problems are 
 possible : 
 
 1. In remodeling a house a room 16 ft. square is changed 
 by lengthening one dimension a certain number of feet and by 
 diminishing the other by twice that number. The area of the 
 resultant room is 296 sq. ft. ; what are its dimensions ? 
 
 2. A triangle has an altitude 2 in. greater than its base, 
 and an area of 32 sq. ft. ; find the length of its base. 
 
 3. A train moving x mi. per hour travels 90 mi. in 15 — x 
 hours. What is its rate per hour ? 
 
 SUMMARY 
 
 I. Definitions. 
 
 1. The square roots of negative numbers are called imaginary 
 numbers. S ec . 137. 
 
 2. In distinction from imaginaries, the rational and irrational 
 numbers hitherto studied are called real numbers. Sec. 139. 
 
 3. A complex number is a binomial, one of whose terms is a 
 real number and the other an imaginary number. Sec. 141. 
 
 II. Processes. 
 
 1. After introducing the symbol i for the imaginary unit 
 V— 1, the operations with imaginary and complex numbers 
 are performed like the operations with real numbers. 
 
 Sec. 142. 
 
 2. Any power of i can be expressed by ± i or ± 1. 
 
 Sec. 145. 
 
 3. The solution of quadratic equations may yield complex 
 numbers. In problems this often indicates the impossibility 
 of the given conditions. Sees. 146, 147. 
 
IMAGINARY AND COMPLEX NUMBERS 
 
 97 
 
 SUPPLEMENTARY WORK 
 
 Graphical Representation 
 
 We have seen that positive integers and fractions can be 
 represented by lines. 
 
 Thus, the line AB represents 3, and °. , 1 — *■ - — i — 
 
 -+-» 
 
 the line BC represents 3£. 
 
 Similarly, we have seen that negative integers and fractions, 
 which for a long time were considered to be meaningless, can 
 be represented by lines. 
 
 Thus, the line BA represents — 3, and 
 
 < i ' ' 1 the line CB represents — 3|. 
 
 Irrational numbers can also be represented 
 by lines. 
 
 Thus, in the right-angled triangle abc, the line db 
 represents the y/2. 
 
 I 
 
 A 
 
 x- 
 
 t=& 
 
 ).!. 
 
 B 
 
 y 
 
 ;+1 
 
 •~X 
 
 Like the negative number 
 the imaginary number re- 
 mained uninterpreted several 
 centuries. But this number 
 also can be represented graph- 
 ically. 
 
 -i 
 
 Thus, if a unit length on the 
 y-axis be chosen to represent V— 1 
 J or i, the negative unit — V— 1 or 
 
 — i should evidently be laid off in 
 the opposite direction. 3V— 1 or 
 3 i would then be represented by 
 OA and — 3 i by OB, as in the figure, and others similarly. 
 
 The reason for placing V — 1 or i on a line at right angles 
 to the line on which real numbers are plotted may be seen in 
 
98 
 
 ELEMENTARY ALGEBRA 
 
 the fact that multiplying 1 by V— 1 twice changes + 1 into 
 — 1. On the graph + 1 can be changed into — 1 by turning 
 it through 180°. If multiplying 1 by V— 1 twice turns the 
 line 1 through 180°, multiplying 1 by V— 1 once should turn 
 + 1 through 90°. 
 
 For example : 
 
 1. Eepresent graphically V — 4: 
 
 V^4 = vTi = 2 i i this is represented by a line 2 spaces long drawn 
 upward on the y-axis. 
 
 2. Eepresent graphically — V— 3 : 
 
 — V— 3 = — V3 i = — 1.7 i (approximately) ; this is represented by 
 a line 1.7 i spaces long drawn downward on the ?/-axis. 
 
 WRITTEN EXERCISES 
 
 Represent graphically : 
 
 1. 3i. 
 
 2. -2L 
 
 3. V^9. 
 
 4. V-16. 
 
 5.-5 i. 
 
 6. 5 i. 
 
 7. V-^3 
 8. 
 
 9. — 5V^4. 
 
 10. —31. 
 
 11. + 2V^3. 
 
 V-12. 
 
 taken 
 
 12. 5V-9. 
 
 Complex numbers 
 may be represented 
 graphically by a modi- 
 fication of the plan 
 used in representing 
 imaginary numbers. 
 
 EXAMPLES 
 
 1. Represent graph- 
 ically 3 + i. 
 
 To do this 3 is laid off 
 on the axis of real num- 
 bers, (vx'), and i upward 
 on the axis of imaginaries 
 As in other graphical work this locates the point Pi which is 
 to represent the complex number, 3 + i. 
 
IMAGINARY AND COMPLEX NUMBERS 99 
 
 The number Va' 2 + b- is called the modulu s of the complex number 
 o + M. As appears from the figure, OPi = V3 2 + 1 2 , and hence OP x 
 represents the modulus of 3 + i. 
 
 2. Represent graphically 3 — i. 
 
 The point P 2 is the graph of the complex number 3 — i, and OPg 
 
 represents its modulus. 
 
 3. Represent graphically — 3 — 5 i. 
 
 The point P 3 is the graph of the complex number — 3 — 5 f, and OP3 
 represents its modulus. 
 
 We have thus interpreted by means of diagrams positive and 
 negative integers, positive and negative fractions, positive and 
 negative irrational numbers, and positive and negative com- 
 plex numbers ; in fact, all of the numbers used in elementary 
 algebra. 
 
CHAPTER VII 
 
 QUADRATIC EQUATIONS 
 
 GENERAL FORM 
 
 148. The general form for a quadratic polynomial with one 
 unknown quantity is ax 2 + bx + c, where a, b, and c denote any 
 algebraic expressions not involving x, and where a is not zero. 
 If a is zero the polynomial is linear. 
 
 For example: 1. 5x 2 — 7x + 8. 
 
 Here a = 5, b = — 7, c = 8. 
 
 » 7m -x* + 3x 5m 
 
 2n + l 2w-l 
 
 Here Q = , 7w , 6 = 3, c = ~ 5m 
 
 2b + 1 2 » - 1 
 
 WRITTEN EXERCISES 
 Put the following expressions into the form ax 2 + &x + c: 
 
 1. 3x + 5x(x-2) + 4:(x 2 -5). 4. (a; + 9) - q(s* - 11). 
 
 2. 7(4 x— 1) + + 3)(x — 2). 5. (a& + &)(ca> + d). 
 
 3. a(6x + c)(2dx + 3e). 6. (x 2 — a) + (a* - b). 
 
 7. (x + g)(x+p)-(x-g)(2x-^). 
 
 _ fx,l\* 2/9 x 16V 
 8 - ^ + 3j~3(,T~"3j 
 
 9. x 2 + a& — ax — b(a + x + x 2 ). 
 
 10. x(x-2)(x-4)-x-°(x-5). 
 
 11. (2x + l) 2 -(3x + l) 2 +(4x + l) 2 . 
 
 12. (x - 1)0 - 2)0 - 3) -0 + 1)0 + 2)0 + 3). 
 
 100 
 
QUADRATIC EQUATIONS 101 
 
 149. Similarly, every quadratic equation can be put into the 
 form ax 2 + bx + c = by transposing all terms to the left mem- 
 ber and then putting the polynomial which constitutes the left 
 member into the form ax 2 + bx + c. 
 
 EXAMPLES 
 
 1. (3x + 5)(2x-7) = 3x 2 -4, 
 then 6x 2 - 11 x - 35 = 3 a 2 - 4, 
 
 or, 3x 2 -llx-31 = 0. 
 Here a = 3, b = - 11, c = - 31. 
 
 2. (??ix + 3 a) 2 = mx 2 - 5 (amx - 2), 
 
 then m 2 x 2 + 6 ami + 9o 2 = mx 2 — 5 amx + 10, 
 
 or, (m 2 - m)x 2 + 11 amx + 9 a 2 — 10 = 0. 
 Here a = m 2 — m, & = 11 am, c = 9 a 2 — 10. 
 
 WRITTEN EXERCISES 
 Put the following equations into the form ax 2 + bx + c = : 
 
 1. (a-l) 8 = (aH-l) 3 . 3. (7x-±) 2 = 3z + 2. 
 
 2. a; 2 + ex = /a + g(x + e). 4. ( 5 a ~ ^ = 18 a - 2 a. 
 
 X , \ 2 / X ^ 2 
 
 +a = —a 
 
 a+1 J \a— 1 
 _ a-8x a — 4 « a — 5 a; 
 
 b. — — = • • 
 
 a + 6x a — 3x a + 5x 
 7. (x + l)(x 2 -l) = (x 2 + 1) (a; + 2). 
 
 8 ( - 3a\ 2 _f x 3a\fx 4' 
 V2« 4/\2& Ty\5 9 
 
 _ 2a + 5&4-3z « + & 3a-6 + 2a; 
 y. = • • 
 
 3a — 5 6 + 3 # a— & 2a — b + 3x 
 
 10. (2 a - 4 + a) 2 + 4(a 4- 4 + a;) 2 = (3 a + 6 + 2 x)\ 
 
 11. (2a;-|-4&-3) 2 +(2® + 2& + ll) 2 
 
 = (a; + 3 b - 8) 2 + (3 x + 3 b + 8) 2 . 
 
102 ELEMENTARY ALGEBRA 
 
 METHODS OF SOLUTION 
 
 150. General Solution. By solving the general quadratic 
 equation ax 2 + bx + c = 0, general formulas for the roots are 
 obtained. 
 
 Solve: ax 2 + bx + c = 0. (1) 
 
 Dividing by a, which is not 0, X 2 + — + - = 0. (~) 
 
 a a 
 
 Adding -^— to complete the 
 
 4fl2 hr 7)2 /,2 ,. 
 
 square and subtracting the x 2 +— + — - V - = 0. (5) 
 
 same, a 4 ,/2 4 f< 2 a 
 
 Writing the second term as 
 the square of its square , ? , 2 / y/frl _ 4 ac \ 2 
 
 Factoring (5), 
 
 
 \ 2a 2a j\ 2a 2a ) K ' 
 
 2a V 2a / 2a \ 2a J K l 
 
 Denoting these roots by r x and r 2 : 
 
 -f> + V& 2 -4ac 
 ri = 2a 
 
 — b— V& 2 — 4 ac 
 
 **2 = ~ 
 
 2a 
 By substituting in these formulas the values, including the 
 signs, that a, b, c have in any particular equation, the roots of 
 that equation are obtained. This is called solution by formula. 
 
 EXAMPLE 
 
 Solve : 3 x 2 - 9 x + 5 = 0. (1) 
 
 Here, a = 3, b = - 9, c = 5, (2) 
 
 _ 9 ± V81 - m _ ± V2T ... 
 
 6 (j 
 
QUADRATIC EQUATIONS 103 
 
 WRITTEN EXERCISES 
 
 Solve by formula : 
 
 1. x 2 - x -1 = 0. 9. x 2 -x + 6 = 0. 
 
 2. ar + 3.r + l = 0. 10. 2a 2 — as + 2 = 0. 
 
 3. ar + 2.r-l = 0. 11. 3.C 2 - 2 a; + 1 = 0. 
 
 4. x 2 - 4 a 4- 4 = 0. 12. 7^ + 6a;-4 = 0. 
 
 5. or 9 - 5 x + 6 = 0. 13. 4 a,- 2 -12 x + 9 = 0. 
 
 6. ^-3.r + 2 = 0. 14. 3x 2 + 5x-2 = 0. 
 
 7. a- 2 -13a- + 9 = 0. 15. 5 a 2 - 4x + 6 = 0. 
 
 8. 2x 2 -7x-3 = 0. 16. 7x 2 + 5x-8 = 0. 
 
 151. Literal Quadratic Equations. When any of the coeffi- 
 cients of a quadratic equation involve letters, the equation is 
 called a literal quadratic equation. 
 
 Such equations are solved in the usual way. 
 
 EXAMPLES 
 
 1. Solve: x 2 4- 6 mx + 8 = 0. (1) 
 
 x 2 + 6mx=-8 (£) 
 
 Completing the square, X 2 + 6 mx + 9 m 2 = 9 m 2 — 8. (3) 
 
 .-. (x + 3 m) 2 = 9 ro a -8. (4) 
 
 .-. x + 3ct = ± V 9m 2 -8. (5) 
 
 .-. x = - 3 m± V9 m 2 — 8. (6) 
 
 2. Solve : £ 2 + gt + /i = 0. (i) 
 
 Here = 1, 6 = g, C = h. (#) 
 
 Hence, bv Sec. 150, t - ~9 ± ^9 2 ~ + h . 
 
 2 
 
 (3) 
 
 3. Solve: gt 2 + 2vt = 2s. (1) 
 
 gt 2 + 2vt-2s = 0. (2) 
 
 Here a = g, b = 2 w, c = — 2 s. (3) 
 
 Hence, by Sec. 150, t = - — ± — V(2 I') 2 - 4 gr f-2s) . (4) 
 
 2<7 2g 
 
 = - - ± - Vv 2 + 2gs. (5) 
 
 a g 
 
 = -(-v±Vv 2 + 2gs). (£) 
 
104 
 
 ELEMENTARY ALGEBRA 
 
 WRITTEN EXERCISES 
 
 Solve : 
 
 1. f-+at = 7c. 
 
 2. u 2 4- ku + 1 = 0. 
 
 3. v 2 + mo = 1. 
 
 4. ax 2 + bx + c = 0. 
 
 5. a: 2 4-ax + o = 0. 
 
 6. m 2 ^ 2 + 2 ma; = — 1. 
 
 7. x~ + 2px-l = 0. 
 
 8. 4 a^-4ax +16=0. 
 
 9. aV 4- 2 ax 4- 5 = 0. 
 10. m¥ + 4mx-6=0. 
 
 11. x 2 -4ax = 9. 
 
 12. £ 2 -8*4-24d = 9d' 2 . 
 
 13. 5aar J 4-3fre + 2& 3 = 0. 
 
 14. ay 2 -(a-b)y-b = 0. 
 
 15. 6 2 a^ - 2 to = ac - 1. 
 
 16. w 2 4- 4 aw 4- a 2 = 0. 
 
 17. .T 2 -3«x-(-10a 2 = 0. 
 
 18. v 2 — 4; amv = (a 2 — m 2 ) 2 . 
 
 19. 2ar J -3a = a(3-4a;). 
 
 20. w 2 -a 2 =2b(a- w). 
 
 152. Collected Methods. We have used three methods of 
 solving quadratic equations : 
 
 1. Factoring. 
 
 Equation 
 
 ^-3x4-2 = 0. 
 
 x 2 — (a + b)x + ab = 0. 
 
 Factors 
 
 (x~2)(x-l). 
 (x — a){x — b). 
 
 Roots 
 X = 2, X = 1. 
 
 x = a, x = b. 
 
 2. Completing the square. 
 
 Equation Solution 
 
 cc 2 + x + 2 = 0. See First Course, 
 
 Sec. 285, p. 225. 
 
 ax 2 + bx + c = 0. See Sec. 150. 
 
 Roots 
 
 £C = 
 
 X: 
 
 -l±V-7 
 
 &±V& 2 -4 
 
 ac 
 
 3. Formula. 
 
 Equation 
 
 3ar + 2ic-7 = 0. 
 
 Solution 
 
 See Sec. 150. 
 
 ax 2 + bx + c = 0. See Sec. 150. 
 
 iC = 
 
 # = 
 
 2a 
 
 Roots 
 
 -1±V22 . 
 
 — ■ ■ • 
 
 3 
 
 — b± -y/b 2 — 4 ac 
 2a~~ 
 
QUADRATIC EQUATIONS 105 
 
 WRITTEN EXERCISES 
 
 6. 
 
 a^_a;_30 = 0. 
 
 7. 
 
 x 2 + x-12 = 0. 
 
 8. 
 
 x 2 - 3 x + 2 = 0. 
 
 9. 
 
 x 2 + 11 x + 30 = 0. 
 
 0. 
 
 x 2 - 7 x + 12 = 0. 
 
 Solve by factoring : 
 
 1. x 2 -x-6 = 0. 
 
 2. x 2 -x-2 = 0. 
 
 3. £ 2 + £-2 = 0. 
 
 4. k 2 + jc-6 = 0. 
 
 5. ar 9 + 3£ + 2 = 0. 
 
 Solve by completing the square : 
 
 11. x* + x + l = 0. 15. x 2 - 5x + 10 = 0. 
 
 12. ar + 3£+l=0. 16. x*- 16 x + 60 = 0. 
 
 13. o 2 -^ -|-1 = 0. 17. ar' + f ic -h i = 0. 
 
 14. ^-.9x-+.5 = 0. 18. a? + 1.5 a; -3.5 = 0. 
 
 Solve by formula : 
 
 19. 3£ 2 + £ + 5 = 0. 24. x 2 + l = 0. 
 
 20. 2^ -5 £-3 = 0. 25. x 2 + 15 x + 56 =0. 
 
 21. 4 a; 2 + 3 x -1 = 0. 26. x 2 + 8 x + 33 = 0. 
 
 22. 5ar + 2z + 6 = 0. 27. a 2 - 10 x + 34 = 0. 
 
 23. £ 2 + £ + l = 0. 28. 2 a^ + 3 £-27 = 0. 
 
 Solve and test, using whichever of the methods in Sec. 152 
 seems most convenient : • 
 
 29. 
 
 9 y 2 - 4 = 0. 
 
 30. 
 
 Qx 2 - 13 £ + 6 = 
 
 31. 
 
 5 as 2 — 4 £ + 4 = 0. 
 
 32. 
 
 f + 11 1 + 30 = 0. 
 
 33. 
 
 6«*-5s-6 = 0. 
 
 34. 
 
 6r 2 -2r-4 = 0. 
 
 35. 
 
 w s + 4 w> - 3 = 0. 
 
 36. 
 
 2£-l 2.7- + 1_ 
 
 2£+l 2£-l 
 
 
 8 
 
 37. 
 
 £2-2x + 3 = 0. 
 
 38. 
 
 x 2 - 0.3 a» + 0.9 = 0. 
 
 39. 
 
 aJS_ l.i x + 1.2 = 0. 
 
 40. 
 
 11 0^ + 1 = 4(2- £) 2 . 
 
 41. 
 
 x 2 +(a + b)x + ab = 0. 
 
 42. 
 
 £ 2 -(6 + c)£ + 6c = 0. 
 
 43. 
 
 2a£ +(a-2)£-l = 
 
 44. 
 
 9 2 
 
 6 + £ 6 — £ 
 
106 ELEMENTARY ALGEBRA 
 
 45. -2+-^ + _*<). 
 
 x — 4 a; — 6 x — 2 
 
 46. i(x-l)(x-2) = (x-2%)(x-l§). 
 
 47. The product of two consecutive positive integers is 306. 
 Find the integers. 
 
 Solution. 
 
 1. Let x be the smaller integer. 
 
 2. Then x + 1 is the larger. 
 
 3. .-. x(x + 1) is their product. 
 
 4. .-. x(x + 1) = 306, by the given conditions. 
 
 5. .-. x 2 + x - 306 = 0, from (4). 
 
 6 . ... x = -l±Vl + 1^4 = -1±35 = 17? or _ 18> golving (5) 
 
 Since the integers are to be positive, the value — 18 is not admissible. 
 
 * = 17, /. x + 1 = 18, and the integers are 17 and 18. 
 
 Test. 17 • 18 = 306. 
 
 48. There is also a pair of consecutive negative integers 
 whose product is 306. What are they ? 
 
 49. If the square of a certain number is diminished by the 
 number, the result is 72. Fiud the number. 
 
 50. A certain number plus its reciprocal is — 2. What is the 
 number ? 
 
 51. A certain positive number minus its reciprocal is f-. 
 What is the number ? What negative number has the same 
 property ? 
 
 • 52. One perpendicular side of a certain right triangle is 31 
 units longer than the other; the square of their sum exceeds the 
 square of the hypotenuse by 720. Find the sides. 
 
 53. The perimeter of the rectangle 
 x shown in the figure is 62 in. Find the 
 sides. 
 
 54. In a right triangle of area 60 sq. ft. the difference 
 between the perpendicular sides is 7. Find the three sides. 
 
 55. The sum of the hypotenuse and one side of a right 
 triangle is 162, and that of the hypotenuse and the other side 
 is 121. What are the sides '.' 
 
QUADRATIC EQUATIONS 107 
 
 RELATIONS BETWEEN ROOTS AND COEFFICIENTS 
 
 153. Relation of Roots to Coefficients. By adding and multi- 
 plying the values found for the roots (Sec. 150), we obtain 
 respectively, 
 
 fi + r, — jL 
 
 c 
 
 Applying this result to the equation x?+px-\- q = 0, we 
 
 have: 
 
 n + r 2 = -p, 
 
 »i r 2 = q- 
 
 In words : 
 
 In the equation x 2 +px + q = 0, the coefficient of x ivith its sign 
 changed is the sum of the roots, and the absolute term is their 
 product. 
 
 Every quadratic equation can be put into the form x' 1 + px + q = by 
 dividing both members by the coefficient of x' 1 . 
 
 154. By means of Sec. 153 a quadratic equation may be 
 written whose roots are any two given numbers. 
 
 EXAMPLES 
 
 1. Write an equation whose roots are 2, — 3. 
 
 _^ = n + r 2 = 2+(— 3) = — 1. .-.p = l 
 
 q = nr 2 = 2(— 3) = — 6. 
 .-. x 2 + x — 6 = is the equation sought. 
 
 2. Write an equation whose roots are \ + V — 3, ^ — V — 3. 
 
 -p = n + r 2 = a + V^=3) + (£- V^3)= 1. .-.p = - 1 
 q = IV , = (J + V^3) (i - V^3) = i -(- 3) = V- 
 .• . a:' 2 — x + A£ = is the equation sought. 
 
 WRITTEN EXERCISES 
 Write the equations whose roots are : 
 
 1. 4, 5. 3. 24, 30. , 5. a, -b. 
 
 2. f, f . 4. 8|, 10. 6. 8, - 40. 
 
108 ELEMENTARY ALGEBRA 
 
 7. 7, -If 10. —5. —20. 13. a — bi, a + bi. 
 
 8 . _4, + 4. 11. -f±iV5. 14. 1 + 2/, 1-2 1. 
 
 9. |±V=6. 12. |±|V^47. 15. i-V2, i+V2. 
 
 155. Testing Results. The ultimate test of the correctness 
 of a solution is that of substitution ; but this is not always 
 convenient, especially when the roots are irrational. In such 
 cases, the relations between the roots and coefficients are of use. 
 
 For example : Solving 2 a;' 2 — 5 x + 6 = 0, 
 
 or x 2 — | x + 3 = 0, the roots are 
 
 n = | + \ V- 23 and r 2 = f - \V^ 23. 
 
 Adding, — (r\ + r 2 ) = — - ¥ ° = — f , the coefficient of x. 
 
 Multiplying, n r 2 = (f) 2 - Q V- 23)* = §f + ff = 3, 
 the absolute term. 
 
 Therefore, the roots are correct. (Sec. 153.) 
 
 156. In what follows, the coefficients a, b, c, are restricted 
 to rational numbers. 
 
 157. Character of the Roots. By examining the formula for 
 
 the roots, =-^- — — — — , it appears that the character of the 
 
 2 a 
 
 roots as real or imaginary, rational or irrational, equal or un- 
 equal, depends upon the value of the expression b 2 — 4 ac. 
 
 1. If b 2 — 4 ac is positive, the roots are real. 
 
 Thus, in x 2 + 4 x - 3 = 0, b 2 — 4 ac = 16 + 12, or 28, .-. the roots are 
 real and unequal. 
 
 2. If 6 2 — 4 ac is a perfect square, the indicated square root can be 
 extracted, and the roots are rational. 
 
 Thus, in x 2 - 4 x + 3 = 0, b 2 - 4 ac = 16 — 12, or 4, .-. the roots are 
 rational and unequal. 
 
 3. If b 2 — 4 ac is not a perfect square, the indicated root cannot be 
 extracted and the roots are irrational. 
 
 Thus, in x 2 + 5x + 1 = 0, b 2 - 4 ac = 25 - 4 = 21, .-. the roots are 
 irrational. 
 
QUADRATIC EQUATIONS 109 
 
 4. If 6 2 — 4 fflc = 0, the radical is zero, and the two roots are equal. 
 
 Thus, x 2 - lOx + 25 = 0, b 2 - 4 ac = 100 - 4 • 25 = 0, .-. the roots are 
 equal. 
 
 5. If b 2 — 4 ac is negative, the roots are imaginary. 
 
 Thus, in 2 x 2 — x + 1 = 0, b 2 — 4 ac = 1 — 8, or — 7, .*. the roots are 
 complex numbers. 
 
 Consequently, it is merely necessary to calculate b 2 — 4 ac to know in 
 advance the nature of the roots of a quadratic equation. 
 
 158. Discriminant. Because its value determines the char- 
 acter of the roots, the expression b 2 — 4 ac is called the dis- 
 criminant of the quadratic equation. 
 
 ORAL EXERCISES 
 Without solving the equations, find the nature of the roots of : 
 
 l. a? + x-20 = 0. 9> l_+ lsaXm 
 
 4:X 
 
 2. x* + x-3 = 0. 
 
 3. 2x 2 -x + 2 = 0. 
 
 4. 3x 2 -x-\-3 = 0. 
 
 10. 7ar> + 3a;-4 = 0. 
 
 11. _4 cc + 8x 2 + l = 0. 
 
 12. 5 + 4x- 2 -3a: = 0. 
 
 5. 2s» + 2s~4 = 0. 13 7aj+6 + aj2 = . 
 
 6 . 5aj 2 -3a; + 6=0. 14 _6o; + 9^ + 3 = 0. 
 
 7. 3x- 2 -4z + 5 = 0. 15. £c 2 -6a; + 4 = 0. 
 
 8. 6 z 2 + a- 1 = 0. 16. 5^-1 + ^ = 0. 
 
 159. The relation x 2 -\-px + q = x 2 — (i\ + r 2 ) x + rfa may be 
 written : 
 
 (1) x 2 + px + q = (x — n) (fl5 — r 2 ). 
 
 And since x*+px+ q = °^ + bx + c [ n ^oh. p = & ? q __£. 
 
 a a a 
 
 we have 
 
 (2) ax 2 + to + c = a (a — ?-,) (a — r 2 ). 
 
 160. The solution of a quadratic equation, therefore, enables 
 us to factor every polynomial of either form (1) or (2). 
 
HO ELEMENTARY ALGEBRA 
 
 Since i\ and r 2 involve radicals : 
 
 1. Tlte factors mil generally be irrational. 
 
 2. Tlie factors will be rational when r a and r 2 are so ; that is, 
 when b 2 — 4 ac is a perfect square. 
 
 3. Tlie tiro factors involving x icill be the same when the roots 
 are equal ; that is, when b 2 — 4 ac = 0. 
 
 In the last case the expressions are squares and 
 
 (1) becomes (x — i\) 2 , and 
 
 (2) becomes [-\/a(x — r^] 2 . 
 
 EXAMPLES 
 
 Trinomial W - 4 ac Nature of Factors 
 
 1. 3x 2 — 7x + 2 49 — 4.3-2=25 rational of 1st degree. 
 
 2. 3 x 2 - 7 x + 3 49 - 4 • 3 • 3 = 13 irrational. 
 
 3. 2x 2 -8x + 8 64-4-2-8 = equal. 
 
 ORAL EXERCISES 
 
 By means of the above test, select the squares ; also the 
 trinomials with rational factors of the 1st degree : 
 
 1. 8 a 2 -8 a; + 2. 5. a 2 + 3 a; -2. 9. 6 a- 2 + 5 a: -4. 
 
 2. ^ + 4« + 12. 6. a?x 2 + 2ax+l. 10. 6 x 2 — 5x + 9. 
 
 3. 3^ + 3flJ + l. 7 - 4ar° + 4x- + l. n. 4ar-4a;-3. 
 
 4. 3z 2 + 2z + 12. 8. a-- 2 -8 a; + 15. 12. 8X 2 -9 x + 3. 
 
 161 . The actual factors of any quadratic trinomial of the form 
 ax 2 + bx + c can be found by solving the quadratic equation : 
 
 ax 2 + bx + c = 0, 
 and substituting the roots in the relation : 
 
 ax 2 + bx + c = a (x — r 2 ) (as — r 2 ). 
 
 EXAMPLE 
 Factor: 6 a; 2 + 5 a; -4. CO 
 
 Solving 6*2 + 5* -4 = 0, X = — f, (~) 
 
 Substituting -I for r, a ( x _ ri )(a; - r 2 ) = 6(x + f)(x - £)• (5) 
 5 for r 2 and ti for a. 
 
 Therefore, 6 X 2 4- 5 X - 4 = 6 (x + f) (x - J) . (4) 
 
QUADRATIC EQUATIONS 
 
 111 
 
 Factor : 
 
 1. 3 a: 2 -2 a; -5. 
 
 2. 9 a 2 -3 a; -6. 
 
 3. Qf+y-1. 
 
 WRITTEN EXERCISES 
 
 5. 10w 2 -12w+2. 9. 6 a 2 -7 x + 3. 
 
 6. 9u 2 -17u-2. 10. 5 x 2 - 40 a; + 6. 
 
 7. 6 x 2 + 25 a; + 14. 11. a 2 "* -2 a" -3. 
 
 4. 15 y*-4Ly -35. 8. 2z 2 + 5z + 2. 12. c 4 - 13 c 2 + 36. 
 
 GRAPHICAL WORK 
 
 162. Preparatory. 
 
 1. By counting 
 spaces read the length 
 of EF in the figure. 
 
 2. Is it the square of 
 the length of OE ? 
 
 3. Answer similar 
 questions for GH and 
 OH. 
 
 Every point of the 
 curve is so located that 
 the length of its ordinate 
 is the square of its ab- 
 scissa. 
 
 163. Quadratic ex- 
 pressions may be repre- 
 sented graphically. 
 
 For example : 
 
 The curve in the figure 
 is the graph of y = x?. 
 That is, the length of CD 
 is the square of that of 
 OC ; the length of AB is the square of that of OA ; etc. 
 
 i 
 
 1 1 ] 
 
 / 
 
 
 
 ! 
 
 ■••; 
 
 
 I L..16 
 
 
 
 
 IS J 
 
 1 -; 
 
 
 1 
 
 
 / : 
 
 1 :1 j 
 
 : : 
 
 
 
 \ 
 
 / ■■ 
 
 I ;■■ I 
 
 
 ; 
 
 
 
 -••! f 
 
 
 9 
 
 
 
 [F 
 
 ••••j i 
 
 \ 
 
 
 
 -] I 
 
 
 4 
 
 
 /b 
 
 
 ! ; 
 
 * - : : 
 
 1 
 
 
 
 ■4 ! ■•■- 
 
 
 \ 1 
 
 h 
 
 
 
 
 x 1 ' : 
 
 
 y 
 
 c 
 
 A 
 
 E 
 
 !h L 
 
 
 
 
 1 
 
 2 
 
 i 
 
 14 
 
 
 
 
 
 
 
 
 y 
 
NlWIHEK 
 
 SQUARE 
 
 - 5 
 
 25 
 
 -4 
 
 16 
 
 -3 
 
 9 
 
 -2 
 
 4 
 
 - 1 
 
 1 
 
 -0 
 
 
 
 1 
 
 1 
 
 2 
 
 4 
 
 3 
 
 9 
 
 4 
 
 16 
 
 5 
 
 25 
 
 112 ELEMENTARY ALGEBRA 
 
 WRITTEN EXERCISES 
 
 1. Construct on a large sheet of squared paper the points 
 corresponding to this table of squares. 
 
 2. Then sketch a smooth curve through the 
 points beginning with — 5, 25. 
 
 The work should be carefully done, and the result pre- 
 served for later use. As there are no negative values 
 of x 2 , the x-axis should be taken near the lower edge of 
 the paper. The unit should be chosen quite large ; for 
 example, 10 spaces. Then the table might include squares 
 of numbers increasing by tenths: 1, 1.1, 1.3, etc. The 
 curve will be a graphical table of squares and square 
 roots. 
 
 3. Read to one decimal place from the graph 
 
 V2; V3; V5; V6; V7; V8 
 
 Every ^-distance is the square of the corresponding x-distance ; and 
 every x-distance is the square root of the corresponding ?/-distance. We 
 see that for every y-distance there are two corresponding x-distances, 
 one plus and the other minus, corresponding to the two square roots. 
 Thus the points of the curve for which y = 4 are those whose values 
 of x are 2 and — 2 respectively, i.e. Vi = ± 2. 
 
 164. Graphical Solution of Quadratic Equations. Any value of 
 x which satisfies the system 
 
 \y = -px-q 
 
 makes x 2 equal to — px — q, or x 2 -\-px + q = 0. 
 
 The values of x satisfying the system may be read from the 
 graph of y = x 2 . 
 
 EXAMPLE 
 
 Solve graphically x 2 — x — 6 = 0. 
 
 1. Construct the graph of y = x 2 . (As in Sec. 163.) 
 
 2. Construct the graph of y = x + 6. 
 
 3. They intersect at points for which x = — 2 and + 3. 
 . •• the roots of x 2 — x — 6 = are — 2, 3. 
 
QUADRATIC EQUATIONS 
 
 113 
 
 Notes. 1. Step 2 may be done by simply noting two points of the 
 graph of y — x + 6 and laying a ruler connecting them. The roots can 
 
 be read while the ruler is in position, and thus the same graph for y = x 2 
 can be used for several solutions. 
 
 2. The equation must first be put in the form x 2 + px + q = 0, if not 
 so given. 
 
 WRITTEN EXERCISES 
 
 Solve graphically : 
 
 1. x 2 -5x + § = 0. 
 
 2. x 2 + 3 x + 2 = 0. 
 
 3. .r-2x-3=0. 
 
 4. ^+2^-3=0. 
 
 5. ar-3x-40 = 0. 
 
 6. z 2 + 4;c+4 = 0. 
 
 7. 2^-^-1 = 0. 
 
 8. 3a 2 -2a-l=0. 
 
 9. x 2 + x + ± = 0. 
 
 10. x 2 + x—2=0. 
 
 11. 4f + 4.r + l=0. 
 
 12. a 2 - 9 = 0. 
 
 13. The path of a projectile fired horizontally from an eleva- 
 tion, as at in the figure on p. 114, with a given velocity may 
 
114 
 
 ELEMENTARY ALGP^BRA 
 
 be represented by the graph of the equation y = f— , where 
 
 2v 2 ' 
 
 g = o2 and v is the initial velocity of the projectile in feet 
 per second. Let v = 16 ft. per second and compute the num- 
 bers to complete the table of values of x and y. 
 
 Table 
 
 1 7 ;3 ;4 :5 16 [7 \8 \9 MO 
 
 X 
 
 y 
 
 
 
 
 
 1 
 
 ( ) 
 
 4 
 
 ( ) 
 
 8 
 
 ( ) 
 
 9 
 
 ( ) 
 
 16 
 
 ( ) 
 
 24 
 
 ( ) 
 
 32 
 
 ( ) 
 
 48 
 
 ( ) 
 
 Read from the graph of this table the horizontal distance 
 traveled by the projectile when it is 4 ft. below the starting 
 point. 
 
 14. Construct similarly the path of a projectile whose initial 
 velocity is 32 ft. per second. 
 
 15. A cannon of a fort on a hill is 300 ft. above the plane 
 of its base. The cannon can be charged so as to give the 
 projectile an initial velocity of 100 ft. per second. What range 
 does the cannon cover ? 
 
 16. The enemy is observed at a point known to be 1\ mi. 
 from the foot of the vertical line, in which the cannon stands. 
 With what initial velocity must the ball be discharged to strike 
 the enemy ? 
 
QUADRATIC EQUATIONS 115 
 
 CERTAIN HIGHER EQUATIONS SOLVED BY THE AID OF 
 QUADRATIC EQUATIONS 
 
 165. We have found the general solution of linear and 
 quadratic equations with one unknown. Equations of the 
 third and the fourth degree can also be solved generally by 
 algebra, and certain types of equations of still higher degree 
 as well ; but these solutions do not belong to an elementary 
 course. We shall take up only certain equations of higher 
 degree whose solution is readily reduced to that of quadratic 
 equations. 
 
 CO 
 
 (*) 
 
 (3) 
 (4) 
 (5) 
 
 («) 
 
 The four values, of x are the four roots of the given equation of the 
 fourth degree. Test them all by substitution. 
 
 2. Solve: x 6 - 3 ^-4 = 0. (1) 
 
 Let y= a-s, then y 2 — 3 y — 4 = 0. (2~) 
 
 Solving (2), y — 4, 
 
 and y =" - 1. ' (5) 
 
 .'. by the substitution in (2), X s = 4, or X 3 — 4 = 0, 
 
 and X' - — 1, or X 3 + 1 = 0. {4) 
 
 Factoring (A), X s - 4 = (x - ^I)(X 2 + Vi ■ X + V¥) =0. (5) 
 
 and X 3 + 1 = (X + 1)(X 2 — X+I)=0. (6) 
 
 Solving (j), x=</i, or #4(_.| + jV^3), or ^4(-i-£V=3). (7) 
 
 Solving (6), x=-l, or ±— |\/^3, or i + i\/^3). (5) 
 
 3. Solve: (ar 8 - 3 a; + l)(ar J - 3 x + 2) = 12- CO 
 
 This may be ( x 2 _ 3 x + nr( x 2 _ 3 X + 1) + 1] = 12. (2) 
 
 written 
 If y is put for 
 
 sb* — S.a>+1, the 
 
 equation becomes 2/(2/ + 1) = 12, (o) 
 
 or J/ 2 + y - 12 = 0. (4) 
 
 
 EXAMPLES 
 
 Solve : 
 
 .r 4 - 
 
 9 x- 2 + 8 = 0. 
 
 Let y = .v 2 ; then the given 
 
 equation becomes, 
 Solving for y, 
 
 ,,2 
 
 -92/+8 = 0. 
 
 y = 8 or 1. 
 
 Therefore 
 
 
 x 2 = 8. 
 
 Or, 
 
 
 x 2 = l. 
 
 Solving (A), (5), 
 
 
 x = ± V8, ± 1 
 
116 ELEMENTARY ALGEBRA 
 
 Solving, 
 
 y = 3 or ■ 
 
 -4. 
 
 (5) 
 
 Then from (3), 
 
 x- - 3 * + 1 = 3, 
 
 
 (6) 
 
 and 
 
 x 2 -3x + 1 =-4. 
 
 
 (7) 
 
 Solving (6), 
 
 x = S±VTf - 
 2 
 
 
 (*) 
 
 Solving (7), 
 
 x 3±V-11 
 
 
 (9) 
 
 These are the four roots of the given equation of the fourth degree. 
 
 WRITTEN EXERCISES 
 Solve as above : 
 
 1. x«-7a?+6 = 0. 9. a- 3 = l-a: 6 . 
 
 2. rf-8-M-2=0. 1Q> ^ + 5= 5 
 
 3. a; 10 -5. ^ + 6 = 0. ^ + 3 
 
 4. x 4 + 13x 2 + 36 = 0. 11. x 2n -Ax n -5 = 0. , 
 
 5. *»-3a?+l = 0. 12. 2a: 6 + 5ar ! + 2 = 0. 
 
 6. 12 - x* = 11 x- 2 . 13. se* + aa?-8a 2 = 0. 
 
 7. af" - 6ic n + c = 0. 14. (a; 2 + 4) 2 -4(a: 2 +4)+4=0. 
 
 8 - -it—. r + -A^ = V^ - 15. a 2 +3a;=l 
 
 z 2 + l a 2 + 2 a 2 + 3 -r". - g^ + Sx + 1 
 
 16. (a?-3x + l)(a? — 3x + 2) = 12. 
 
 17. (a; 2 -l) 2 + 2(x- 2 -l) + l = 0. 
 
 18. (x 2 + 5x — l)(ar + 5a + l)= — 1. 
 
 166. Binomial Equations. Equations of the form x n ±a = 
 are called binomial equations. The simpler cases admit of being 
 solved by elementary processes. 
 
 EXAMPLES 
 
 1. Solve : x> - 1 = 0, or x 3 = 1. (7) 
 
 Factoring ai»- 1, (a; — 1) (x 2 + X + 1) = 0. (0) 
 
 Finding equations . . 
 
 equivalent to (2), X = 1, X 2 + £ + 1 = 0. (3) 
 
 _ 1 _ ■«/_ 3 
 Solving (3), X = 1, X = (4) 
 
 Thus we have found the three numbers such that the cube of each is 1, 
 or the three cube roots of unity. 
 
 Verify this statement by cubing each number in step (4). 
 
QUADRATIC EQUATIONS 117 
 
 Note that ( - I + i V^3) 2 = ~ 1 ~ V ~ 3 = ~ 1 ~' V§ , 
 
 2t 2t 
 
 1 *i 4. f i i / — 5\o — 1 + V— 3 — 1 + i V3 
 also that (-i~iv — 3) 2 = ±- = ^ 
 
 Hence, if w stands for one of the complex cube roots of unity, w 2 is the 
 other. 
 
 Every number has three cube roots ; for example, the cube roots of 8 
 are 2, and 2 a> and 2 ufl. 
 
 Verify this by cubing 2, 2 w, and 2 w 2 . 
 
 2. Solve: x 4 + 1 =0, or x A = -1. (1) 
 
 Factoring a*+ 1, (x 2 - l) (x 2 + t') = 0. (2) 
 
 Solving (3), X 2 = I, X 2 = - i. (5) 
 
 Solving (3), X - ± a/7, X = ± V— 7. (4) 
 
 These are the -four numbers, each of which raised to the fourth power 
 equals — 1, or the four fourth roots of — I. 
 
 WRITTEN EXERCISES 
 
 1. Find the 3 cube roots of — 1 by solving X s + 1 = 0. 
 
 2. Find the 4 fourth roots of 1 by solving x* — 1 = 0. 
 
 3. Find the 6 sixth roots of 1 by solving 
 
 (aje - 1) = (ar 3 - l)(x 3 + 1) = 0. 
 
 4. Find the 4 fourth roots of 16 by solving x 4 — 16 = 0. 
 
 5. Find the 3 cube roots of 8 by solving x 3 — 8 = 0. 
 
 6. Show that the square of either irrational cube root of 
 - 1 is the negative of the other irrational cube root. 
 
 7. Show that the sum of the three cube roots of unity is 
 zero ; also that the sum of the six sixth roots of unity is zero. 
 
 SUMMARY 
 I. Forms and Definitions. 
 
 1. A general form of the quadratic equation is : 
 
 ax 2 + bx + c = 0. Sec. 148. 
 
 2. The general forms of the roots of this equation are : 
 
 -6+V6 2 - 
 
 -4 
 
 ac 
 
 2a 
 
 -&-V6 2 - 
 
 -4 
 
 ac 
 
 r 2 = — » — Y ' Sec. 150. 
 
 2a 
 
118 
 
 ELEMENTARY ALGEBRA 
 
 3. The discriminant of the quadratic equation is b 2 — 4 ac. 
 
 If 5-' _ 4 ac is greater than 0, the roots are real and unequal. 
 
 If },2 _ 4 ac i s equal to 0, the roots are real and equal. 
 
 If &2 _ 4 ac i s less than 0, the roots are complex numbers. 
 
 If i,i _ 4 m is a perfect square, the roots are rational. 
 
 If ip. _ 4 ac is not a perfect square, the roots are irrational. 
 
 Sees. 157, 158. 
 
 4. The discriminant of the quadratic equation enables us 
 to rind by inspection the nature of the factors of a quadratic 
 expression. Sec. 160. 
 
 5. In the quadratic equation sc 2 + px + q = 0, the coefficient 
 of x with its sign changed is the sum of the roots, and the 
 absolute term is their product. Sec. 153. 
 
 6. Equations of the form x" ± a = are binomial equations. 
 
 Sec. 166. 
 
 7. The three cube roots of unity are 1, w, and w 2 , where 
 
 iV3 
 
 -l + <-V3 or _-l 
 
 Sec. 166. 
 
 2 2 
 
 1 1. Processes. 
 
 1. Methods of solution: By factoring, by completing the 
 square, by formula. Sec. 152. 
 
 2. Certain higher equations may be solved by the methods 
 of quadratic equations. Sec. 16o. 
 
 REVIEW 
 WRITTEN EXERCISES 
 
 Solve : 
 
 1. x- = 6 x — 5. 
 
 2. to- — w — 1 = 0. 
 
 3. a-2 _ 6 x - 7 = 0. 
 
 4. v- + 2 v + 6 = 0. 
 
 5. x 2 - 5 x + 1 = 0. 
 
 6. 2 ar - .t + 3 = 0. 
 
 7. 3 .r - x + 7 = 0. 
 
 8. x 2 - 5x + 11 = 0. 
 
 9. a 2 — 14 a> + 5 = 0. 
 
 10. 3x*-9x-% = 0. 
 
 H. a; 2 -4a?- 9021 = 0. 
 
 12. a- 2 - 4 a + 9021 = 0. 
 
 13 . a 2 + 4 a - 9021 = 0. 
 
 14 . a? + 4 .r + 9021 = 0. 
 
 15. a 2 + 30 .r + 221 =0. 
 
 16. x 2 - 30 x - 221 = 0. 
 
17. 
 
 9 9 
 
 X b 
 
 18. 
 
 ex'- + foe + a = 0. 
 
 19. 
 
 («-2)(aj + 3) =16. 
 
 £0. 
 
 x 2 = 6 x + 16. 
 
 21. 
 
 24 -lO.r = .T 2 . 
 
 22. 
 
 (x - 1)- = a; + 2. 
 
 23. 
 
 5 a; + x- 2 + 6=0. 
 
 24. 
 
 as» _ 9 x + 14 = 0. 
 
 25. 
 
 a .2 + 3 a, _ 70 _ o. 
 
 26. 
 
 4s 2 -4a;-3 = 0. 
 
 27. 
 
 3a 2 -7<c + 2 = 0. 
 
 28. 
 
 or - 10 cc + 21 = 0. 
 
 
 
 QUADRATIC EQUATIONS 119 
 
 29. a 2 - 10 x + 24 = 0. 
 
 30. x 2 + 10 x + 24 = 0. 
 
 31. a,- 2 = 9 x 2 - (a- + l) 2 . 
 
 32. 9 x 2 + 4 x - 93 = 0. 
 
 33. 4 x 2 + 3 x- - 22 = 0. 
 
 34. 6 x 2 - 13 a? + 6 = 0. 
 
 35. (x+ l)(a?4-2) = x + 3. 
 
 36. — - - + - = c. 
 
 O + X — x 
 
 37. — *— —^ = aa- + 6 2 - 
 a -|-a; 
 
 38 8 + H « + 12 = 1 
 a; + 2 2(a? + 19) 2* 
 
 i)^2,+ 3 ^- 2 ). 
 
 40 
 
 x-3 2 
 
 « — a a; — & _ (a — Iff 
 
 x — b x — a (x — a)(x—b) 
 
 41. (as - l)\x + 3) = *(* + 5) (a; - 2). 
 
 42. x 2 - 6 aca- + a 2 (9 c 2 - 4 b 2 ) = 0. 
 
 State what can be known by means of the discriminant 
 and without solving, concerning the factors of the following 
 trinomials : 
 
 43. 3x 2 -2x + l. 45. 5 v 2 + 20 y + 20. 
 
 44. 4ar + lla;-l. 46. 2x 2 -x + 3. 
 
 Similarly, what can be known about the roots of : 
 
 47. * 2 -9 = 0? 49. z 2 -*-l = 0? 
 
 48. 5x 2 + a,- + 2 = 0? 50. 5a 2 = 9a;? 
 
 How must a be chosen in order that : 
 
 51. The roots of x 2 + ax -f- 5 = shall be imaginary ? 
 
 52. The roots of ax 2 + 6 x + 1 = shall be real ? 
 
120 ELEMENTARY ALGEBRA 
 
 53. The roots of x 2 + 4 x + 2 a = shall be real and of oppo- 
 site signs ? 
 
 54. The roots of (a -f 1) x 2 + 3 x — 2 = shall be imaginary ? 
 
 55. The roots of -Ix 2 — ax + 2 = shall be real and both 
 positive ? 
 
 Find the values of ra for which the roots of the following 
 equations are equal to each other. What are the correspond- 
 ing values of x? 
 
 56. x 2 — 12aj + 3m = 0. . 58. 4x 2 + mx + x + 1 = 0. 
 
 57. mx 2 + 8 x + m = 0. 59. mx 2 -f 3 mx — 5 = 0. 
 
 60. A number increased by 30 is 12 less than its square. 
 Find the number. 
 
 61. The product of two consecutive odd numbers is 99. 
 What are the numbers ? Is there more than one set ? 
 
 62. Find two consecutive even numbers the sum of whose 
 squares is 164. 
 
 63. Find a positive fraction such that its square added 
 to the fraction itself makes f . 
 
 64. If a denotes the area of a rectangle and p its perimeter, 
 show that the lengths of the sides are the roots of the equation 
 
 x 2 — -zX-\-a = 0. 
 
 65. The diagonal and the longer side of a rectangle are to- 
 gether equal to 5 times the shorter side, and the longer side 
 exceeds the shorter by 35 meters. Find the area of the rec- 
 tangle. 
 
 66. A rug 9 ft. by 12 ft. covers \ the floor of a room, and 
 can be laid so that the uncovered strip of floor about the rug 
 is of the same breadth throughout. Find the dimensions of the 
 room. 
 
 67. A company of soldiers attempts to form in a solid 
 square, and 56 are left over. They attempt to form in a 
 square with 3 more on each side, and there are 25 too few. 
 How many soldiers are there ? 
 
 68. It took a number of men as many days to dig a ditch as 
 
QUADRATIC EQUATIONS 
 
 121 
 
 there were men. If there had been G more men the work 
 would have been done in 8 days. How many men were there ? 
 
 69. Solve: 2x 2 + Qx + c = 0. 
 
 What value has c if the two values of x be equal ? If they be 
 reciprocal ? 
 
 70. A public library spends $180 monthly for books. In 
 June the average cost per book was 15 ^ less than in May, and 
 60 books more were bought. How many books were bought 
 in May ? 
 
 71. When water flows from an orifice in a tank the square 
 of the velocity (v) equals 2 g times the height 
 (h) of the surface above the orifice. Write 
 the equation that denotes this fact, g is 
 the " constant of gravity " and may be taken 
 as 32. 
 
 72. What does the square of the velocity, 
 (v 2 ), at A in Fig. 1 equal ? Find this velocity. Fig. l. 
 
 73. What would be the velocity of the water if an opening 
 were made halfway up the tank shown in the figure ? 
 
 74. Find the velocity with which water rushes through an 
 opening at the base of a dam against which the water stands 
 25 ft. high. 
 
 75. The distance on the level from the bottom of the vertical 
 wall to the point where the stream reaches 
 the ground is called the range. The range 
 in Fig. 2 is a. If t is the number of sec- 
 onds taken by the water to reach the ground 
 after leaving o, then a = vt. Take a to be 
 10 ft., and find t after computing v as above. 
 
 76. Water flowing from a leak in a dam reaches the plane of 
 the base 20 ft. from the dam. The opening is 10 ft. below 
 the surface of the water. How long after leaving the dam 
 does the water strike the base? 
 
 77. Water flowing from an orifice in a standpipe reaches the 
 level ground 40 ft. from the base in 3 seconds, 
 the column of water above the orifice ? 
 
 9 
 
 K----CL--* 
 
 Fig. 2. 
 
 How high is 
 
122 ELEMENTARY ALGEBRA 
 
 SUPPLEMENTARY WORK 
 ADDITIONAL EXERCISES 
 
 Solve for x : 
 
 1.1 1 
 
 a a 4- x a 2 — x 2 
 
 = 0. 
 
 Ill 1 
 
 2- - + T H 7 = 0. 
 
 a b x a -\-b-\-x 
 
 3. (2 a - 5 - a;) 2 + (3 a - 3 a-) 2 = (a + 5 - 2 a-) 2 . 
 
 4. (3£-4a + 3&) 2 + (2a + a) 2 =(>-4a+&) 2 
 
 4-(2.e-3a4-4&) 2 . 
 
 5. (7a + 3&+a) 2 +(4a-&-8a) 2 -(4a + 3& + 4a) 2 
 
 = (7 a + & - 6 a) 2 . 
 
 6. (2a + 4& + 6z)(3a-9& + a) = (2a-5& + 5x) 2 . 
 
 7. (a; + a)(5 a - 3 a - 4 &) = (a + a - 2 6) 2 . 
 
 8. (5 x + 4 a + 3 &)(10 aj - 6 a + 8 6) = (5 x + a 4- 7 6) 2 . 
 
 9. (26a; + a 4- 22 6) (14 a?+ 13 a - 26) = (16 x + lla + S b) 2 . 
 10. (8c4-10 4-4x)(18c4-160 4-24z) = (12c + 40 + lla;) 2 . 
 
 14/4-16 2/4-8 2tf 
 
 11 
 12 
 
 21 8/ -11 
 
 a 2 + x 2 (b + c) 2 
 
 a 2 (b - c) 
 
 2 
 
 i« 3a- 2 -27 , 90 + 4X 2 
 13 - ^ + 3 +-^-^9-- 7 ' 
 
 .t 2 +x-2 ^ + 1_ Q 
 x* + 2x-3 a>-3 
 
 1 111 
 
 15. — -= = -+r 
 
 a-\-b—x a b x 
 
 , n / sA 3a + 3aA _ l-2a, Q x l + « 
 
 16. (a-x)^l--_ r j-2 = r -^(c-3)--_ I . 
 
 2c 
 
QUADRATIC EQUATIONS 123 
 
 Quadratic Equations involving Radicals 
 
 If the equation involves but one radical, the method of 
 Sec. 96, p. 48, can be used. 
 
 When the equation resulting from squaring is not directly 
 solvable, substitution is sometimes useful, 
 
 EXAMPLE 
 
 Solve: 5x?-3x + ^5x i -3x + 2 = 18. (1) 
 
 Putting 5x 2 -3x=jr, 
 the given equation be- 
 comes 
 
 y + Vy + 2 = 18. (2) 
 
 Subtracting y, \/y + 2 = 18 — y. (3) 
 
 Squaring, y + 2 = 324 - 36 y + y 2 . (4) 
 
 Rearranging, y 2 — 37 y + 322 = 0. (5) 
 
 » 37±V372-4-322 ^ 
 
 Solving, y = — (6) 
 
 = 23 or 14. (7) 
 
 = 23, (*) 
 
 and 5x 2 -3x = 14. (9) 
 
 Solving these, X = — — , (10) 
 
 and X — 2. or — 1.4. (11) 
 Test. Substituting, it appears that 2 and — 1.4 satisfy the equation. 
 
 The values ° ± ^ satisfy 
 10 
 
 Hence the values of x 
 
 are determined from OX" — o X — Z.3, (A; 
 
 6x 2 - 3x - V5x 2 - 3x + 2 = 18. 
 Sometimes successive squaring is necessary. 
 
 EXAMPLE 
 
 Solve: V2x + 6+V3x + l=8. (1) 
 
 Rearranging, V2x + 6 = 8 - V3x + 1. (£) 
 
 Squaring, 2 x + 6 = 64-16 V3x-fl +3 x+1. (3) 
 
 Collecting, 16V3X+1 = X + 59. (4) 
 
 Sq conecting gain aDd x 2 - 650 X + 3225 = 0. (5) 
 
 Solving, x = 5, or 645. (6) 
 
 Test. Trial shows that the first of these values satisfies the given equa- 
 tion ; and it is obvious on inspection that the second cannot satisfy the 
 equation. 
 
124 ELEMENTARY ALGEBRA 
 
 Sometimes it is best first to transform the given ex- 
 pression. 
 
 EXAMPLE 
 
 Solve: 2x + VI¥+9 = - 2x + 1 + ^ 5x + 6 
 
 2 x - V4 ar + 9 
 
 CO 
 
 Clearing of 
 
 fractions, = 2 X + 1 + V5 X + 6. (2) 
 
 Rearranging, 8 — 2 X — y/b X + 6. (3) 
 
 Squaring and 
 collecting, 4 X z - 37 £ + 58 = 0. (4) 
 
 Hence, x = 2, or 7J. (5) 
 
 Test. By trial, 2 is seen to satisfy the given equation. To avoid the 
 complete work of substituting 7^, we note that every root of (2) must 
 satisfy (3). If x = 7 J, the left member of (3) is negative and the right 
 member positive. Hence 7^ is not a root. It would satisfy (5) if the 
 radical had the negative sign. 
 
 WRITTEN EXERCISES 
 Solve : 
 
 1. a-V3o; + 10 = 6. 3. v^r-Vaj-15 = l. 
 
 2. 6 x- «*- 12 = 0. 4. x%-x$-6 = 0. 
 
 5. x 2 + x + VaF+ x + 1 = — 1. 
 
 6. x- + a + V# 2 + 2 a = — a. 
 
 7. x -2 _j_ 6 a _ 9 = Vx 2 + 6a + 7. 
 
 8. 4 Var + 7 a; + 10 = Vo • Var 9 + 5 a; + 10. 
 9. Var' + 6 a; - 16 + (x + 3) 2 = 25. 
 
 10. 2*-5»* + 2 = 0. is. V^W5 + — 4== 5 - 
 
 11. af* — af*— 6=0. Va?— 5 
 
 12. 3 x~% - 4 a?"* = 7. 14. a; + 2 + (a? + 2)* = 20. 
 
 15. ar 5 + 3.x-3 + Va; 2 + 3a;+17 = 0. 
 
 16. Va7+7+V3a?-2 = - ^±^- * 
 
 V3a;-2 
 
 17. ,,; + y,^T8 = 3a; - V ^-^. 
 
 a; — Var* — 8 
 
 18. Vx(a + .r) + Vx(a — x) = 2 Vax. 
 
QUADRATIC EQUATIONS 125 
 
 Factoring applied to solving Equations 
 Factor Theorem. The relation 
 
 y? +px + q = (a — n) — »») 
 
 is a special case of an important general theorem, known as 
 the factor theorem : 
 
 If any polynomial in x assumes the value zero when a is sub- 
 stituted for x, then x — a «s a factor of the polynomial. 
 
 We prove this first for the particular polynomial 
 
 x* - 3 x 3 + 7 x 2 - 2 x - 10. 
 
 By substituting 2 for x we find that the polynomial assumes the value 
 zero. 
 
 Suppose the polynomial to be divided by x — 2, and denote the quo- 
 tient by Q and the remainder by B, the latter being numerical. 
 
 Then x* - 3 x s + 7 x 2 - 2 x - 16 = (x - 2) § + B. 
 
 In this equation substitute 2 for x ; the left number becomes zero as 
 just seen ; 2 — 2 is also 0, and times Q is 0, no matter what value Q 
 may have ; hence the result of substitution is, 
 
 = + B, or B = 0. 
 
 Consequently x 4 - 3 x 3 + 7 x 2 — 2 x — 10 = (x — 2) Q, or x — 2 is a 
 factor of x 4 — 3 x 3 + 7 x 2 — 2 x — 16. This may now be verified by actual 
 division, but the value of the theorem lies in the fact that by it we know 
 without actual division that x — 2 is a factor of the polynomial. 
 
 Notation for Polynomials. It is convenient to have a symbol to repre- 
 sent any polynomial involving x. For this purpose the symbol P(x) 
 is used. It is read " P of x." P(2) means the same polynomial when 2 
 is substituted for x, P(— V5) means the polynomial when — V5 is put for 
 x ; and P(a) means the polynomial when a is put for x. 
 
 General Proof: Let P(x) denote the given polynomial. 
 
 Suppose P(x) to be divided by x — a. There will be a certain quotient, 
 call it Q(x), and a remainder, B. This remainder will not involve x, 
 
 otherwise the division could be continued. 
 
 » 
 
 We have then : P(x) = (x - a) Q (x) + B. (1) 
 
 In this equation, put a in place of x : 
 
 P(a) = (a-a)Q(a) + B. (8) 
 
12G ELEMENTARY ALGEBRA 
 
 By hypothesis, P(a) = 0, a — a = 0, and R remains unchanged when 
 x is replaced by a, for there is no x in R. Hence 
 
 = + R, or R = 0. (3) 
 
 Substituting this value of R in (1), 
 
 P(x) = (x-a)Q(x). (4) 
 
 That is, x — a is a factor of P (x) , as was to be proved. 
 
 WRITTEN EXERCISES 
 
 In each polynomial substitute the values given for x, and 
 use the factor theorem, when applicable, to determine a factor 
 of the polynomial : 
 
 Polynomial Values fok » 
 
 1. x* + 3 x 3 + 4 x 2 - 12 x - 32 1, 2, - 2. 
 
 2. x* - 7 x 3 + 14 x 2 + x - 21 1, - 1, 2, 3. 
 
 3. 2x 4 + 5x 3 -41x 2 -64x + 80 4,-4,5,-5. 
 
 By use of the factor theorem, prove that each polynomial 
 has the factor named: 
 
 Polynomial Factor 
 
 4. x 3 + 12x2 + 31 x- 20 x + 5. 
 
 c a^-l a 8 -l 
 
 5. x— a. 
 
 x — 1 a— 1 
 
 6 . ^_2x- 2 + — -fm 3 -2m 2 + — ^ a-m. 
 
 m \ x y 
 
 7. (x-a) 2 + (x-^) 2 -(a-6) 2 x-Z>. 
 
 8. x 3 + 2x 2 + 3x + 2 x + 1. 
 
 9. x 3 " + 2x 2n + 3x" + 2 x n + l. 
 
 10. x 3 + ax 2 — aV — a 8 x — a 2 . 
 
 11. 9x 5 +(3a 2 -12a)x 4 -4a 3 x 3 +3a 2 x+a 4 x + 
 
 a 2 
 3' 
 
 Roots and Factors. The factor theorem may be worded 
 thus : 
 
 If a is a root of the equation P(x) = 0, then x - a f s a factor 
 of the polynomial P(x). 
 
QUADRATIC EQUATIONS 127 
 
 It is also evident that : 
 
 If x — a is a factor of P(x), then a is a root o/P(x) = 0. 
 
 For, let P(x) = (x— a) Q (*). 
 
 Putting a for x, P(a) = (a — a)(j)(rt). 
 
 = 0. 
 
 Thus there is a complete correspondence between root of 
 equation and linear factor of 'polynomial. As in quadratic 
 polynomials, so also in all higher polynomials : To every linear 
 factor there corresponds a root, and vice versa. If we know a 
 linear factor, we can at once determine a root ; if we know a 
 root, we can at once write a linear factor. 
 
 This property enables us to write out any equation of which 
 all the roots are known. 
 
 EXAMPLE 
 
 Write the equation whose roots are 2, 3, — 1, 0. 
 
 The factors are x - 2, x — 3, x — (— 1), and, x — 0. 
 The polynomial is (x — 2) (x — 3) (x + l)x. 
 The equation is x 4 — 4 x 3 + x 2 + 6 x — 0. 
 
 WRITTEN EXERCISES 
 
 Write the equations whose roots are : 
 
 1. 3, — 4, 0. 5. a, b, c. 
 
 2. 2, — 2, V5, — V5. 6. a, — b, c. 
 
 3. 4, 1 + V2, 1 - V2. 7. a, - a, b, - b, 0. 
 
 4. 6, - 1, i, - i. 8. 2 - V3 i, 2 + V3 i, 4. 
 
 When a root is known, we thereby know a factor of the 
 given polynomial, and this may aid in finding the other roots 
 of the equation. 
 
 EXAMPLE 
 
 Find the roots of the equation x 3 — 3 x l — 4 x + 12 = 0. 
 
 The equation x 3 - 3x 2 -4x+ 12 =0 (i) 
 
 has the root — 2. (Verify by substitution.) Hence the polynomial has 
 the factor x — (— 2) or x + 2. Finding the other factor by division, the 
 equation may be written, 
 
 (x + 2)(x 2 -5x + 6) = 0. (JS) 
 
128 ELEMENTARY ALGEBRA 
 
 The left member will be zero, either if 
 
 x + 2 = 0, (3) 
 
 or x 2 -5x + G = 0. (4) 
 
 Equation (3) has the root — 2, and equation (4) has the roots 2 and 3. 
 Hence the roots of (2) are 2, — 2, and 3. 
 
 WRITTEN EXERCISES 
 
 Each of the following equations has oue of the six numbers, 
 ± 1, ± 2, ± 3, as a root. Find one root by trial, and then solve 
 completely as in the example : 
 
 1. x s + 5x 2 — 4 : x-20 = 0. 5. ar 3 + a 2 + a- + l = 0. 
 
 2. x s + 3x 2 -16x-±8 = 0. 6. x 9 + 3x 2 -5x-10 = 0. 
 
 3. ^-3^ + 5^-15 = 0. 7. 2a 3 -2a 2 -17a + 15 = 0. 
 
 4. x s + 2x 2 -ax-2a = 0. 8. a- 3 + 3 a- 2 -4a-- 12 = 0. 
 
 9. x 3 - (2 b + 3) x 2 + (b 2 + 6 & - c) x + 3 & 2 - 3 c 2 = 0. 
 
 10. By finding the h. c. f. of their first members, find the 
 two roots that are common to the equations 
 
 a j*_4o 3 + 2<B 2 + a; + 6 = and 2a- 3 -9a- 2 + 7a- + 6 = 0. 
 
 Whenever an equation has the right member zero, any 
 factors that may be given are aids to the solution. The roots 
 corresponding to factors of the first degree can at once be 
 written out. Factors of higher degree determine equations of 
 corresponding degree to be solved. 
 
 WRITTEN EXERCISES 
 Solve by equating each factor to zero : 
 
 1. a(a--l)(a- + 2) = 0. 4. x(x + 2i)(x-7 V§) = 0. 
 
 2. (a + 2)(a--5)(a + V3) = 0. 5. (« - 4) (a 8 - 4) (x + 4) = 0. 
 
 3. (a- 2 -l)(x - 7)(x + 5) = 0. 6. x(x - ai)(x* - 6 2 ) = 0. 
 
 7. x (x - 7) (x + V7)(a-° + 7) = 0. 
 
 8. (a- 2 + 5.T + 6)(a 2 -9a- + 14) = 0. 
 
 9. (a 2 -7a- + ll)(.r + 2a + 5) = 0. 
 
QUADRATIC EQUATIONS 
 
 129 
 
 10. x(x 2 -4)(x 2 + 7x-3)=0. 
 
 11. (x 2 - 25) (a? + 3) (a? - 6 x + 13) = 0. 
 
 12. (x — a) (x + 6) .r (ar — 2 ex + c 2 ) = 0. 
 
 13. (ar + 2 ox- + 6) (x 2 - a 2 b) (5 x - 1) = 0. 
 
 Graphical Work 
 
 Graphical Representation of Solutions. The solutions of ax 2 + 
 hx -f- c = may be represented graphically by plotting the curve 
 corresponding to y = ax 2 -+- bx + c. The solutions of the equa- 
 tion will be represented by the point where the curve cuts the 
 iK-axis. 
 
 EXAMPLE 
 
 Represent the solutions of x 2 -f- x — 2 = 0. 
 
 To draw the curve y — x 2 + 
 x — 2, we make a table of a 
 sufficient number of pairs of 
 values of x and y, plot the cor- 
 responding points, and sketch a 
 smooth curve through them. 
 
 -4 
 
 10 
 
 -3 
 
 4 
 
 -2 
 
 
 
 -1 
 
 -2 
 
 3 
 
 -2& 
 
 1 
 
 2 
 
 -21 
 
 1 
 
 4 
 
 _ 9 3 
 
 
 
 -2 
 
 1 
 
 
 
 2 
 
 4 
 
 3 
 
 10 
 
 4 
 
 18 
 
 Note. When the integral 
 values do not satisfactorily out- 
 line the shape of the curve, frac- 
 tional values of x must also be 
 used as above. 
 
130 ELEMENTARY ALGEBRA 
 
 This graph not only represents the solutions of ar + x— 2 = 0; 
 but, what is far more important, it represents the values and 
 variation of the polynomial x 2 + x — 2 for all values of x 
 (within the limits of the figure). Thus EF represents the 
 value of the trinomial for x = \, and CD that for x=— 2\. 
 The figure also tells us that x 2 + x — 2 is positive when x is 
 negative and numerically greater than— 2 (algebraically less 
 than — 2). That when x is algebraically less than —2 and in- 
 creasing, the trinomial is positive and decreasing, reaching 
 the value zero when x=—2. As x increases, the trinomial 
 continues to decrease, becoming • negative, reaching its least 
 value for x = — \. It then begins to increase, through negative 
 values, reaching zero when x = l. As x increases further, the 
 trinomial continues to increase, becoming positive and continu- 
 ing to increase as long as x increases. 
 
 The graph enables us to read off approximately not only the 
 roots of x 2 + x — 2 = 0, but also those of any equation of the 
 form x 2 + x — 2 = a, if they are real. 
 
 For example : 
 
 1. The roots of x 2 + x — 2 = 3 are indicated by the points where the 
 line drawn parallel to the x-axis through the point 3 on the y-axis cuts the 
 graph. 
 
 2. The line parallel to the x-axis through the point — 3 does not cut 
 the graph at all. This tells us that there are no real values of x which 
 make the trinomial x- + x — 2 equal to —3. The equation x' 2 +x— 2 
 = — 3 has imaginary roots. 
 
 The graph shows (1) that the equation x 2 + x — 2 = a has two real and 
 distinct roots whenever a is positive ; also when a ranges from zero to a 
 little beyond — 2 ; (2) that it has imaginary roots when a is less than a 
 certain value between — 2 and — 3 ; (3) at a certain point the line paral- 
 lel to the x-axis just touches the graph. This corresponds to the case of 
 equal roots of the equation. The value of a in this case may be read 
 from the graph as — 2\. This means that the equation x 2 + x — 2 = — 2J, 
 or x' 2 + x + \ = 0, has equal roots. This may be verified by solving the 
 equation or by applying the discriminant (Sec. 384, p. 349). 
 
 Note. Since even the best of drawing is not mathematically accurate, 
 results read off from a graph are usually only approximately correct. 
 The closeness of the approximation depends on the degree of accuracy in 
 the drawing. 
 
QUADRATIC EQUATIONS 131 
 
 WRITTEN EXERCISES 
 Treat each trinomial of Nos. 1 to 9 below as follows : 
 
 (1) Draw the graph of the trinomial. 
 
 (2) From the graph discuss the variations of the trinomial. 
 
 (3) From the graph read off approximately the roots of the 
 equation resulting from equating the trinomial to zero. 
 
 (4) If the trinomial be equated to a, read from the graph 
 the range of values of a, for which the roots of the equation 
 are, (i) real and distinct, (ii) real and equal, (iii) imaginary. 
 
 (5) Last of all, verify those of the preceding results which 
 relate to roots of the equations by solving the equations or by 
 use of the discriminant. 
 
 x 2 + 4 x — 5. 7. x 2 + 2 x + 3. 
 
 3a 2 - 2x. . 8. x*-2x-l. 
 
 ar-4. 9. 8 + 2X-X 2 . 
 
 10. Draw the graph of the function 2 x* + 5& — 3, and 
 state how it varies as x varies from a negative value, numer- 
 ically large at will, through zero to a large positive value. For 
 what values of x is the function positive ? For what values 
 negative ? 
 
 11. What is the graphic condition that ax 2 + bx + c shall 
 have the same sign for all values of x ? What must therefore 
 be the character of the roots of ax 2 + bx + c = 0, if the tri- 
 nomial ax*-\-bx + c has the same sign for all values of x? 
 
 Determine m so that each of the following trinomials shall 
 be positive for all values of x : 
 
 12. x 2 + mx + 5. 13. 3X 2 — 5x + m. 14. mx 2 + 6 x + 8. 
 
 1. x 2 + 5x — 6. 
 
 4 
 
 2. x* + 3x + 2. 
 
 5 
 
 3. 2a? — 5aj + 2. 
 
 6 
 
CHAPTER VIII 
 SYSTEMS OF QUADRATIC AND HIGHER EQUATIONS 
 
 SIMULTANEOUS QUADRATIC EQUATIONS 
 
 167. Two simultaneous quadratic equations with two un- 
 knowns cannot in general be solved by the methods used in 
 solving quadratic equations, because an equation of higher 
 degree usually results from eliminating one of the unknoAvns. 
 But many simultaneous quadratic equations can be solved by 
 quadratic methods, and certain classes of these will be taken 
 up in this chapter. 
 
 168. Class I. 
 
 A system such that substituting from one equation into the other 
 produces a quadratic equation. 
 
 1. A system of equations composed of a linear equation and a 
 quadratic equation can always be solved by substitution. 
 
 
 EXAMPLE 
 
 
 
 J3a; + 42/ = 24. 
 
 CO 
 
 r e : 
 
 { x> + y* = 25. 
 
 (2) 
 
 From (i), 
 
 3 x - 24 - 4 y. 
 
 (3) 
 
 From (3), 
 
 x = 8 ^« 
 
 3 
 
 (4) 
 
 Substituting (4) in (2), 
 
 (8-^) 2 +2/ 2 = 25. 
 
 {5) 
 
 Simplifying (5), 
 
 25 2/2-192?/ + 351 =0. 
 
 (G) 
 
 Factoring (6), 
 
 O/-3)(25.y-ll7)=0. 
 
 (7) 
 
 Solving (7), 
 
 y = 3 and 2/ = — • 
 9 J 26 
 
 (8) 
 
 From (/,), 
 
 44 
 
 x = 4 and x = 
 
 25 
 
 (9) 
 
 132 
 
SIMULTANEOUS QUADRATIC EQUATIONS 133 
 
 Test. 3 . 4 + 4 . 3 = 24 and ?Li* + i^l = 24. 
 
 25 25 
 
 32 + 42 _ 25. 
 
 It should be noted that to every x there corresponds only one particular 
 value of y, and vice versa. The proper correspondence may be seen by 
 noticing which value of one unknown furnishes a given value of the other 
 in the process of solution. In the example above, y = 3 produces x = 4. 
 
 2. When both equations are quadratic, substitution is appli- 
 cable if the result of substitution is an equation having the quad- 
 ratic form. 
 
 EXAMPLE 
 
 • (x 2 + y 2 = 25. (1) 
 
 1 ay = 12. {2) 
 
 Solve 
 
 12 
 
 From (2), X = (3) 
 
 y 
 
 (10\2 
 — j + y 2 — 25. (4) 
 
 Simplifying (U), we obtain an „ 4 _ 05 u 2 1 1 44 _ n (k\ 
 
 equation in quadratic form, J y ~ • \ J 
 
 Factoring (5), (y* - 16) (^ - 9) = 0. (6) 
 
 Solving (6), y = ± 4, and ±3. (7) 
 
 Substituting (7) in (3), X = ± 3, and ± 4. (5) 
 
 Test. Taking both values to be positive, or both to be negative, 
 
 (±3)2 + (±4)2 = 25. 
 
 (±3) -±4 = 12. 
 
 There are frequently various methods of solving the same problem. 
 Thus, in the last example, multiply (#) by 2, add it to or subtract it from 
 (1) ; the resulting equations (x + y)' 2 — 49, and {x — y) 2 = 9, can be 
 solved by extracting the square roots, and adding and subtracting the 
 results. 
 
 „ , _ WRITTEN EXERCISES 
 
 Solve and test : 
 
 1. x 2 + y* = 13, 3. 3 x 2 - 2 xy = 15, 
 2x + 3y = 13. 2x+3y = 12. 
 
 2. 2x 2 -y 2 = U, 4. x 2 -xy + y 2 = 3, 
 3x + y = ll. 2x + 3y = 8. 
 
134 ELEMENTARY ALGEBRA 
 
 5. x 2 + y 2 = 2xy, 11. x — y = l, 
 
 x + y = 8. x 2 — y 2 = 16. 
 
 6. ar + y 2 = 25, 12. 3xy + 2x + y = 4S5, 
 x + y = l. 3x-2y = 0. 
 
 7. x — y = 1 £-, 13. .r -}->/ = a, 
 a# = 20. . c 2 + ?/ 2 = &. 
 
 8. z 2 -?/ 2 = 16, 14. x — y = b, 
 x-y = 2. xy = a 2 . 
 
 9 ?L = ^_i^ 15 - 2x = x2 -y 2 > 
 
 ' y 2 ~ 9 ?/ ' 2 a = 4 xy. 
 
 x-y = 2. ± 4 
 
 16. -=P-, 
 
 10. 2z + ?/=7, 2/ ooj 
 
 ar + 2?/ 2 = 22. z 2 -?/ 2 = 81. 
 
 169. Class II. 
 
 A system in which one equation has only terms of the second 
 degree in x and y can be solved by finding x in terms of y, or 
 vice versa, from this equation and substituting in the other. 
 
 Solve 
 
 
 ( x 2 — 5 xy + 6 y 2 = 0. 
 
 CO 
 
 r e: 
 
 a; 2 -/ = 27. 
 
 (*) 
 
 Dividing (J) by y 2 , 
 
 x 2 Say °y 2 = o, 
 J/ 2 2/ 2 y 2 
 
 (5) 
 
 Simplifying (3), 
 
 (iM;) +6=0 - 
 
 (4) 
 
 Factoring (4), 
 
 (H(H= ft 
 
 (5) 
 
 Solving (5), 
 
 ? = 2, - = 3. 
 
 2/ y 
 
 (e) 
 
 From (6), 
 
 x = 2 y, x = 3 y. 
 
 (7) 
 
 Substituting x = 1y in (2), 
 
 (2 y) 2 - ,/-• = 27. 
 
 (*) 
 
 Solving (*■), 
 
 y=±3. 
 
 (0) 
 
SIMULTANEOUS QUADRATIC EQUATIONS 135 
 
 x = ±6. 
 
 Substituting (9) in x=2y, 
 Substituting se= 3 y in (2), 
 
 Solving (11), 
 
 Substituting (12) in x=Zyl 
 
 Test. 
 
 Taking both values to f (±6) 2 -5(± 3) (±6) + 6(± 3) 2 = 0. 
 
 be positive or both -{ 
 
 negative, I (±6) 2 - (±3) 2 = 27. 
 
 (10) 
 (11) 
 (12) 
 
 (13) 
 
 Taking the signs as 
 before, 
 
 ± 9 \/6\ 2 
 
 -^m) + °m=°- 
 
 {m°-m=*- 
 
 Notes. 1. When the equation in -, as in the fourth step, cannot he 
 
 y 
 
 factored hy inspection, the formula for solving the quadratic equation is 
 used. Putting - — s, a quadratic equation is obtained whose solution 
 
 y 
 
 will lead to the results of step (7) above. 
 
 2. When an equation has the right member zero, it is unnecessary to 
 divide by x 2 or y 2 , if the left member can be factored by inspection. 
 Thus, in (1) above, (x — 3 y) (x — 2 y) = 0, hence x = 3 y and x = 2 y, as 
 
 in (7). 
 
 Solve : 
 
 WRITTEN EXERCISES 
 
 1. x 2 -2xy-3y°- = 0, 
 xr + 2y 2 = 12. 
 
 2. a? + Xy + tf = 0, 
 
 ^ + ^ = -1. 
 
 3. 6 x 2 4- 5 xy + y 2 = 0, 
 
 y*-. X -y = 32. 
 
 4. 2a? — 5xy + 2y 2 = 0, 
 4aj 8 -4/ + 3« = 330. 
 
 5. x(x + ?/) = 0, 
 
 a 2 _ ^ + y- = 27. 
 
 6. x 2 — ?/ 2 = 0, 
 
 ar 6 
 
 3xy + y = 105. 
 
 7. 3x 2 -2xy-?/ 2 = 0, 
 
 8. 4 or + 4 £?/ + ,?/'-' = 0, 
 x + 3y 2 -2x = 195. 
 
 9. x 2 + 3a;-4 ?/ + xy = 33, 
 x 2 + 7 x?/ + 10 if = 0. 
 
 10. iB 2 4-aV = 0, 
 
 a 2 + ?/-/ = - a 2 . 
 
136 ELEMENTARY ALGEBRA 
 
 170. Class III. 
 
 A system of two simultaneous quadratic equations whose terms 
 are of the second degree in x and y ivith the exception of the ab- 
 solute terms can be solved by reducing the system to one of 
 Class II. 
 
 This can be done in two ways : 
 
 1. Make their absolute terms alike, and subtract. The result- 
 ing equation has every term of the second degree in x and y. 
 
 EXAMPLE 
 
 Solve: f^ + ^ = 66, (i) 
 
 \ rf - f = 11. (2) 
 
 Multiplying (2) by 6, 6 X 2 — 6 y 2 = 66. (3) 
 
 Subtracting (i) from (3), 5 X 2 — xy — 6 y 2 — 0. (4) 
 
 Factoring (4), (5 X — 6 y) (X + y) = 0. (5) 
 
 Expressing x in terms of y, X — § y and X = — y. (6) 
 
 Substituting as=§ y in (2), §| y 2 — y 2 — 11. (7) 
 
 Solving (7), ?/ = ± 5. (5) 
 
 Substituting i/=± 5 in a; = | y, X = ± 6. (9) 
 
 Similarly, substituting as=- y in (2), y 2 — y 2 = 11. (10) 
 
 But this leads to = 11. (11) 
 
 (Zi) being impossible, the solution is 
 
 x = ± 6, y = ± 6. (12) 
 
 TEST. Taking the values [ (±6) 2 + ( ±6)( ± 5) = 66. 
 to be both positive < f +6Y 2 — f + 5V 2 = 11 
 
 or both negative, { \^ / \ ^ / 
 
 2. Substitute vxfor y throughout the equations and solve for v. 
 
 EXAMPLE 
 
 2^-3^ + ^ = 4, CO 
 
 l x 2 -2x 2 / + 3r = 9. (0) 
 
 Putting y=vx in (1) and (2), 2 x 2 — 3 UX 2 + V 2 X 2 = 4, (5) 
 
 and, x 2 - 2 VX 2 + 3 t? 2 X 2 = 9. (4) 
 
 Factoring (3) and (4), X 2 (2 - 3 V + t' 2 ) = 4. (5) 
 
 x 2 (l -2t7 + 3u 2 )=9. (6) 
 
SIMULTANEOUS QUADRATIC EQUATIONS 
 
 137 
 
 Equating the values of X 2 in 4 _ 9 
 
 (5) and (6), 2 — 3 V + V* 1 - 2 v + 3 v 2 
 
 • (?) 
 
 Clearing (7) of fractions, 3 V 2 + 19 V — 14 = 0. 
 
 (*) 
 
 Solving (S), V -— 7, §. 
 
 (5) 
 
 Since y = vx, V — ~ ' X, 
 
 (20) 
 
 2x 
 and, 2/ — -^ - " 
 
 U-0 
 
 From U), 2 x 2 + 21 x 2 + 49 x 2 = 72 x 2 = 4. 
 
 (12) 
 
 '.Solving (12), * = ± T ^2- 
 
 6 
 
 (13) 
 
 From (10), 2/ = =F 7 ^ 2 - 
 
 6 
 
 (H) 
 
 Similarly, from (11) and (1), X = ± 3, 
 
 (15) 
 
 From (15) and (1/), # = ± 2. 
 
 (16) 
 
 Test as usual. 
 
 
 „ . WRITTEN EXERCISES 
 Solve : 
 
 
 1. a- 2 + ?/ 2 = 41, 9. aj 2 -ajy = 54, 
 a*/ = 20. a*/ - y 2 = 18. 
 
 
 2. x 2 + xy = a?, 10 - ^ + ^ = 12, 
 tf + xy = b 2 . tf + xy = 24;. 
 
 o 2 , , * 7 11- 4x 2 + 3?/ 2 = 43, 
 
 3. x 2 + xy + y = 7, J 
 
 4. z 2 + ^-2</ 2 = 4, 12. E±* + *=2'=|, 
 
 „ n x—y x+y 4 
 
 x 2 -3xv + 2 = 0. 2 i 2 on 
 
 ? ' ar + ?r = 20. 
 
 5. x 2 +6ajy + 2^ = 133, 13 o x 2 + 3 xy + f- = 80, 
 
 x 2 -y 2 = lG. X y-x 2 =6. 
 
 6. jb 2 - 12 a# + 119=0, 14 ^ 2 _47/ 2 = 20, 
 2^-2^ + 24 = 0. ^ = 12 . 
 
 7. a 2 aj 2 + 6Y = c a , 15. a: 2 - ay + ?/ 2 = 21, 
 x*-tf = c?d?. 2xy-y 2 = 15. 
 
 8. ar + ?/ 2 = 13, 16. 4 ^ + 3/ = 43, 
 xy = 6. 3ar-2/ 2 -3 = 0. 
 
 10 
 
138 ELEMENTARY ALGEBRA 
 
 171. Class IV. 
 
 A system in which each equation is unaltered when x and y are 
 interchanged can be solved by letting x = u + v and y = u — v. 
 
 EXAMPLE 
 
 Solve: l*+t-.-f-1», (1) 
 
 { xy + x + y = S9. («) 
 
 Let x = u+ v, y=u - v, then (1) becomes 
 
 (u + v) 2 +(u-v) 2 -(u + v )-(u-v) = 78, (S) 
 
 and (2) becomes 
 
 (u + v)(u — v) + (u + v) + (u - v)=39. (4) 
 
 Simplifying {3), 2 U 2 + 2 V 2 — 2 U = 78. (5) 
 
 Simplifying (4), M 2 — V 2 + 2 U = 39. (6) 
 
 Dividing (5) by 2, and sub- „ „ '„ _ fl . 
 
 trading (6) from the result, ^ » — ou — u. ^/; 
 
 Expressing u in terms oft 1 , W — rLil— f^") 
 
 3 v J 
 
 Substituting (8) in (6), i^! + ^L - 39. (of) 
 
 9 3 W 
 
 Simplifying (9), 4 tf* + 3 V 2 - 9-39 = 0. (10) 
 
 Solving (JO), » 2 = — ^o (-7-0 
 
 and, « 2 = 9. (12) 
 
 Solving (ii) and (IS), V = ± £ V— 39. (i5) 
 
 Substituting (i3) and UA) in (S), v = ± 3, (i\f) 
 
 Then, « = -¥, or 6. (i5) 
 
 -13±V-39 ,_>. 
 
 Also X = U+V = ^ (16) 
 
 = 6 ± 3 = 9 or 3, (i7) 
 
 - 13TV-39 ,, ON 
 
 and, y = u — V = — (18) 
 
 Also =6 =F3 = 3 or 9. (19) 
 
 Such equations may often be solved by some of the previous methods, 
 in which case it is usually preferable to do so. 
 
1. 
 
 xy — (x + y) — l = 0, 
 
 
 xy = 2. 
 
 2. 
 
 y x 2 
 
 
 M-4 
 
 SB ?/ 
 
 3. 
 
 ^ + 2 ,2 = 39, 
 
 
 y — » = 3. 
 
 4. 
 
 v*-xy + y i - r ? = Q, 
 
 
 x — y + l = 0. 
 
 SIMULTANEOUS QUADRATIC EQUATIONS 139 
 
 , WRITTEN EXERCISES 
 
 Solve : 
 
 5. x 2 — xy + tf = 12, 
 
 x- + xy + f = 4. 
 
 6. zy = 3(z4-y) } 
 
 a* + f = 160. 
 
 7 . ^ + ?/ 2 + ^ + 2/ = 188, 
 a-?/ = 77. 
 
 8. x 2 -xy + y* = 19, 
 
 xy = 15. 
 
 9. 2x- + 2f-(x-y)=9, 
 tf = l. =1. 
 
 172. The foregoing classes of simultaneous quadratic equa- 
 tions are applied in the following problems. 
 
 WRITTEN EXERCISES 
 
 1. Two square floors arte paved with stones 1 ft. square ; 
 the length of the side of one floor is 12 ft. more than that of 
 the other, and the number of stones in the two floors is 2120. 
 Find the length of the side of each floor. 
 
 Solution. 
 
 Let x be the length in feet of a side of the smaller floor and y be the 
 
 length of that of the other, then 
 
 x = y—12. (i) 
 
 and by the given conditions, X 2 + y 2 = 2120. (~) 
 
 Substituting (i) in (2), (y - 12) 2 + y 2 = 2120. (8) 
 
 Simplifying (3), y 2 - 1 2 y - 988 = . (4) 
 
 Solving (4), ?/ = 38 and — 26, (5) 
 
 Substituting (5) in (l), x = 26 and - 38. (<?) 
 
 The negative values not being admissible, the squares are 26 ft. and 38 ft. on 
 
 a side. (7) 
 
 Test. 26 2 + 38 2 = 2120. 
 
140 ELEMENTARY ALGEBRA 
 
 2. The sura of the sides of two squares is 7 and the sum 
 of their areas is 25. Find the side of each square. 
 
 3. The hypotenuse of a certain right triangle is 50, and the 
 length of one of its sides is f that of the other. Find the sides 
 of the triangle. 
 
 4. The difference between the hypotenuse of a right triangle 
 and the other two sides is 3 and 6, respectively. Find the sides. 
 
 5. A number consists of two digits ; the sum of their 
 squares is 41. If each digit is multiplied by 5, the sum of 
 these products is equal to the number. Find it. 
 
 6. The difference between two numbers is 5 ; their product 
 exceeds their sum by 13. Find the numbers. 
 
 7. In going 120 yd. the front wheel of a wagon makes 6 
 revolutions more than the rear wheel ; but if the circumference 
 of each wheel were increased 3 ft., the front wheel would make 
 only 4 revolutions more than the rear wheel in going the same 
 distance. Find the circumference of each wheel. 
 
 8. The diagonal of a rectangle is 13 in. ; the difference be- 
 tween its sides is 7 in. Find the sides. 
 
 9. The diagonal of a rectangle is 29 yd., and the sum of its 
 sides is 41 yd. Find the sides. 
 
 10. The sum of the perimeters of two squares is 104 ft. ; the 
 sum of their areas is 346 sq. ft. Find their sides. 
 
 11. The difference between the areas of two squares is 231 
 sq. in. ; the difference between their perimeters is 28 in. Find 
 their sides. 
 
 12. Two steamers set out simultaneously from San Fran- 
 cisco and Honolulu, 2100 mi. apart, and travel toward each 
 other ; they meet 840 mi. from Honolulu in 1\ da. less than 
 the difference between their rates in miles per hour. Find the 
 rate of each per hour. 
 
 13. Two trains leave New York simultaneously for St. 
 Louis, which is 1170 mi. distant ; the one goes 10 mi. per hour 
 faster than the other and arrives 9| hr. sooner. Find the 
 rate of each train. 
 
SIMULTANEOUS HIGHER EQUATIONS 141 
 
 SIMULTANEOUS HIGHER EQUATIONS 
 
 173. Certain systems containing higher equations can be 
 solved by the methods of this chapter. 
 
 EXAMPLES 
 
 1. Solve: 
 
 f ar 5 + y 3 =18 xy, 
 \x +y =12. 
 
 GO 
 
 {2) 
 
 Put a; = u + v, and 
 
 y = 11 — w, 
 
 
 Then, from {1) 
 
 (?< + v) 3 + (u — v) 3 =-18(?< + fl)(M — v), 
 
 (3) 
 
 and from (2), 
 
 (u + v) + (u -v) = 2u = 12. 
 
 (4) 
 
 Combining, (U) 
 and (3), 
 
 216+ 18w 2 = 9(36-« 2 ). 
 
 (-5) 
 
 Solving (5), 
 
 » = ±2. 
 
 (C) 
 
 •'• 
 
 a; = u + v = 8 and 4, 
 
 (7) 
 
 and 
 
 y = tt — v — 4 and 8. 
 
 (-5') 
 
 Test as usual. 
 
 
 
 2. Solve: 
 
 (x* + y 4 = 706, 
 U-y =2. 
 
 00 
 
 09) 
 
 Put a? = u+t>, 
 
 
 
 y =u — V, 
 Then from (1) 
 
 (m+v) 4 +(m-») 4 =706, 
 
 (3) 
 
 and from (2), 
 
 (?« + ») — (« — v)= 2. 
 
 00 
 
 Simplifying (&), 
 
 0=1. 
 
 (5) 
 
 From (5) and (3), 
 
 (« + l) 4 + (?<-l) 4 = 706. 
 
 (6) 
 
 Simplifying (6), 
 
 M 4 + 6 ifi - 352 = 0. 
 
 (7) 
 
 Solving (7). 
 
 u 2 = - 22. 
 
 (*) 
 
 and 
 
 u 2 = 16. 
 
 (5) 
 
 • 
 • • 
 
 rt= -fcV-22, ±4. 
 
 (20) 
 
 x = u+v=±V— 22+1. (22) 
 k = ± 4 + 1 = 5, - 3. (2£) 
 
 and, y = u— f»= ± V-22— 1. (25) 
 
 t/=±4-l = -5, +3. {14) 
 
 Test as usual. 
 
142 ELEMENTARY ALGEBRA 
 
 g , . WRITTEN EXERCISES 
 
 1. x 3 + y 3 = lS9, 4. x 4 + y* = 81, 
 x + y = 9. x + y = 5. 
 
 2. x 3 + y* = 72, 5. x*+y* = 2, 
 x + y = 6. x + y = 2. 
 
 3. x* + y* = l&9, 6. x 4 + y 4 = 10,001, 
 tfy + xy 1 = 180. x — y = 9. 
 
 SUMMARY 
 
 1. Certain classes of simultaneous quadratic equations can 
 be solved by quadratic methods : Sec. 167. 
 
 (1) Class I. A system of equations such that substituting 
 from one equation into the other produces an equation of 
 quadratic form. Sec. 168. 
 
 (2) Class II. A system in which one equation has only 
 terms of the second degree in x and y can be solved by finding 
 x in terms of y, or vice versa, from this equation and substitut- 
 ing in the other. Sec. 169. 
 
 (3) Class III. A system of two simultaneous quadratic 
 equations whose terms are of the second degree in x and y, 
 with the exception of the absolute terms, can be solved by 
 reducing the system to one of Class II. 
 
 This can be done in two ways : 
 
 Make their absolute terms alike and subtract. The result- 
 ing equation has every term of the second degree in x and y. 
 
 Substitute vx for y throughout the equations and solve 
 for v. Sec. 170. 
 
 (4) Class IV. A system in which each equation is unaltered 
 when x and y are interchanged, can be solved by letting 
 x = u + v and y = u— v. Sec. 171. 
 
 2. Certain simultaneous higher equations can be solved by 
 quadratic methods. Sec. 173. 
 
QUADRATIC AND HIGHER EQUATIONS 143 
 
 REVIEW 
 
 _ , WRITTEN EXERCISES 
 
 Solve : 
 
 1. x + y = 7.5, 13. af 5 — y 3 = 665, 
 xy = 14. x — y = 5. 
 
 2. 3x-2y = 0, 14. ^ + 0^ + ^=19, 
 xy = 13.5. a;?/ = 6. 
 
 3. x + y = 7, 15. (a + 2)(z-3) = 0, 
 ar J -?/ 2 = 21. x 2 + 3xy + y 2 = 5. 
 
 4. a; — ?/ = 5, 16. a; 2 — / = 3, 
 
 x 2 + y 2 = 37. x 2 + 2/ 2 -^ = 3. 
 
 5. a; -y =1, 17 xj\-y + x-ji = 5 
 3x 2 + y 2 = 31. x-y x + y 2' 
 
 a?+V = 90. 
 
 tf + xy + y 2 -- 
 
 x* + x*tf+y i = 21. 
 
 x 2 + y 2 -l = 2xy, 
 xy(xy + 1) = 8190. 
 
 6. x—y =5, 
 
 x 2 + 2xy + y 2 = 75. 18. a^ + agf + y* : = 7, 
 
 7. x + y = 7(x-y), 
 x 2 + y 2 = 225. 19- ar ! + ?/ 2 --l=2a;y, 
 
 8. 5(x 2 -y 2 ) = 4(x 2 +y 2 ), 
 x + y = 8. 
 
 9. x + y = 9, 
 
 ary 10 
 
 Vzy vo 
 
 10. ^L = 4, 
 aj* = 20. 
 
 11. x + y = a, 
 x 3 + y s = b 3 . 
 
 12. x — y = a, 
 tf-f = b\ & + tf = 5(x + y)+2< 
 
 20. 
 
 a + # = 5, 
 
 
 1+1 = 5. 
 
 x y 6 
 
 21. 
 
 A/ 9 2 
 
 x + y = - = xr — yr 
 2/ 
 
 22. 
 
 » 2 -y 2 =144, 
 
 
 x —y — 8. 
 
 23. 
 
 x 2 + z# = j^, 
 
 
 •*# + f = T8 • 
 
 24. 
 
 
144 ELEMENTARY ALGEBRA 
 
 25. a? + tf + 2(x + y) = 12, 28. x 2 + 2 xy + 7 / = 24, 
 xy-(x±y) = 2. 2x 2 -xy-y 2 = 8. 
 
 26. a? + a# + y 2 = 21, 29. (2a>-3)(3#-2)=0, 
 
 a — V^ + ^ = 3. 4 : x 2 + 12xy — 3y 2 = 0. 
 
 Suggestion. In Ex. 26 divide (1) Suggestion. In Ex. 29, from equa- 
 by (2), obtaining x + y + y/xy = 7. tion (1), x = f. The correspond- 
 Add this equation to (2) and find ing value of y is found by 
 x in terms of y. substituting this value of a; in 
 
 equation (2). 
 
 27. .r 2 + to + £* = 133, 30. x(x-y) = 0, 
 
 t + x — -yjtx = 7. x 2 + 2xy + y 2 = 9. 
 
 31. x—y — -\/x — y = 2, 
 
 a? -3/* = 2044. 
 
 Soggestion. In the first equation put Vx — y = 3, and solve for z, 
 finding 
 
 x — y = 1, or 4. 
 
 Then from the second equation, 
 
 x 2 + xy + y 2 = 2044, or 511. 
 
 Put y — x — 1 in x 2 + xy + y 2 = 2044, and solve for x. 
 
 Similarly in the case of x 2 + xy + y 2 = 511. 
 
 32. 3x 2 -4y 2 =8, 
 5(aj-&)-4t/ = 0. 
 
 For what values of k are the solutions real ? Imaginary ? 
 
 Equal ? 
 
 33. Vic — Vy = 2, 
 
 (■y/x — -y/y) 'Vxy = 30. 
 
 34. Two men, A and B, dig a trench in 20 days. It would 
 take A alone 9 days longer to dig it than it would B. How 
 long would it take A and B each working alone ? 
 
 35. A man spends $539 for sheep. He keeps 14 of the 
 flock that he buys, and sells the remainder at an advance of $2 
 per head, gaining $28 by the transaction. How many sheep 
 did he buy and what was the cost of each ? 
 
QUADRATIC AND HIGHER EQUATIONS 
 
 145 
 
 SUPPLEMENTARY WORK 
 
 Graphs of Simultaneous Quadratic Equations 
 Preparatory. 
 1. Iu the same diagram construct graphs to represent the 
 
 equations : 
 
 x- 2 + f = 25, 
 x — y=— L 
 
 Compare the result with this figure. 
 
 IMI>> ]i luAauluia) 
 
 
 
 
 
 
 * "='1 
 
 ■" 
 
 ( 1 1 1 
 
 i 
 
 
 5 
 
 / 
 
 
 ; | | 
 
 .. 
 
 
 
 
 
 
 
 — ■ | : / 
 
 j 
 
 
 
 
 
 
 
 / \ 
 
 
 1 
 
 
 
 | 
 
 \ 
 
 — ■' 
 
 
 
 
 
 
 I 
 
 
 
 
 
 
 
 
 
 
 
 - \ ■ ./ 
 
 
 : 
 
 
 
 
 ' ! :•• 
 
 i j 
 i i 
 
 • 
 
 
 
 
 
 
 
 L/j 1 
 
 : 
 
 / 
 
 j 
 
 LX; 
 
 
 
 
 
 : 
 
 ...... 
 
 
 \ \ 1 
 
 
 2. Solve the system of equations in Exercise 1. 
 Compare the values of x and y with the coordinates of the 
 intersections of the graphs. 
 
 In how many points does the straight line intersect a circle ? 
 How many solutions has the given system of equations ? 
 
146 
 
 ELEMENTARY ALGEBRA 
 
 3. Construct in one diagram the graphs of the aquation 
 
 \xy=l2. 
 Compare the result with this figure. 
 
 .1 
 
 I I I' If f I 
 
 - f [■■■ 
 
 •"X"] ] : T" 
 
 1 J 
 
 1 i 
 
 \ r™ 
 
 : \ ' 
 - \ ? 
 
 : ; •< 
 
 n \ 3 3 
 
 I - - 
 
 
 \/ 
 
 ]- 
 
 
 
 1 1 '■ 
 
 -/ 
 
 ^ V**^^ " 
 
 I [ r i r 
 
 ^^p>*' 
 
 \ i / 
 
 1 » ! 
 
 o 
 
 l : i 
 
 '' \ 
 
 
 F 
 
 ;•- 
 
 >d [ 
 
 
 
 i |" 
 
 XX \ 
 
 ■■••; j \ 
 
 X 
 
 \ \ 
 
 4. Solve the system of equations in Exercise 3. 
 Compare the values of x and y with the coordinates of the 
 intersections of the graphs. 
 
 Graphical Solution of Two Simultaneous Equations. Every 
 point of the graph of one of the equations has coordinates, x 
 and y, that satisfy that equation ; the points of intersection 
 are points of both graphs and therefore have coordinates, x and 
 y, that satisfy both equations. Hence, to solve two simul- 
 taneous equations graphically, draw their graphs and read the 
 coordinates of their intersections. The coordinates of each 
 point of intersection correspond to a solution. If the graphs 
 do not intersect, the system of equations has no real roots. 
 
i^R 
 
 QUADRATIC AND HIGHER EQUATIONS 147 
 
 WRITTEN EXERCISES 
 Solve graphically, and test by computing x and y. 
 
 1. 2a 2 + y=l, 7. x i + y 2 = 25, 
 x-y = 2. aj 2 - 2 / 2 = 7. 
 
 2. a- 2 -?/ 2 = 25, 8. x 2 -3xy + 2y 2 = 0, 
 x + y = l. ^ + ^=16. 
 
 3. x 2 + y 2 = 10, 9. 4z 2 + 9?/ 2 = 36, 
 a;y = 3. x 2 + y 2 = 25. 
 
 4. a- 2 + 2/ 2 =13, 10. 4ar° + 9/ = 36, 
 xy = 6. 2x-3y = 5. 
 
 5. a^ + r = 13, 11. 4z 2 -92/ 2 = 30, 
 a? + 2y = l. a 2 + ^ = 16. 
 
 6. v?-tf = 5, 12. 4ar 9 -9/ = 36, 
 3 a? - y = 7. a?y = 18. 
 
CHAPTER IX 
 
 PROPORTION 
 
 174. Proportion. An equation between two ratios is called 
 a proportion. 
 
 Thus, - = - is a proportion. 
 
 b d 
 
 The older form of writing this proportion is a : b :: c:d ; whence b 
 and c are called the means and a and d the extremes. When so written, 
 the proportion is read " a is to b as c is to d." At present, it is more 
 
 customary to use the form - = - and to read it "a over b equals c over d." 
 
 b d 
 
 ft O 
 
 175. Fourth Proportional. In the proportion - = -, the 
 fourth number, cl, is called the fourth proportional. 
 
 176. Third Proportional. In the proportion - = -, the third 
 
 b c 
 number, c, although in the fourth place, is called the third 
 proportional to a and b. 
 
 177. Mean Proportional. In the proportion - = -, b is called 
 the mean proportional between a and c. 
 
 178. Relation to the Equation. A proportion is an equation 
 and is to be treated in accordance with the properties of 
 equations. 
 
 EXAMPLES 
 
 1. Find the fourth proportional to the numbers 6, 8, 30. 
 
 Let '• lie the fourth proportional, then, hy Sec. 175, - = — . (i) 
 
 8 x 
 
 Multiplying by S x, 6 x = 8 • 30. (2) 
 
 Dividing by fi, £ = 40. (3) 
 
 Therefore 40 is the fourth proportional to 6, 8, 30. 
 
 148 
 
PROPORTION 149 
 
 2. Find the third proportional to 5 and 17. 
 
 Let x be the third proportional, then, by Sec. 176, — = — • (7) 
 
 Multiplying both members by 17 x, 5 X = 17 2 . (,£) 
 
 Solving (2), 5C = if2 = 57|. (5) 
 
 Therefore 57f is the third proportional to 5 and 17. 
 
 WRITTEN EXERCISES 
 
 1. Write m, n, and £> so that p shall be the third propor- 
 tional to m and n. 
 
 2. Write m, w, and p so that m shall be the mean propor- 
 tional between m and p. 
 
 Find x in each of the following proportions : 
 
 3. 
 
 _5_1 
 
 x 4 
 
 
 
 
 7. 
 
 1.21 x 
 x .09 
 
 4. 
 
 ; IB 7 
 11 1331 
 
 
 
 
 8. 
 
 1_42 
 6 x 
 
 5. 
 6. 
 
 x--16 = i 
 - 24 -=- x = 
 
 :2 
 
 a; -T- 
 
 -12. 
 
 9. 
 
 75 -a; 
 — x 3 
 
 10. Find the third proportional to 8 and V— 5. 
 
 11. Find the third proportional to — 6 and V3. 
 
 12. Find the mean proportionals between the following 
 numbers : 9 and 16 ; — 25 and — 4 ; V— 3 and. V— 7 ; 1 
 and —1. 
 
 13. If a sum of money earns S48 interest in 5 yr., how 
 much will it earn in 16 yr. at the same rate per cent ? 
 
 14. A city whose population was 40,000 had 2500 school 
 children ; the total population increased to 48,000, and the 
 number of children of school age increased proportionally. 
 How many children of school age were there then ? 
 
 15. What number must be added to each of the four num- 
 bers, 5, 29, 10, 44, to make the results proportional ? 
 
150 
 
 ELEMENTARY ALGEBRA 
 
 16. A lever need not be straight, although it 
 must be rigid. Thus, the crank and the wheel and * 
 axle are varieties of the lever, and the law of the 
 lever applies to them. Thus, in Fig. 1, 
 
 W p' 
 Find W if P = 14, p = 16, and w = 4. 
 
 17. Find the unknown number: 
 
 Fig. 1. 
 
 
 (1) 
 
 (2) 
 
 (3) 
 
 p= 
 
 3a 
 
 __ 
 
 a — b 
 
 w= 
 
 66 
 
 5p 
 
 (a + by 
 
 p = 
 
 2c 
 
 8p 
 
 a + b 
 
 w = 
 
 — 
 
 2p 
 
 — 
 
 18. If an axle is 6 in. in diameter, what must be the 
 diameter of the wheel in order that a boy exerting a force of 
 50 lb. may be able to raise 800 lb. weight ? 
 
 19. A brakeman pulls with a force of 150 lb. 
 on a brake wheel 16 in. in diameter. The force 
 is communicated to the brake by means of an 
 axle, A, 4 in. in diameter. What is the pull on 
 the brake chain ? 
 
 20. Fig. 3 represents a weight W acting at P on an 
 inclined plane, whose rate of slope is a vertical units to b 
 horizontal units. It is known that the weight W acts in 
 two ways : a force N pressing directly against the surface 
 and tending to produce friction, and a force F parallel to 
 the plane and tending to cause the ^ 
 
 weight to slide down the plane. It is known 
 that these various quantities are related to 
 each other thus : ^--' 
 
 Z- 
 w 
 
 a 
 
 i 
 
 W 
 
 Find F and ^V, if TP= 15 lb., a = 20 in., b = 21 in., I = 29 in. 
 
PROPORTION 
 
 151 
 
 21. Find I and N, if F=W lb., TT=34 lb., a = 4 ft., 6 = 
 1\ ft. 
 
 22. Find 6 and I, if F=66 lb., ^=112 lb., 17=130 lb., 
 a = 33 in. 
 
 23. Find IF and a, if i^=200 lb., N=45 lb., & = 9, Z = 41. 
 
 24. It is known that the pressure P 
 exerted by a letter press on a book is 
 connected with the force F and the other 
 dimensions indicated in Fig. 1 by the fol- 
 lowing relation : 
 
 F_ 
 
 h 
 
 2 trr 
 
 Fig. 1. 
 
 Using ^ 2 - as it, what pressure 
 is exerted by a force of 20 lb. ap- 
 plied to the wheel of a letter press 
 16 in. in diameter and the threads 
 of whose screw are ^ in. apart ? 
 
 25. It is known that if a circular arch of span s is raised h 
 units in the middle, as indicated in Fig. 2, then h, s, r satisfy 
 the proportion : 
 
 h: 
 
 s 
 2 2 
 
 :2r-h. 
 
 If a bridge in the form of a circular 
 arch is to have a span of 200 ft. and be 
 raised in the middle 6 ft. above the level 
 of the end points, what is the radius of 
 the circle ? 
 
 26. Find the unknowns : 
 
 --%- 
 
 V 
 
 Fig. 2. 
 
 
 (1) 
 
 (2) 
 
 (3) 
 
 (4) 
 
 h = 
 
 s = 
 
 r = 
 
 10 
 60 
 
 rr 
 
 i 
 
 287 
 
 176 
 
 187 
 
 1470 
 18015 
 
1.52 
 
 ELEMENTARY ALGEBRA 
 
 27. When a force P is applied 
 at H, the dimensions indicated in 
 the figure and the weight W are 
 known to be related as follows : 
 
 P_ m 
 
 W~ 
 
 dr 
 '2irlR 
 
 If 72=18 in., r=10 in., 1=12 in., 
 d = \ in., and - 2 T 2 be used for ■*, 
 what weight can a man raise 
 
 who turns the handle H with a force of 90 lb. ? 
 
 28. In the formula of Exercise 27 find P, if W= 1000 lb., 
 
 B = 21 in., I = 15 in., r = 11 in., d = § in. 
 
 179. 7?t am/ proportion the product of the means equals the 
 product of the extremes. 
 
 For, - = - , and multiplying both members by bd, ad = bc. 
 b d 
 
 180. Conversely, 
 
 If the product of two numbers equals the product of two other 
 numbers, the four numbers can be arranged in a proportion, the 
 two factors of one product being the means, and the two factors 
 of the other product the extremes. 
 
 For, if ad 
 
 b d 
 
 be, divide both members by bd, and - = -■ 
 
 b d 
 
 181. If - = c ,then 
 
 b d a c 
 
 For, if two numbers are equal, we know that their reciprocals are 
 equal. See First Cocrse, Sec. 193, p. 141. 
 
 The older form of statement that is still sometimes used is: If four 
 numbers form a proportion, they are in proportion by inversion. 
 
 182. If - = -, then - = -• 
 
 b d c d 
 
 For, multiplying both members by 
 
 canceling b and c, 
 
 6 
 
 c 
 
 a _b 
 b~ c ' 
 a _b 
 c d 
 
 c 
 
 or, 
 
 The older form of statement is : If four numbers form a proportion, 
 they are in proportion by alternation. 
 
PROPORTION 153 
 
 183. If ^,then^±- 6 =^. 
 
 b d b d 
 
 For, adding ± 1 to both members, - ± 1 = - ± 1. 
 
 b <I 
 
 Therefore, performing the processes, a ^ - = c ^ • 
 
 6 a" 
 
 The older form of statement is : If four numbers form a proportion, 
 they are in proportion by composition or by division. 
 
 184. If ^ = ^ , then «±& = ^±I?. 
 
 b d a — b c — d 
 
 For, from Sec. 183, £±-$ = £+-£. 
 
 ft d 
 
 a- 6 c - «" 
 
 and 
 
 d 
 
 Dividing these equations, member for member, 
 
 a + b _ c + d 
 a — b c — d 
 
 The older form of statement for this is : If four numbers form a pro- 
 portion, they are in proportion by composition and division. 
 
 185. The mean proportional between two numbers is the 
 square root of their product. 
 
 For, from - = -, Sec. 403, we find b = Vac. 
 b c 
 
 186. The properties of proportion are ^^seful in solving cer- 
 tain problems. 
 
 00 
 
 (*) 
 
 (3) 
 
 
 
 EXAMPLES 
 
 olve: Va 
 
 + bx ■ 
 + bx- 
 
 ■f Va — bx 1 
 
 Va 
 
 — Va — bx % 
 
 By Sec. 1S4, 
 
 2Va + bx 3 
 
 
 2Va — bx - l 
 
 Simplifying (2), 
 11 
 
 — Va + bx = SVa — bx. 
 
154 ELEMENTARY ALGEBRA 
 
 Squaring (3), Ct + bx =9<Z — 9 &2C. (4) 
 
 Solving (U), 10bx = 8 a, (5) 
 
 and CC = — - • (6) 
 
 56 
 
 ~ . a c i 4.1 j. « 4- & Va 2 4- b 2 , yx 
 
 2. Given - = -, show that — !— = - • (J) 
 
 6 d c + cZ Vc 2 4-d 2 
 
 By Sec. 183, ■ 7" = — ^ (*) 
 
 o a 
 
 .-. by Sec. 182, — = - • (5) 
 
 c + rf a 
 
 Squaring both members of the ?L = £_. (/) 
 
 given equation, ^-2 ( p 
 
 Applying Sec. 183 to (4), 
 
 g 2 + & 2 _ C 2 + ^2 
 & 2 d 2 
 
 a 2 + & 2 _ & 2 
 
 From (6)i 
 
 y q a + &a _ & 
 Vc 2 + d 2 d 
 
 c + d v c 2 + <7 2 
 
 WRITTEN EXERCISES 
 
 a & \ fw a2 + 6 ' 2 » 2 - &2 
 1. Given - = -, show that - - = — 
 
 be a+ c a — c 
 
 /-i • a & i, 4.1, 4. a 2 — & 2 b 2 — c 
 2. Given - = -, show that 
 
 ■> 
 
 be a 
 
 3. Given | = c -, show that ^^- 2 = ^^ • 
 
 (S) 
 
 Applying Sec. 182 to (5), g2 + ^ = J 2 ' ( 6 ) 
 
 (7) 
 
 Equating values of £ in (3) and «±& - ^'±j! . (5) 
 
PROPORTION 155 
 
 Solve the equations: 
 
 4. V# + y/b : V# — s/1) = a : b. 5. -— = f- = — - — • 
 
 27 9 x — y 
 
 6. A passenger on an express train observes that a train 
 going in the opposite direction passes him in 2 sec. If it 
 had been going the same way, it would have passed in 30 sec- 
 onds. What is the ratio of the rates of the. two trains ? 
 
 187. The following theorem concerning ratios is sometimes 
 used : 
 
 In a series of equal ratios (a continued proportion), the sum 
 of the numerators divided by the sum of the denominators is 
 equal to any one of the ratios. 
 
 Proof : Let - = - = — = ... be the equal ratios and r be their common 
 value. b d f (i) 
 
 Then - = r,-= r, - = r, and so on. (2) 
 
 b d f w 
 
 /. a = br, c = dr, e = fr, and so on. (5) 
 
 .-. adding the equations in (<?), (a + c + e •••) = (& -j- d + /--) r - (4) 
 
 .-. solving (4) for r, a + c + e "' = r . r 5 \ 
 
 v ' b+d+f- ^ ' 
 
 a + c + p ■■• the sum of the numerators a c , 
 
 .-. —^ — — = ■ = -, or -, and so on, since 
 
 b + d+f'-' the sum of the denominators b d 
 
 r is any of these ratios. (6) 
 
 188. This theorem may be applied to certain equations. 
 
 EXAMPLES 
 Solve : 
 
 Applying the theorem, 
 
 Simplifying (2), 
 
 Solving (3), 
 and 
 
 Test by substitution. 
 
 2 x - 2 2 
 
 CO 
 
 4-9z 9a; 
 
 2x_2_. 
 4 ~9x 
 
 (.*) 
 
 x_ 2 
 
 2 ~ 9 x 
 
 (3) 
 
 9x 2 =4, 
 
 (4) 
 
 x=±l 
 
 
156 
 
 ELEMENTARY ALGEBRA 
 
 WRITTEN EXERCISES 
 
 2 
 
 1. Given - = - = —, show that — !— — — = -■ 
 
 b d f b+d-f d 
 
 „ ~, . a c e i , , . a 2 + c 2 + e 2 e 
 
 2 . Given- ^- = ? show that y + (P+/i = ? 
 
 Va 2 + c 2 a 
 
 3. Also show that - = r • 
 
 V& + d ° 
 
 4. Given— ==£ = -, show that-— ^ ; = — . 
 
 n q s 2n — Sq -f- 5 s n 
 
 5. Given x = b -=°, show that » = f i^±A 
 
 a c d c + d 
 
 n v a ■ x — 7 5x + l 
 
 6. Find x in = • 
 
 x + 2 3x — 2 
 
 SUMMARY 
 I. Definitions. 
 
 1. An equation between two ratios is called a proportion. 
 
 2. The numbers b and c are called the means, and a and 
 d the extremes. Sec. 174. 
 
 3. In the proportion - = -, the fourth number d is called the 
 fourth proportioned. Sec. 175. 
 
 4. In the proportion - = -, the third number c is called the 
 third proportional. Sec. 176. 
 
 5. In the proportion - = -, b is called the mean proportional 
 
 b c 
 
 between a and c, and is the square root of their product. 
 
 Sees. 177 and 185. 
 
 II. Properties. 
 
 1. A proportion is an equation and therefore subject to the 
 laws of equations. Sec. 178. 
 
 2. In any proportion the product of the means equals the 
 product of the extremes. Sec. 17 ( J. 
 
PROPORTION 157 
 
 3. If the product of two numbers equals the product of 
 two other numbers, the four numbers can be arranged in a 
 proportion, the two factors of one product being the means, 
 and those of the other product the extremes. Sec. 180. 
 
 4. If - = c , then - = - (inversion). Sec. 181. 
 
 b d a c v 
 
 5. If - = -, then - = - (alternation). Sec. 182. 
 
 b d c d 
 
 6. If - = -, then c ±±l = < LA d (composition or division). 
 
 6 d h d Sec. 183. 
 
 7. If - = -, then ^_±A=^±_!? (composition and division). 
 
 6 d a ~ b c ~ d Sec. 184. 
 
 8. In a series of equal ratios the sum of the numerators 
 divided by the sum of the denominators equals any one of the 
 ratios. Sec. 187. 
 
 REVIEW 
 
 ORAL EXERCISES 
 
 3 5 
 
 1. What is the fourth proportional in - = -? 
 
 3 5 
 
 2. What is the third proportional in - = -? 
 
 3. What is the mean proportional between 4 and 9 ? 
 
 4. State the principle according to which it follows from 
 
 a c that a + 6 - c+d 
 
 d b d 
 
 5. What is the value of x in - = -? 
 
 a x 
 
 WRITTEN EXERCISES 
 Find the proportional to : 
 1. x, y, ab. 2. p 2 , q, m 2 q. 3. x 2 , y 2 , z 2 . 
 
158 ELEMENTARY ALGEBRA 
 
 Find the mean proportional between : 
 4. a\a 2 b 2 . 5. -6, 4a 2 6 2 . 6. 27x>f,3y. 
 
 (I c 
 
 When : = - show that : 
 o a 
 
 ac _ a~ 
 
 
 9 9 
 
 a- a- — 
 
 b 2 
 
 c 2 <?- 
 
 d 2 
 
 a + 2b_ 
 2b 
 
 _c+2d 
 2d 
 
 10. 
 
 2a-3b 2c-3d 
 
 8 . " = " ~" . 11. 
 
 9. ^-r^^ = ^r-". 12. 
 
 4« 
 
 3b 
 
 i 2 - 
 
 5b 2 
 
 3d 
 4c 2 - 
 
 5d 2 
 
 
 5 b 2 
 -b 
 
 _Vc, 
 
 5d 2 
 
 
 a 
 
 ; 2 -6 2 
 
 
 c-cl Vc 2 - d 2 
 
 13. The ratio of two numbers is 3 to 5 and the sum of their 
 squares is 544. Find the numbers. 
 
 14. The lengths of the sides of a triangle are in the ratio of 
 3 to 4 to 5 and their sum is 120 ft. What is the length of 
 each side ? 
 
PROPORTION 159 
 
 SUPPLEMENTARY WORK 
 
 Adding the same positive number to both numerator and 
 denominator of a ratio, 
 
 (1) Increases the absolute value of the ratio if the numerator 
 is less than the denominator ; and 
 
 (2) Diminishes the absolute value of the ratio if the numerator 
 is greater than the denominator. 
 
 For, let - be the given ratio and x the given number. 
 6 
 
 To determine whether - or a + x is the greater, subtract the second 
 
 ratio from the first. 
 
 Then - — a ~^~ x — a ^ ~*~ ax ~ a ^ ~ ^ x — ( a ~ ^) x 
 ' b b + x~ 6(6 + a;) ~ b(b + x) ' 
 
 (1) If a > b, the last ratio is a positive number, therefore the 
 
 ft 
 
 ratio - would be diminished by adding x to each of its terms, 
 b 
 
 which proves (2). 
 
 (2) If a < b, the last ratio is a negative number, therefore 
 the ratio - would be increased by adding x to each of its terms, 
 
 which proves (1). 
 
 (3) If a = b, the last ratio is zero, therefore the ratio - would 
 be unchanged by adding x to each of its terms. 
 
 ORAL EXERCISES 
 
 1. How is the value of | changed by adding 4 to each term ? 
 
 2. How is the value of | changed by adding 5 to each term ? 
 
 3. How is the value of 4 changed by adding a to each term ? 
 
 4. How is the value of | changed by adding b to each term ? 
 
 5. What is the effect on £ of adding c to each term ? 
 
CHAPTER X 
 
 VARIATION 
 
 189. Direct Variation. When two variable quantities vary 
 so as always to remain in the same ratio, each is said to 
 vary directly as the other. Each increases or decreases at the 
 same rate that the other increases or decreases. 
 
 Consequently, if x and y are two corresponding values of the 
 variables and k the fixed ratio, then 
 
 - = k and x — ky. 
 V 
 
 For example, at a fixed price (k) per article, the total cost (x) of a 
 
 number of articles of the same sort varies directly with the number of 
 
 x 
 articles (y). That is, - = k. 
 
 Likewise in the case of motion at a uniform rate (r), the distance tra- 
 versed (d) varies as the time of motion (t) . That is, - = r. 
 
 190. A symbol still occasionally used for " varies as " is oc 
 
 Tims, u x varies as y " is written x oc y, 
 and " d varies as £" is written dazt. 
 
 191. Relation of Variation to Proportion. When one vari- 
 able varies directly as another, any pair of values of the vari- 
 ables forms a proportion with any other pair. 
 
 For, — = rand — = r; .\ — = — , which is a proportion. 
 
 yl yll yl yll 
 
 192. Expressions for Direct Variation. We have thus seen 
 that the relation x varies directly as y may be expressed in any 
 one of three ways : 
 
 (a) x = ky, by use of the equation. 
 
 (b) xcxy, by use of the symbol of variation. 
 
 (c) — = — . , by use of the proportion ; x', y' and x", y" being 
 
 y y 
 
 any two pairs of corresponding values of the variables. 
 
 160 
 
VARIATION 161 
 
 WRITTEN EXERCISES 
 
 1. Write the statement "v varies as w" in the form of an 
 equation, also in the form of a proportion. 
 
 2. Write the statement x = ky by use of the symbol oc ; 
 also in the form of a proportion. 
 
 t' t" 
 
 3. Write — = 7^77 by use of the symbol oc. 
 
 4. The weight (w) of a substance varies as the volume (v) 
 when other conditions are unchanged. Express this law by use 
 of the equation. By use of the symbol oc. Also in the form of 
 a proportion. 
 
 5. In the equation w = kv, if to = 4 and v = 2, what is the 
 value of k ? Using this value of k, what is the value of w 
 when v = 25? 
 
 6. When 1728 cu. in. of a substance weigh 1000 ounces, 
 what is the ratio of the weight (w) to the volume (v) ? What 
 volume of this substance will weigh 5250 ounces ? 
 
 7. The cost (c) of a grade of silk varies as the number of 
 yards (n). Find the ratio (r) of c to n when c is $ 7.00 and 
 
 c i i c' 
 
 8. If in Exercise 7, - =r, what does — equal ? Given that 
 
 u n 
 
 40 yd. of silk cost $ 60, find the cost of 95 yd. by means of the 
 
 c' c 
 proportion — = - . Also by means of the equation c = nr. 
 
 193. Inverse Variation. A variable x is said to vary inversely 
 
 as a variable y, if it varies directly as -■ 
 
 Inverse variation means that when one variable is doubled the other 
 is halved ; when one is trebled the other becomes \ of its original value, 
 and so on. 
 
 194. Expressions for Inverse Variation. The relation "x 
 varies inversely as y " may be expressed : 
 
 (a) By the equation, x = Jc(-\ or x = -, .-. xy = k. 
 
162 ELEMENTARY ALGEBRA 
 
 (b) With the symbol of variation, y oc -■ 
 
 (c) As a proportion ~ = ^- T . 
 
 WRITTEN EXERCISES 
 
 1. Write the statement "v varies inversely as w"in the 
 form of an equation. Also in the form of a proportion. 
 
 2. Write t = - by use of the symbol oc, also in the form of 
 a proportion. 
 
 3. In a bicycle pump the volume (v) of air confined varies 
 inversely as the pressure (p) on the piston. Write the rela- 
 tion between v and p in three ways. 
 
 4. In Exercise 3, if v = 18 (cu. in.) and p = 15 (lb.), what 
 is k in v — - ? What is the pressure ( p) when v = 1 (cu. in.) ? 
 
 5. In an auditorium whose volume (c) is 25,000 cu. ft. 
 there are 2000 persons (p). What is the number (h) of cubic 
 feet of air space to the person ? What will be the number 
 when 1000 more persons come in ? 
 
 6. The area of a triangle varies as the base times the alti- 
 tude. If the area is 12 when the base is 8 and the altitude 3, 
 what is the area of a triangle whose base is 40 and altitude 20 '.' 
 
 7. The area of a circle varies as the square of its radius. 
 The area of a circle of radius 2 is 12.5664; what is the area of 
 a circle whose radius is 5.5 ? 
 
 8. The volume of a sphere varies as the cube of its radius. 
 If the volume of a sphere whose radius is 3 is 113.0976, what 
 is the volume of a sphere whose radius is 5 ? 
 
 9. If x oc y and x = 6 when y = 2, find x when y = S. 
 
 Suggestion, x = ky. .-. 6 = k • 2 and k = 3. Substitute y — 8 in 
 the equation x = 3 y. 
 
 10. Determine k in xccy, if .r = 10 when y = 20. Also if 
 x = 1 when y = 5. If x = 100 when y = 10. 
 
 11. If x oc iv and y cc w, prove that x 4- y x w. 
 
VARIATION 
 
 163 
 
 12. If x oc w and w oc y, prove that xy oc iv 2 . 
 
 13. If x oc ii and *o oc z, prove that — oc^ • 
 
 w z 
 
 14. If xccy-z and a = l when ?/ = 2 and z = 3, find the con- 
 stant A;. 
 
 15. Given y = z-\-iv, zccx and miock; and that sc = l when 
 w = 6, and that x = 2 when z = 20. Express y in terms of x. 
 
 Suggestion, y = kx + k'x ; determine k and k'. 
 
 16. Given z<xx + y and yccx 2 , and that ic=|when y=^ 
 and z = \. Express z in terms of x. 
 
 GRAPHICAL WORK 
 
 195. The relation "oj varies as y" has been expressed by the 
 linear equation x = ky, and has been represented by a straight 
 line (see Eirst Coukse, Sees. 212 and 213, pp. 168-170). 
 
 The relation " x varies inversely as y " has been expressed by 
 
 k 
 the equation x = - • The graph of this equation is a curve. 
 
 y 
 
 2 
 Suppose, for illustration, that k = 2. Then x = - • For this equation : 
 
 y 
 
 The table is 
 
 X 
 
 2/ 
 
 
 5 
 
 2 
 5 
 
 
 4 
 
 1 
 2 
 
 
 3 
 
 2 
 3 
 
 
 2 
 
 1 
 
 
 1 
 
 2 
 
 
 i 
 
 2 
 
 4 
 
 
 -1 
 
 — 
 
 2 
 
 -2 
 
 — 
 
 1 
 
 - 3 
 
 — 
 
 2 
 3 
 
 -4 
 
 — 
 
 1 
 2 
 
 -5 
 
 — 
 
 2 
 
 5 
 
 The graph is 
 
 y 
 
 j- 
 
 
 r j" 
 
 
 
 i 
 
 
 
 
 
 4- 
 
 
 
 j 4 
 
 
 
 
 « •■ 
 
 
 
 j 
 
 j 3 
 
 
 
 ..; 
 
 ■4 ; 
 
 
 [ |" 
 
 
 1 2 
 
 1 
 
 
 
 : 
 
 
 - 
 
 
 : 
 
 
 
 -4: -3; 
 
 -2) - 
 
 ii 
 
 1 
 
 :2 
 
 13 
 
 |4 ; 
 
 
 ! [•■ 
 
 
 1 -1 
 
 
 
 
 ...1 3 
 
 
 
 
 \-2 
 V 3 
 
 
 
 
 
 
 j h 
 
 
 T'-4 
 
 ■■- 
 
 
 
 -j | 
 
 
 • 
 
 
 
 
 _J_ 
 
 '"1 
 
 
 y 
 
164 
 
 ELEMENTARY ALGEBRA 
 
 WRITTEN EXERCISES 
 Kepresent graphically : 
 
 1. x = -, or xy = 1. 
 
 y 
 
 _ 2 
 
 2. a; = — -, or xy = — 2. 
 
 3. # = -, or a-?/ = 4. 
 
 4. 3»= ^— , or 3xy = — 1. 
 
 2/ 2/ 
 
 5. Under standard conditions the volume (V) of a confined 
 
 gas varies inversely as the pressure (jp). That is, v = -. This 
 
 is known as Boyle's Law. Suppose when the piston is at a in 
 
 B the figure, the volume of the 
 gas in part A is 1 cu. ft., 
 
 P P 
 
 and the pressure at P is 
 
 5 lb. When the pressure is 
 
 37 1 10 lb. the piston moves to b, 
 
 A and the volume of the gas B 
 
 becomes \ cu. ft. What will be the volume of the gas when 
 
 P=20 lb.? 
 
 k 
 
 Taking k = 5 in the equation v = -, the table of values is 
 
 P 
 
 p = 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 G 
 
 7 
 
 8 
 
 9 
 
 10 
 
 15 
 
 20 
 i 
 
 4 
 
 V = 
 
 5 
 
 2* 
 
 l 3 
 
 n 
 
 1 
 
 5 
 6 
 
 5 
 7 
 
 5 
 8 
 
 5 
 9 
 
 1 
 
 i 
 
 3 
 
 The graph of the table is : 
 
 
 I v 
 
 i r r 
 
 5 
 
 J 4 
 
 
 - : - : ~.: :-.^ > .- ;- .. ■ = ; 
 
 3 
 
 
 i 
 
 {.nil 
 
 2 
 
 
 ...|.^vi. : 1 | ■ 1 1 i ! j i L ! | ! i ..j 
 
 |_1 
 
 
 \^_____L_L : ! 
 
 i ° 
 
 I 
 
 i 
 
 .2 .3 |4 |6 6 :7 8 jB 10 11 12 13 ; 14 .15 15 1? 13 19 20 ' 
 
 : 
 
 
 ...j ! | L ,., ;..;....! | \ | \ 
 
 I i i 
 
 Read from the graph the pressure on the piston necessary to hold this 
 volume of gas at | cu. ft. ; at | cu. ft. ; at 5 cu. ft. ; at 2 cu. ft. 
 
VARIATION 
 
 165 
 
 6. The attraction or " pull " of the earth on bodies in its neighborhood 
 is the cause of their weight. The law of gravitation states that the weight of 
 a given body varies inversely as the square of its distance from the center 
 of the earth. This law may be expressed by the equation 
 
 k 
 w = — • 
 d? 
 
 To construct the graph which shows the nature of the relation between 
 to and d as they vary, k may be taken to be 1. 
 
 Fill the blanks in the table : 
 
 w 
 
 ±1 
 
 ±5 
 
 ±i 
 
 ±.25 
 
 ±•2 
 
 1 
 
 ( ) 
 
 ( ) 
 
 ( ) 
 
 ( ) 
 
 7. Plot the graph for the table in Exercise 6. 
 
 SUMMARY 
 
 I. Definitions. 
 
 1. When two variable quantities vary so as always to 
 remain in the same ratio, each varies directly as the other. 
 
 Sec. 189. 
 
 2. When one variable varies directly as another, any pair of 
 values of the variables forms a proportion with any other pair. 
 
 Sec. 191. 
 
 3. A variable x varies inversely as a variable y, if it varies 
 
 directly as - • 
 
 Sec. 193. 
 
 II. Processes. 
 
 1. Processes involving direct variation are performed by 
 reference to the equation x = ky. Sec. 192. 
 
 2. Processes involving inverse variation are performed by 
 
 reference to the equation x = 
 
 Sec. 194. 
 
 y 
 
 3. Direct and inverse variation may be represented graphi- 
 cally. Sec. 195. 
 
166 
 
 ELEMENTARY ALGEBRA 
 
 SUPPLEMENTARY WORK 
 
 A large number of problems in science may be solved by the 
 following plan : 
 
 EXAMPLES 
 
 1. The "law of gravitation" states that the weight of a given 
 body varies inversely as the square of its distance from the 
 center of the earth. What is the weight of a body 5 mi. above 
 the surface of the earth, which weighs 10 lb. at the surface 
 (4000 mi. from the center) ? 
 
 Method. There are two variables in the problem, the weight (w) and 
 the distance (d). There are also two parts or cases in the statement, one 
 in which the value of one variable is unknown, and one in which the 
 values of both variables are given. 
 
 Arrange the data as follows : 
 
 
 10 
 
 d 
 
 1st case 
 2d case 
 
 X 
 
 10 
 
 4005 
 4000 
 
 To this table apply the law of variation expressed in the physical law. 
 
 Since the law is : w varies inversely as the square of d, the values of d 
 
 must be squared, and the ratio x : 10 equals the inverse ratio of 4005 2 to 
 
 4000 2 ; that is, 
 
 x _ 4000 2 
 
 10 _ 4005 2 ' 
 .-. x = 9.96, and the weight is 9.96 lb. 
 
 2. The squares of the times of revolution of the planets 
 about the sun vary directly as the cubes of their distances 
 from the sun. The earth is 93,000,000 mi. from the sun, 
 and makes a revolution in approximately 365 da.; what is 
 the distance of Venus from the sun, taking its time of revolu- 
 tion to be 226 da. ? 
 
VARIATION 10" 
 
 Solution 
 
 
 t = time of re vol. 
 
 d = distance from sun 
 
 1st case 
 2d case 
 
 365 
 
 226 
 
 93,000,000 
 
 X 
 
 According to the astronomical law the times must be squared and the 
 distances cubed ; then, since the law is that of direct variation : 
 
 365' 2 _ 93,000,000 3 
 226 2 ~ X s 
 
 \2 
 
 .-. x = 93,000, 000-^/ f—\ =68,900,000, and the distance of Venus 
 from the sun is approximately 69,000,000 miles. 
 
 WRITTEN EXERCISES 
 
 1. The intensity of light from a given source varies in- 
 versely as the square of the distance from the source. If the 
 intensity (candle power) of the light from an electric lamp 
 is 4 at a distance of 150 yd., what is its intensity at a dis- 
 tance of 25 yd. ? 
 
 2. According to the first sentence of Exercise 1, how much 
 farther from an electric light must a surface be moved to 
 receive only i as much light as formerly? 
 
 3. The time of oscillation of a pendulum varies directly as 
 the square root of its length. What is the length of a pendu- 
 lum which makes an oscillation in 5 sec, a 2-second pendulum 
 being 156.8 in. long? 
 
 4. According to Exercise 3, what is the time of oscillation 
 of a pendulum 784 in. long ? 
 
 5. The distance through which a body falls from rest varies 
 as the square of the time of falling. A body falls from rest 
 576 ft. in 6 sec. ; how far does it fall in 10 sec. ? 
 
 6. Volumes of similar solids vary as the cubes of their 
 linear dimensions. The volume of a sphere of radius 1 in. is 
 4.1888 cu in.; what is the volume of a sphere whose radius is 
 5 in. ? 
 
168 
 
 ELEMENTARY ALGEBRA 
 
 7. According to Exercise 6, what is the radius of a sphere 
 whose volume is 33.5104 cu. ft. ? 
 
 8. According to Exercise 6, if a flask holds \ pt., what is 
 the capacity of a flask of the same shape 4 times as high ? 
 
 9. In compressing a gas into a closed receptacle, as in pump- 
 ing air into an automobile tire, the pressure varies inversely 
 as the volume. If the pressure is 25 lb. when the volume is 
 125 cu. in., what is the pressure when the volume is 115 cu. in. ? 
 
 10. According to Exercise 9, if the pressure is 50 lb. when 
 the volume is 250 cu. in., what is the volume when the pres- 
 sure is 10 lb. ? 
 
 11. It is known that if one gear wheel turns another as in the figure, 
 
 the number of revolutions of the two 
 are to each other inversely as their 
 number of teeth. That is, if the first 
 has C\ teeth and makes Bi revolutions, 
 and the second has C'z teeth and makes 
 in the same time i? 2 revolutions, 
 
 then 
 
 Find R, if d = 25, C, = 15, and R, = 6. 
 Find the numbers to fill the blanks : 
 
 R\ Co 
 
 R-2 C\ 
 
 
 (i) 
 
 (2) 
 
 (3) 
 
 (4) 
 
 Ci = 
 
 42 
 
 60 
 
 5 n 
 
 — 
 
 a = 
 
 — 
 
 50 
 
 3 n 
 
 40 
 
 #i = 
 
 12 
 
 — 
 
 21 
 
 8n 
 
 R> = 
 
 9 
 
 15 
 
 — 
 
 6 n 
 
 12. A diamond worth $ 2000 was broken into two parts, to- 
 gether worth only $ 1600. If the value of a diamond is propor- 
 tional to the square of its weight, into what fractions was the 
 original diamond broken ? (Find result to nearest hundredth.) 
 
 13. If the volume of a sphere varies as the cube of its radius, 
 find the radius of a sphere -whose volume equals that of the sum 
 of two spheres whose radii are respectively 6 ft. and 3.5 ft. 
 
VARIATION 169 
 
 14. The number of vibrations (swings) made by two pen- 
 dulums in the same time are to each other inversely as the 
 square roots of their lengths. If a pendulum of length 
 39 in. makes 1 vibration per second (seconds pendulum), about 
 how many vibrations will a pendulum 10 in. long make ? How 
 long must the pendulum be to make 10 vibrations per second ? 
 
 15. Two towns join in building a bridge which both will use 
 and agree to share its cost ($ 5000) in direct proportion to 
 their populations and in inverse proportion to their distances 
 from the bridge. One town has a population of 5000 and is 
 2 mi. from the bridge; the other has a population of 9000 
 and is 6 mi. from the bridge. What must each pay ? 
 
 16. If a : b =p : q, prove that 
 
 
 a 2 . h2 . « _ „2 i_ Q 2 . P . 
 
 
 a 1 u . — p -\- (/ . 
 
 a + b P+ C l 
 
 Suggestion. 
 
 The given proportion may be written : 
 
 
 a _b 
 P~ 1 
 
 Let 
 
 - = r, then a = pr, b = qr. 
 
 p 
 
 The proportion to be proved may be written : 
 
 (a + h)(a* + ftg) = (p +g)(;> 2 + 9 2) 
 a 3 p 3 
 
 Substituting the values of a and b above, the left member readily 
 reduces to the right. 
 
 Note. A good method for proving such identities is to begin with the 
 required relation and transform it into the given relation, or to transform 
 both the given and the required relation until they reduce to the same 
 thing. 
 
 17. Solve, using the principle of composition and division, 
 
 (a - V2bx + x 2 ) : (a - b) = (a + V2 bx -f- x 2 ) : (a + b). 
 
 18. If «1 = «i = «? = . . . = On prQve that 
 
 &i & 2 b 3 b n 
 
 q 1 + « g + «3-l \-a H _ <h § 
 
 h+h +&3+ "• + &„ &/ 
 
 19. Given x + -y/x : x — V# : : 3 Vcc + 6 : 2 Vai, to find a;. 
 12 
 
CHAPTER XI 
 
 SERIES 
 
 196. Series. If a set of numbers is specified according to 
 some law or system, the set of numbers is called a series. 
 
 197. Terms. The numbers constituting the series are called 
 its terms, and are named from the left, 1st term, 2d term, etc. 
 
 The following are examples of series : 
 
 1. 1,2,3,4,5,.-.. 7. 1, 1, |, j, |, .... 
 
 2. 1, 3, 5, 7, 9, ••.. 8. 1, 3, 9, 27, 81, 243, .... 
 
 3. 1,5,9,13,17,.-.. 9. 2,f,f,^, .... 
 
 4. 3,6,9,12,15,-... 10. 1,1,1,1,1,1,.... 
 
 5. 1, li 2, 21 3, .... 11. V2, V3, V4, V5, V6, V7, •••• 
 
 6. 2, 4, 8, 16, 32, .... 12. 100, 99, 98, 97, 96, 95, .... 
 
 ORAL EXERCISES 
 1-12. State the next five terms of each series above. 
 
 198. When the laiv of a series is known, any term may be 
 found directly. 
 
 EXAMPLES 
 
 1. The law of the second series in Sec. 197 is that each term is two 
 more than the preceding. To get the tenth term we start from 1 and 
 add 2 for each of 9 terms. That is, the tenth term is 1 + 9 • 2 = 19. 
 
 Similarly, the 12th temi is 1 + 11 • 2 = 23, 
 the loth term is 1 + 14 • 2 = 29, 
 the 47th term is 1 + 46 . 2 = 93, 
 the nth term is 1 + (n — 1)2 = 2 n — 1. 
 
 2. The law of the eighth series is that each term after the first is 3 
 times the preceding term. To get the ninth term we start at 1 and multiply 
 by 3 eight times, or by 3 s . That is, the ninth term is 1 • 3 8 = G561. 
 
 Similarly, the 12th term is 1 • 3" = 177,147, 
 the nth term is 1 • 3"- 1 = 3»-i. 
 
 170 
 
SERIES 171 
 
 WRITTEN EXERCISES 
 
 1-4. Select four of the series in Sec. 197 that can be treated 
 like the first example and write the 10th, 12th, 15th, and 47th 
 terms of each. 
 
 5-8. Select four of the series in Sec. 197 that can be treated 
 like the second example and write the 8th, 10th, and ?ith terms 
 of each. 
 
 9. Write similarly the 7th, 11th, 20th, 47th, and «th term 
 for any 6 of the above series. 
 
 199. We shall give only two types of series, the arithmetical 
 and the geometric, the laws of which are comparatively simple. 
 
 ARITHMETICAL SERIES 
 
 200. Arithmetical Series. A series in which each term after 
 the first is formed by adding a fixed number to the preceding 
 term is called an arithmetical series or arithmetical progression. 
 
 201. Common Difference. The fixed number is called the 
 common difference, and may, of course, be negative. 
 
 For example : 
 
 1. 7, 15, 23, 31, 39,. ••• is an arithmetical series having the common dif- 
 ference 8. 
 
 2. 16, 14^, 13, 111, io, ... is an arithmetical series having the common 
 difference — f. 
 
 ORAL EXERCISES 
 
 1. Name all the arithmetical series in the list, Sec. 197. 
 
 2. State the common difference in each of these series. 
 
 3. State the 10th term in each of these series. 
 
 WRITTEN EXERCISES 
 
 1. Write the nth term in each arithmetical series in the 
 list, Sec. 197. 
 
 2. Beginning with 2 find the 100th even number. 
 
 3. Beginning with 1 find the 100th odd number. 
 
172 ELEMENTARY ALGEBRA 
 
 4. Beginning with 3 find the 200th multiple of 3. 
 
 5. A city with a population of 15,000 increased 600 persons per 
 year for 10 yr. What was the population at the end of 10 yr. ? 
 
 202. A general form for an arithmetical series is : 
 
 a, a + d, a + 2 d, a + 3 d, ■■-, a +(n - T)d, •••, 
 where a denotes the first term, 
 
 d denotes the common difference, and 
 n denotes the number of the term. 
 
 203. Last Term. If the last term considered is numbered n 
 and denoted by I, we have for the last of n terms the formula : 
 l = a+{n — l)d. 
 
 204. The Sum of an Arithmetical Series. The sum of n terms 
 of an arithmetical series can be found readily. 
 
 EXAMPLE 
 Find the sum of the first 6 even numbers. 
 
 1. Let 8 = 2 + 4 + 6 + 8 + 10+12. 
 
 2. We may also write s- 12 + 10 + 8 + 6 + 4 + 2. 
 
 3. Adding (1) and (2), 
 
 2s = (2 + 12) + (4 + 10) + (6 + 8) +(8 + 6) + (10 + 4) 
 
 + (12 + 2) 
 = 6(2 + 12) ; for each parenthesis is the same as 2 + 12. 
 
 4 . ., s = 6(2_±i2) =42 . 
 
 WRITTEN EXERCISES 
 
 Find similarly the sum of : 
 
 1. The first 6 odd numbers. 
 
 2. The first 6 multiples of 3. 
 
 3. The first 5 multiples of 7. 
 
 4. The first 4 multiples of 8. 
 
 205. General Formula for the Sum. The general form of the 
 series may be treated in the same way. 
 
 If I denotes the last of n terms, the term before it is denoted 
 
SERIES 173 
 
 by / — d, the next preceding by 1 — 2 d, and so on. Hence, the 
 sum of n terms may be written : 
 
 s=a +(a+d)+(a+2d) + ••• + (/ -2 d) + (l - d) + l. 
 And also, s= I + (I -d) + (1-2 d) + — + (a+2d) + (a+d)+a. 
 
 Whence, adding, 2s=(a + l) + (a + l) A \-(a+l) = n(a+l). 
 
 Therefore, s = w ^ a ' • 
 
 Or, in words, 
 
 T%e sum of any number of terms of an arithmetical series is one, 
 half the sum of the first and the last terms times the number of 
 terms. 
 
 By using the value of I in Sec. 203, s = n ^ 2a + fa- 1 )* 1 ') . 
 
 This permits the calculation of s without working out separately the 
 value of I. 
 
 WRITTEN EXERCISES 
 
 For each series in the following list find : First, the sum of 
 10 terms. Second, the sum of n terms. 
 
 1. 1,2,3,4,5,—. 4. 3,6,9,12,15,—. 
 
 2. 1,3,5,7,9,-... 5. 1, li 2, 2|, 3,.-.. 
 
 3. l,5j 9,13,17,.-.. 6. 100, 99, 98, 97, 96, 95,- ••. 
 
 7. A man invests $100 of his earnings at the beginning of 
 each year for 10 yr. at 6%, simple interest. How much has 
 he at the end of 10 yr. ? 
 
 Solution. 
 
 1. The last investment bears interest 1 yr. and amounts to $106 ; the 
 next to the last bears interest 2 yr. and amounts to $112, etc. ; the first 
 bears interest 10 yr. and amounts to $ 160. 
 
 2. Hence, a=$ 106, d = $6, and n = 10. 
 
 3. Therefore, I = 106 + 9 • 6 = 160. 
 
 4. Therefore, s = V°- (106 + 160) = 1330. 
 
 5. The man has $1330. 
 
 8. If $ 50 is invested at the beginning of each year for 20 yr. 
 at 5 % simple interest, what is the amount at the end of 20 yr. ? 
 
174 ELEMENTARY ALGEBRA 
 
 9. If a body falls approximately 16 ft, the first second and 
 32 ft. farther in each succeeding second, how far does it fall 
 in 5 sec. ? 
 
 206. Collected Results. The three chief formulas of arith- 
 metical series are : 
 
 1. l = a+(n-l)d. 
 
 2. s- 2 
 
 " 3 _ s = n[2a + (n-l)<Z\ . 
 
 GEOMETRIC SERIES 
 
 207. Geometric Series. A series in which each term after the 
 first is formed by multiplying the preceding term by a fixed 
 number is called a geometric series, or a geometric progression. 
 
 208. Common Ratio. The fixed multiplier is called the com- 
 mon ratio, and may be negative. 
 
 For example : 
 
 1. 2 is the common ratio in the geometric series 2, 4, 8, 16, •••. 
 
 2. — i is the common ratio in the geometric series, 27, —9, 3, —1, J, •••. 
 
 ORAL EXERCISES 
 
 1. Name all the geometric series in the list, Sec. 197. 
 
 2. State the common ratio in each of these series. 
 
 3. State the 6th term of each of these series. 
 
 WRITTEN EXERCISES 
 
 1. The series 1, 3, 9, 27, 81, •••, whose ratio is 3, is the same 
 as 1, 3 1 , 3 2 , 3 3 , 3 4 , •••. Write by use of exponents the 6th term 
 of this series ; the 8th term ; the 10th term ; the 15th ; the 
 25th; the 100th. 
 
 2. The series 3, -#, f, — f, ••■, whose ratio is — %, is the 
 same as 3, 3(-£), 3(-£) 2 , 3(-i) 3 , ••-. Write by use of ex- 
 ponents the 5th term of this series ; the 8th term ; the 10th ; 
 the 25th ; the 50th. 
 
SERIES 175 
 
 209. A general form for the geometric series is: 
 
 ,.n-l 
 
 "j 
 
 '> 
 
 where a denotes the first term, 
 
 r denotes the common ratio, and 
 n denotes the number of the terms. 
 
 210. Last Term. If the last term is numbered n, and de- 
 noted by I, then we have for the last of n terms the formula, 
 
 l = ar n -\ 
 
 211. The Sum of a Geometric Series. The sum of n terms of a 
 geometric series can readily be found. 
 
 EXAMPLE 
 
 Find the sum of 5 terms of the series 2, 6, 18, 54, 162. 
 
 Solution. 
 
 Let s = 2 + 6 + 18 + 54 + 162. (/) 
 
 ■"aSS & the 3s = 6 + 18 + 54 + 162 + 486. (« ) 
 
 Subtracting (1) from (2), 3 S - S = 486 — 2, (3) 
 
 or, 2s = 484. (4) 
 
 Dividing by 2, S = 242. (5) 
 
 WRITTEN EXERCISES 
 Find similarly the sum of 5 terms of each of these series: 
 
 1. 6,30,150,-. 3. i,^,^,.... 
 
 2. 7,-14,28,.-.. 4. i -i, iV, •••• 
 
 212. General Formula for the Sum. The general form of the 
 series may be treated in the same way. If I denote the last 
 
 of n terms, the term before it is denoted by -, the next preceding 
 
 r 
 
 by—, and so on. Hence, the sum of n terms may be written : 
 
 1. s — a + ar + «r + --\ \-l. 
 
 r r 
 
176 ELEMENTARY ALGEBRA 
 
 2. Then rs = ar + ar 2 + -\ \- I + Ir. 
 
 r 2 r 
 
 3. Subtracting, s — rs = a — Ir. 
 
 4. Or, (1 — r) s = a — Zr. 
 
 a — Ir Ir — a 
 
 ' .*.s = - = -• 
 
 1 — r ?* — 1 
 
 In words, 
 
 T7te s?*m o/ any number of terms of a geometric series is the 
 ratio times the last term diminished by the first term and divided 
 by the ratio less 1. 
 
 By using the value of I (Sec. 210), 
 
 s = 
 
 ar" 1 • r — a ar n — a 
 
 r-1 r-1 
 
 Thus, s may be found without first computing I. 
 
 WRITTEN EXERCISES 
 
 1. Find the sum of 6 terms of the series ^, \, \, •••• 
 
 2. Find the sum of 10 terms of the series \, — \, \, •••. 
 
 3. Find the sum of 8 terms of the series 1, .25, .0625, •••. 
 
 4. Find the sum of 12 terms of the series 27, — 9, 3, — 1,—. 
 
 5. An air pump exhausted the air from a cylinder containing 
 1 cu. ft. at the rate of j 1 ^ of the remaining contents per stroke. 
 What part of a cubic foot of air remained in the cylinder after 
 25 strokes ? 
 
 6. The population of a town increased from 10,000 to 
 14,641 in 5 yr. If the population by years was in geometric 
 series, what was the rate of increase per year ? 
 
 213. Collected Results. The three chief formulas of geo- 
 metric series are : 
 
 1. l = ar n ~\ 
 
 Ir — a 
 
 2. s = 
 
 3. s = 
 
 r-1 
 
 ar' 1 — a 
 
 r-1 
 
SERIES 
 
 177 
 
 1st 
 
 8100 (1.05) 
 
 2d 
 
 •$100 (1.05) 2 
 
 3rd 
 
 $100 (l.os) 3 
 
 4th 
 
 $100 (1.05)* 
 
 WRITTEN EXERCISES 
 
 1. $100 is placed on interest at 5%, compounded annually. 
 
 (1) What is the amount at the end of the first year ? 
 
 (2) What is the principal for the second year ? 
 
 (3) What is the amount at the end of the second year ? 
 
 Notice that the amounts appear in 
 the right-hand column of the table. 
 
 (4) Indicate similarly the amount 
 of $100 at the end of 5 yr. ; 10 yr. ; n 
 yr. Which formula of geometric series 
 expresses the amount for n yr. ? 
 
 2. Indicate the amount of $100 at 6%, compounded an- 
 nually, at the end of 1 yr.; 2 yr. ; 5 yr. ; 10 yr. ; n yr. 
 
 3. Many savings banks pay interest at the rate of 3 %, 
 compounded semiannually. 
 
 Indicate the amount of $100 under the above conditions 
 at the end of 6 mo. ; 1 yr. ; 18 mo. ; 2 yr. ; 10 yr. ; n yr. 
 
 Note. The numerical value of these expressions can be computed 
 readily by logarithms. 
 
 Solve by the use of logarithms : 
 
 4. What is the amount of $ 1 at 1 % compound interest for 
 8 yr. ? 
 
 5. A man deposits $100 in a bank paying -i</ interest, com- 
 pounded annually, on the first day of each year for 5 yr. 
 How much will he have on deposit at the end of 5 yr. ? 
 
 6. Determine similarly the amounts of : 
 
 
 Deposit at Beginning 
 or Each Tear 
 
 Kate of Interest Com- 
 pounded Annually 
 
 Number of Tears 
 
 (1) 
 
 $25 
 
 5 
 
 8 
 
 (2) 
 
 10 
 
 6 
 
 15 
 
 (3) 
 
 43.20 
 
 4 
 
 20 
 
 w 
 
 39.87 
 
 3 
 
 20 
 
178 ELEMENTARY ALGEBRA 
 
 MEANS 
 
 214. Means. . Terms standing between two given terms of a 
 series are called means. 
 
 215. Arithmetical Mean. If three numbers are in arith- 
 metical series, the middle one is called the arithmetical mean 
 between the other two. 
 
 The arithmetical mean between a and b is found thus : 
 
 1. Let A be the mean and d the common difference. 
 Then the terms may he written A — d and A + d. 
 
 2. Whence, A — d = a and A + d = b. 
 
 3. Adding, 2 A = a + 6 and A = ^±-5 ■ 
 
 216. The arithmetical mean between two numbers is one half 
 their sum. 
 
 217. Geometric Mean. If three numbers are in geometric 
 series, the middle one is called the geometric mean between the 
 other two. 
 
 The geometric mean between a and b is found thus : 
 
 1 . Let g be the geometric mean. 
 
 2. Then 2=1 
 
 a g 
 
 3. .-. g- = ab and g — Vab. 
 
 218. Tlie geometric mean between two numbers is the square 
 root of their product. There are really two geometric means, one 
 negative and one positive. 
 
 The geometric mean between two numbers is the same as 
 their mean proportional. 
 
 ORAL EXERCISES 
 
 State the arithmetical mean between : 
 
 l. 8,12. 2. (5,3. 3. 4,-10. 4. 5a, 13a. 
 
 State the geometric mean, including signs, between : 
 
 5. 8,6. 6. 3,12. 7. a, a 5 . 8. 2 a?, 32 a T . 
 
SERIES 179 
 
 219. Any number of means may be found by use of 
 formulas already given. 
 
 EXAMPLES 
 
 1. Insert 5 arithmetical means between 4 and 12. 
 
 1. In this case a — 4, I = 12, and n = 7. 
 
 2. .'. I = a + O — l)'i becomes 12 = 4 + (7 - l)d. 
 
 12 — 4 
 
 3. Solving for c?. d = — - — = 1J. 
 
 b 
 
 4. Adding ]i to4, and 1^ to that result, and so on, the means 
 
 are found to be 5|, 6|, 8, 9i, and lOf. 
 
 2. Insert 4 geometric means between — 27 and \. 
 
 1. In this case a = — 27, / = |, and n = 6. 
 
 2. .-. Z = ar 11 - 1 becomes \ — — 27 r 5 . 
 
 3. Solving for r, ?' 5 = = — — • 
 
 & 9 • 27 3 5 
 
 Therefore, r =—\. 
 
 4. Multiplying — 27 by — \, and multiplying this result by — i, 
 
 and so on, the means are found to be 9, — 3, 1, and — \. 
 
 WRITTEN EXERCISES 
 
 1. Insert 3 arithmetical means between 6 and 26. 
 
 2. Insert 10 arithmetical means between — 7 and 144. 
 
 3. Insert 3 geometric means between 2 and 32. 
 
 4. Insert 4 geometric means between — ^ and 31. 
 
 OTHER FORMULAS 
 
 220. Arithmetical Series. By means of the formulas of Sec. 
 206, any two of the five numbers, a, n, I, d, s, can be found when 
 the other three are given. 
 
 EXAMPLES 
 
 1. Given n = 6, s — 18, I — 8, find a, d. 
 
 1. For these values, formulas (1) and (2) become : 
 
 8 = a + (<S - l)d, 
 
 18= 6(« + 8) 
 2 
 
18 q ELEMENTARY ALGEBRA 
 
 2. We have thus two equations to determine the two num- 
 
 bers, a, d. 
 From the second equation, a = — 2. 
 
 3. Using this value in the first equation, d = 2. • 
 
 2. Given a = 4, I = 12, s = 56, find w and d. 
 
 1. For these values formulas (1) and (2) become: 
 
 12 = 4 + (» - l)d. 
 
 66 = *C 4+12 >. 
 2 
 
 2. .-. from the second equation, n = 7. 
 
 3. Substituting in the first, 12 = 4 + 6 d, 
 
 4. therefore d = f . 
 
 3. Given n = 12, s = 30, I = 10, find a, d. 
 
 1. Formulas (1) and (3) become : 
 
 10 = a + 11 <Z. 
 30= 12(2q+lld) = 12 a + 6r , fZ< 
 
 2 
 
 2. Solving these equations for a and d : 
 
 a = — 5, and d = |f . 
 The same results would be found by using formulas (1) and (2), since 
 (3) is only another form of (2). 
 
 221. The same problems can also be solved generally; that 
 is, without specifying numerical values. 
 
 EXAMPLE 
 
 Regarding n, s, d as known, find a, I. 
 
 1. From (1), Sec. 206, I - a =(n - Y)d. 
 
 2. From (2), Sec. 200, a + I = — • 
 
 n 
 
 3. Adding (1) and (2), 2 I = (»- 1) d + — • 
 
 Or, 
 
 ; = ( "-!>* | s . 
 
 9 e r(ti \}d s~\ 
 
 L Substituting in (2) above, a = "- — J — + - 
 
 n L 2 "J 
 
 _ .s (w-l)d . 
 ~n 2 
 
SERIES 
 
 181 
 
 WRITTEN EXERCISES 
 By use of the formulas in Sec. 206, find, the following 
 
 l. 
 2. 
 
 3. 
 
 5. 
 6. 
 
 7. 
 
 8. 
 
 9. 
 
 10. 
 
 11. 
 
 12. 
 
 13. 
 14. 
 15. 
 16. 
 
 17. 
 
 18. 
 
 19. 
 20. 
 
 Find 
 
 I 
 I 
 
 I 
 I 
 
 a 
 
 a 
 
 a 
 a 
 
 d 
 d 
 d 
 d 
 
 n 
 
 n 
 n 
 
 In Terms of 
 
 ad n 
 ads 
 
 a n's 
 1 1 n s 
 
 ad n 
 a d I 
 
 an I 
 
 d » I 
 
 dn I 
 d n s 
 
 d I s 
 n I s 
 
 an I 
 an s 
 a I s 
 n I s 
 
 ad I 
 ads 
 
 a I s 
 
 d I s 
 
 Result 
 
 I =a + (w — \)d 
 I 
 
 I = - 
 
 i [ - d ± V8 ds + (2 a - d) 2 } 
 2s 
 
 I = 
 
 n 
 
 s Q- \)d 
 n 2 
 
 % n[2a + (n- l)rf] 
 1+ a f - a? 
 
 2d 
 
 = \ n [2 I- (n- l)d] 
 
 a 
 
 a 
 
 a 
 a 
 
 I- (« - \)d 
 s_ (». - \)d 
 n 2 
 
 = \ \_d ± V(2 I + d) 2 - 8 ds] 
 
 2 s 
 
 d 
 d 
 d 
 d 
 
 1 — a 
 ii-l 
 
 2 (s — an) 
 11(11 — 1) 
 
 I 2 - a 2 
 2 s — I — a 
 2(nl- s) 
 11(11 — 1) 
 
 1 — a , ■. 
 n = h 1 
 
 d 
 
 ii 
 
 d-2a± V (2 a - d) 2 + 8 ds 
 
 n =■ 
 
 n = 
 
 2d 
 
 2 s 
 
 1 + a 
 
 2 1 + d ± V(2Z + c?) 2 - 
 
 2d 
 
 8ds 
 
 Note, a, I, d, s may have any values, but n must be a positive integer. 
 Hence, when n is one of the unknowns, not all the solutions that satisfy 
 the equations will correspond to a possible arithmetical series. 
 
182 ELEMENTARY ALGEBRA 
 
 222. Geometric Series. By means of the formulas of Sec. 
 U13, any two of the five numbers, a, n, I, r, s, can be found 
 when the other three are given. 
 
 EXAMPLES 
 
 1. Given s = 1024, r = 2,a = 2, find I. 
 
 1. For these values formula (2) becomes: 
 
 1024 = 2-^? = 2 (7-1). 
 2-1 v J 
 
 2. .-. Z = 513. 
 
 2. Given r = 3, n = 5, s = 363, find a. 
 
 1. For these values formulas (1) and (2) become; 
 
 l=.a-3*. 
 31 -a 
 
 363 = ' 
 
 2 
 
 2. Eliminating 1, 363=^! — 11. 
 
 3. Therefore, 363 = q \ 242 , and a = 3. 
 
 3. Given s = 363, a = 3, r = 3, find n. 
 
 1. For these values formula (3) becomes: 
 
 a . s n _ a 
 363 = 5_2 2. 
 
 2 
 
 2. Therefore, 3" - 1 = 242, and 3" = 243. 
 
 3. By factoring 243, n is seen to be 6. 
 
 Note. In finding n it may not be possible to factor as in the case of 
 243 above. In this case logarithms may be applied. 
 
 223. The same problems can be solved generally, that is, 
 without specifying numerical values. 
 
 EXAMPLE 
 Express I in terms of a, n, and s. 
 
 1. From formula (2) r= s -^^, or, r»-i= ^~^"" . 
 
 s-l is -I)"- 1 
 
 Cl (g ryV>-l 
 
 2. .-. substituting in (1 ) I = — ^ ' ' • 
 
 (.s — i) n 
 
 3. .-. l(s - n_1 - a (« - a)"" 1 = 0. 
 
 This equation is of a degree higher than 2 in I when n > 3. But for 
 ?i equal to or less than 3 it can be solved by methods already explained. 
 
SERIES 
 
 183 
 
 WRITTEN EXERCISES 
 By use of the formulas in Sec. 213, find the following: 
 
 Note. In Exercises 3, 12, and 16, only the equation connecting the 
 unknown numbers with the given ones can be found : 
 
 1 
 2. 
 3. 
 4. 
 
 5. 
 6. 
 
 7. 
 
 8. 
 
 10. 
 
 11. 
 12. 
 
 13. 
 14. 
 15. 
 
 16. 
 
 Find 
 
 a 
 a 
 
 In Terms of 
 
 a r n 
 ars 
 
 a n s 
 r n s 
 
 a r n 
 a rl 
 
 a n I 
 r n I 
 
 r n I 
 
 r ns 
 
 r Is 
 
 n 1 s 
 
 an! 
 
 a n s 
 
 a 1 s 
 
 nls 
 
 Result 
 
 I = ar"' 1 
 _ a + (r-l).t 
 r 
 
 l(S- O"" 1 -rt(S -«)"-! = 
 
 (r — l)s?-"- 1 
 r n — 1 
 
 s = 
 
 s = 
 
 «Q"-1) 
 
 r-1 
 rl — a 
 r-1 
 
 n—li— n— 1/ — 
 
 VY" - Va» 
 
 '-Vl-"-lTa 
 lr n - I 
 
 yWI y.H-1 
 
 a = 
 
 i.n-1 
 
 a = 
 
 (r - 1) s 
 
 r" — 1 
 a = rl — (r — l)s 
 
 a (s - a)"" 1 - I (s - I)"- 1 = 
 
 n-\ a 
 
 r= X 
 
 s s — a rt 
 
 r n — ?• + - - = 
 a a 
 
 s — a 
 
 s - I 
 s 
 
 r n - — r n ~ l +- — - = 
 s — / s — / 
 
184 ELEMENTARY ALGEBRA 
 
 SUMMARY 
 
 I. Definitions. 
 
 1. A set of numbers specified according to some law is called 
 a series. The numbers constituting a series are called its terms. 
 
 Sees. 196, 197. 
 
 2. If each term of a series after the first is found by adding 
 a fixed number {common difference) to the preceding term, the 
 series is called an arithmetical series, or arithmetical progression ; 
 if each term after the first is found by multiplying the preceding 
 term by a fixed number {common ratio), the series is called a geo- 
 metric series, or geometric progression. Sees. 200, 201, 207, 208. 
 
 3. If three numbers are in arithmetical (or geometric) series, 
 the middle one is called the arithmetical (or geometric) mean 
 between the other two. Sees. 440-444. 
 
 II. Notations. a = first term. 
 
 d = common difference. 
 
 r = common ratio. 
 
 n = number of a term, or number of terms 
 
 considered. 
 I = last (of n terms). 
 s = sum (of n terms). 
 
 III. Important Formulas. 
 
 Arithmetical Series 
 
 Geometric Series 
 
 I = a -f (n — l)d 
 
 n (a + I) 
 S= 2 
 
 o 
 
 2 
 
 Z = crr"- 1 
 
 Ir — a 
 
 a,r" — a 
 
 S - r-1 
 
 REVIEW 
 
 WRITTEN EXERCISES 
 
 1. Find the 47th multiple of 7. 
 
 2. Find the sum of the first 12 multiples of 4. 
 
SERIES 185 
 
 Find the 20th term, and the sum of 12 terms of each series-. 
 
 4. 8, 11, 14, 17, .... 
 
 5. 2 9 , 2 6 , 2 s , .... 
 
 6. a + &, a — b, a — 3 b, a — 5 b. 
 
 Find the eighth term, and the sum of 8 terms : 
 
 7. 1, 4,16, -.. 10. 1, -2, 2 2 , -2 3 , .... 
 
 8. 3,6,12,.-.. 11. ii, T L -. 
 
 9. 2, -4, 8, -16, .... 12. 100, -40, 16, .... 
 
 Find the twelfth term, and the sum of 12 terms : 
 
 13. 2,4,6, -•. 16. f,^,^, .... 
 
 14. -5, -3, -1, ••-. 17. 4, -3, -10, ••.. 
 
 IK 1 6 5 ... 1Q 1 —1 _1_1 ... 
 
 ID. -L, 7 , j, .". J.O. 2 , 3 , g , 
 
 19. Find three numbers whose common difference is 1 and 
 such that the product of the second and third exceeds that of 
 the first and second by ^. 
 
 20. The first term of an arithmetic series is n 2 — n — 1, the 
 common difference is 2. Find the sum of n terms. 
 
 21. In Italy the hours of the day are numbered from 1 to 
 24. How many strokes would a clock make per day in strik- 
 ing these hours ? 
 
 22. How many strokes does a common clock striking the 
 hours make in a day ? 
 
 23. A man leases a business block for 20 years under the 
 condition that, owing to estimated increase in the value of the 
 property, the rental is to be increased $50 each year. He 
 pays altogether $19,500. What was the rental of the first 
 year ? The last ? 
 
 24. A railroad car starting from rest began to run down 
 an inclined plane. It is known that in such motion the dis- 
 tances passed over in successive seconds are in arithmetical 
 progression. It was observed that at the end of 10 sec. the car 
 had passed over 570 ft. and at the end of 20 sec. 2340 ft. 
 
 13 
 
186 ELEMENTARY ALGEBRA 
 
 from the starting point. How far did it run the first 
 second? How far from the starting point was it at the 
 end of 15 sec. ? 
 
 25. It is known that if a body falls freely, the spaces passed 
 over in successive seconds are in arithmetical progression, and 
 that it falls approximately 16 ft. in the first second and 48 ft. 
 in the next second. To determine the height of a tower, a ball 
 was dropped from the top and observed to strike the ground 
 in 4 sec. Find the height of the tower. 
 
 26. An employee receives a certain annual salary, and in 
 each succeeding year he receives $72 more than the year 
 before. At the end of the tenth year he had received alto- 
 gether $ 10,440. What was his salary the first year ? The 
 last? 
 
 27. The 14th term of an arithmetical series is 72, the fifth 
 term is 27. Find the common difference and the first term. 
 
 28. A man is credited $100 annually on the books of a build- 
 ing society as follows : At the beginning of the first year he 
 pays in $100 cash. At the beginning of the second year 
 he is credited with $6 interest on the amount already to his 
 credit ; and he is required to pay $94 in cash, making his 
 total credit $200. At the beginning of the third year he is 
 credited with $12 interest, and pays $88 in cash, and so on. 
 How much is his payment at the beginning of the tenth year? 
 What is his credit then ? How much cash has he paid 
 altogether ? 
 
 29. At each stroke an air pump exhausts f of the air in the 
 receiver. What part of the original air remains in the receiver 
 after the 8th stroke? 
 
 30. At the close of each business year, a certain manufac- 
 turer deducts 10 <f from the amount at which his machinery 
 was valued at the beginning of the year. If his machinery 
 cost $ 10,000, at what did he value it at the end of the fourth 
 year ? . 
 
 31. By use of logarithms, find its valuation at the end of the 
 20th year. 
 
SERIES 187 
 
 SUPPLEMENTARY WORK 
 
 Special Series 
 
 Finite Series. So far we have treated only series with a fixed 
 number of terms. A series which comes to an end is called a 
 finite series. 
 
 Infinite Series. A. series whose law is such that every term 
 has a term following it is called an infinite series. 
 
 For example : 
 
 2, 5, 8, 11, •••, 239 as here written ends with 239. But the law of the 
 series would permit additional terms to be specified. In the above ex- 
 ample, the next following terms would be 242, 245, etc. It is obvious 
 that however many terms may have been specified, still more can be 
 made by adding 3. The series is thus unending. 
 
 Similarly, all of the series so far considered might have been con- 
 tinued by applying their corresponding laws. 
 
 The term " infinite 1 ' comes from the Latin infinities, and is here used 
 with the meaning, unending. 
 
 If the coefficients of the binomial expansion be regarded as a series, 
 
 , (»- 1) n(n - l)Q-2) n(n - 1)Q - 2)(n - 3) 
 ' '^"i - '" 3! ' ' 4! 
 
 they furnish, when n is a positive integer, instances of series that come to 
 an end according to the law of the series. If n — 3, the series has 4 
 terms, and if n = 10, the series has 11 terms ; for the positive integer n, 
 it has Ti + 1 terms. This is true because the factors n, n — 1, n — 2, and 
 so on, will finally in the (n + 2)nd term contain n — n or zero. There- 
 fore the series has n + 1 terms. 
 
 But if n is a negative integer or a fraction, none of the factors, n, 
 n — 1, n — 2; and so on, become zero, and the series can always be ex- 
 tended farther. That is, if n is a negative integer or any fraction, the 
 series is unending or infinite. 
 
188 ELEMENTARY ALGEBRA 
 
 Infinite Geometric Series. The subject of infinite series is of 
 great importance, but is far too difficult to be taken up here. 
 We mention simply a few properties of infinite geometric 
 series whose ratio is numerically less than 1. 
 
 The following are examples of such series : 
 1 4 '2 1 i i ... 
 
 9 3 3 3 3 
 
 z> D 5 5' 25' T251 • 
 
 3. .5, .05, .005, .0005, •••. 
 4 1_ii—i i- _ i 
 
 rx ' A ' 2~' ?' ?) T6' 32' 
 
 State the ratio and the next three terms of each series. 
 
 I. The terms become numerically smaller and smaller. Each 
 term is numerically smaller than the one preceding it, for it is 
 a proper fraction of it. 
 
 II. Tlie terms become numerically small at will. That is, 
 however small a number may be selected, there are terms in 
 the series smaller than it, and when r is numerically less than 
 1, the term ar n ~ l may be made numerically small at will, by 
 taking n sufficiently large. 
 
 This seems obvious from the consideration of the series given above as 
 examples. It is not difficult to prolong these series until their terms are less 
 than T ^ say, or -j-^g^, and from this it seems plausible to think that the 
 terms would become less than one millionth, or one billionth, or any other 
 number, if a sufficient number of terms are taken. As a matter of fact 
 this is true, but the proof is too difficult to be given here. 
 
 III. We have proved that, if s n denote the sum of the first 
 
 n terms of a geometric series, 
 
 * P 
 _ a — ar n 
 
 1 — r 
 
 This may be written : s n = ar n ~ x [ V 
 
 1 — r \1 — rj 
 
 By taking n sufficiently large, the product of ar n ~ x and the 
 
 fixed number can be made as small as desired. As more 
 
 1 — r 
 
 and more terms of the series are added, the sum differs less. 
 
SERIES 189 
 
 and less from ; and if sufficiently many terms are taken, 
 
 1 — r 
 
 the sum comes as close as we please to 
 
 1 — r 
 
 The number is called the limit of the sum of n terms, 
 
 1 — r 
 
 as n is increased without bound. Denoting this limit by s, we 
 
 have : 
 
 a 
 s = - . 
 
 The number s is not the sum of all the terms of the series, for since the 
 terms of the series never come to an end, the operation of adding them 
 cannot be completed. We cannot end an unending process. The num- 
 ber s is simply the number to which the sum of the first n terms of the 
 series approximates more and more closely as n increases. 
 
 For example : 
 
 1 4 
 
 When a = 4, and r = -, then s = — — = 8. 
 
 2' 1-| 
 
 To test this, we form successive values of s n . 
 
 S2 = 6. 
 
 s 3 = 7. 
 Si = 7J. 
 s 5 = 7f . 
 
 s 6 = 7|. 
 
 It appears that the values of s n approximate more and more closely to 
 8 as n is increased. 
 
 WRITTEN EXERCISES 
 Find the limit of the sum of the series : 
 1. 1 + 1+1-+.... 3. 5 + | + | + ^.... 
 
 2- W+W+-. 4. 8-I+A-A+-- 
 
 5. Test the results of the preceding exercises by finding 
 successive values of s n . 
 
 6. In an infinite geometric series s = 2 and r = i ; find a. 
 
 7. Find the fraction which is the limit of .333333 •••, or 
 .3 + .03 + .003+ •-.. 
 
190 ELEMENTARY ALGEBRA 
 
 8. Find the limit of .23232323 • • • or .23 + .0023 + .000023 + .... 
 
 , r- r-~- r— r- — r-r- 1 9. Triangles are drawn in a 
 
 Z ^IfiiiTm^ lilirr^ itiV Ik I rectangle of dimensions indi- 
 a b d e c cated, B being the midpoint ot 
 < 12 | Ns - -> A q D that of BCf E that of 
 
 DC, and so on. What limit does the sum of the areas of the 
 triangles approach as more and more triangles are taken ? 
 
 ADDITIONAL EXERCISES 
 
 1. Find the sum of 16 terms of the series, 
 
 27, 221, 18, 131 .... 
 
 2. Find the sum of 18 terms of the series, 
 
 3. The difference between two numbers is 48. The arith- 
 metic mean exceeds the geometric mean by 18. Find the 
 numbers. 
 
 4. Express as a geometric series the decimal fraction 
 
 .0373737-.. 
 What is its limiting value? 
 
 5 If — - — — , are in arithmetical progression, show 
 
 b — a 2 b b — c 
 
 that a, b, c are in geometric progression. 
 
 Suggestion. The supposition means that 
 1111 
 
 b - a 2b 2f> b-c 
 This reduces to b 2 — ac. 
 
 6. Find the amount in n years of P dollars at r per cent 
 per annum, interest being compounded annually. 
 
 7. During a truce, a certain army A loses by sickness 14 
 men the first day, 15 the second, 16 the third, and so on ; 
 while the opposing army B loses 12 men every day. At the 
 end of fifty days the armies are found to be of equal size. 
 Find the difference between the two armies at the beginning 
 of the truce. 
 
SERIES 191 
 
 8. A strip of carpet one half inch thick and 29f feet long 
 is rolled on a roller four inches in diameter. Find how many 
 turns there will be, remembering that each turn increases the 
 diameter by one inch, and taking as the length of a circum- 
 ference --J-- times the diameter. 
 
 9. Insert between 1 and 21 a series of arithmetic means 
 such that the sum of the last three is 48. 
 
 (1 C 
 
 10. If - = - , prove that ab + cd is a mean proportional be- 
 
 (J Cv 
 
 tween a 2 + c 2 and b 2 + d 2 . 
 
 11. The sum of the first ten terms of a geometric series 
 is 244 times the sum of the first five terms; and the sum of 
 the fourth and the sixth term is 135. Find the first term and 
 the common ratio. 
 
CHAPTER XII 
 ZERO: INTERPRETATION OF RESULTS 
 
 ZERO AND ITS PROPERTIES 
 
 224. Definition of Zero. Zero may be defined as the result 
 of subtracting a number from itself. 
 
 6 — 6 = 0; a — a = 0. 
 
 225. Addition. By definition of zero, a +0=a+b — b = a, 
 since to add b and immediately to take it away again leaves 
 the original number a. 
 
 226. Subtraction. Similarly, a — = a — (b — b) = a — b + 
 b = a, since to take away b, then at once to replace it, leaves 
 the original number a. 
 
 To add or subtract zero does not alter the original number. 
 
 227. Multiplication. By definition of zero, 
 
 • a — (b — b)a = ba — ba = 0. 
 That is, if one factor is zero, the product is zero. 
 
 Multiplication by zero simply causes the multiplicand to vanish. 
 
 228. Division. We recall that division is the process of 
 finding a number (quotient) which when multiplied by a given 
 number (divisor) shall have a given product (dividend). 
 
 12 -=- 3 or ±£ simply proposes the problem : By what must 
 3 be multiplied to produce 12 ? The proof that 12 -f- 3 = 4 is 
 the fact that 3 x 4 = 12. 
 
 Likewise a-=-0, or -, simply proposes the problem, By what 
 
 must zero be multiplied to produce a ? 
 
 Let x denote the desired number. Then • x = a. 
 
 192 
 
ZERO. INTERPRETATION OF RESULTS 193 
 
 But we know that zero times any number is zero. If a is 
 not zero, there is no number x that satisfies the above equation. 
 
 That is, - = no number, since there is no number whose prod- 
 uct with is a. 
 
 If a is zero, every number x satisfies the equation. That is, 
 - = any number, since times any number = 0. 
 
 Division by zero is therefore either entirely indefinite or im- 
 possible. In either case it is not admissible. 
 
 229 If we divide one literal expression by another, there 
 is no guarantee that the result is correct for those values of 
 the letters that make the divisor zero. 
 
 EXAMPLE 
 
 
 
 
 Let 
 
 
 
 a = b. 
 
 
 CO 
 
 Multiplying both members by a, 
 
 
 
 a 2 = ab. 
 
 
 (*) 
 
 Subtracting b 2 from both members, 
 
 
 
 «2 _ jy2 = a b- 
 
 ■b 2 . 
 
 (3) 
 
 Factoring, 
 
 (a 
 
 + b)(a- b)=b(a- 
 
 -6). 
 
 (4) 
 
 Dividing both members by a — l>, 
 
 
 
 a + b = b. 
 
 
 {5) 
 
 Substituting the value of a from (i), 
 
 
 
 6 + 6 = 6. 
 
 
 (.6) 
 
 Or, 
 
 
 
 2b = b. 
 
 
 (7) 
 
 Dividing by b, 
 
 
 
 2 = 1. 
 
 
 (*) 
 
 The work is quite correct to equation (4) inclusive. But by dividing 
 equation (4) by an expression that, according to the conditions of the 
 problem, is zero, we find as result an incorrect equation, 
 
 230. In all divisions, therefore, we must assure ourselves 
 that the divisor is not zero. If a literal divisor is used, 
 the result can be depended upon only for such values of the 
 letters as do not make the divisor zero. 
 
 EXAMPLES 
 
 1. The town B is d miles distant from A; two trains leave 
 A and B simultaneously, going in the same direction (that 
 from A towards B), at the rate of m mi. per hour (train from A) 
 
194 ELEMENTARY ALGEBRA 
 
 and q mi. per hour (train from B). At what distance from B 
 will the two trains be together ? 
 
 Solving this problem by the usual method, we find as the result —2 
 
 m — q 
 If el =£ (read " d is not equal to "'), and if m = q, the result assumes 
 
 the form 2L, an indicated division by zero. This means that the problem 
 
 is impossible under these conditions. This is evident also from the mean- 
 ing of m and q in the problem. If the two trains go in the same direction 
 at the same rate, the one will always remain d miles behind the other. 
 
 If, however, d = 0, and m = q, the result assumes the form - , which 
 
 equals any number whatever. This also agrees with the conditions of the 
 problem. If d is zero, B and A are coincident, and the two trains are 
 together at starting. If m=q, they both run at the same rate, and always 
 remain together. They are therefore together at every distance from B. 
 
 2. We have solved (First Course, Sec. 267, p. 222) the 
 
 equations 
 
 ax + by = e, 
 
 ex + dy =f. 
 
 •4-1,4.1 14. de — bf af—ec 
 with the result x = ■- ; y = -± • 
 
 ad — be ad — be 
 
 I. Let us give the letters a, b, c, d, such values that ad — be 
 =0; for example, a=2, 6=1, c=4, d=2. And let us give e and 
 / such values that de — bf is not 0; for example, e = 5, /= 4. 
 
 Then the above results become : 
 
 6 12 
 x = - : y = • 
 
 0' 9 
 
 The indicated division by zero means that the problem is impossible. 
 There is no pair of values that satisfies both equations. This appears 
 readily also by substituting the values of a •••/ in the given equations, 
 which then become : 2 a; + ?/ = 5 
 
 4 x + 2 y = 4. 
 Dividing the second equation by 2, the system becomes : 
 
 2x + y = 6, 
 
 2x + y = 2, 
 
 and it is obvious that no set of values of x and y can make 2 x + y equal 
 to 5 and also equal to 2. 
 
 The two equations are called incompatible or contradictory. 
 
ZERO. INTERPRETATION OF RESULTS 
 
 195 
 
 This condition can be illustrated graphically : 
 
 Drawing the graphs of 
 2 x + y = 5 and 2 x + y = 2, 
 the two lines appear to be 
 parallel. That two parallel 
 straight lines do not intersect is 
 the geometric condition corre- 
 sponding to the fact that a system 
 of two incompatible equations 
 has no solution. 
 
 II. Retaining the values 
 of a ••• d above, let us give 
 e and / such values that 
 the numerators of the result 
 both become zero ; for example, e = 5, /= 10 
 
 The result assumes the form : 
 
 
 
 
 X" 
 
 •••j 
 
 
 ■■ 
 
 
 
 \ 
 
 5 
 
 r 
 
 
 
 ;•■■■ 
 
 
 
 4 
 
 V 
 
 
 
 
 
 
 2 
 1 
 
 , \ 
 
 
 -■CC 
 
 
 
 
 
 
 l\ 2 
 
 \ 3 
 
 
 
 x = - ; y 
 
 0' J 
 
 
 0* 
 
 This indicates that x may have any value ; and also that y may have 
 any value. 
 
 Substituting the values of a ■•• f in the given equations, they become : 
 
 2x + y = 5 
 4x + 2y = 10. 
 
 It appears that the second equation is twice the first, and hence equiva- 
 lent to it. Any values of x and y that satisfy the first, will also satisfy 
 the second. 
 
 We can choose arbitrarily any value for x and then determine a value 
 of y to go with it by means of the first equation. For example, choosing 
 x = 3, then 2 • 3 + y = 5, which gives y = — 1. These values of x and y 
 satisfy both equations. Similarly, any value can be chosen for y, and a 
 value of x can be determined such that the pair of values satisfies the given 
 system. That is, any value of x is a root ; likewise any value of y is a 
 root, in agreement with the meaning of the form, -, assumed by the 
 result of the general solution. 
 
 The two equations are dependent. Every solution of one is 
 a solution of the other. The conditions can be illustrated 
 graphically. 
 
196 
 
 ELEMENTARY ALGEBRA 
 
 If we undertake to make the 
 graphs of the two equations as 
 given, we find that they lead to 
 the same straight line. The 
 two graphs are coincident ; 
 every point of the straight line 
 is a common point of the two 
 graphs. Any abscissa x is the 
 abscissa of a common point of 
 the graphs ; any ordinate y is 
 the ordinate of a common point 
 of the graphs. 
 
 Note. The study of expres- 
 sions which may assume the 
 exceptional forms mentioned above, especially those which may assume 
 the form - , is very important, both from the point of view of later mathe- 
 matics and the physical sciences ; but what has been said above will suffice 
 for the needs of the present work. 
 
 231. We have thus seen that systems of two linear equa- 
 tions in two unknowns may be classified as follows : 
 
 1. Independent (the ordinary case, admitting one solution). 
 
 2. Contradictory (admitting no solution). 
 
 3. Dependent (admitting a boundless number of solutions). 
 
 WRITTEN EXERCISES 
 
 Construct the graphs of each of the following systems and 
 classify them according to Sec. 231 : 
 
 1. 3x + y = 2, 
 x + y = 0. 
 
 2. 2x-y = l, 
 Ax-2y = 2. 
 
 3. s — t = 6, 
 
 4. x +2z = 10, 
 x -f- 3 z = 11. 
 
 5. a = 25, 
 y = 10. 
 
 6. lQx + hy = 25, 
 2x + y = 5. 
 
 7. 7x + Uy = 7, 
 x + 2y = 2. 
 
 8. 12x-3y = 8, 
 3y—x =4. 
 
ZERO. INTERPRETATION OF RESULTS 197 
 
 232. The discussions above are instances of what may be 
 called interpretation of results. That is, after the conditions 
 of a problem have been expressed by equations, and the equa- 
 tions solved, the result must be examined to s"ee whether it is 
 admissible under the conditions of the problem; the various 
 possible combinations of the literal expressions given must be 
 discussed ; and exceptional or noteworthy sets of values pointed 
 out. 
 
 EXAMPLES 
 
 1. Find three consecutive integers such that their sum shall 
 be equal to 3 times the second. 
 
 Solution. 1. Let x = the first. 
 
 2. Then x + 1 = the second, 
 
 3. and x + 2 = the third. 
 
 4. .-. x + (x + 1) + (x + 2) = 3(sc -f 1), by the condi- 
 
 tions of the problem. 
 
 5. ... (3 _ 3) ( X + 1) = 0, or (:>: + 1) = 0. 
 
 Interpretation of the Result. The equation determines no particu- 
 lar value of x ; it exists for every value of x. Consequently, every three 
 consecutive integers must satisfy the given conditions. 
 
 2. Find three consecutive integers whose sum is 57, and the 
 sum of the first and third is 40. 
 
 Solution. 1. Let x = the first. 
 
 2. Then x + 1 = the second, 
 
 3. and x + 2 = the third. 
 
 4. Then, x + (x + 1) + (x + 2) = 57, 
 
 5. and x + (x + 2) = 40, by the given conditions. 
 
 6. From (4), x = 18. 
 
 Interpretation of the Result. This result for x will not satisfy 
 equation (5) ; therefore no three consecutive integers satisfy the condi- 
 tions of the problem. 
 
 WRITTEN EXERCISES 
 
 Solve and interpret the results : 
 
 1. Fifteen clerks receive together $150 per week; some 
 receive $ 8 and others $ 12 per week. How many are there 
 receiving each salary ? 
 
198 ELEMENTARY ALGEBRA 
 
 2. A train starts from New York to Richmond via Phila- 
 delphia and Baltimore at the rate of 30 miles an hour, and 
 two hours later another train starts from Philadelphia for 
 Richmond at the rate of 20 miles an hour. How far beyond 
 Baltimore will the first train overtake the second, given that 
 the distance from New York to Philadelphia is 90 miles and 
 from Philadelphia to Baltimore 96 miles ? 
 
 3. The hot-water faucet of a bath tub will fill it in 14 
 minutes, the cold-water faucet in 10 minutes, and the waste 
 pipe will empty it in 4 minutes. How long will it take to fill 
 the tub when both faucets and the waste pipe are opened ? 
 
 4. If the freight on a certain class of goods is 2 cents per 
 ton per mile, together with a fixed charge of 5 cents per ton 
 for loading, how far can 2000 tons be sent for $ 80 ? 
 
 5. Find three consecutive integers whose sum equals the 
 product of the first and the last. 
 
 I 
 
SUPPLEMENT 
 
 GEOMETRIC PROBLEMS FOR ALGEBRAIC SOLUTION 
 
 The problems in the following list may be used as supple- 
 mentary work for pupils that have studied plane geometry. 
 In the body of the Algebra numerous problems have been given 
 applying geometric facts which the pupil has learned in the 
 study of mensuration in arithmetic. In the following list 
 the problems contain the application of other relations and 
 theorems of geometry, and typical solutions have been inserted 
 to suggest to the pupil the method of attack. 
 
 LINEAR EQUATIONS. ONE UNKNOWN 
 
 1. In a given triangle one angle is twice another, and the 
 third angle is 24°. Find the unknown angles. 
 
 Solution. Let x be one of the unknown angles, 
 
 then 2 x is the other. (i) 
 
 Because the sum of the angles of a triangle = 180°, 
 
 x + 2 x + 24° = 180°. 0?) 
 
 Solving equation (2), 
 
 x = 52° and 2 x = 104°. (3) 
 
 The angles of the triangle are 52°, 104°, 24°. 
 
 2. In a certain triangle one angle is three times another, 
 and the third angle is 36°. Find the unknown angles. 
 
 3. In a given right-angled triangle one acute angle is f the 
 other. Find the angles. 
 
 4. In a certain isosceles triangle the angle opposite to the 
 base is 18°. Find the angles at the base. 
 
 5. The three angles A, B, and C of a given triangle are in 
 the ratio of 2, 3, and 5. Find the angles. 
 
 199 
 
200 
 
 ELEMENTARY ALGEBRA 
 
 6. Given an angle A such that a point B situated on one 
 side 11 in. from the vertex is 8 in. distant from the other side. 
 Find a poiut C on the same side as B, and 
 equidistant from B and the other side of the 
 angle. 
 
 Solution. Let A be the given angle, then the 
 figure represents the conditions of the problem. 
 From the similar triangles A CE and ABD, we have 
 
 AC^AB 
 CE Brf 
 
 11- a; _ 11 
 
 8 ' 
 
 or, 
 
 From (1), 
 Then, 
 and 
 
 x 
 
 88-8x= 11 a\ 
 88 = 19 x, 
 
 X : 
 
 GO 
 
 8 8 
 
 T5' 
 
 (3) 
 (4) 
 
 The distance CB is f f in., or 4£f . 
 
 7. Solve problem 6, if the point B is 9 in. from A and 6 in. 
 from side AD. 
 
 8. Solve problem 6, if the point B is a in. from the vertex 
 of the angle and b in. from the other side of the angle. 
 
 9. Two points A and B are 8 in. apart. Parallels are 
 drawn through A and B ; on these parallels the points A' and 
 B' are located on the same side of the straight line through AB 
 and at distances 6 in. and 5 in. from A and B, respectively. 
 Determine the point where the line A'B' cuts the line AB. 
 
 s^b 
 
 Solution. Let C be the desired point and let BC — x. 
 Then, by similar triangles, 
 
 BC AC 
 
 or 
 
 BB> A A' ' 
 
 x _ x + 8 
 
 (0 
 GO 
 
GEOMETRIC PROBLEMS FOR ALGEBRAIC SOLUTION 201 
 
 Hence, 
 
 The point is 40 in. from B. 
 
 6x 
 
 x 
 
 5 x + 40. 
 40. 
 
 (5) 
 (-4) 
 
 10. Solve the same problem if yl'and B'lie on opposite sides 
 of AB. 
 
 11. Solve the same problem if the distance AB is d, and 
 the points A', B' lie on the same side of AB, and at distances a 
 and 6 from A and .B respectively, with a>b. 
 
 12. Solve the preceding problem if the points A' and B' lie 
 on opposite sides of AB. 
 
 13. The three sides of a triangle are 11, 9, 12. A perpen- 
 dicular is dropped on the side of length 11 from the opposite 
 vertex. Find the lengths of the segments into which the foot 
 of the perpendicular divides that side. 
 
 Solution. Using the notations of the figure, 
 
 A>=12>-a* 
 and fc2 = 92-(ll~a;)2. 
 
 From (i) and (2), 12 2 - x 2 = 9 2 - (11 - x) 2 . 
 
 Rearranging (3) , 12 2 - 9 s + ll 2 = 22 x. 
 
 Solving (4), « = ff 
 
 The other segment is 11 
 The segments are ff and ff . 
 
 x, or ff . 
 
 00 
 
 (*) 
 
 (5) 
 
 (4) 
 
 (5) 
 
 (6) 
 
 14. Solve the preceding problem if the sides are 4, 7, 9, and 
 the perpendicular is dropped on the side of length 7. 
 
 15. Solve the same problem if the sides of the triangle are 
 a, b, c, and the perpendicular is dropped on the side of length a. 
 
 14 
 
202 
 
 ELEMENTARY ALGEBRA 
 
 16. The lower base of a trapezoid is 12, the upper base is 10, 
 and the altitude is 4. Determine the altitude of the triangle 
 formed by the upper base and the prolonga- 
 tion of the two non-parallel sides until they 
 meet. 
 
 E 
 r* 
 
 /» 
 
 10 
 
 Df 
 
 l i 
 l l 
 
 i i 
 i \ 
 i • 
 
 or 
 
 Av_ 
 
 or 
 
 JB 
 
 12 
 
 Solution. Using the notations of the figure, 
 
 EF^BC 
 EG A& 
 
 x _10 
 
 z + 4~12' 
 
 Hence, 12 x = 10 x + 40, 
 
 x = 20. 
 The altitude is 20. 
 
 (-0 
 
 (3) 
 
 00 
 
 17. Solve the same problem if the lower base is a, the upper 
 base b, and the altitude h. 
 
 LINEAR EQUATIONS. TWO UNKNOWNS 
 
 18. A rectangle 5 in. longer than it is wide is inscribed in a 
 triangle of base 12 in. and altitude 9 in., the longer side resting 
 on the base of the triangle. Find the dimensions of the 
 
 rectangle. 
 Solution. 
 Then, 
 
 Let x denote the longer side and y the shorter. 
 x — y = 5. 
 
 W 
 
 E< 
 
 /' 
 
 ft 
 
 L 
 
 F \ 
 
 tP 
 
 / 
 
 \ 
 
 
 i 
 
 
 1 
 
 V 
 
 V 
 
 X 
 
 G 
 
 \ 
 
 m j 
 
 or 
 
 12 
 
 In the similar triangles ABC and AED, 
 
 AF = ED 
 AG BC' 
 
 0— y _ x 
 9 12* 
 
 (3) 
 
GEOMETRIC PROBLEMS FOR ALGEBRAIC SOLUTION 203 
 
 From (3), 
 
 From (1) and (£), 
 
 108 - 12 y = 9 x. 
 108 -12 ?/ = 9Q/ + 5), 
 or 21 y = 63. 
 
 y = s. 
 
 From (7) and (i), x = S. 
 
 The dimensions of the rectangle are 3 in. and 8 in. 
 
 (.4) 
 (5) 
 (6) 
 (?) 
 (*) 
 
 19. Solve the preceding problem if the shorter side rests on 
 the base of length 12 in. 
 
 20. Solve Problem 18 if the difference of the sides is d, the 
 length of the base of the triangle is a, the altitude is h, and 
 the longer side of the rectangle rests on the base of the triangle. 
 
 21. Solve the preceding problem if the shorter side of the 
 rectangle rests on the given base of the triangle. 
 
 22. A rectangle similar to a rect- 
 angle whose sides are 5 and 8 is 
 inscribed in a triangle of base 32 
 and altitude 20. The longer side 
 of the rectangle rests on the given 
 base of the triangle. Find the di- 
 mensions of the rectangle. 
 
 Solution. Let x and y denote the sides of the inscribed rectangle. 
 Then from the similar triangles EDC and ABC, 
 
 DE _ CF 
 
 AB~ CG' 
 
 x _ 20-y 
 
 32 _ 20 
 From the similarity of the rectangles, 
 
 x 8 
 
 or 
 
 a) 
 
 (3) 
 (4) 
 
 (6) 
 (7) 
 
 (•'0 
 
 23. Solve the preceding problem if the shorter side of the 
 rectangle rests on the given side of the triangle. 
 
 
 y 5 
 
 From (5), 
 
 x = %y. 
 
 From (;?), 
 
 20 x = 640 - 32 y. 
 
 From (4) and (;7), 
 
 32 y = 640-32?/, 
 
 
 64 y = 640. 
 
 
 y = io. 
 
 From (S) and (A), 
 
 x = 16. 
 
-04 ELEMENTARY ALGEBRA 
 
 24. Solve the same problem if the given side of the triangle 
 is of length a, and the altitude on it is of length h, and the 
 given rectangle has dimensions 1 and m, provided the inscribed 
 rectangle has its side corresponding to the side I of the given 
 rectangle resting on the given base of the triangle. 
 
 25. The bisector of an angle of a given triangle divides the 
 side opposite to the angle into two segments of lengths 4 in. 
 and 7 in. The difference between the other two sides of the 
 triangle is 5 in. Find the perimeter of the triangle. 
 
 Solution. Let ABC be the given triangle, AD the bisector of angle 
 A, and x and y the required sides. 
 
 Then, x — y = 5, given in the problem, • (l) 
 
 x 7 
 and - = -, by geometry the bisector divides the opposite side into segments 
 
 proportional to the adjacent sides. rg\ 
 
 4x = 7y, from (,?). (5) 
 
 7 y - 4 y = 20, from (3) and 4 times (2)". (4) 
 
 Then, y = 6§, solving (4). (5) 
 
 Then, .,• = llf, from (5) and (/). (6) 
 
 The perimeter is 11 in. + 6| in. + llf in. = 29i in. 
 
 26. Solve the preceding problem, if the segments of the base 
 are I and m and the difference between the sides is d. 
 
 27. The sides of a triangle are 8 ft., 12 ft., and 15 ft, and 
 the angle between the sides 8 and 12 is bisected by a line cut- 
 ting the side 15. What is the length of each segment of the 
 line 15? 
 
GEOMETRIC PROBLEMS FOR ALGEBRAIC SOLUTION 205 
 
 LINEAR EQUATIONS. THREE UNKNOWNS 
 
 28. The points A, B, and O are situated so that AB = 8 in., 
 BC=6 in., .10=5 in. Find the 
 radii of three circles having the 
 three points as centers and each 
 tangent to the other two externally. 
 
 Solution. Let .r, ?/, z, be the radii of 
 the three circles as indicated in the figure. 
 
 Then, 
 
 x + y = 8, 
 x + z — 5, 
 V + z = 6. 
 
 Adding (i), (J), and (<?), and dividing the result by 2, 
 
 x + y + z - ■*£. 
 
 From (5) and (4), 
 From (,J) and (4), 
 From (i) and (^), 
 The radii are 3£ in., 41 in., and H in. 
 
 y = h 
 
 z = 
 
 (4) 
 (-5) 
 
 (7) 
 
 29. Solve the same problem if AB = 12, 50=16, and 
 .10=20. 
 
 30. Solve Problem 28, if BC= % AC= b 
 a 
 
 AB = c. 
 
 31. A triangle .ISO is circumscribed 
 about a circle and its sides are tangent at 
 points E, D, F. The sides of the triangle 
 are a = 8 in., b = 15 in., and c = 12 in. Find 
 the segments into which points E, D, F 
 divide the sides. 
 
 Solution. The tangents from an external point 
 are equal, hence in the figure, 
 
 x + y = 8, (1) 
 
 x + z = 15, (..') 
 
 II + Z = 12. (J) 
 
 Solving these equations, the segments are x 
 
 y 
 
 z = 
 
206 
 
 ELEMENTARY ALGEBRA 
 
 32. Solve the same problem, if a = 24 ft., b = 10 ft., and 
 c = 24 ft. 
 
 33. A man owned a triangular, unfenced field of sides 600 yd., 
 
 700 yd., and 800 yd. He sold a tri- 
 angular piece cut off by a straight 
 line parallel to the side of length 
 800 yd. and found that 1800 running 
 yards of fence would be required to 
 inclose what remained. Find the 
 
 \) B lengths of the sides of the portion 
 sold. 
 
 Solution. Using the notations of the figure, 
 
 AB + BC + AC + 2 DE = AB + BE + EC + CD + DA + 2 DE 
 
 = (AB + BE + ED + DA) + (EC + CD + DE). 
 
 2100 + 2x = 1800 +( x + y + z). 
 
 CD = DE 
 CA AB' 
 
 Hence, 
 But 
 
 and 
 
 or 
 
 and 
 
 Hence, 
 
 From (7) and (8), 
 
 From (9) and (2), 
 Hence, 
 
 CE . 
 CB 
 
 DE 
 ~ AB' 
 
 y 
 
 700 
 
 % 
 ~800' 
 
 z 
 600 
 
 _ X 
 
 ~800" 
 
 y 
 
 7x 
 
 ~ 8 ' 
 
 z 
 
 _'6x 
 ' 4 ' 
 
 
 21* 
 
 x + y + z = 
 
 8 
 
 2100+2 2 = 1800 + 
 
 21 x 
 
 ox 
 
 8 
 
 = 300. 
 
 From (12) and (7), 
 From (12) and (8), 
 
 The lengths of the sides are 
 CD = 420 yd. 
 
 x = 480. 
 y = 420. 
 z = 360. 
 
 00 
 
 (*) 
 
 (3) 
 
 (4) 
 (5) 
 (6) 
 (7) 
 (*) 
 
 (IS) 
 
 (U) 
 
 Z>£' = 480yd., #(7 = 360 yd., and 
 
GEOMETRIC PROBLEMS FOR ALGEBRAIC SOLUTION 207 
 
 34. Solve Problem 33 if the division line runs parallel to the 
 side of length 700 yd. 
 
 35. Solve Problem 33 if the division line runs parallel to the 
 side of length 600 yd. 
 
 36. Solve Problem 33 if the lengths of the sides are a, b, c, 
 with the division line parallel to a, and the length of fence 
 required is 2 p. 
 
 QUADRATIC EQUATIONS 
 
 37. The sides of a triangle are AC = T, BC = 9, and AB = 10. 
 Calculate the length of the altitude 
 on the side 10, and of the two seg- 
 ments into which the altitude divides 
 that side. 
 
 Solution. Using the notations of the 
 figure, 
 
 V) 
 0*) 
 
 (3) 
 
 (4) 
 (5) 
 
 («) 
 
 (7) 
 
 
 h 2 = 7 2 - x 2 , 
 
 and 
 
 A2 = 92 _ (10 _ jc)2. 
 
 Then, 
 
 72_x 2 = 9 2 -(10-x) 2 , 
 
 or 
 
 72 = 9 2 - 100 + 20 x 
 
 Then, 
 
 68 = 20x, 
 
 and 
 
 x = V ; 
 
 also, 
 
 10 - x = A*. 
 
 By (i), 
 
 /i2 = 72_(l_r-,2. 
 
 Then, 
 
 V7' 2 • 5 2 - 17 2 
 
 _ V(35 + 17) (35- 17) 
 5 
 
 Vo2 • 18 
 
 V'2-20-2-9 6a/26 
 
 5 
 
 5 
 6 a/26 
 
 17 S3 
 
 The segments are — and - - and the altitude is 
 
 5 6 5 
 
 (9) 
 
208 ELEMENTARY ALGEBRA 
 
 38. Calculate similarly the length of the altitude on the 
 side of length 7, and of the segments into which the altitude 
 divides that side. 
 
 39. Calculate similarly the length of the altitude on the 
 side of length 9, and of the segments into which the altitude 
 divides that side. 
 
 40. If the lengths of the sides of a triangle are a, b, c, calcu- 
 late the lengths of the segments into which each side is divided 
 by the altitude on that side. 
 
 41. In a circle of radius 10, a 
 chord is drawn at distance 6 from 
 the center. Find the radius of a 
 circle that is tangent to the circle, 
 to the chord, and to a diameter per- 
 pendicular to it. 
 
 Solution. Using the notations of the figure : 
 
 EC' 2 = EF' 2 + FC' 2 . (i) 
 
 (10 - x) 2 = x 2 + (6 + x)*. {2) 
 
 100 - 20 x + x 2 = x 2 + 36 -)- 12x + x 2 . (3) 
 
 x 2 + 32 x - 64 = 0. (4) 
 
 x = _32W32 2 +4Tg (5) 
 
 _ 32±16V4!TT 
 
 - 2 {6) 
 
 = — 16±8V5. (7) 
 
 The negative value of x being inadmissible under the geometric condi- 
 tions, we have : 
 
 x=-16 + 8Vo. (5) 
 
 42. Solve the same problem, if the radius of the given circle 
 is 12 ft. and the chord is 4 ft. from the center. 
 
 43. Solve Problem 41, if the radius of the given circle is 
 /• and the chord is at d distance from the center. 
 
GEOMETRIC PROBLEMS FOR ALGEBRAIC SOLUTION 209 
 
 44. A point P is selected on a diameter of a circle of radius 
 6, at the distance 1 from the 
 center. At P a perpendicular 
 is erected to the diameter in 
 question, and a tangent is 
 drawn to the circle such that 
 the point of contact of the 
 tangent bisects the segment of 
 the tangent lying between the 
 perpendicular and the diame- 
 ter produced. Find the dis- 
 tance from the center to the 
 point where the tangent cuts 
 the diameter produced. 
 
 Solution. Let 
 
 CE = x. 
 
 Then in the right triangle CDE, 
 
 DE 2 = x 2 - 6 2 . 
 
 From the similar triangles DC'E and DGE, 
 
 DE _ GE 
 CE DE' 
 
 or DE 2 =CE-GE 
 
 = x ■ GE. 
 PE 
 
 Since D bisects FE, 
 
 GE = 
 
 2 
 x — 1 
 
 or 
 
 From (2), (5), and (7), x(x ~ l) = x 2 - 6 2 , 
 
 x 2 - x = 2 x 2 - 72. 
 .-. x 2 + x - 72 = 0. 
 
 x = — 
 
 1 ± Vl + 288 
 
 :-0, 8. 
 
 C-0 
 
 (*) 
 
 (3) 
 
 (4) 
 (5) 
 
 (6) 
 
 (7) 
 
 (^) 
 
 (9) 
 (10) 
 
 {11) 
 (12) 
 
 The negative root also indicates a solution. It means that a second 
 tangent satisfying the required conditions cuts the diameter produced on 
 the opposite side from F, at the distance 9. 
 
210 
 
 ELEMENTARY ALGEBRA 
 
 45. Solve the same problem if the radius is 12 and the point 
 lies at the distance 2 from the center. 
 
 46. Solve the same problem if the radius is 15, and the dis- 
 tance of P from the center is 35. 
 
 47. Solve the same problem if the radius is r and the dis- 
 tance from the center is d. 
 
 48. The tangent to a circle is a mean proportional between 
 the segments of the secant from the same point. Find the 
 length of the tangent, if the segments of the secant are 4 ft. 
 and 9 ft. 
 
 49. Two chords AB and CD intersect at within the circle. 
 The product of OA and OB equals the product of OC and OD. 
 Given OA = 4, OB = 8, and CD = 12, find OC and OD. 
 
 50. The owner of a triangular lot whose sides are 70, 88, 
 and 140 rd. in length wishes to divide it by a straight fence 
 
 into two parts that shall be 
 equal in area and also have the 
 same perimeter. If the fence 
 connects the sides of length 70 
 and 140 rd., how must it be 
 placed ? 
 
 Solution. Let DE be the desired 
 position of the fence. 
 
 Then, 
 
 AABC = 2ADBE. 
 
 Since the triangles have one angle in common, 
 A ABC 70-140 
 
 or, by (1), 
 
 £\DBE 
 
 2: 
 
 xy 
 70-140 
 
 or 
 
 xy 
 xy = 35- 140. 
 
 By the conditions of the problem, 
 
 BD + BE + DE = DA + AC + CE + ED. 
 
 GO 
 
 (J) 
 
 (5) 
 (4) 
 
 (5) 
 
 Subtracting DE from both members and replacing the other lines by 
 their values, 
 
GEOMETRIC PROBLEMS EOR ALGEBRAIC SOLUTION 211 
 
 x + y = 70 - y + 1-40 - x + 88, 
 
 (6) 
 
 2(x + y) = 298, 
 
 (?) 
 
 x -f- y - 149. 
 
 (*) 
 
 From (4), 4 xy = 4 • 35 • 140 
 
 
 = I40 2 . 
 
 (9) 
 
 Squaring (8) and subtracting (9) from the result, 
 
 
 (x - y) 2 = 149 2 - 140 2 
 
 (10) 
 
 = (149+ 140)(149-140) 
 
 (U) 
 
 = 289 • 9. 
 
 (12) 
 
 Then, x — y = ± 17 ■ 3 
 
 
 = ±51. 
 
 (13) 
 
 From (8) and (13), x = 100, 49. 
 
 (U) 
 
 y = 49, 100. 
 
 (15) 
 
 These values satisfy the algebraic equations, but in the concrete prob- 
 lem y — 100 is inadmissible, since y lies on the side of length 70. Hence, 
 in the concrete problem, the result is x — 100, y = 49. 
 
 51. Solve the same problem if the fence connects the sides 
 of length 70 and 88. 
 
 52. Solve the same problem if the fence connects the sides 
 of length 88 and 140. 
 
 53. Solve the same problem if the sides are of length a. b, 
 c, and the fence connects the sides of lengths a 
 and c. 
 
 54. Find the sides of a right-angled triangle, 
 given its area 25, and its perimeter 30. 
 
 Solution. Let x, y, z denote the sides of the triangle, 
 z being the hypotenuse. 
 
 Then, 
 
 and 
 
 x 2 + y 2 = z 2 , 
 x + y -f z = 30, 
 
 ^ = 25. 
 2 
 
 Multiplying both members of (3) by 4, 
 
 2xy = 100. 
 
 Adding (4) and (1), x 2 + 2 xy + y 2 = z 2 + 100, 
 or (x + y) 2 = z 2 + 100. 
 
 From (2), x + y = 30 — z, 
 
 or t» + y) 2 = C30-») 2 . 
 
 (4) 
 
 (5) 
 
 (6) 
 (7) 
 
 (8) 
 
212 ELEMENTARY ALGEBRA 
 
 From (8) and (6), (30- z) 2 = z- + 100. (3) 
 
 (i0) 
 
 From (7), x + y = 5 ^, (i$) 
 
 or 
 
 
 (30- 
 
 -*>» 
 
 = :- + 100. 
 
 900 
 
 -6O2 
 
 + Z 2 
 
 = s 2 + 100. 
 
 
 
 6O2 
 
 = 800. 
 
 
 
 z 
 
 — 3 • 
 
 
 : 
 
 '■ + y 
 
 — 5_a 
 
 3 ' 
 
 X 2 
 
 + 2 ay + y2 : 
 
 — ipOO 
 
 — 9 • 
 
 Multiplying (4) by 2 and subtracting the result from (14), 
 
 X 2 -2.nj + y* = 1<1Q, (25) 
 
 Adding (13) and (16) and dividing the result by 2, 
 
 2o±5V7 
 
 .'■ = 
 
 3 
 
 Subtracting (16) from (13) and dividing the result by 2, 
 
 (17) 
 
 y = ^^- m 
 
 The sides are 25 + 5 ^, . 25-5V7 and 40 
 3 3 ' 3 
 
 55-. Solve the same problem if the area of the triangle is 64. 
 and the perimeter 48. 
 
INDEX 
 
 The numbers in the Index refer to pages in the Book 
 
 Absolute Terms, 26. 
 Addition, 1. 
 
 Commutative Law of, 2. 
 
 Associative Law of, •>. 
 
 of Fractions, 16. 
 
 of Radical Expressions, 44. 
 
 Method of, 31. 
 Algebraic Expressions, 8. 
 Alternation, 152. 
 Associative Law, 3, 12. 
 
 Base, 79. 
 
 Binomial Equations, 116. 
 
 Characteristic, 79, 80. 
 Common Difference, 171. 
 Common Ratio, 174. 
 Commutative Law, 2, 11, 12. 
 Completing the Square, 36. 
 Complex Numbers, 91. 
 Composition, 153. 
 and Division, 153. 
 
 Degree of Equations, 26. 
 Difference, 4. 
 Discriminant, 109. 
 Distributive Law, 12. 
 Division, 14. 
 
 Law of Exponents in. 54. 
 
 of Fractions, 19. 
 
 of Imaginaries, 93. 
 
 Equations, 25. 
 Identical, 25. 
 Conditional, 25. 
 Solving, 26, 27, 31, 32, 36. 
 Roots of, 25. 
 of One Unknown, 25. 
 of Two Unknowns, 31, 132. 
 Systems of, 31. 
 Simultaneous, 31, 132. 
 Equivalent, 26. 
 Degree of, 26. 
 Linear, 26. 
 Quadratic, 26, 36. 
 Higher, 26, 141. 
 with Three or More Unknowns, 
 
 no 
 OO. 
 
 Radical, 48. 
 Binomial, 116. 
 Incompatible, 194. 
 Contradictory, 194. 
 Dependent, 195. 
 Exponents, 53. 
 Laws of, 53. 
 Fractional, 56, 65. 
 Negative, 63, 65. 
 Zero, 65. 
 
 Factoring, 20, 109. 
 
 Applied to Equations, 125. 
 Factors, 11. 
 Factor Theorem, 125. 
 
 213 
 
214 
 
 ELEMENTARY ALGEBRA 
 
 Forms, 26, 36. 
 
 Linear, 26. 
 
 Quadratic, 36. 
 Formulas, 32, 102, 175. 
 Fourth Proportional, 148. 
 Fractions, 15. 
 
 Numerator and Denominator 
 of, 15. 
 
 Terms of, 15. 
 
 Reduction of, 16. , 
 
 Processes with, 16-19. 
 
 Complex, 19. 
 
 Graphs, 1, 2, 3, 5, 11, 12, 15, 32, 
 97, 111, 129, 145, 163, 195. 
 
 Imaginary Numbers, 90. 
 
 Processes with, 91. 
 
 Powers of, 94. 
 
 as Roots of Equations, 94. 
 Interpretation of Results, 197. 
 Irrational Numbers, 42. 
 
 Logarithms, 74. 
 Linear Equations, 26. 
 Linear Forms, 26. 
 
 Mantissa, 79. 
 
 Mean Proportional, 148, 153. 
 
 Means, 178. 
 
 Arithmetical, 178. 
 
 Geometric, 178. 
 Multiplication, 11. 
 
 Commutative Law of, 11. 
 
 Law of Exponents in, 53. 
 
 of Relative Numbers, 13. 
 
 Distributive Law of, 12. 
 
 of Fractions, 17. 
 
 of Radicals, 45. 
 
 of Imaginaries, 92. 
 
 Negative numbers, 5. 
 
 Notation of Positive and Nega- 
 tive Numbers, 6. 
 
 of Polynomials, 125. 
 Numbers, 5, 15. 
 
 Relative, 5, 7, 8. 
 
 Positive, 5. 
 
 Negative, 5. 
 
 Rational, 42. 
 
 Irrational, 42. 
 
 Real, 90. 
 
 Imaginary, 90. 
 
 Complex, 91. 
 
 Order of Operations, 3. 
 
 Parentheses, 9. 
 
 Notation of, 125. 
 Positive Numbers, 6. 
 Powers, 5. 
 
 Law of Exponents for, 54. 
 Product, 11. 
 Progression, 171. 
 
 Arithmetical, 171. 
 
 Geometric, 174. 
 Proportion, 148. 
 Proportional, 148. 
 
 Fourth, 148. 
 
 Third, 148. 
 
 Mean, 148. 
 
 Quadratic Equatious, 36, 100. 
 Complete, 36. 
 Incomplete, 36. 
 General Form of, 100. 
 General Solution of, 102. 
 Literal, 103. 
 Simultaneous, 132. 
 
 Radical Expressions, 42. 
 Properties of, 42. 
 Addition of, 44. 
 Subtraction of, 44. 
 
INDEX 
 
 215 
 
 Multiplication of, 45. 
 
 Square root of, 51. 
 
 In Equations, 48, 123. 
 Radicals, 42. 
 Ratio, 148. 
 
 Common, 174. 
 Rational Numbers, 42. 
 Rationalizing Factors, 47. 
 
 the Denominator, 47. 
 Real Numbers, 90. 
 Reciprocal, 18. 
 Relative Numbers, 5, 7, 8. 
 Roots of Equations, 25. 
 
 Relation of, to Coefficients, 107. 
 
 Series, 170. 
 
 Arithmetical, 171, 179. 
 
 Sum of, 172, 17-">. 
 
 Geometric, 174, 182. 
 
 Finite, 187. 
 
 Infinite, 187. 
 Signs, 6. 
 
 of Operation, 6. 
 
 of Character, 6. 
 Square Roots, 45, 46. 
 
 Substitution, 31. 
 
 Method of, 31. 
 Subtraction, 4. 
 
 Method of, 31. 
 Subtraction, of Fractions, 16. 
 
 of Radicals, 44. 
 
 of Imaginaries, 92. 
 Surds, 42. 
 
 Terms, 170. 
 
 Absolute, 26. 
 
 in Series, 170, 172, 175. 
 Testing, 27, 108. 
 Third Proportionaly- 448. ■ — 
 
 Unknowns, 26. 
 
 Value, 5. 
 
 Absolute, 6. 
 Variables, 160. 
 Variation, 160. 
 
 Direct, 160. 
 
 Relation of, to Proportion, 160. 
 
 Inverse, 161. 
 
 Zero, 192. 
 
 (2) 
 
THIS BOOK IS DUE ON THE LAST DATE 
 STAMPED BELOW 
 
 AN INITIAL FINE OF 25 CENTS 
 
 WILL BE ASSESSED FOR FAILURE TO RETURN 
 THIS BOOK ON THE DATE DUE. THE PENALTY 
 WILL INCREASE TO 50 CENTS ON THE FOURTH 
 DAY AND TO $1.00 ON THE SEVENTH DAY 
 OVERDUE. 
 
 
 f 
 
 22 1941 
 
 Ml 
 
 *WF S 
 
 ^Trr 
 
 MAR 19 WO 
 
 SEP *'?1941 
 
 OCT 14 1946 
 
 ■j ■ 
 
 FEB 14 1942 
 
 
 APR m m2 
 
 n« 
 
 NOV 13 1944 
 
 ^EC 
 
 mov 301944 
 
 1AB1 B5-1P1 
 
 QSAD 
 
 00m-7,' 
 
° MSK (ROM WHICH BORROWED 
 
 LOAN DIPT. 
 
 LD62A-50m.7,'65 
 (i , o/o6slO)94 12 A 
 
 IT n GeneralLib '-ary 
 U n ivers«yofCal^ rnia 
 Berkeley