LIBRARY OF THE University of California. OTF^T OF" Class mmmmmmmf- 'M ^ •sL-' »sL^ 'sL^ sL-" «\L^ "sX^ 'sX^ ■sj^ •sj/^ 'vXo'' •^ .^ »^ •jx. ^N, v^ .^ pj^;; Pj^c ^pysr":;:^ Class * ^ Pi Accession No./r/.d.4/^ [KM aiARf Ij 113 n .A. n. T^- . GIFT OF REY.JJl.BENT8N,DJ. '• ^ 'sj^ ^'vj^ "sL^ -sj^ «sX^ 'vX ^ «J^ •\L^ '\1^ "•'^ *^^ •^^^^ •'Tm v-']^. P^ i-^ *M^ ' •'TSi •/TSi " r^ Digitized by tine Internet Archive in 2008 witin funding from IVIicrosoft Corporation http://www.arcliive.org/details/firstlessonsinalOOdavirich FIRST LESSONS ALGEBRA, EMBRACING THE ELEMENTS SCIENCE. BY CHARLES DAVIES, ITJTHOR OF MENTAL AND PRACTICAL ARITHMETIC, ELEMENTS OF SURVKYIKG, BI.EMENTS OF DESCRIPTIVE AND ANALYTICAL GEOMETRY, ELEMENTS OP DIFFERENTIAL AND INTEGRAL CALCULUS, AND A TREATISE ON SHAPES, SHADOWS AND PERSPECTIVE. PUBLISHED BY A. S. BARNES & Co, Hartford.— WILEY & PUTNAM; COLLINS, KEESE & Co, New-York.— PERKINS & MARVIN, Boston.— THOMAS, COWPERTHWAIT & Co, Philadelphia.— GUSHING & SONS, Baltimore. 1839. DAVIES' COURSE OF MATHEMATICS. DAVIES' MENTAL and PRACTICAL ARITHMETIC— Designed for the use of Academies and Schools. It is the purpose of this work to explain, in a brief and clear manner, the properties of numbers, and the best rules in their various applications. DAVIES' KEY— To Mental and Practical Arithmetic. DAVIES' FIRST LESSONS in ALGEBRA— Being an introduction to the Science. DAVIES' BOURDON'S ALGEBRA— Being an abridgment of the work of M. Bourdon, witli the addition of practical examples. DAVIES' LEGENDRE'S GEOMETRY and TRIGONOMETRY — Being an abridgment of the work of M. Legendre with the addition of a treatise on Mensuration of Planes and Solids, and a table of Logarithms and Logarithmic sines. DAVIES' SURVEYING— With a description and plates of, the Theo- dolite. Compass, Plane-Table and Level, — also. Maps of the Topo- graphical Signs adopted by the Engineer Department, and an explana- tion of the method of surveying the Public Lands. DAVIES' ANALYTICAL GEOMETRY— Embracing the Equations of the Point and Straight Line — of the Conic Sections — of the Line and Plane in Space — also, the discussion of the General Equation of the second degree, and of surfaces of the second order. DAVIES' DESCRIPTIVE GEOMETRY— V^^ith its applications to Spherical Projections. DAVIES' SHAD0V7S AND LINEAR PERSPECTIVE. DAVIES' DIFFERENTIAL AND INTEGRAL CALCULUS— With numerous applications. Entered according to the Act of Congress, in the year one thousand eight hundred and thirty-eight, by Charles Davies, in the Clerk's Office of the District Court of the United States, for the Southern District of New York. Stereotyped by Henry W. Rees, 32 Ann Street, New York. PREFACE. Although Algebra naturally follows Arithmetic in a course of scientific studies, yet the change from num- bers to a system of reasoning entirely conducted by letters and signs is rather abrupt and not unfrequently discourages and disgusts the pupil. In the First Lessons it has been the intention to form a connecting link between Arithmetic and Algebra, to unite and blend, as far as possible, the reasoning on numbers with the more abstruse method of analysis. The Algebra of M. Bourdon has been closely fol- lowed. Indeed, it has been a part of the plan, to furnish an introduction to that admirable treatise, which is justly considered, both in Europe and this country, as the best work on the subject of which it treats, that has yet appeared. 111869 4 PREFACE. This work, however, even in its abridged form, is too voluminous for schools, and the reasoning is too elaborate and metaphysical for beginners. It has been thought that a work which should so far modify the system of Bourdon as to bring it within the scope of our common schools, by giving to it a more practical and tangible form, could not fail to be useful. Such is the object of the First Lessons. It is hoped they may advance the cause of education, and prove a useful introduction to a full course of mathematical studies. HiRTFORD, September^ 1838. CONTENTS. CHAPTER I. PRELIMINARY DEFINITIONS AND REMARKS. ARTICLESf! Algebra — Definitions — Explanation of the Algebraic Signs, - 1 — 23 Similar Terms — Reduction of Similar Terms, - - • 23 — 26 Addition— Rule, -----.-. 26 — 28 Subtraction — Rule — Remark, - - - - - - 28 — 33 Multiplication — Rule for Monomials, ----- 33 — 36 Rule for Polynomials and Signs, ------ 36 — 38 Remarks — Properties Proved, ------ 38 — 43 Division of Monomials — Rule, ------ 42 — 45 Signification of the Symbol ao, - - ^ - - - 45 — 46 Of the Signs in Divison, ------- 46 — 47 Division of Polynomials, ------ 47_49 CHAPTER H. ALGEBRAIC FRACTIONS. l)efinitions — Entire Quantity — Mixed Quantity, • - 49—53 To Reduce a Fraction to its Simplest Terms - - . 62 To Reduce a Mixed Quantity to a Fraction, - - - 63 To Reduce a Fraction to an Entire or Mixed Quantity, - - 64 To Reduce Fractions to a Common Denominator, - - - 65 To Add Fractions, • - 56 6 CONtENTS. To Subtract Fractions, To Multiply Fractions, To Divide Fractions, - ARTICLES. 57 59 CHAPTER III EQUATIONS OP THE FIRST DEGREE. Definition of an Equation — Properties of Equations, Transformation of Equations — First and Second, - - - Resolution of Equations of the First Degree — Rule, Questions involving Equations of the First Degree, Equations of the First Degree involving Two Unknown Quantities, - - - - - - -- Elimination — By Addition — By Subtraction — By Comparison, - Resolution of Questions involving Two or more Unknown Quantities, --------- 60- -66 66- -70 70 71- -73 72 73- -76 76—79 CHAPTER IV. OF POWERS. Definition of Powers, -------- 79 To raise Monomials to any Power, ----- 80 To raise Polynomials to any Power, ----- 81 To raise a Fraction to any Power, ----- 82 — 83 Binomial Theorem, -------- 84 — 90 CHAPTER V. Definition of Squares — Of Square Roots — And Perfect Squares, 90—96 Rule for Extracting the Square Root of Numbers, - - 96 — 100 Square Roots of Fractions, ------ lOO — 103 Square Roots of Monomials, 103 — 107 Calculus of Radicals of the Second Degree, - - • 107—109 CONTENTS. 7 ARTICLES. Addition of Radicals, .---.•- 109 Subtraction of Radicals, - - - - - - - 110 Multiplication of Radicals, - - - - - - 111 Division of Radicals, - - - - - - - 112 Extraction of the Square Root of Polynomials, - . - 113 — 116 CHAPTER VI. Equations of the Second Degree, - - - y - 116 Definition and Form of Equations, - - - - 116 — 118 Incomplete Equations, ------- us — 122 Complete Equations, ------- 122 Four Forms, 123—127 Resolution of Equations of the Second Degree, - - - 127 — 128 Properties of the Roots, - - - - . - 128—134 CHAPTER VIL Of Progressions, ------.-• 135 Progressions by Differences, ------ 136 — 138 Last Term, 138—140 Sum of the Extremes — Sum of the Series, - ' - - 140 — 141 The Five Numbers — To Find any Number of Means, - 141 — 144 Geometrical Proportion and Progression, - - - - 144 Various Kinds of Proportion, ------ 144 — 166 Geometrical Progression, -- 166 Last Term— Sum of the Series, 167 — 171 Progressions having an Infinite Number of Terms, ' - - 171 — 172 The Five Numbers— To Find One Mean - - - - 172—173 FIRST LESSONS IN ALGEBRA CHAPTER I. Preliminary Definitions and Remarks. 1. Quantity is a general term embracing every thing which can be increased or diminished. , 2. Mathematics is the science of quantity. 3. Algebra is that branch of mathematics in which the quantities considered are represented by letters, and the ope- rations to be performed upon them are indicated by signs. These letters and signs are called symbols. 4. The sign +, is called plus ; and indicates the addition of two or more quantities. Thus, 9 + 5, is read, 9 plus 5, or 9 augmented by 5. If we represent the number nine, by the letter «, and the number 5 by the letter h, we shall have a+ 5, which is read, a plus h ; and denotes that the number represented by a is to be added to the number represented by b. 5. The sign—, is called minus ; and indicates that one Quest. — 1. What is quantity 1 2. What is Mathematics 1 3. What is Algebra 1 "WHiat are these letters and signs called 1 4. What does the sign plus indicate 1 6. What does the sign minus indicate ? 10 FIRST LESSONS IN ALGEBRA. quantity is to be subtracted from another. Thus, 9— 5 is read, 9 minus 5, or 9 diminished by 5. In like manner, a—b, is read, a minus h, or a diminished by h. 6. The sign x , is called the sign of multiplication ; and when placed between two quantities, it denotes that they are to be multiplied together. The multiplication of two quantities is also frequently indicated by simply placing a point between them. Thus, 36 x 25, or 36.25, is read, 36 multiplied by 25, or the product 36 by 25. 7. The multiplication of quantities, which are represented by letters, is indicated by simply writing them one after the other, without interposing any sign. Thus, ah signifies the same thing as axh, or as a,h ; and ahc the same as axhxc, or as a.b.c. Thus, if we suppose a=:36, and Z>r=25, we have a5 = 36x25=: 900. Again, if we suppose ^=2, b=3 and c=4, we have ahc=z2x3x4=z24. It is most convenient to arrange the letters of a product in alphabetical order. 8. In the product of several letters, as abc, the single let- ters, a, bj and c, are called factors of the product. Thus, in the product ah, there are two factors, a and b ; in the product abc, there are three, a, 5, and c. Quest. — 6. What is the sign of multiplication 1 What does the sign of multiphcation indicate 1 In how many ways may multiplication be expressed 1 7. If letters only are used, how may their multiplication be expressed! 8. In the product of several letters, what is each letter called 1 How many factors in db 1 — In abc ? — In abed 1 — In ahcdfl DEFINITION OF TERMS. 11 9. There are three signs used to denote division. Thus, a-^b denotes that a is to be divided by h. - denotes that a is to be divided by h. o a\b denotes that a is to be divided by b. 10. The sign =, is called the sign of equality^ and is read, is equal to. When placed between two quantities, it denotes that they are equal to each other. Thus, 9 — 5=4 : that is, 9 minus 5 is equal to 4 : Also, a-\-bz:z.c, denotes that the sum of the quantities a and b is equal to c. If we suppose a = 10, and b = 5, we have a-{-b=zCj and 10 + 5=c = 15. 1 1. The sign >, is called the sign of inequality, and is used to express that one quantity is greater or less than another. Thus, « > Z> is read, a greater than b ; and a<^b is read, a less than b ; that is, the opening of the sign is turned towards the greater quantity. Thus, if a =9, and 5=4, we write, 9>4. 12. If a quantity is added to itself several times as a-\-a+a-\-a+a-\-a, we generally write it but once, and then place a number before it to show how many times it is taken. Thus, a-\-a-\-a-\-a-{-a=i^a. Quest. — 9. How many signs are used in division 1 What are they 1 10. What is the sign equality \ When placed between two quantities, what does it indicate! 11. For what is the sign of inequahty usedl Which quantity is placed on the side of the opening "? 12. What is a co- efficient 1 How many times is ah taken in the expression ah ] In 3ai ! In ^ah 1 In bah 1 In 6a5 \ If no co-efficient is written, what co-efficient is understood'^ 12 FIRST LESSONS IN ALGEBRA. The number 5 is called the co-efficient of a, and denotes that a is taken 5 times. When the co-efficient is 1 it is generally omitted. Thus, a and \a are the same, each being equal to «, or to one a. 13. If a quantity be multiplied continually by itself, as axaxaxaxa, we generally express the product by writing the letter once, and placing a number to the right of, and a little above it : thus, aXaXaxa Xa=:a^. The number 5 is called the exponent of a, and denotes the number of times which a enters into the product as a factor. For example, if we have a^, and suppose a = 3, we write, a^ = axaXa=3^=:3xSx3=z27. If a=i4, a3 = 43=:4x4x4 = 64, and for ariz5, a^=5^=:5x5x 5 = 125. If the exponent is 1 it is generally omitted. Thus, a^ is the same as a, each expressing a to the first power. 1 4. The power of a quantity is the product which results from multiplying the quantity by itself. Thus, in the example a3— 43 = 4x4x4 = 64, 64 is the third power of 4, and the exponent 3 shows the degree of the power. 15. The sign V , is called the radical sign, and when Quest. — 13. What does the exponent of a letter showl How many times is a a factor in a2 1 \n a^\ In a^ 1 In as '? If no exponent is written, what exponent is understood 1 14. What is the power of a quantity '\ What is the third power of 2 ] Express the 4th power of a, 15. Express the square root of a quantity. Also the cube root. Also the 4th root. DEFINITION OP TERMS. 13 prefixed to a quantity, indicates that its root is to be ex- tracted. Thus, '^oTov simply -/tTdenotes the square root of a. -y/o'denotes the cube root of a, -t/o" denotes the fourth root of a. The number placed over the radical sign is called the in- dex of the root. Thus, 2 is the index of the square root, 3 of the cube root, 4 of the fourth root, &;c. If we suppose a = 64, we have 16. Every quantity written in algebraic language, that is, with the aid of letters and signs, is called an algebraic quantity, or the algebraic expression of a quantity. Thus, is the algebraic expression of three times the number a ; -25^^ ^^® algebraic expression of five times ( the square of a ; 3 i ia < r is the algebraic expression of seven times 7a^'^ ^ the product of the cube of a by the square ( of 6; q _ci ( is the algebraic expression of the difference I between three times a and five times b ; 2a2 — 3a5+452^ is the algebraic expression of twice the square of «, diminished by three times the product of a by 5, augmented by four times the square of b. 1. Write three times the square of a multiplied by the cube of b. Ans. 3a^\ Quest. — 16. What is an algebraic quantity 1 Is 5ab an algebraic quantity 1 Is 9a 1 Is^l IsZb — x] 2 14 FIRST LESSONS IN ALGEBRA. 2. Write nine times the cube of a multiplied by h, dimin- ished by the square of c multiplied by d. Ans, ^a%—c^d. 3. If «=i2, Z>iiz3, and cz=zb, what will be the value of 3^2 multiplied by V^ diminished by a multiplied by h multi- plied by c. We have ^a%'^—ahc = ^ X 22 X 32-2 X 3 X 5=78. 4. If cz=4, 5 = 6, c=7, cZ=8, what is the value of 9a?+hc—ad1 Ans. 154. 5. If fl=7, Z>=3, c=7, (?=1, what is the value of Gad-^W^c — Ad'^'^. Ans, 227. 6. If a = 5, 5 = 6, c=6, t?=5, what is the value of 9ahc — Sad-{-Ahc'i Ans. 1564. 7. Write ten times the square of a into the cube of h into c square into d^, 17. When an algebraic quantity is not connected with any other, by the sign of addition or subtraction, it is called a monomial^ or a quantity composed of a single term, or sim- ply, a term. Thus, 3«, 5a2, 7a352^ are monomials, or single terms. 18. An algebraic expression composed of two or more parts, separated by the sign + or — , is called d, polynomial^ or quantity involving two or more terms. For example, 3a-55 and 2a^—'ich+W^ are polynomials. 19. A polynomial composed of two terms, is called a hi" nomial ; and a polynomial of three terms is called a trinomial. Quest. — 17. What is a monomial 1 Is ^ab a monomial 1 18. What is a polynomial 1 Is 3a — & a polynomial 1 19. What is a binomial] What is a trinomiall DEFINITION OF TERMS. 15 20. Each of the literal factors which compose a term is called a dimension of this term : and the degree of a term is the number of these factors or dimensions. Thus, 1 is a term of one dimension, or of the first IS second degree. degree. { - , ^ is a term of two dimensions, or of the 7 3^ 2 — 7 h ( is of six dimensions, or of the sixth "~ I degree. 2 1 . A polynomial is said to be homogeneous, when all its terms are of the same degree. The polynomial ^a—2h+c is of the first degree and homogeneous. —Aah-\-h'^ is of the second degree and homogeneous. ^a^c—4:c'^+2cH is of the third degree and homogeneous. 8a3+4«6+c is not homogeneous. 22. A vinculum or bar — , or a parenthesis ( ), is used to express that all the terms of a polynomial are to be considered together. Thus, a+h-fcxh, or {a+h+c)xh, denotes that the trinomial a+h+c is to be multiplied by b ; also, a-\-b+cXc-\-d-\-f, or {a+h+c)x{c-\-d+f), denotes that the trinomial a+h+c is to be multiplied by the trinomial c+d+f. When the parenthesis is used, the sign of multiplication is usually omitted. Thus, (a+h+c)y.h is the same as [a+h + c)h. Quest. — 20. What is the dimension of a termi What is the degree of a term 1 How many factors in ^ahc 1 Which are they 1 What is its degree "! 21. When is a polynomial homogeneous 1 Is the polyno- mial 2a36+3a262 homogeneous'? Is ^a^h — ^3] 22. For what is the vinculum or bar used 1 Can you express the same with the parenthesis 1 16 FIRST LESSONS IN ALGEBRA. !23. The terms of a polynomial which are composed of the same letters, the same letters in each being affected with like exponents, are called similar terms. Thus, in the polynomial lah + ^ah —AaW + 5a^% the terms 7ab, and Sab, are similar : and so also are the terms —4a^^ and da^b"^, the letters and exponents in both being the same. But in the binomial 8a^-\-7ab^, the terms are not similar ; for, although they are composed of the same letters, yet the same letters are not affected with like exponents, 24. When an algebraic expression contains similar* terms, it may be reduced to a simpler form. 1. Take the expression 3ab + 2aby which is evidently equal to 5ab. 2. Reduce the expression 3ac-\-9ac-\-2ac to its simplest form. Ans. 14ac. 3. Reduce the expression abc-\-4abc-\-5abc to its sim- plest form. In adding similar terms together we abc take the sum of the coefficients and 4abc annex the literal part. The first term, 5abc abc, has a coefficient 1 understood, lOabc (Art. 12). 25. Of the different terms which compose a polynomial, some are preceded by the sign +, and the others by the sign — , The first are called additive terms, the others, subtractive terms. Quest. — 23. What are similar terms of a polynomiall Are Sasja and 6a2&2 similar 1 Are 2a2i2 and 2a3j2 ] 24. If the terms are positive and similar, may they be reduced to a shnplerformt In what way' DEFINITION OF TERMS. 17 The first term of a polynomial is commonly not preceded by any sign, but then it is understood to be affected with the sign +. 1 . John has 20 apples and gives 5 to William : how many has he left ? Now, let us represent the number of apples which John has by a, and the number given away by h : the number he would have left would then be represented by a—b. 2. A merchant goes into trade with a certain sum of money, say a dollars ; at the end of a certain time he has / gained b dollars : how much will he then have ? Ans. a-{-b dollars. If instead of gaining he had lost b dollars, how much would he have had 1 Ans. a — b dollars. Now, if the losses exceed the amount with which he began business, that is, if b were greater than a, we must prefix the minus sign to the remainder to show that the quantity to be subtracted was the greatest. Thus, if he commenced business with $2000, and lost $3000, the true difference would be —1000 : that is, the subtractive quantity exceeds the additive by $fOOO. 3. Let a merchant call the debts due him additive, and the debts he owes subtractive. Now, if he has due him $600 from one man, $800 from another, $300 from another, and owes $500 to one, $200 to a second, and $50 to a third, how will the account stand? Ans. $950 due him. 4-. Reduce to its simplest form the expression 3a^ -{-5a'^b — 3a^ + 4a^ — Ga^J _ a^, Quest. — 25. What are the terms called which are preceded by the sign + ? What are the terms called which are preceded by the sign — . If no sign is prefixed to a term, what sign is understood ? If some of the terms are additive and some subtractive, may they be reduced if similar ? Give the rule for reducing, them, . Does the reduction affect the expo- nents, or only the coefficients ? 2* 18 FIRST LESSONS IN ALGEBRA. Additive terms. Suhtractive terms. + ^a?h — Za^h + ba% — Qa?h 4- ^a^h — a^h Sum +\2a% Sum — lOo^^'. But, \2a%-\0a?h=z2a?b. Hence, for the reduction of the similar terms of a polyno- mial we have the following RULE. I. Form a single additive term of all the terms preceded hy the sign plus ; this is done by adding together the coefficients of those terms ^ and annexing to their sum the literal part. II. Form, in the same manner, a single suhtr active term. III. Subtract the less sum from the greater, and prefix to the result the sign of the greater. Remark. — It should be observed that the reduction affects only coefficients, and not the exponents. • EXAMPLES. 1. Reduce to its simplest form the polynomial -\-2a%c'^ — A a^c^ + Sa^c^- — 8a^c^ + 1 1 a^c'^. Find the sum of the additive and subtractive terms sepa- rately, and take their difference : thus. Additive terTns. Subtractive terms. + 2a^c'^ — 4a^c^ + 6a^c^ ' — Sa^c'^ + lla^ Sum — I2a^c^ Sum -\-l9o^ Hence, the given polynomial reduces to 1 9a^bc^ — 1 2u^c^ = 7a^bc^. ADDITION. 19 2. Reduce the polynomial 4a26— Sa^i— Oa^^+lla^j to its simplest form. Ans. —2a^b, 3. Reduce the polynomial 7abc'^—abc'^—7abc^-\-8abc^ + 6abc'^ to its simplest form. Ans. ISabc^. 4. Reduce the polynomial 9cb^—Sac^-\-l5cb^-\-8ca-{-9ac'^ — 24cb^ to its simplest form. Ans. ac^-^-Sca. The reduction of similar terms is an operation peculiar to algebra. Such reductions are constantly made in Algebraic Addition, Subtraction, Multiplication, and Division. ADDITION. 26. Addition in Algebra, consists in finding the simplest equivalent expression for several algebraic quantities, con- nected together by the sign plus or minus. Such equivalent expression is called their sum. 1. What is the sum of 3ax-\-2ab and —2ax + ab. — ^- 3ax~{-2ab We reduce the terms as in Art. 25, —2ax-\- ab and find for the sum ax-{-3ab 2. Let it be required to add together \ -, the expressions : i The result is 3a-\-bb-\-2c an expression which cannot be reduced to a more simple form. Quest. — 26. What is addition in Algebra ? What is such simplest and oquivalent expression called ? ^0 FIRST LESSONS IN ALGEBRA. Again, add together the monomials ? 2a^P The result after reducing (Art. 25), is . . ISa^b^ r 2a^-'Aah 3. Let it be required to find the sum \ q2_QA4_7,a of the expressions 2ah~b}p' Their sum, after reducing (Art. 25) is . ba^—bah—Al)^ 27. As a course of reasoning similar to the above would apply to all polynomials, we deduce for the addition of algebraic quantities the following general RULE. I. Write down the quantities to he added so that the similar terms shall fall under each other, and give to each term its proper sign. II. Reduce the similar terms, and annex to the results the terms which cannot he reduced, giving to each term its respec* tive sign. EXAMPLES. 1. What is the sum of Sax, 5ax, ■—2ax and ISax. ? A?is. I9ax. 2. What is the sum of 4ah-\-8ac and 2a5— 7ac+<^. ? Ans. 6ah-}-ac-{-d. 3. Add together the polynomials, 3a'^ — 2h^—iah, 5a'^ — h^-\-2ah, and 3ah-^3c^^2h^. The term 3a^ being similar to 5a2, we write Sa^ for the result of the reduction of these two terms, ^ at the same time slightly crossing them, as in the first term. QuBST. — 37. Give the rule for the addition of Algebraic quantities. -\-3(f^h—2b^—3c^ 8a24. ab — 5b^—3c^ ADDITION. 21 Passing then to the term —4ah, which is similar to-\-2ab and -{-Sab, the three reduce to -\-ab, which is placed after 8a^, and the terms crossed like the first term. Passing then to the terms involving b^, we find their smn to be — 5b'^, after which we write — Sc^. The marks are drawn across the terms, that none of them may be overlooked and omitted. (4) (5) (6) a 6a 5a a ba 5b 2a Tla 5a-\-5b (7) Sab (8) Sac 5ab Sab Sac llac (11) 12a— 6c — 3a— 9c (9) (10) 7abc-{-9ax Sax-{-Sb — Sabc—Sax 5ax — 95 Aabc-\-6ax \Sax — 6b 9a — 15c Note. — If az=:5, 5=4, c=:2, a?=l, what are the values of the several sums above found. (12) (13 ) (14) 9a+/ 6ax — 8ac 3«/+ g -\-m 6a-\-g -7ax— 9ac ag—Saf—m 2a-/ ax -\- 17 ac ab— ag-\-Sg ^-\-g ab-{-4g (15) (16) 7x-{-Sab+ 3c 8a;2+ 9acx-\-lSa%'^c^ Sx — Sab — 5c —7x^— ISacx+lAa^b'^c'^ 5x~9ab — 9c -4a;2+ 4acx—20aH^c^ 9x—9ab — llc — 3a;^+ + 7a^'^c^ (17) (18) 22A-3c-7/+3^ l9ah^-\-Sa^*-Sax^ — 3A+8c— 2 /— 9^4- 5^ — 17aA2— 9a35 ^ + 9ax^ 19A4 5c-9/'-6^+5^ 2ah^—6a^^-\- ax^ 29 FIRST LESSONS IN ALGEBRA. (19) (20) 7a?— 9y+5;^+3— g Sa+ h — 0? — 3y —8—- ^ 2a— 5+ c — a;+ y—^z+\ + lg — 3«+ J +2df —2a?+6y+ 3-^—1— g —6J— 3c+-3— c + 2J+2^. ^n^. — a— 9J + 8c— 3(^— 2^. 11. From . . —3a-\-b-'Sc+7e—5f+3h—7x—l3y take k-{-2a--9c+8e—7x+7f—7/--'3l^k. Ans. — 5a+54-c— c— 12/+3A— 12y+3Z. 12. From a-{-b take a— Z>. ^?i^. 2b. 13. From 2a:— 4a-.2^> + 5 take 8 — 5b-{-a-i-6x. Ans. —ix—5a+3b—3. 14. From 3a-\-b-\-c—d—lO take c+2«— c?. JiW-y. Gf+6 — 10. 15. From 3a+b+c—d—l0 take 5— 19 + 3«. ^?i5-. c— J4-9. 16. From 2ab+b^—ic + bc—b take 3«2— c+Z*2, ^w^. 2ab-^3a^—3c-^bc'-b, 17. From a^+3i2c-|.«j2_^j^ take b^-^-ab^-^abc. Ans. a^+3b-C'^b^. 18. From 12a;+6«-.4i+40 take 4Z'— 3a+4a?+6J--10. Ans. 8x-^9a—8b--ed+50, 19. FroHi2a;— 3«+4^+6c— 50 take 9a + a:+6Z>— 6c— 40. ^W5. ic— 12a— 25+12c— 10. 20. From 6a— 45— 12c+12a? take 2a:— 8a+45— 6c. Ans. 14a— 85— 6c+ 10a:. 21. From 8a5c— 1253a+6ca:— 7a:y take 7ca:— o-y— IS^^a. Ans. Sabc-i-b^a-^cx—dxy, SUBTRACTION* 27 31. By the rule for subtraction, polynomials may be subjected to certain transformations. For example, becomes . . . In like manner becomes or, again, Also, . . . becomes Also, . . . becomes 6a2— 3ab + 2b^ — 2^c, 6a'^—{3ah — 2b^ -\-2bc), 7^3 _ Sa'^b— 462c +652, 7a^'^{8a^b+ 4b'^c—6b'^), 7^3— 8a^b-(4b^c-6b^). 8a3— 7*2 + c —J, 8^3 _ (752 _ c +d). 9^3 _ a + 3a2 —d, 953 _ (^ _ 3^2 _|_^), 32. Remark. — From what has been shown in addition and subtraction, we deduce the following principles. 1st. In algebra, the words add and sum do not always, as in arithmetic, convey the idea of augmentation; for a—b, which results from the addition of —b to a, is properly speaking, a difference between the number of units ex- pressed by a, and the number of units expressed by b. Consequently, this result is numerically less than a. To distinguish this sum from an arithmetical sum, it is called the algebraic sum. Thus, the polynomial 2a'^—3a'^b-^3b^c is an algebraic sum, so long as it is considered as the result of the union Quest. — 31. How may you change the form of a polynomial ? 32. In algebra do the words add and sum convey the same idea as in arithme- tic ? What is the algebraic sum of 9 and — 4? Of 8 and — 2? May an algebraic sum ever be negative ? What is the sum of 4 and — 8 1 Do the words subtraction and difference in algebra always con- vey the idea of diminution ? What is the algebraic difference between 8 and — 4? Between a and — hi 28' FIRST LESSONS IN ALGEBRA. of the monomials 2c^^ —^oF'b, +3^2^, with their respec- tive signs ; and, in its proper acceptation, it is the arithmeti- cal difference between the sum of the units contained in the additive terms, and the sum of the units contained in the subtr active terms. It follows from this, that an algebraic sum may, in the numerical applications, be reduced to a negative number, or a number affected with the sign — . 2nd. The words subtraction and difference do not always convey the idea of diminution ; for, the numerical difference between -\-a and — h being a-\-h, exceeds a. This result is an algebraic diff^erence, and can be put under the form of a—{ — b)z=za-{-h. MULTIPLICATION. 33. If a man earns a dollars in one day, how much will he earn in 6 days ? Here it is simply required to re- peat the number a, 6 times, which gives Qa for the amount earned. 1. What will ten yards of cloth cost at c dollars per yard 1 Ans. 10c dollars. 2. What will d hats cost at 9 dollars per hat ? Ans. 9d dollars. 3. What will b cravats cost at 40 cents each ? Ans. 405 cents. 4. What will b pair of gloves cost at a cents a pair ? Quest. — 33. What is the object of multiplication in algebra 1 If a man earns a dollars in one day, how much will he earn in 4 days 1 In 6 days 1 In 6 days 1 MULTIPLICATION. 89 Here it is plain that the cost will be found by repeating h as many times as there are units in a : Hence, the cost is ah cents. Ans. ah cents. Note. — If we suppose «— 6, c=4, and c?z=3, w^hat would be the numerical values of the above answers ? 5. If a man's income is 2a dollars a week, how much will it be in 4^ weeks. Here we must repeat 3a dollars as many times as there are units in Ah weeks ; hence, the pro- duct is equal to 2axAh = l2ah, If we suppose a=z4: and 5 = 3 the product will be equal to 144. 34. Remark.— It is plain that the product I2ah will not be altered by changing the arrangement of the factors ; that is, \2ah is the same as «5xl2, or as hax\2, or as ax 12x5 (See Arithmetic, ^ 22). 35. Let us now^ multiply 3a252 ^y 2a%, which may be placed under the form 3a252 X 2a% = 3 X 2aaaahhh ; in which a is a factor four times, and h a factor three times: hence (Art. 13). 3^252 X 2a25 = 3 X 2aaaahhh=z6a'^h^, in which, we multiple/ the co-efficients together and add the exponents of the like letters. Quest. — 34. Will a product be altered by changing the arrangement of the factors 1 Is Sab the same as Sba 1 Is it the same as aXSbl As 5 X 3a T 35. In multiplying monomials what do you do with the co- efficients 1 What do you do with the exponents of the common letters ? If a letter is found in one factor and not in the other, what do yo^i do 7 3* 80 FIRST LESSONS IN ALGEBRA. Hence, for tlie multiplication of monomials, we have the following RULE. I Multiply the co-effccients together. II. Write after this product all the letters which are com- mon to the multiplicand and multiplier, affecting each letter with an exponent equal to the sum of the two exponents with lohich this letter is affected in the two factors. III. If a letter enters into but one of the factors, write it in the product with the exponent with which it is affected in the factor. EXAMPLES. 1. Sa^bc"^ X Habd"^ = 56a^'^c^^d\ 2. 21 a^^cd X 8abc^ =z 1 6Sa^b^cM. 3. 4abcx7df = 2Sabcdf (4) (5) Multiply 3a^ 12a^-x by 2a^ \2x'^y (6; Qxy z ay'^z Ga'^b'^ 144a'^x^y Saxy^z^, (7) (P'xy 2xy'^ 2a^x'^y^ (8) 3«Z»2c3 ^a^b^c (9) 87 ax'^y 3^*y3 10. Multiply 5aWx^ by 6c V. 11. Multiply lOa'^b^c^ by 7acd. 12. Multiply 9a%xy by 9a^xy. 13. Multiply 36a%'^c^d^ by 20abVd\ 14. Multiply 27 axyz by 9a%'^c'^d'^xyz. Ans. 15. Multiply IZc^h^c by Sahxy. 261 ab^xY- Ans. 30^352^5^8^ Ans. 70a^¥c^d. Ans. 8la%^x^y'^. Ans. 720a%^c^d\ Ans \04a^l^cxy MULTIPLICATION. 31 16. Multiply 20a^h^cd by 12a^x^y. Ans. 2\0d^¥cdx^y. 17. Multiply lAa^hH^y by 20a?c^xhj. Ans. 2S0a'¥cH^xh/. 18. Multiply Sa^Z^y by la^hxi/. Ans. bQa^h^xy^. 19. Multiply Ibaxyz by ba^hcdx^y'^ . Ans. 37da%cdx^y^z, 20. Multiply 51«2y2^2 by 9a?hc^x^y. Ans. 459a'^bc^x'^y^, 36. We will now proceed to the multiplication of poly- nomials. Take the two polynomials a-\-b-{-c, and d-\-fj composed entirely of additive terms ; the product may be presented under the form {a-{-b-]-c) {d+j). It is now re- quired to take the multiplicand as many times as there are units in d s,ndf. Multiplicand a-{-b-{-c Multiplier d-^f taken d times ad-\-bd-\-cd taken/times +af-\-bf-j-cf entire product .... ad-^-bd-^-cd-^-af-j-bf-i-cf. Therefore, in order to multiply together two polynomials composed entirely of additive terms : Multiply successively each term of the multiplicand by each term of the multiplier, and add together all the products, EXAMPLES. 1. Multiply 3fl2+ 4:ab + b^ by 2a + 5b 6a3+ Sa%'\-2ab'^ The product, after reducing, -{'\ba%-\-20ab'^+bb^ becomes .... Qa^-^23a:^b + 22ab'^-\-bbK Quest. — 36. How do you multiply two polynomials composed of additive terms 1 33 FIRST LESSONS IN ALGEBRA. (2) (3) x^+y^ a;^+ ocy^ +7 ax X +y CLX -\-bax x^+xy'^ ax^+ ax'^y^-\-ld^x'^ a?34-a?y2^_-jn2y_|_y3 6aa;^+6aa?y-j-42a^a?2. 4. Multiply x'^-\'2ax+d^ by x-\-a. Ans. x^-\-Zax'^+^a?x+a^ 5. Multiply x^+y^ hj x+y. Ans. x^-\-xy'^-\-x'^y-\-y^ 6. Multiply 3«52+66fV by Sai^+Sa^c^. Ans. 9a2J4_(. 27^3^,2^2 _|_i8aM 7. Multiply a25^+c2(^ by a+h. Ans. a?b'^+ac^d+a%'^-{-hc^d, 8. Multiply Zax'^+9ah'^-\'Cd^ by Ga^cS. 9. Multiply 64a3a;3 + 27a2a:+9a5 by Sa?cd. Ans. bl2a^cdx^+2\Qa^cdx + l[2a^hcd, 10. Multiply a'^+2ax-\-x'^ by a+o:. 11. Multiply a^+^aH+^ax'^-{-x^ by «+«?. Ans. a!^-\-Aa'^x+Qa?'x'^+Aax^-{-x^, 37. To explain the most general case, multiply a—h by c—d. The required product is equal to a —h a—b taken as many times as there c —d are units in c—d. If then we mul- ac — bc tiply by c, which gives ac — bc, we —ad+bd have got too much by a—b taken ac—bc—ad-j-bd. d times ; that is, we have ad—db MULTIPLICATION. 33 too much. Changing the signs, and subtracting this from the first product (A.rt. 30), we have (« — b) (c — d)=zac — be — ad-\-bd. Let us suppose (2=10, bz^iQ, <: = 5, and d=^\\ in which case we find the product {a — b) (c — d)=zac — be — ad-\-bd=i\Q. Hence, we have the following rule for the signs. When two terms of the multiplieand and multiplier are ajfected with the same sign, the corresponding product is affect- ed with the sign -\- ; and when they are affected with contrary signs, the product is affected with the sign — . Therefore we say in algebraic language, that + multi- plied by -|-, or — multiplied by —, gives +5 —multi- plied by +) 01* + multiplied by — , gives — . Hence, for the multiplication of polynomials we have the following RULE. Multiply all the terms of the multiplicand by each term of the multiplier, observing that like signs give plus in the pro- duct, and unlike signs minus. Then reduce the polynomial result to its simplest form. EXAMPLES- 1. Multiply .... 2ax — ^ah by ^x — b. The product Qax"^— 9abx becomes after .... — 2a:bx-^3ab^ reducing 6ax^~llabx-^3ab^. QoEST. — 37. What does + multiplied by + give? + multiplied by — 1 — multiplied by + ? — multiplied by — ? Give the rule for the multiplication of polynomials ? 84 FIRST LESSONS IN ALGEBRA. 2. Multiply a*—2P by a — h. Ans. a^—2aP—a^h+2h\ 3. Multiply a;2--3a?— 7 by x—2. Ans. a?^— 50:^— a:-|-l4. 4. Multiply 3a^-^5ab-\-2b^ by a^ — 7ab, Ans. 3a'^—26a^-\-37a^^—Uab\ 6. Multiply b^+¥+¥ by b^—1. Ans. b^—b^. 6. Multiply x'^—2x^y+4xh/ — 8xy^-{-l6i/ by a:+2y. JLw5. a:^-f 32y^. 7. Multiply 4a;^— 2y by 2y. Ans, Sx^i/ — 4i/^, 8. Multiply 2a?+4y by 2a?— 4y. Ans. Ax'^—ldj/'^. 9. Multiply x^+x^y+xj/^+y^ by a:— y. 10. Multiply x^-{-xy+y^ by a?^— a:y+y^- ^71^. a:*+a?y+y*- IL Multiply 2^—3ax+4x^ by 5a2_6^.p_2a;2^ Ans. 10a^—27a^x-^3ia'^x^-'l8ax^—8x\ 12. Multiply 3a?2— 2a;y+5 by x^+2xy—3. Ans. 3x^ + 4x^y — 4a?2 — 4a?2y^ + 1 6xy-- 1 5. 13. Multiply 3x^+2x^y^ + 3y^ by 2a;3_3a:y + 5y3. . ( 6a:6 — 5a;5y2_6a:^y4_^6a:3y24. ^^* < 15a:3y3-9a^y+10a:2y5-M53^. i 14. Multiply 8ax — 6ab—c by 2aa; + aJ+c. Ans, 1 6fl52aj2 __ 4gj2^-j. _ 6a2Z>2 + 6«ca; — 7a5c — c\ 15. Multiply 3a2— 552+3c2 by a^ — b^. Ans. 3a*-8a252+3a2c2+5i*~35V. 16. 3a^ — 5bd+ cf — 5g24-4g>^— 8c/: PK).red. ~ 1 5a^4- 37aHd--29a'^cf—20b'^d'^-h44bcdf~ 8c'^p. MULTIPLICATION. ./ 35 38. To finish with what has reference to algebraic mul- tiplication, we will make known a few results of frequent use in Algebra. Let it be required to form the square or second power of the binomial {a-\-b). We have, from known principles, {a+b)^={a+b) (a+b)=za^+2ab+b\ That is, the square of the sum of two quantities is equal to the square of the first, plus twice the product of the first by the second, plus the square of the second, 1. Form the square of 2fl+35. We have from the rule {2a +Uf = 4a2 + \2ab -h 9R 2. {bab+^acf =z2^a%'^+ ZOa'^bc+ ^aW 3. (pa^+Ba%f z=:25a^ + SOa^b +Ua^b\ 4. (6«a;+9a2a:2)2 = 36a2a;2_f.i08a3a;3+81a%*. ^' 39. To form the square of a difference a— 5, we have (a-bf^ia-b) {a^b)-a^-'2ab+b\ That is, the square of the difference between two quantities is equal to the square of the first, minus twice the product of the first by the second, plus the square of the second, 1 Form the square of 2a— b. We have (2a-Z>)2=:4a2~.4a5+R 2. Form the square of 4ac— ic. We have (4«c - bcf = 1 6a2c2 - Sabc'^ + b'^c'^, 3. Form the square of 7a^b'^—l2ab^. We have (7^252 _i2a53)2_ 49^4^,4 _i68a365+ 144^2^6. Quest. — 38. What is the square of the sum of tw6 quantities equal to t 39. What is the square of the difference of two quantities equal to 1 86 FIRST LESSOKS IN ALGEBRA, 40. Let it be required to multiply «-f6 by a—b. We liave {a+b)x(a-b) = a^-P. Hence, the sum of two quantities, multiplied by their differ' ence, is equal to the difference of their squares. 1. Multiply 2c+b by 2c— b. We have (2c+b) X (2c— Z>) = 4c2— Z>2. , 2. Multiply 9ac-f 35c by 9ac—3bc. We have {9ac+3bc){9ac — 3bc) = Sla'^c'^ — 9b^c^. 3. Multiply Sa^+7ab^ by 8a^—7ab'^. We have (8a^+7ab^){Sa^—7ab^)z=z6ia^—i9a'^b\ 41. It is sometimes convenient to find the factors of a polynomial, or to resolve a polynomial into its factors. Thus, if we have the polynomial ac-\-ab-\-ad, we see that a is a, common factor to each of the terms : hence, it may be placed under the form a{c+b-\-d). 1. Find the factors of the polynomial a'^b^-\-a^d-\-a'^f Ans. a'^{b'^-\-d-\-f). 2. Find the factors of Sa^ + Ga^^-^b^d, Ans. b{3a^-{-Ga^b + bd). 3. Find the factors of 3a'^b+9a'^c-{-l8a'^xi/. Ans. 3^2(5 -f 3c +6a?y). QuERT. — 40. What is the sum of two quantities multiplied by their difference equal to 1 DIVISION. 37 4. Find the factors of 8a^cx—lSacx^+2ac^i/'^30a^c^x, Ans. 2ac(4cfa;—9a;2+c*y— ISa^c^ir), 5. Find the factors of a^+2ah+b^. Ans, {a+b)x{a+h). 6. Find the factors of o^—bK Ans, {a+b)x{a—by 7. Find the factors of a^—Hab+b^ Ans. (a--5)x(a— 5). S. DIVISION. 42. Algebraic division has the same object as arithmeti- cal, viz : having given a product, and one of its factors, to find the other factor. We will first consider the case of two monomials. The division of 72a^ by 8a^ is indicated thus : It is required to find a third monomial, which, multiplied by the second, will produce the first. It is plain that the third monomial is 9a^ ; for by the rules of multiplication Sa^x9a'^=72a^ Hence, we have =9g^, a result which is obtained by dividing the coefficient of the dividend by the coefficient of the divisor, and subtracting the exponents of the like letter. Quest. — 42. What is the object of division iti Algebra 1 Give the rule for dividing monomials 1 4 30 FIRST LESSONS IN ALGEBRA. Also, I^^^=^5a^-ih^'^c = 5a^c, for, 7ah x da^c = 35a^^c, • Hence, for the division of monomials we have the foUovring I. Divide the coefficient of the dividend hy the coefficient of the divisor. II. Write in the quotient, after the coefficient, all the letters common to the dividend and divisor, and affect each with an exponent equal to the excess of its exponent in the dividend over that in the divisor. III. Annex to these, those letters of the dividend, with their respective exponents, which are not found in the divisor. From these rules we find ^Ba%^c'^d , ,, , IbOa^h^cd"^ ^ ^,, , =:4a'^oca: :; — - — zzzocrb^cd. 12ab^c ' SOa^'d^ 1. Divide I6x^ by 8a?. Ans. 2x. 2. Divide I5axy^ by Say. Ans. 5xy^. 3. Divide S4aPx by I2b^. Ans. 7abx. 4. Divide 36a^b^c^ by 9a^^c. Ans. 4ab^c. 5. Divide 88^352^ by Sa'^b. Ans. Uabc. 6. Divide 99a^b^x^ by lla^^x^ Ans. 9ab^x. 7. Divide I08x^y^z^ by bix^z. Ans. 2xy^z^. 8. Divide 64x'^y^z^ by 16x^y^z^. Ans. Axyz. 9. Divide 9QaWc^ by \2a%c. Ans. Qa^¥c^, 10. Divide bid^c^d^ by 27 acd. Ans. 2a^c^d^. 11. Divide ZQa'^¥d^ by 2a%H. Ans. \9abd\ DIVISION. 39 12. Divide 42a^b^c^ by 7ahc. Ans. dahc, 13. Divide 64a^¥c^ by 32a*5c. Ans, 2aPc\ 14. Divide 12Sa^x^i/ by IQaxy^, Ans, Qa^x^y^. 15. Divide U2hd^p by 2#/. An^. 6656^/5. 16. Divide 256^4^,9^8^7 by IQa^hc^, Ans, IQab^c^dP, 17. Divide 200a^m?n^ by 50aPmn. Ans. 4amn, 18. Divide SOOx^y^z^ by 60a:y2^. ^;^^^ bx^y^z, 19. Divide 27 a^b^c^ by 9a5c. ^w^. 3«*5c. 20. Divide 64a^y^z^ by 32fl5y5;j7^ ^^^^^ 2a2y;3'. 21. Divide SSa^b^c^ by lla^Z^^c^. Ans. Qa%'^c\ 43. It follows from the preceding rule, that the division of monomials v^^ill be impossible, 1st. When the coefficients are not divisible by each other. 2nd. When the exponent of the same letter is greater in the divisor than in the dividend. 3rd. When the divisor contains one or more letters which are not found in the dividend. When either of these three cases occurs, the quotient re- mains under the form of a monomial fraction ; that is, a monomial expression, necessarily containing the algebraic sign of division, but which may frequently be reduced. Take, for example, I2a'^b^cd, to be divided by 8a^bc\ which is placed under the form na'^b^cd ^ Sa?bc^ ' Quest. — 43. What is the first case named in which the division of monomials will not be exact 1 What is the second 1 What is the third 1 If either of these cases occur, can the exact division be made 1 Under what form will the quotient then remain 1 May this fraction be often reduced to a simpler form 1 40 FIRST LESSONS IN ALGEBRA. this may be reduced by dividing the numerator and denomi- nator by the common factors 4, «^, h, and c, which gives Sa%c^ 2c 44. Hence, for the reduction of a monomial fraction we have the following RULE. I. Suppress the greatest factor common to the two co- efficients. II. Subtract the less of the two exponents of the same letter from the greater, and write the letter affected with this differ- ence in that term of the fraction corresponding with the greatest exponent. III. Write those letters which are not corrmion, with their ^ respective exponents ^in the term of the fraction which contains them. From this new rule, we find, (1) (2) ASaWcd? 4ad2 , 37a Pc^d 37b^c and dea^^c^de'' 3bce ' 6a^c^d^~~ da'^d ' (3) (4) 7a'^b I , 4a2Z>2 2a also — ...070 = A . * and Ua^b^ '~ 2ab ' Qa¥ 3b^ 7bc^ 5. Divide 49a^^c^ by I4a^bc\ Ans. — — 2a 6. Divide 6amn by 3abc. Ans. -^ — . •^ be 7. Divide }8c^b^mn^ by I2a^¥cd. Ans. tt^-j- •^ 2a^b^^cd Quest. — 44. Give the rule for the reduction of a monomial fraeticMx DIVISION. 41 8. Divide l^a^h^c^d^ by \^db^cd?m. Arts. 9. Divide 12a?c^W' by Yla^c^ly^d, Ans, 10. Divide \^^a%^xmn by 25a^^d, Ans, 4b^m 6 Aa^hxmn 11. Divide 9Qa^h^c^df by IbaHxy, Ans, — |r— — • 12. Divide ^^mVfx^y'^ by Ibam^nf, Ans. — — ^J^. 127 13. Divide I27d^xh/ by 16cZ^;ry, An^. TT^-g-^. 45. If v^e have an expression of the form a c^ a^ a^ a^ — , or — , or —, or — , or — , &c^ a a? a^ a^ aP and apply the rule for the exponents, we shall have But since any quantity divided by itself is equal to 1, it follows that — =aO = l, ^=02-2:^^^1, &c, a a^ or finally, if we designate the general exponent by m, we have or that is, any power of which the exponent is is equal to 1. Quest. — 45. What is afi equal to? What is Jfi equal tol What is the power of any number equal to, when the exponent of the power is Ot 4* 42 FIRST LESSONS IN ALGEBRA. 2. Divide Ba^b^c^d by 2a^b^d. "ZaWd = 3a2 -2^2-2^4^ -i^3e*. 3. Divide 8a*5V(?5 by \a%'^d^d^. Ans. 2a^, 4. Divide I6a%^d^ by S^ejsj, Ans. 2d^. 5. Divide 32m^n^x'^y^ by 47n3w3a!:^. j1»^. 8a?y, e. Divide dda'^b^d^c^ by 2U^¥d^cK Ans. 4bd^. SIGNS IN DIVISION. 46. The object of division, is to find a tliird quantity called the quotient, which, multiplied by the divisor, shall produce the dividend. Since, in multiplication, the product of two terms having the same sign is affected with the sign + , and the product of two terms having contrary signs is affected with the sign — , we may conclude, 1st. That when the term of the dividend has the sign +, and that of the divisor the sign of + y the term of the quo- tient must have the sign + . 2nd. When the term of the dividend has the sign -f , and that of the divisor the sign —, the term of the quotient must have the sign — , because it is only the sign — , which, multiplied with the sign —, can produce the sign + of the dividend. Quest.— 46. What will the quotient, multiplied by the divisor, be equal to ? If the multiplicand and multiplier have like signs, what will be the sign of the product? If they have contrary signs, what will be the sign of the product ? When the term of the dividend and the term of the divisor have the same sign, what will be the sign of the quotient ? When they have diiferent signs, what will be the sign of the quotient ? DIVISION. 43 3rd. When the term of the dividend has the sign — , and that of the divisor the sign + , the quotient must have the sign — . Again we say for brevity, that, + divided by +? and — divided by — , give + ; — divided by +, and -f divided by — , give — . EXAMPLES. 1. Divide iax by —2a. Ans. —2x, Here it is plain that the answer must be —2x\ for, —2a X — 2a?= + 4«a?, the divisor 2. Divide 3. Divide 4. Divide 5. Divide 6. Divide 7. Divide 8. Divide 9. Divide 10. Divide 11. Divide 12. Divide 13. Divide 14. Divide 15. Divide 16. Divide 17. Divide 18. Divide 36a3a;2 by —I2a^x, -^5Sa-b^c'^(P by 29a^^c. — 84«4Z>5j3 by —42a^b'^d. 64c^d^x^ by IGc^dx. — 885Vy6 by '-24Pcdx^, 77aY2^ by -ll«y^^. 84a'^b^c'^d by '-42a'^b'^c^d. —QOd?¥c^d by —I2a%''c^d'^, — 88a%'c^ by Sa^b^c^, I6x^ by —8a?. — 15a2a?y3 by 2ay. — Siab^x by — 12R — 96a^^c^ by I2a^c. — lUa^^c^d^ by —SGa^b^c^d, 256a^c^x^ by — 16a^cx'^. —SOOa^b^c^x^ by SOa'^bh'^x, dOOa^^c^ by -lOOa^JV. Ans. +5- Ans. — Sax, Ans. —2abcd^. Ans. 2a%W. Ans. Ad'^x'^. Ans +H^' ^'''' + 2cd ' Ans. —7. Ans. —2. 1_ abed' Ans. —Wab. Ans. — 2x. Ans. —baxy'^. Ans. labx. Ans. — Sabc^. Ans. Aa^b^cd"^. Ans. — 16abcx. Ans. — lOabcx. Ans. — 5abc^. 44 FIRST LESSONS IN ALGEBRA. 19. Divide —6ia^b^c'' by '~Sa^b''c^. Ans. Sabc. 20. Divide +96^5^,4^9 by —24M^d, Ans. —4ab^d\ 21. Divide 72a^Pd^ by —Sa^b'^d, Ans. —9abd^. I a — X Quotieiit. Division of Polynomials, FIRST EXAMPLE. 47. Divide a'^~2ax-{-x'^ by a—x. It is found most convenient, Dividend. Divisor. in division in algebra, to place a^ — 2ax-{-x'^ the divisor m the right of the a^ — ax dividend, and the quotient di- — ax-^-x"^ rectly under the divisor. — ax-j-x"^ We first divide the term a^ of the dividend by the term a of the divisor : the partial quotient is a, which we place under the divisor. We then multiply the divisor by a, and subtract the product a^—ax from the dividend, and to the remainder bring down x'^. We then divide the first term of the remainder, — ax by a, the quotient is — x. We then multiply the divisor by —x, and, subtracting as before, we find nothing remains. Hence, a—x is the exact quotient. In this example, we have written the terms of the dividend and divisor in such a manner that the exponents of the same letter shall go on diminishing from left to right. This is what is called arranging the dividend and divisor with reference to a certain letter. By this preparation, the first term on the left of the dividend, and the first on the left of the divisor, are always the two which must be divided by each other in order to obtain a term of the quotient. Quest. — 47. What do you understand by arranging a polynomial with reference to a particular letter ? DIVISION. 45 48. Hence, for the division of polynomials we have the following RULE. I. Arrange the dividend and divisor with reference to. a cer- tain letter, and then divide the first term on the left of the dividend hy the first term on the left of the divisor, the result is the first term of the quotient ; multiply the divisor hy this term, and subtract the product from the dividend. II. Then divide the first term of the remainder hy the first term of the divisor, which gives the second term of the quotient ; multiply the divisor hy the second term, and suhtract the pro* duct from the result of the first operation. Continue the same process until you ohtain for a remainder ; in which case the division is said to he exact. SECOND EXAMPLE. Let it be required to divide b\a%'^+\Oa^—ABa% — \^h^+iah^ by 4ah—5a'^+3h\ We here arrange with reference to a. Dividend. Divisor. l0a^—48a^-\-5la'^h^-\- 4ah^ — l5h\ -\-lOa^— Sa^~ ^aW — 40a3Z>+57«2Z,2_^ 4a63— IS^^'t — 40a3Z>+32a2^>2^24«53 — 5a2-|-4a64-352 — 2a2+8a5— 562 Quotient. 25a'^h^~20ah^ — l5h^ 25a2J2_20«^»3_i5^,4 Quest. — 48. Give the general rule for the division of pol)momials 1 If the first term of the arranged dividend is not divisible by the first term of the arranged divisor, is the exact division possible 1 If the first term of any partial dividend is not divisible by the first term of the divisor, is the exact division possible 1 46 FIRST LESSONS IN ALGEBRA. Remark. — When the first term of the arranged dividend is not exactly divisible by that of the arranged divisor, the complete division is impossible ; that is to say, there is not a polynomial which, multiplied by the divisor, will produce the dividend. And in general, we shall find that a division is impossible, when the first term of one of the partial dividends is not divisible by the first term of the divisor. GENERAL EXAMPLES. 1. Divide ISac^ by 9a?. Ans. 2x. 2. Divide 10a?y by —bx^y. Ans. —2y, 3. Divide —^ax^y'^ by ^x'^y. Ans. —ay. 4. Divide —Sx'^ by — 2a?. Ans. +4a?. 5. Divide \0ah-{-l5ac by 5a. Ans. 25+ 3c. 6. Divide Wax—bAx by 6a?. Ans. 5«— 9. 7. Divide \Ox'^y—\by'^—^y by 5y. Ans. 2x'^-3y — l, 8. Divide l2a-\-3ax—18ax'^ by 3a. Ans. 4:-\-x—6x^. 9. Divide 6ax'^-{-9a'^x-{-a'^x^ by ax. Ans. 6x-\-9a-\-ax. 10. Divide a'^-\-2ax+x^ by a+x. Ans. a-\-x, 11. Divide a^ — 3a^y-\-3ay'^—y^ by a—y. Ans. a^ — 2ay-{-y^. 12. Divide 2A.a% — l2aHh'^ — Qah by —Qah. Ans. — 4«+2a2cZ>+l. 13. Divide 6a?^— 96 by 3a;— 6. Ans. 2a;3+4a:2+8a;+16. 14. Divide . . . a^~^a'^x+\0a^x'^—\0aV-{-5ax^—x^ by a^ — 2ax-\-x'^. Ans. a^ — 3(i^x-^3ax^ — a?^. 15. Divide 48a:3— 76aa:2 — 64a2a:+105a3 by 2x — 3a. Ans. 2ix^—2ax — 35a^, DIVISION* 4t 16. Divide y^ — ZyV+^yH^—x^hyy^ — ^y'^x+Zyx'^—x^. ^ Ans, y^+Sy^aj+Syic^+a?^. 17. Divide Q4.a'^h^ —2ba?h^ by SaW-{-bah\ Ans. Sa?b'^ — bah^. 18. Divide Qa^-\-2^a%+22ah'^+b¥ by ^a?-\-Aah-^h'^. Ans. 2a+5b, 19. Divide 6ax^+6ax'^y^+42a^x'^ by ax+5ax. Ans. x^-i-xy^-{-7ax. 20. Divide . . - 1 5a^-{- 37 a^-bd-29a'^cf—20bhP+44bcdf ~8c2/2 by 3a^-—5bd-{-cf. Ans. —ba^+Ud—Scf. 21. Divide aj^-f aiy-f-y* by x^—xy-{-y^. Ans. x'^-{-xy-\-y^. 22. Divide a?*— y* by a;— y. u4.w^. a;^+a;2y+a?y2 4-y3. 23. Divide 3a^—Sa'^b^ + 3a'^c'^+5b^ — 3b^c^ by a2_j2^ ^w^. 3a2_5^>2 + 3c2. 24. Divide . . Ga:^— 5a;^y2 — Gic^y^+Gaj^y^+lSaj^yS — ga^^yi + 10a;2y5+15yS by 3a;3+2a;2y2-j-3y2^ Ans. 2a;^ — 3a:2y3-f5y3. 25. Divide . — c^+lGa^oj^— 7a6c— 4a2Z>a;— Ga^J^-j-eaca? by Sax— Gab— c. Ans, 2ax+ab + c, /, 26. Divide .... 3a:*+4a;3y-4a;2_4<^2y2_^16:ry-15 by 2a:y+a:2_3. Ans. 3x'^—2xy-\-o. 27. Divide x^+32y^ by a:+2y. Ans. x^—2x^y-\-4:X^y^—Sxy^-\-l6y'^, 28. Divide 3a^ -26 a^-Uab^-h 37 a^^ by 2b'^-5ab -j_3a2 ^715. d^—7ab* 48 FIRST LESSONS IN ALGEBRA* CHAPTER IL Algebraic F7^actions. 49. Algebraic fractions should be considered in the same point of view as arithmetical fractions, such as |^, W ; that is, we must conceive that the unit has been divided into as many equal parts as there are units in the denominator, and that one of these parts is taken as many times as there are units in the numerator. Hence, addition, subtraction, mul- tiplication, and division, are performed according to the rules established for arithmetical fractions. It will not, therefore, be necessary to demonstrate those rules, and in>their application we must follow the procedures indicated for the calculus of entire algebraic quantities. 50. Every quantity which is not expressed under a fractional form is called an entire algebraic quantity. 51. An algebraic expression, composed partly of an entire quantity and partly of a fraction, is called a mixed quantity. Quest. — 49. How are algebraic fractions to be considered 1 What does the denominator show ? What does the numerator show ? How then are the operations in fractions to be performed '? 60. What is an entire quantity 1 61. What is a mixed quantity? ALGEBRAIC FRACTIONS. 49 To reduce a fraction to its simplest terms. 52. The rule for reducing a monomial fraction to its low- est terms has already been given (Art. 44). With respect to polynomial fractions, the following are cases which are easily reduced. 1. Take, for example, the expression a?—2ah-\-h'^' This fraction can take the form {a+h) {a-^h) {a -by (Art. 39 and 40). Suppressing the factor a— 5, which is common to the two terms, we obtain a-{-b a — b' 2. Again, take the expression 5a^—l0a^-\-5ab^ 8a^—8a^ This expression can be decomposed thus : 5a{a'^—2ab-i-b^) 8a^{a-b) ' ba{a—bY or, Sa\a—by Quest. — 52. How do you reduce a fraction to its simplest terms ^ 5 I 50 FIRST LESSONS IN ALGEBRA. Suppressing the common factors a{a—h), the result is Sa Hence, to reduce any fraction to its simplest terms, we sup- press or cancel every factor common to the numerator and denominator. Note. — Find the factors of the numerator and denomina- tor as explained in (Art. 41). EXAMPLES. 1. Keduce — ^ ^ ^ to its simplest terms. Ans. 4a2+2ac2* „ „ ^ 15«5c+25a9J . . , 2. Keduce ^^ ^ . ^^ ^ to its simplest terms. ^a^c+baU Ans. 3. Reduce , ^,- „ „ to its simplest terms. A ^^ Ans. 3^7 4. Reduce ^ ^ ■ to its simplest terms. ^•^^■^"•^^ 637^336^- ^""-Ti-iTia- 6. Reduce -_ - to its simplest terms. Ans. —8. „ ^ , 2A¥—3Qah^ ^ 4b-6a 7. Reduce t^-— ,- — ^rTr-rr^. ^n^. 48a^6*—66a5^>6* ' Sa^-^lla^b^. ALGEBRAIC FRACTIONS. 51 ^ CASE II. 53. To reduce a mixed quantity to the form of a fraction. RULE. Multiply the entire part hy the denominator of the fraction ; then connect this product with the terms of the numerator hy the rules for addition, and under the result place the given denominator, EXAMPLES. 1. Reduce Q\ to the form of a fraction 43 6x7=42: 42 + 1=43: hence, 6j=y. 2. Reduce x ^ — to the form of a fraction. X d^—x^ x^—{o^—x^\ 2aj2 — cP- ^ X = ^= . Ans, ax~\~ tK 3. Reduce x to the form of a fraction. 2a . ax^x'^ Ans. 2a 2a?—7 4. Reduce 5 -\ • to the form of a fraction. 3a; 17a;-7 Ans, — . 3a? 5. Reduce 1 to the form of a fraction. a 2a—x-\-l Ans. . Quest.—- 63. How do you reduce a mixed quantity to the form of a fraction ] 6* FIRST LESSONS IN ALGEBRA. ^ 3 6. Reduce l+2a? — to the form of a fraction. 10^2_}_4^_|_3 i*>^ Ans. - 5x 3c+4 7. Reduce 2€t+b — to the form of a fraction. iea+Sb — 3c—4: Ans, g . Qct cc ■""■ clu 8. Reduce 6ax-\-h to the form of a fraction. 4a 18a^x-{-5ab Ans. . 4a 8 -|- 6a2 J2^4 9. Reduce 8+3ah ■■ , , — to the form of a frac- 96abx^+30a'^b'^x^—8 tion. Ans, ,^ - , . I2abx'^ 3 52 _ 8c* 10. Reduce 9H r- — to the form of a fraction. a—b^ 9«_-652-8c* Ans. j^ . CASE in. 54. To reduce a fraction to an entire or mixed quantity. RULE. Divide the numerator by the denominator for the entire part, and place the remainder, if any, over the denominator for the fractional part. Quest. — 64. How do you reduce a fraction to an entire or mixed quantity ? ALGEBRAIC FRACTIONS. 63 EXAMPLES. , T. :. 8966 1. Reduce — - — to an entire number. 8 8)8966( 1120 ... 6 rem. Hence, 1120f=: Ans. 2. Reduce to a mixed quantity. X Ans, a- X ax — x^ . . , 3. Reduce to an entire or mixed quantity. Ans, a—x. ^ , ah— 20^ . , 4. Reduce r to a mixed quantity. 2^2 Ans. a — 5. Reduce to an entire quantity. Ans, a + x, a — x 6. Reduce — to an entire quantity. X — y Ans. x'^+xy + y^. 7 Reduce to a mixed quantity. DX 3 Ans. 2x—l-{-—-. 5x 8. Reduce to a mixed quantity. . o r. 32a2ir Ans. 4:X^—S-] — . fgl FIRST LESSONS IN ALGEBRA. CASE IV. 55. To reduce fractions having different denominators to equivalent fractions having a common denominator. RULE. Multiply each numerator into all the denominators except its own, for the new numerators, and all the denominators together for a common denominator. EXAMPLES. 1. Reduce ^, J, and f, to a common denominator. 1x3x5 = 15 the new numerator of the 1st. 7x2x5^:70 „ „ „ 2nd. 4x3x2=24 „ „ „ 3rd. and 2 X 3 X 5=30 the common denominator. Therefore, J^, J§, and f ^, are the equivalent fractions. Note. — It is plain that this reduction does not alter the values of the several fractions, since the numerator and denominator of each are multiplied by the same number. 2. Reduce — and — to equivalent fractions having aXcz=:ac > hxb=h^y a common denominator. the new numerators, and hxc=:bc the common denominator. Quest. — 56. How do you reduce fractions to a common denominator! ALGEBRAIC FRACTIONS. 65 Hence, — and — are the equivalent fractions. 3. Reduce — and to fractions having a com- , . ^ ac ^ ab-^-W' mon denominator. Ans, 7— and — ; . DC be 4. Reduce — , — , and J, to fractions having a Z(X oC common denominator. Ans, -r — , --; — , and 6ac 6ac 6ac ^ T^ 1 3 2a; . 2x . . , . 5. Reduce — , — , and a-\ , to tractions havmg a common denominator. 9a Sax , 12a'^+24x Arts. -— — , -— — , and 12a ' 12a ' 12a 1 a a I /K" 6. Reduce — , — , and — — — , to fractions .4 o a~\~ X having a common denominator. 3a+3a; 2a^^2aH Ga^+Gic^ 6a+6a?' 6a +60; ' Qa -\-Qx ' ft Qny' rfl — QQ^ 7. Reduce 7-^, — - — , and = — to a common Jo DC a denominator. 5acd \Sahdx , \^a%c — Ibhcx"^ Ans. ' ^, , , -T-rT~r» and l^hcd ' Ibhcd ' \bhcd 8. Reduce — , , and -, to a common de- ^a c a-\-b nominator. ac^-\-c% 5a^ — 5ab^ , 5ac^ Ans, -—z — , -— -r — , and ba^c-\'babc ' baH+babc * ba^c+babc 66 FIRST LESSONS IN ALGEBRA. CASE V.- 56. To add fractional quantities together, RULE. Reduce the fractions, if necessary, to a common denomina- tor ; then add the numerators together, and place their sum over the common denominator. EXAMPLES. 1. Add f, f, and | together. By reducing to a common denominator, we have 6x3x5=90 1st numerator. 4x2x5 = 40 2nd numerator. 2x3x2 = 12 3rd numerator. 2x3x5 = 30 the denominator. Hence, the fractions become 90 40 12 142 30 ' 30 ' 30~" 30 ' which, by reducing to the lowest terms become 4^. 2. Find the sum of -^, -r-, and -pr.. d j Here axdxf—adf\ cxhx f=cbf > the new numerators. ex bxd=:ebd 5 And bxd xf= bdf the common denominator. adf cbf . ebd adf-\-cbf-\-ebd Quest. — 56. How do you add fractions. ALGEBRAIC FRACTIONS. S? 3. To cf r— add b-\ Ans. a+o-\ he 4. Add — , — and — together. Ans, x+ 5. Add — -— and — together. Ans. — — . 6. Add x-\ — — ■ to 3x-\ — . Ans, Ax-\ — — . 5iK X I Of 7. It is required to add 4x, —- — , and together. Ans, 4a: +" 2ax _ _ I __, _, _ __ 8. It is required to add — , — , and — - — together. 49a;+12 Ans. 2x+- 60 9. It is required to add 4a?, — , and 2+— together. 44a:+90 Ans, 4x-\-' — . 45 2x Sx 10. It is required to add 3x+— and x — -- together. o y Ans. 3x-\- 11. Required the sum of ac—— - and 1 — -, oa a Sa^cd—ebd - j-Sad—Sac '^'*^- Fad 45 58 FIRST LESSONS IN ALGEBRA. 57. To subtract one fractional quantity from another. RULE. I. Reduce the fractions to a common denominator, II. Subtract the numerator of the fraction to he subtracted from the numerator of the other fraction, and place the dif- ference over the common denominator. EXAMPLES. 3 2 1. What is the difference between --- and — . 7 8 Z^^ 2 _24 14_10___5_ Y""T"~56~56~56~28* ^^* 2. Find the difference of the fractions —-7-- and — . 26 6c TJ^^z^ (^-" «)x3c=:3ca;— 3ac ) , •n^re, ^ . s «, . , ^, J- the numerators And, 25 X 3c = 65c the common denominator. _- 3ca? — 3ac Aah — Shx 3ca? — 3ac — 4:ah-{-Shx ^''"''' —637 66^-= 633 • ^"*- 3. Required the difference of -— - and — . Ans. 7 5 • 35 ' 4. Required the difference of 5y and -^. Ans. — -^. o 8 5. Required the difference of — and — . Ans. --^^. Quest. — 57. How do you subtract fractions ? ALGEBRAIC FRACTIONS. 59 6. Required the difference between — 7 — and -j. Ans. 3a; 4-^ 7. Required the difference of — — — and dx-\-ad — he hd * 2a;+7 Ans 5b 8 24x+8a—l0bx—35b 40^ • 8. Required the difference of 3a?+— and oc . cx-\-bx — ab Ans, 2x-\- hc CASE VII. 58. To multiply fractional quantities together. RULE. ♦ If the quantities to be multiplied are mixed, reduce them to a fractional form ; then multiply the numerators together for a numerator and the denominators together for a denominator. t a EXAMPLES. 1. Multiply ori. We first reduce the com- pound fraction to the sim- ple one ^2? ^^^ then the mixed number to the equiva- lent fraction ^^ ; after which, we multiply the numerators and denomina- tors together. by 8^- Operation. , 3 3 ^^ Y=42' 3 25 75 25 Hence, — x — = = — . ' 42 3 126 42 Ans. 25 42* 60 FIRST LESSONS IN ALGEBRA. 2. Multiply a-\ by -r- Fi^^st, cH =— ^ . ad a a Hence, .... X— r= 5 — . Ans, a a ad 3. Required the product of — and — . Ans. — ^r-. 4. Required the product of — and 5 2a 3a; 3 Ans. 5a 5. Find the continued product of — , and — =-. ^ a' c 2b Ans. 9ax. 6. It is required to find the product of b-\ and — . . ah-\-hx Ans. . X ^2 J2 a:^_|_J2 7. Required the product of — 7 and -j— — . x^-b^ Ans. b^c+bc^ 8. Required the product of x-\ -, and r- a a-\-b . ax" — ax-\-x'^ — 1 ^ns. ^— — . a^-\-ao (P' — X^ a — 0? "'' X -{-x^ ' ;4 — «2.-v2 9. Required the product of a-\ by Ans. ax + ax^ — aj2 — x' Quest. — 58. How do you multiply fractions together? ALGEBRAIC FRACTIONS. 64 CASE VIII. 69. To divide one fractional quantity by another. RULE. Reduce the mixed quantities, if there are any, to a fractional form; then invert the terms of the divisor, and multiply the fractions together as in the last case, EXAMPLES. 10 , 5 1. Divide. • • • 24 ^ "¥* If the divisor vrere 5, the Operation, quotient would be ■^^-^. But, ^ _& ^ since the divisor is |- of 5, 8 "~ 8 the true quotient must be 8 10 10 times -^1^-^, for the eighth of 24 ~ "~ 120 a number will be contained 10 80 2 v^ Q „ in the dividend 8 times more 120 ~ 120 "" 3 * than the number itself. In this operation we have actually multiplied the numerator of the dividend by 8 and the denominator by 5 ; that is, we have inverted the terms of the divisor and multiplied the fractions together. X.. ., ^ , / 2. Divide . . a? + 62 -^ a?— ^> * Ans. x-\ — -. 10. Divide 6a2 + _ by c^-^-J?. 5 -^ 2 60a2 4.25 J.n^. IL Divide ISc^— a:4-4- by a^— A 6 -^ 5 10c2—5a;+5a* 90^2— 55a;4-5a 5a^b — b^ 12. Divide 20a:2-4^ by x'^-^. 20dcYx^--8abf dc^x^-dc^b-hdc^' EQUATIONS OF THE FIRST DEGREE. 63 CHAPTER III. Of Equations of the First Degree, 60. An Equation is the expression of two equal quanti- ties with the sign of equality placed between them. Thus, a?=(z+5 is an equation, in which x is equal to the sum of a and 5. 6 1 . By the definition, every equation is composed of two parts, separated from each other by the sign = . The part on the left of the sign, is called the first member ; and the part on the right, is called the second memher. Each mem- ber may be composed of one or more terms. Thus, in the equation x-=za+h^ x is the first member, and a-\-h the second. 62^ Every equation maybe regarded as the emmciation, in algebraic language, of a particular question. Thus, the equation a: +3?= 30, is the algebraic enunciation of the following question : Quest. — 60. What is an equation 1 61. Of how many parts is every equation composed 1 How are the parts separated from each other? What is the part on the left called '? What is the part on the right called \ May each member be composed of one or more terms 1 In the equation a; =a+ 5, which is the first member? Which the second ? How many terms in the first member ? How many in the second 1 I 64 FIRST LESSONS IN ALGEBRA. To find a number which being added to itself, shall give a sum equal ^o 30. Were it required to solve this question, we should first express it in algebraic language, which would give the equation By adding x to itself, we have 2a;r=30. And by dividing by 2, we obtain Hence we see that the solution of a question by algebra consists of two distinct parts. 1st. To express algebraically the relation between the knovm and unknown quantities. 2nd. To find a value for the unknown quantity, in terms of those which are known. This latter part is called the solution of the equation. The given or known parts of a question, are represented either by numbers or by the first letters of the alphabet, a, b, c, &c. The unknown or required parts are repre- sented by the final letters, x, y, z, 2_453c2^_5a6_j_2a2j2c2_3a353, Second Transformation, 69. When the two members of an equation are entire polynomials, to transpose certain terms from one member to the other. 1. Take for example the equation 5a:— 6 = 8 + 20?. If, in the first place we subtract 2x from both members, the equality will not be destroyed, and we have 5x—6—2x=:S. Whence we see that the term 2x, which was additive in the second member becomes subtractive in the first. Quest. — 69. What is the second transformation? What do you understand by transposing a term 1 Give the rule for transposing from one member to the other. 70 FIRST LESSONS IN ALGEBRA. In the second place, if we add 6 to both members, the equality will still exist, and we have 5a?— 6~2a?+6 = 8+6. Or, since ~6 and +6 destroy each other, we have 5x-'2x=S+6, Hence the term which was subtractive in the first mem- ber, passes into the second member with the sign of addition. 2. Again, take the equation ax+h=zd-—cx. If we add ex to both members and subtract h from them, the equation becomes ax-\-b-}-cx—b=d—'Cx-\-cx—b. or reducing ax+cx=d-^b. When a term is taken from one member of an equation and placed in the other, it is said to be transposed. Therefore, for the transposition of the terms, we have the following RULK. An7/ term of an equation may be transposed from one mem^ her to the other by changing its sign, TO. We will now apply the preceding principles to the resolution of equations. 1. Take the equation 4x—3=2a;+5. EQUATIONS OP THE FIRST DEGREE. 71 By transposing the terms — 3 and 2ir, it becomes 4a?— 2a?=5+3. Or, reducing 2a: =8. Q Dividing by 2 af=— =4. Verification. If now, 4 be substituted in the place of x in the given equation 4a:— 3 = 2a;4-5, it becomes 4 X 4—3=2 x 4+5. or, 13 = 13. Hence, the value of x is verified by substituting it for the unknown quantity in the given equation. 2. For a second example, take the equation 5a; 4a; 7 13ar 12 3 8 6 By making the denominators disappear, we have 10a;— 32a?-312=21— 52a:, or, by transposing 10a:-32a;+52a;=21 + 312 by reducing 30a; =333 333 111 a result which may be verified by substituting it for x in the given equation. 3. For a third example let us take the equation (3a— a;)(a— ^) + 2«a;=45(a:+a). 72 FIRST LESSONS IN ALGEBRA. It is first necessary to perform the multiplications indica- ted, in order to reduce the two members to two polynomials, and thus be able to disengage the unknown quantity x, from the known quantities. Having done that, the equation becomes, 3a'^—ax—3ab+bx-{-2axz=z4:bx+4ab, or, by transposing —ax-\-bx-\-2ax—4bx ^ziab-^Sab — da^f by reducing ax—Sbx =z7ab — 3a^. Or, (Art. 41). {a-3b)xz=:7ab-3a'^. Dividing both members by a— 3b we find _ 7ab — 3a^ a — 36 Hence, in order to resolve an equation of the first degree, we have the following general RULE. I. If there are any denominators, cause them to disappear, and perform, in both members, all the algebraic operations indicated; we thus obtain an equation the two members of which are entire polynomials. n. Then transpose all the term^ affected with the unknown quantity into the first member, and all the known terms into the second member. HI. Reduce to a single term all the terms involving x : this term will be composed of two factors, one of which will be X, and the other all the multipliers of x, connected with their respective signs. IV. Divide both members by the number or polynomial by which the unknown quantity is multiplied. Quest. — 70. What is the first step in resolving an equation of the first degree 1 What the second 1 What the third ? What the fourth ? EQUATIONS OF THE FIRST I>EGREE. 73 EXAMPLES. 1. Given 3a:— 2+24 = 31 to find x, Ans. xz=:3, 2. Given x-\~l8=z3x~5 to find x. Ans, a;==ll— . 3. Given 6— 2a;+10=20--3a?— 2 to find x. Ans, a? =2. 4. Given x-{-—x-\-—x=:ll to find x. Ans, x=z6. 1 R 5. Given 2x — — a:+l=5a:— 2 to find x. Ans,xz=z-=-, Z I s \ 6. Given 3«a?+— — 3 = 5ir— a to find x, 6— 3a Ans. xz '6a-26 7. Given — - — h-^=20 — to find x, 2 3 2 Ans. a?=23— -. 4 8. Given — - — {-—-=4 : — to find x. o6 Ans. a? =3-—. 13 9. Given —+x=:-- — 3 to find x. 4 2 8 10. Given ; 4=:f to find x. c d -^ Ans, x=4. cdf+4cd Ans. x=:-—^ — -7—. 3ad--2b€ Y4 FIRST LESSONS IN ALGEBRA. 11. Given !f^-i^=4-6 to find .. 56 + 95— 7c Ans. 37=- 16a X x—2 . X 13 12. Given — —^~=.— ta find x. Ans. a; = 10. X X X X r J. n ^ 13. Given r+ ^=/ ^"^ ^^^ ^• a o c a abcdf ■^^^' ^^J^d-acd-^ahd-ahc' Note. — ^What is the numerical value of x, when a— I, 5=2, c=3, d=\, 5 = 5, and/=6. 14. Given ^-.^-^=-12if to find .-. Ans. a?=14. 15. Given x-^^+^^=«+l to find a.. An5. 07 = 6. 16. Given x+^+~^='ioo-AZ to find x. 4 5 D ^715. a;=60. 17. Given 2ar — = — -^ — ^o find x. Ans. a?=3. 18 Given 2x-\ — ^——x+a to find a;. ^a+d Ans. x=z- 6 + 5 ax—b , a bx bx—a - , 19. Given — p-+~=y — - to find x. 35 Ans. x=- '3a-2b' EQUATIONS OF THE FIRST DEGREE. 76 80. Find the value of x in the equation {a+h){x—h) 4ab—b^ a^—hx 7 3a — 7-7 2x-\ . a — o a-jro o Ans, dcz=- 2b{2a^+ab—b^) Of Questions producing Equations of the First Degree involving but a single unknoivn quantity. 71. It has already been observed (Art. 62), that the solution of a question by algebra consists of two distinct parts: 1 St. To express the conditions of the question algebrai- cally ; and 2-d. To disengage the unknown from the known quantities. We have already explained the manner of finding the value of the unknown quantity, after the question has been stated ; and it only remains to point out the best methods of putting a question in the language of algebra. This part of the algebraic solution of a question cannot, like the second, be subjected to any well defined rule. Sometimes the enunciation of the question furnishes the equation immediately ; and sometimes it is necessary to dis- cover, from the enunciation, new conditions from which an equation may be formed. Quest. — 71. Into how many parts is the resolution of a question in algebra divided 1 What is the first step 1 What the second 1 Which part has already been explained ? Which part is now to be considered ? Can this part be subjected to exact rules ? Give the general rule for stating a question 76 FIRST LESSONS IN ALGEBRA. Ill almost all cases, however, we are enabled to discover the equation by applying the following RULE. Consider the 'problem solved, and then indicate, hij means cf algebraic signs, upon the known and unknown quantities, the same operations which it would he necessary to perform, in order to verify the unknown quantity, had it been known. QUESTION'S. 1 . To find a number to which if 5 be added, the sum will be equal to 9. Denote the number by x. Then by the conditions x+b-9. This is the statement of the question. To find the value of x, we transpose 5 to the second member, which gives a; = 9 — 5 = 4. Verification, 4 + 5 = 9. 2. Find a number such, that the sum of one half, one third, and one fourth of it, augmented by 45, shall be equal to 448. Let the required number be denoted by x. Then, one half of it will be denoted by — . one third „ „ by -— . one fourth , „ by — . 3 X EQUATIONS OF THE FIRST BEGREB T7 And by the conditions, This is the statement of the question. To find the value of x, subtract 45 from both members : this gives By clearing the terms of their denominators, we obtain 6x-j-4x-i-3x=iS36, or 13a; = 4836. IT 4836 ^^„ Hence, a;=— — =372. Verijlcalion. 372 372 ^72 -7— +-t;-+—— +45 = 186 + 124+93+45=448. <* o 4 3. What number is that whose third part exceeds its fourth by 16, Let the required number be represented by x. Then, ---0;= the third part, -— a;= the fourth part. And by the question This is the statement. To find the value of x, we clear the terms of the denominators, which gives 4a?— 3a?=192. and a; =192. 7* 1 78 FIRST LESSONS IN ALGEBRA. Verification, 192 192 z=z64~48=zl6. 3 4 4. Divide $1000 between A, B and C, so that A shall have $72 more than B, and C $100 more than A. Let ' x= B's share of the $1000. Then x-\- 72= A's share, and 07+172= C's share, their sum is 3a;+ 244 = $1000. This is the statement. By transposing 244 we have 3a;=1000-244=756 756 and 0;=-— — =252= B's share. Hence, x+ 72 =252+ 72 = $324= A's share. And a;+ 172 =252+172 = $424= C's share. Verification. 252 + 324 + 424 = 1000. 5. Out of a cask of wine which had leaked away a third part, 21 gallons were afterwards drawn, and the cask being then guaged, appeared to be half full : how much did it hold? Suppose the cask to have held x gallons. Then, — what leaked away. o And -^+ 21= what had leaked and been drawn. o X 1 Hence, ---+ 21=-— a: by the question, o 2 This is the statement. EQUATIONS OF THE FIRST DEGREE. 79 To iiiid X, we have 2x-\-l26=i3^, and — X =: — 126, or X =: 126, by changing the signs of both members, which does not destroy their equality. Ver if cation. ^^'^-+21=42+21=63=i^ 3 6. A llsh was caught whose tail weighed 9lb., his head weighed as much as his tail and half his body, and his body weighed as much as his head and tail together ; what was the weight of the fish ? Let 2x=z the weight of the body. Then, 9 + a:=: weight of the head ; and since the body weighed as much as both head and tail, 2a:=:9 + 9 + a:, which is the statement. Then, 2a;— a:=18 and a?=18. Verification. 2x=z36lb.=z weight of the body. 9-{-xz=::27lb.z=: weight of the head. dlb.z:^ weight of the tail. Hence, 72lb. = weight of the fish. 80 FIRST LESSONS IN ALGEBRA. 7. The sum of two numbers is 67 and their difference 1 9 : what are the two numbers ? Let xzzz the least number. Then, x-\-19=z the greater. and by the conditions of the question 2a;+19 = 67. This is the statement. To find X, we first transpose 19, which gives. 2a; = 67-19 = 48; 48 hence, x=z—z=24, and a?+ 19=43. Verification, 43+24=67, and 43-24=19. Another Solution. Let X represent the greater number : tnen, aj— 19 will represent the least, and, 2a;— 19 = 67, whence 2a; = 67+l9; 86 therefore, x= — =43. and consequently a;— 19=43 — 19=24. General Solution of this Problem. The sum of two numbers is a, their difference is b. What are the two numbers ? EQUATIONS OF THE FIRST DEGREE. 81 Let X be the least number, x+b will represent the greater. Hence, 2x+bz:^a, whence 2x=:a—b; , f. a — b a b therefore, X'=.—- — == — , ' 2 2 2' and consequently, x-{-bz=i— —+£>=—+—. 2 to find a', y and 2^. ^«5. a?=:r8, y=:9, ;^— 12. 2a; -f 4y— 3^=22"] 2. Given ^ 4a;— 2y4- 5^^=18 ^ to find a?, y and z, I 6x+ 7ij— z=:63j Arts. a?=i:3, y=7, ^ = 4. ^+-2-y+Y^=:32 3. Given ^ — a:-h— -3/+— 2^— 15 ^ to find a?, y and z. AnS. Xz^\2, y=:20, ;?r=:30. 4. Divide the number 90 into four sucli parts that the first increased by 2, the second diminished by 2, the third multiplied by 2, and the fourth divided by 2, shall be equal to each other. This question may be easily solved by introducing a new unknown quantity. Let X, y, z, and u, be the required parts, and desig- nate by m the several equal quantities which arise from the conditions. We shall then have '<4 EQUATIONS OF THE FIRST DEGREE. 113 From which we find m x=:m — 2, yr=m+2, z=:—, u=z2m. And by addmg the equations, 771 x-\-i/-\-z-\-u=^m-\-m-\ — —-\-2m=z4^m. And since, by the conditions of the question, the first member is equal to 90, we have 4j77?=z90, or |w=90; hence mr=20. Having the value of m^ we easily find the other values : viz. a;=18, y:=22, z—\^, w=40. ^^ 5. There are three ingots composed of different metals mixed together. A pound of the first contains 7 ounces of silver, 3 ounces of copper, and 6 of pewter. A pound of the second contains 12 ounces of silver, 3 ounces of cop- per, and 1 of pewter, A pound of the third contains 4 ounces of silver, 7 ounces of copper, and 5 of pewter. It is required to find how much it will take of each of the three ingots to form a fourth, which shall contain in a pound, 8 ounces of silver, 3f of copper, and 4^ of pewter. Let ir, y, and z represent the number of ounces which it is necessary to take from the three ingots respectively, in order to form a pound of the required ingot. Since there are 7 ounces of silver in a pound, or 16 ounces, of the first ingot, it follows that one ounce of it contains ^-^ of an ounce of silver, and consequently in a number of ounces Ix denoted by aj, there is -— ounces of silver. In the same 16 10* 114 FIRST LESSONS IN ALGEBRA. manner we would find that -— ^ and -— , express the num- Ib lb ber of ounces of silver taken from the second and third, to form the fourth ; but from the enunciation, one pound of this fourth ingot contains 8 ounces of silver. We have, then, for the first equation, 16^ 16 16 or, making the denominators disappear, 7a;+12y + 4^==128. As respects the copper, we should find 3x+37/+7z=z60, and with reference to the pewter 6x+y+5z=:68, As the coefficients of y in these three equations, are the most simple, it is most convenient to eliminate this un- known quantity first. Multiplying the second equation by 4, and subtracting the first, we have DX-{-2iZz=:ll2. Multiplying the third equation by 3, and subtracting the second from the product, 150^4-80 = 144. Multiplying this last equation by 3, and subtracting the preceding one from the product, we obtain 40a? =: 320, whence a? =8. EQUATIONS OF THE FIRST DEGREE. 115 Substitute this value for x in the equation 150^+8^=144 ; it becomes 120-f-8^ = 144, whence zz=^3. Lastly, the two values x=8, z — d, being substituted in the equation Qx+y + ^z — QS, give ^ 484-y+15=:::68, whence y = ^- Therefore, in order to form a pound of the fourth ingot, we must take 8 ounces of the first, 5 ounces of the second, and 3 of the third. Verification. If there be 7 ounces of silver in 16 ounces of the first ingot, in 8 ounces of it, there should be a number of ounces of silver expressed by 7x8 . 16 • In like manner, 12x5 , 4x3 and % 16 16 will express the quantity of silver contained in 5 ounces of the second ingot, and 3 ounces of the third. Now, we have 7X8 12x5 . 4x3 128 ^ 16 ' 16 ' 16 16 ~ ' therefore, a pound of the fourth ingot contains 8 ounces of silver, as required by the enunciation. The same condi- tions may be verified relative to the copper and pewter. 116 FIRST LESSONS IN ALGEBRA. 6. A's age is double B's, and B's is triple of C's, and the sum of all their ages is 140. What is the age of each ? Ans, A's = 84, B's=:42, and C's =14. 7. A person bought a chaise, horse, and harness, for £60 ; the horse came to twice the price of the harness, and the chaise to twice the price of the horse and harness. What did he give for each ? ^ ^13 6s. Sd. for the horse. Ans. < jC 6 I3s. Ad. for the harness. ' jC40 for the chaise. 8. To divide the number 36 into three such parts that i of the first, i of the second, and i of the third, may be all equal to each other. A7is. 8, 12, and 16. 9. If A and B together can do a piece of work in 8 days, A and C together in 9 days, and B and C in ten days ; how many days w^ould it take each to perform the same work alone ? Ans. A 14f|, B 17ff, C 23/i-. 10. Three persons, A, B, and C, begin to play together, having among them all $600. At the end of the first game A has won one-half of B's money, which, added to his own, makes double the amount B had at first. In the second game, A loses and B wins just as much as C had at the beginning, when A leaves off with exactly what he had at first. How much had each at the beginning ? A71S. A $300, B $200, C $100. 11. Three persons. A, B, and C, together possess $3640. If B gives A $400 of his money, then A will have $320 more than B ; but if B takes $140 of C's money, then B and C will have equal sums. How much has each ? Ans. A $800, B $1280, C $1560. 12. Three persons have a bill to pay, which neither alone is able to discharge. A says to B, " Give me the 4th of your money, and then I can pay the bill." B says to C, " Give me the 8th of yours, and I can pay it. But EQUATIONS OF THE FIRST DEGREE. 117 C says to A, " You must give me the half of yours before I can pay it, as I have but $8." What was the amount of their bill, and how much money had A and B ? , j Amount of the bill, $13. ( A had $10, andB $12. 13. A person possessed a certain capital, which he placed out at a certain interest. Another person, who possessed 10000 dollars more than the first, and who put out his capi- tal 1 per cent, more advantageously, had an income greater by 800 dollars. A third person, who possessed 15000 dol- lars more than the first, putting out his capital 2 per cent, more advantageously, had an income greater by 1 500 dollars. Required the capitals of the three persons, and the rates of interest. ^^^ 5 Sums at interest, $30000, 40000, 45000. ■\ Rates of interest, 4 5 6 pr. ct. 14. A widow receives an estate of $15000 from her de- ceased husband, with directions to divide it among two sons and three daughters, so that each son may receive twice as much as each daughter, and she herself to receive $1000 more than all the children together. What was her share, and what the share of each child ? ^ The widow's share, $8000. Ans. < Each son's, 2000. ( Each daughter's, 1000. 15. A certain sum of money is to be divided between three persons. A, B, and C. A is to receive $3000 less than half of it, B $1000 less than one third part, and C to receive $800 more than the fourth part of the whole. What is the sum to be divided, and what does each receive ? r Sum, $38400. Ans. < A receives 16200. B „ 11800. L C „ 10400 118 FIRST LESSONS IN ALGEBRA. CHAPTER IV. Of Powers, 19. If a quantity be multiplied several times by itself, the product is called the power of the quantity. Thus, az=za is the root, or first power of a. ay^arz^a^ is the square, or second power of a, aXaXa — a^ is the cube, or third power of a. aXaXaXa=a^ is the fourth power of a. axaxaxaxaz=:a^ is the fifth power of a. In every power there are three things to be considered • 1st. The quantity which is multiplied by itself, and which is called the root or the first power. 2nd. The small figure which is placed at the right, and a little above the letter. This figure is called the exponent of the power, and shows how many times the letter enters as a factor. 3rd. The power itself, which is the final product, or result of the multiplications. Quest. — 79. If a quantity be continually multiplied by itself, what is the product called 1 How many things are to be considered in every power 1 What are thoy ! OF POWERS. IIP For example, if we suppose « = 3, we have «= 3 the root, or 1st power of 3. «2 — 32--3x3=: 9 the second power of 3. a"5 — 33 = 3 X3 X3=: 27 the third power of 3. a^rr 3^=3 X3 X3x3=: 81 the fourth power of 3. flS — 35 = 3x3 X3x 3 X3=r243 the fifth power of 3. In these expressions, 3 is the root, 1, 2, 3, 4 and 5 are the exponents, and 3, 9, 27, 81 and 243 are the powers. To raise monomials to any potoer, 80. Let it be required to raise the monomial 2a?h'^ to the fourth power. We have (2a352)4=2«3^,3 y^ 2a3J2 X 2^352 X 2a?h-, which merely expresses that the fourth power is equal to the product which arises from writing the quantity four times as a factor. By the rules for multiplication, this pro- duct becomes (2a3Z>2y4^2'^a3 + 3 + 3 + 3j2 + 2 + 2 + 2_2%12^,8. from which we see, 1st. That the coefficient 2 must be raised to the 4th power ; and, " 2nd. That the exponent of each letter must be multiplied by 4, the exponent of the power. As the same reasoning would apply to every example, we have, for the raising of monomials to any power, the following 120 FIRST LESSONS IN ALGEBRA. RULE I. Raise the coejficient to the required power, II. Multiply the exponent of each letter hy the exponent of the power. EXAMPLES. 1 . What is the square of Sce^y^ ? Ans. 9a'^y^. 2. What is the cube of Qa^y-x 1 Ans. 216«'ya;^. 3. W^hat is the fourth power of 2d^y%^ 1 4. What is the square of a-¥'y^ ? Ans. a^h^^y^. 5. What is the seventh power of a%c(P 1 Ans. a^-Wc'd?\ 6. What is the sixth power of a%^c^d 1 Ans. a^%^^c^"d^. 7. What is the square and cube of —2a-lr ? Square. Cube. — 2^253 —2a%- -2^252 ' -2^ -^Aa^¥ -\-Aa^¥ " -2a2^3 By observing the way in which the powers are formed, we may conckide, 1st. When the root is positive, all the powers will he positive. 2nd. When the root is negative, all the even powers tcill be positive and all the odd powers negative. Quest. — 80. What is a monomial 1 Give the rule for raising a monomial to any power. When the root is positive, how will the powers be 1 When the root is negative, how will the powers be 1 OF POWERS. 121 8. What is the square of •—2a^¥ ? Ans. Aa^^^, 9. What is the cube of —^a^y'^cl A7is, —\2^a}^y^c'^» 10. What is the eighth power of —d?xy^ ? Ans. +a2%8yi6^ 11. What is the seventh power of —cP-yx^ ? Ans. — a^^y^x^^, 12. What is the sixth power of 2a¥y^ 1 Ans. Qia%'^^y'^^. 13. What is the ninth power of — cdx^y"^ ? Ans. — c9(Z9a;18y27, 14. What is the sixth power of —'^ah'^d ? Ans. 729 a^^'^d^ 15. What is the square of —lOa^^c^ 1 Ans. WOa'^Mc^, 16. What is the cube of —9a%^d-p ? Ajis. —129a}%^^d^p. 17. What is the fourth pow-er of —Aa^h^c'^d'' 1 Ans. 256a20^,i2ci6^20. 18. What is the cube of —4:a?hVd ? Ans. —QAa%^cH^. 19. What is the fifth power of 2w^lrxy ? Ans. Z2a}^h^^x^y^. 20. What is the square of 20a:^y^c5 ? Ans. AOOx^y^c^^. 21. What is the fourth power of ^a-h\^ ? Ans. Sla%^c^'^. 22. What is the fifth power of —c^d'^x'^y'^ ? Ans. -ciojiVOyio. 23. What is the sixth power of —ac^dfl Ans. a^c^'^d^f^. 24. What is the fourth power of —2aP'c'^d^ 1 Ans. l^a^c^d^^. 11 122 FIRST LESSONS IN ALGEBRA. To raise Polynomials to any power. 8 1 . The power of a polynomial, like tliat of a mono- mial, is obtained by multiplying the quantity continually by itself. Thus, to find the fifth power of the binomial a~\-h, we have a -{- h 1st power. a+ h a^-j- ah + ah+h'^ a^-\-2ah -\-P 2nd power. a -{- b + a^b-}- 2ab'^ + h^ a^+da^-i- 3ab^ + b^ . . . 3rd power. a + b + a^+ 3a^'^-\-3ab^ + b^ a'^-{-ia^-{- 6a^^+ 4ab^ + 5^ 4th power. a + b a^-\-4a^b-{- ea^^+~4^^3 -{- ab^ a^^^a^b+lOa%'^+\OaV)'i+bab^-^b^ Ans. Remark. — 82. It will be observed that the number of multiplications is always 1 less than the imits in the expo- QuEST. — 81. How is the power of a polynomial obtained. OF POWERS 123 nent of the power. Thus, if the exponent is 1, no multipli- cation is necessary. If it is 2, we multiply once ; if it is 3, twice ; if 4, three times, &c. The powers of polyno- mials may be expressed by means of the exponent. Thus, to express that a-^-b is to be raised to the 5th power, we write (a+bY, which expresses the fifth power of a+b. 2. Find the 5th power of the binomial a—b. a ~ b *"" r""^** a^ — ab — ab +62 a^-2ab +5^ . 2nd power. a - b a^-2a^b^ ah"^ - a%^ 2«52 _ -53 a?—3a%^ 3a&2 _ -53 ... 3rd power. a - b a^-Sa% + 3a2^,2_ - ab^ - a?b + 3^252- - 3ab^ + b^ a^—A:a?b + 6^2^.2 _ - Aah^ + ¥ 4th power. a ~ b a^-Aa^b + QaW- - 4^253+ ab^ — a^b + 4a^^- - Qa%^-^4:ab^-b^ ^5 _ 5^45 _|_ 10^352 - -I0a^^+5ab^-b^ Ans, Quest. — 82. How does the number of multiplications compare with the exponent of the power 1 If the exponent is 4, how many multipli- cations 1 124 FIRST LESSONS IN ALGEBRA. 3. What is the square of 5a—2c + d, 5a — 2c + d 5a — 2c -\- d 25«2_l0ac+ 5ad — lOac-i- 4c^ —2cd + 5ad—2cd+ d^ 25a^-20ac+l0ad-\-Ac'^ —A^cd+d^ Ans. 4. Find the 4th power of the binomial 3a— 25. ^a — 2h 1st power. 3a — 2h 9^2 _ Qah — Qah + Ah'^ 9a^— l2ah-\-Ah'^ 2nd power. 3« — 25 27«3-, 36^25+ 12a52 27a3_ 54a25+36«52— 8^3 . . 3rd power. 3a — 2h 81a^-162a354-108a252_24«J3 ■^ 5Aa'^h-\-\0Sa%'^—l[2a¥ -{-!&¥ Sla^— 216a3Z> -1-216^252 — 96a53-f 165^ Ans. 5. What is the square of the binomial a+1 1 Ans. a2_j_2«-|_l. 6. What is the square of the binomial a — 11 Ans. a^—2a-\-l, 7. What is the cube of 9a — 35 ? Ans, 729a3— 729a25+243a52_2753, 8. What is the third power of a— 1 ? Ans. a3_3^2_|_3^_l OF POWERS. 125 9. What is the 4th power of x—yl Arts, a?*— 4a;3y+6a;y— 4a7y3 + y*. 10. What is the cube of the trinomial ipc+y+zl Ans. x^ + Zxhj+2x'^z-\-'^xy'^+^xz'^+'^i/z-\-Zyz'^+Qxyz -{-y^-\-z^. 11. What is the cube of the trinomial 2a'^—'4ab-\-3b^ 1 Ans. Sa^—4Sa^b+l32a'^b'^—208a^^+l9Sa^b^—l0Sab^ + 21¥. To 7^aise a Fraction to any Poioer. 83. The power of a fraction is obtained by multiplying the fraction by itself ; that is, by multiplying the numerator by the numerator, and the denominator by the denominator. Thus, the cube of ^-, which is written b / a \^ a a a a^ is found by cubing the numerator and denominator sepa- rately. 2. What is the square of the fraction 7- — ? b-{-c We have /a — c \2^ {a—cY _ a?'—2ac-\-c^ \b^) ~ [b+cf ~"P+2^+^ ^^' 3. What is the cube of ^- 1 Ans. -~^. 3bc 27^>V QuEST.^-83. How do you find the power of a fraction 1 11* 126 FIRST LESSONS IN ALGEBRA. 4. What is the fourth power of — r— 1 Ans. 1 Qxhj^ ' 5. What is the cube of - — ^ ? x-{-y x"^ — ^x^y + 3a?y2 — y^ x'^-\-^x'^y-\-'ixy'^-\-y^* Q>ctx '^'^ 6. What is the fourth power of ? Ans. Aay 1 Qy"^ 7. What is the fifth power of ? ^;?^. 8. What is the square of IQyz 32y^z^ ax—y^ by — a: dP'x'^ — 2axy-\-y'^ ^y — '^hxy-\-x^ 9. What is the cube of -^—- — ? x-\-2y x'^-\-Qx'^y-\-\2xy'^-\-Sy'^ Binomial Theorem. 8 4. The method which has been explained of raising a polynomial to any power, is somewhat tedious, and hence other methods, less difficult, have been anxiously sought for. The most simple which has yet been discovered, is the one invented by Sir Isaac Newton, called the Binomial Theorem. Quest. — 84. What is the object of the Binomial Theorem 1 Who discovered this theorem \ BINOMIAL THEOREM. 127 85. In raising a quantity to any power, it is plain that there are four things to be considered : — 1st. The number of terms of the power. 2nd. The signs of the terms. 3rd. The exponents of the letters. 4th. The coefficients of the terms. Of the Terms. 86. If we take the two examples of Article 81, which we there wrought out in full ; we have (^+^,)5,^^5_L.5«4^_|_10a3^,3_|_10«253_{_5^j4_^55 . By examining the several multiplications, in Art. 8 1 , we shall observe that the second power of a binomial contains three terms, the third power four, the fourth power five, the fifth power six, &c ; and hence we may conclude — That the number of terms in any power of a binomial, is one greater than the exponent of the power. Of the Signs of the Terms. 87. It is evident that when both terms of the given bi- nomial are plus, all the terms of the poicer will be plus. 2nd. If the second term of the binomial is negative, then all the odd terms, counted from the left, ivill be positive, and all the even terms necrative. Quest. — 85. In raising a quantity to any power, how many things are to be considered'? What are theyl — 86. How many terms are there in any power of a binomial 1 If the exponent is 3, how many terms'? If it is 4, how many terms'? If 5'? &c. — 87. If both terms of the binomial are positive, how are the terms of the power '? If the second term is negative, how are the signs of the terms 1 128 FIRST LESSONS IN ALGEBRA. Of the Exponents. 88. The letter which occupies the first place in a bino- mial, is called the leading letter. Thus, a is the leading letter in the binomials a-\-h, a—h. 1st. It is evident that the exponent of the leading letter in the first term will be the same as the exponent of the power ; and that this exponent will diminish by unity in each term to the right, until we reach the last term, which does not contain the leading letter. 2nd. The exponent of the second letter is 1 in the second term, and increases by unity in each term to the right, until we reach the last term, in which the exponent is the same as that of the given power. 3rd. The sum of the exponents of the two letters, in any term, is equal to the exponent of the given power. This last remark will enable, us to verify any result obtained by the binomial theorem. Let us now apply these principles in the two following examples, in which the coefficients are omitted : — {a+hf . . . a^+a^b-{-a^h'^-^a?P + a^¥ + ah^-{-h\ {a — bf,.. a^—a^h+a^h'^—a?P+a?¥—ah^-\-h^. As the pupil should be practised in writing the terms with- out the coefficients and signs, before finding the coefficients, we will add a few niore examples. Quest. — 88. Which is the leading letter of the binomial 1 What is the exponent of this letter in the first term 1 How does it change in the terms towards the right 1 What is the exponent of the second letter in the second term '\ How does it change in the terms towards the right 1 What is it in the last term 1 What is the sum of the exponents in any term equal to*? BINOMIAL THEOREM. 129 1. {a-^-hy . . a^ + aH-\-ab^^P. 2. (a— by . . a'^—a^b + a'^b'^ — ab^+hK 3. {a+by . . a^+a^b+a^^+a^^+ab^+b^. 4. {a-by . . a^-a%+a^b^-a^b^ + a^^-a'^b^+ab^-b'^. Of the Coefficients. 89. The coefficient of the first term is unity. The co- efficient of the second term is the same as the exponent of the given power. The coefficient of the third term is found by multiplying the coefficient of the second term by the exponent of the leading letter, and dividing the product •♦y 2. And finally — If the coefficient of any term be multiplied by the exponent of the leading letter^ and the product divided hy the number which marks the place of that term from the left, the quotient will be the coefficient of the next term. Thus, to find the coefficients in the example {a-by . . , a''-a^b + a^b^-a^P-\-a^^-a^'+ab'-h'' we first place the exponent 7 as a coefficient of the second term. Then, to find the coefficient of the third term, we multiply 7 by 6, the exponent of a, and divide by 2. The quotient 21 is the coefficient of the third term. To find the coefficient of the fourth, we multiply 21 by 5, and divide the product by 3 : this gives 35. To find the coeffi.cient of the fifth term, we multiply 35 by 4, and divide the product by 4 : this gives 35. The coefficient of the sixth term, found in the same way, is 21 ; that of the seventh, 7 ; and that of the eighth, 1. Collecting these coefficients, we have (a-by = «7_7a6j_2io6j2_35a4j3+35a3j4_2ia3J6+7a5«-i'. 130 FIRST LESSONS IN ALGEBRA. Remark. — We see, in examining this last result, tliat the coefficients of the extreme terms are each unity, and that the coefficients of terms equally distant from the extreme terms are equal. It will, therefore, be sufficient to find the coefficients of the first half of the terms, from which the others may be immediately written. EXAMPLES. 1. Find the fourth power of a-{-h. 2. Find the fourth power of a — b. Ans. a^—4a^+6a'^b^ — Aab^-{-b\ 3. Find the fifth power of a+b. Ans. a^-i-5a'^b-\-10a^'^-i-l0a'^b^-\-5ab^-{-b\ 4. Find the fifth power of a — b. A?is. a^ — ba^b+lOaW—lOa^^-i-^ab^—b^. 5. Find the sixth power of a-\-b. Ans. a^-i-6a^b-\-l5a'^b'^+20a^^-\-l5a'^b^+6a¥ + b^ 6. Find the sixth power of a — b. Ans. a^ — 6a^b+l ba^b^ — 20aW + 1 ba^"^ —6ab^-\- ¥. 7. Let it be required to raise the binomial ^a'^c—2bd to the fourth power. It frequently occurs that the terms of the binomial are affected with coefficients and exponents, as in the above Quest. — 89. What is the coefficient of the first term'! What is the coefficient of the second 1 How do you find the coefficient of the third term % How do you find the coefficient of any term 1 What are the coefficients of the first and last terms 1 How are the coefficients of terms equally distant from the extremes 1 BINOMIAL THEOREM. 131 example. In the first place, we represent each term^of the binomial by a single letter. Thus, we place 3a^c=x, and ~2bd='i/, we then have {jc + y )^ = ^'^ + 4aj^y + 6a;2y2 _|_ 4ajy3 _j_ y4^ But, a;2 — 9a4c3, x^::^27a^c^, x^ = Sla^c^ ; and f == ib-^d^, y^ = - Sb^d^, 1/ = 1 6b^d\ Substituting for x and y their values, we have {3a^c-27jd)^=(3a^-cy-\-4{3a^cy{-2bd) + 6 {3a^cY {-2bdY + A{3a^c) {-2bdf+[-2bd)\ and by performing the operations indicated, {3ci^c-2bdY=zS\a^c^-2lQa^c^d + 2lQa^c%'^d'^-9Qa?c¥d^ -^\Q¥d\ 8. What is the square of 3a — Qb1 Arts. 9a2_36^5_j_36^2 9. What is the cube of 3a: — 6y ? Ans. 27a:3— 162a:2y + 324a:y2_21653. 10. What is the square of x—y ? Ans. x^" — 2x1/ -\-y^. 11. What is the eighth power of 7n-{-n ? A71S. 7n^+8jn/n-{-2Sm^n%56m^n^-{- 70772%^+ 5 6 w^s. + 28m^n^+8?nn^-\-n^, 12. W^hat is the fourth power of a~3b 1 Ans. a^ — 1 2^35 + 54^2^2 __ i Qgab^ + 8 1 R 13. What is the fifth power of c—2d 1 Ans. c^ — \0cH^A0cH'^ — Q0cH^ + SQcd^-^32d^. 14. What is the cube of ba — 3d 1 Ans. 125^3 >-.225aV+135a(/2__27rf3 132 FIRST LESSONS IN ALGEBRA. Remark. The powers of any polynomial may easily be found by the Binomial Theorem. 15. For example, raise a-\-b-\-c to the third power. First, put .... h-\-c=:d. Then, - {a+h-\-cYz={a-\-dy = a^-\-3a^d-{-3ad^+d\ Or, by substituting for the value of d, 3a^c -{-3b^c-{- 6abc -\-3ac"-{-3bc'^ + c^. This expression is composed of the cubes of the three terms, plus three times the square of each term by the first powers of the two others, plus six times the product of all three terms. It is easily proved that this law is true for any polynomial. To apply the preceding formula to the development of the cube of a trinomial, in which the terms are affected with coefficients and exponents, designate each term by a single letter, then replace the letters introduced, by their values, and perform the operations indicated. From this rule, we find that {2o?'—Aab^3b'^fz:^Sa^—AQa^b-{-\32a^b'^—20Sa^b^ + 198a2Z,4_i08a65+2756. The fourth, fifth, &c, powers of any polynomial can be found in a similar manner. 16. What is the cube of a— 2Z>-f c ? Ans. a3_853+c3 — 6ft2^,-f 3a2c+12a&2+12^2c+3ac2 — ebc^-—l2abc. ^ EXTRACT mv OF THE SQUARE ROOT. 1 33 CHAPTER V. Extra*:non of the Square Root of Numbers. Forma- tion of the Square and Extraction of the Square Root of Algebraic Quantities. Calculus of Radicals of the Second. Degree. 90. The square or second power of a number, is the product which arises from multiplying that number by itself once : for example, 49 is the square of 7, and 144 is the square of 12. 91. The square root of a number is that number which, being multiplied by itself once, will produce the given num- ber. Thus, 7 is the square root of 49, and 12 the square root of 144: for, 7x7=^49, and 12x12=: 144. 92. The square of a number, either entire or fractional, is easily found, being always obtained by multiplying this number by itself once. The extraction of the square root of a number is, however, attended with some difficulty, and requires particular explanation. Quest. — 90. What is the square, or second power of a number! — 91. What is the square root of a number 1 12 1 134 FIRST LESSONS IN ALGEBRA* The first ten numbers are. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10; and their squares, 1, 4, 9, 16. 25, 36, 49, 64, 81, 100; and reciprocally, the numbers of the first line are the square roots of the corresponding niunbers of the second. We may also remark that, the square of a number expressed hy a single figure^ will contain no figure of a higher denomination than tens. The numbers of the last line, 1, 4, 9, 16, &c, and all other numbers which can be produced by the multiplication of a number by itself, are called perfect squares. It is obvious that there are but nine perfect squares among all the numbers which can be expressed by one or tAVO figures : the square roots of all other numbers expressed by one or two figures, will be found between two whole numbers differing from each other by unity. Thus 55, which is comprised between 49 and 64, has for its square root a number between 7 and 8. Also 91, which is com prised between 81 and 100, has for its square root a number between 9 and 10. 03. Every number may be regarded as made up of a certain number of tens and a certain number of units. Thus 64 is made up of 6 tens and 4 units, and may be ex pressed under the form 60 + 4. Quest. — 92. What will be the highest denomination of the square of a number expressed by a single figure '? What are perfect^ squares 1 How many are there between 1 and 100 1 What are they ? EXTRACTION OF THE SQUARE ROOT. 135 Now, if we represent the tens by a and the units by h, we shall have a^h =64, and (a+^)^ = (64)2; or a^+2ah + h'^zzzA09Q. Which proves that the square of a number composed of tens and units, contains the square of the tens plus twice the product of the tens hy the units, plus the square of the units. 94. If, now, we make the units 1, 2, 3, 4, &c, tens, or units of the second order, by annexing to each figure a ci- pher, we shall have 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, and for their squares, 100, 400, 900, 1600, 2500, 3600, 4900, 6400, 8100, 10000. From which w^e see that the square of one ten is 100, the square of two tens 400 ; and generally, that the square of tens will contain no figures of a less denomination than hun- dreds, nor of a higher name than thousands. Ex. 1. — To extract the square root of 6084. Since this number is composed of more than two places of figures, its roots wdll contain 60 84 more than one. But since it is less than 10000, which is the square of 100, the root will contain but two figures : that is, units and tens. Now, the square of the tens must be found in the two Quest. — 93. How may every number be regarded as made up 1 What is the square of a number composed of tens and units equal to? — ■ 94. What is the square of one ten equal to 1 Of 2 tens 1 Of 3 tens'! &.C. 60 84 49 1184 1184 136 FIRST LESSONS IN ALGEBRA. left-hand figures, wliicli v/e will separate from the other two by putting a point over the place of units, and a second over, the place of hundreds. These parts, of two figures each, are called periods. The part 60 is comprised between the two squares 49 and 64, of which the roots are 7 and 8 : hence, 7 is the figure of the tens sought ; and the required root is composed of 7 tens and a certain number of units. The figure 7 being found, we write it on the right of the given 60 84 78 number, from which we separate it by a vertical line: then we 7x2 = 141 subtract its square, 49, from 60, which leaves a remainder of 1 1 , q~ to which we bring down the two next figures 84. The result of this operation, 1184, con- tains twice the product of the tens hy the units, plusUhe square of the units. But since tens multiplied by units cannot give a product of a less name than tens, it follows that the last figure, 4, can form no part of the double product of the tens by the units : this double product is therefore found in the part 118, which we separate from the units' place, 4. Now if we double the tens, which gives 14, and then di- vide 118 by 14, the quotient 8 is the figure of the units, or a figure greater than the units. This quotient figure can never be too small, since the part 118 will be at least equal to twice the product of the tens by the units : but it may be too large ; for the 118, beside-s the double product of the tens by the units, may likewise contain tens arising from the square of the units. To ascertain if the quotient 8 ex- presses the units, we write the 8 on the right of the 14, which gives 148, and then we multiply 148 by 8. Thus, we evidently form, 1st, the square of the units; and, 2nd, the double product of the tens by the units. This EXTRACTION OF THE SQUARE ROOT. 137 multiplication being effected, gives for a product 1184, a number equal to the result of the first operation. Having subtracted the product, we find the remainder equal to : hence 78 is the root required. Indeed, in the operations, we have merely subtracted from the given number 6084, 1st, the square of 7 tens, or 70 ; 2nd, twice the product of 70 by 8 ; and, 3d, the square of 8 : that is, the three parts which enter into the composi- tion of the square 70-4-8, or 78 ; and since the result of the subtraction is 0, it follows that 78 is the square root of 6084. 05. Remark. — The operations in the last example have been performed on but two periods, but it is plain that the same reasoning is equally applicable to larger numbers, for by changing the order of the units, we do not change the relation in which they stand to each other. Thus, in the number 60 84 95, the two periods 60 84 have the same relation to each other as in the number 60 84 ; and hence the methods used in the last example are equally applicable to larger numbers. 96. Hence, for the extraction of the square root of numbers, we have the following RULE. I. Separate the given number into periods of two figures each, beginning at the right hand : — the period on the left will often contain but one figure. II. Find the greatest square in the first period on the left, and place its root on the right, after the manner of a quotient Quest. — 95. Will the reasoning in the example apply to more than two periods 1 12* 138 FIRST LESSONS IN ALGEBRA. m division. Subtract the square of the root from the first period, and to the remainder bring down the second period for a dividend. III. Double the root already found, and place it on the left for a divisor. Seek how many times the divisor is contained in the dividend, exclusive of the right-hand figure, and place the figure in the root and also at tlie right of the divisor. IV. Multiply the divisor, thus augmented, by the last figure of the root, and subtract the product from the dividend, and to the remainder bring doicn the next period for a new dividend. But if any of the products should be greater than the dividend, diminish the last figure of the root. V. Double the whole root already found, for a new divisor, and continue the operation as before, until all the periods are brought down. 07. 1st Remark. If, after all the periods are brought down, there is no remainder, the proposed number is a per- fect square. But if there is a remainder, you have only found the root of the greatest perfect square contained in the given number, or the entire part of the root sought. For example, if it were required to extract the square root of 665, we should find 25 for the entire part of the root, and a remainder of 40, which shows that 665 is not a perfect square. But is the square of 25 the greatest per- fect square contained in 665 1 that is, is 25 the entire part of the root ? To prove this, we will first show that, the difference between the squares of two consecutive numbers, is equal to twice the less number augmented by unity. Quest. — 96. Give the rule for extracting the square root of numbers. What is the first step 1 What the second 1 What the third 1 What the fourth 1 What the fifth 1 EXTRACTION OF THE SQUARE ROOT. 139 Let . . az=z the less number, and . . a-j-l = the greater. Then . {a-^\f — d^^+2a-\-\, and . . [af-z^d^. Then* difference is n^ 2a+l as enunciated. Hence, the entire part of the root cannot be augmented, unless the remainder exceeds twice the root found, plus unity. But 25 x2+l=:51>40 the remainder: therefore, 25 is the entire part of the root. 98. 2nd Rejmark. — The number of figures in the root will always be equal to the number of periods into which the given number is separated. EXAMPLES. 1. To find the square root of 7225. Ans, 85. 2. To find the square root of 17689. Ans. 133. 3. To find the square root of 994009. Ans. 997. 4. To find the square root of 85673536. Ans. 9256. 5. To find the square root of 67798756. Ans. 8234. 6. To find the square root of 978121. Ans. 989. 7. To find the square root of 956484. Ans. 978. 8. What is the square root of 3^372961 ? Ans. 6031. 9. What is the square root of 22071204 ? Ans. 4698. 10. What is the square root of 106929 ? Ans. 327. 11. What is the square root of 12088868379025 ? Ans. 3476905. Quest. — 98. How many %ures will you alwajrs find in the roott 140 FIRST LESSONS IN ALGEBRA. 99. 3rd Remark. — If the given number has not an exac^ root, there will be a remainder after all the periods are brought down, in which case ciphers may be annexed, forming new periods, each of which will give one decimal place in the root. 1. What is the square root of 36729 ? In this example there are two periods of decimals, which give two places of decimals in the root. 3 67 29 1 191,64 + . 2 9 267 261 38 1 629 381 382 6 3832 ^ 24800 22956 i 184400 153296 31104 Rem. 2. What is the square root of 2268741 ? Ans, 1506,23 + . 3. What is the square root of 7596796 ' 4. What is the square root of 96 ? 5. What is the square root of 153 ? 6. What is the square root of 101 ? Ans. 2756,22 + . Ans. 9,79795 + . Ans. 12,36931 + . Ans. 10,04987+. Quest. — 99. How will you find the decimal part of the root 1 EXTRACTION OF THE SQUARE ROOT. 141 7. What is the square root of 285970396644 1 Ans. 534762. 8. What is the square root of 41605800625 ? Ans. 203975. 9. What is the square root of 48303584206084 ? Ans. 6950078. Extraction of the square root of Fractions. 1 OO. Since the square or second power of a fraction is obtained by squaring the numerator and denominator sepa- rately, it follows that the square root of a fraction will be equal to the. square root of the numerator divided by the square root of the denominator. For example, the square root of — is equal to — : for a a a^ 1. What is the square root of — ? 9 2. What is the square root of — ? 64 3. What is the square root of ~- 1 81 256 4. What is the square root of 361 5. What is the square root of -— ? 64 Ans. 1 2' Ans. 3 Ans. 8 Ans. 16 19' 1 Ans. 2' Quest. — 100. If the numerator -and denominator of a fraction are perfect squares, how will you extract the square root T 142 FIRST LESSONS IN ALGEBRA. 4096 64 6. What is the square root of ^777x7; ? ^n^- ttttt*- « ^Trt. • !_ r 582169 , , 763 7. Wnat IS the square root oi ? Ans. •^^' 101. If neither the numerator nor the denominator is a perfect square, the root of the fraction cannot be exactly found. We can, however, easily find the approximate root. For this purpose ^ Multiply/ ho^h terms of the fraction hy the denominator, which makes the denominator a perfect square without altering the value of the fraction. Then, extract the square root of the numerator, and divide this root hy the root of the denomi^ tor ; this quotient will he the approximate root. 3 Thus, if it be required to extract the square root of — , ] 5 we multiply both terms by 5, which gives — -. We then have -v/TFnr 3,8729+ : hence, 3,8729+ -^ 5 =: ,7745+ — Ans, 7 2. What is the square root of — ? Ans. 1,32287 + . 14 3. What is the square root of -— ? A?ts. 1,24721 + . n 4. What is the square root of 1 1 77: ? 16 Ans. 3,41869+. Quest. — IGl. If the numerator and denominator of a fraction are net perfect squares, how do you extract the square root \ EXTRACTION OF THE SQUARE ROOT. 143 13 5. What is the square root of 7-- ? Ans. 2,71313 + . 36 15 6. What is the square root of 8— ? Ans, 2,88203+. 5 7. What is the square root of — ? Ans. 0,64549+. 3 8. W^hat is the square root of 10— ? Ans. 3,20936+. 10!3. Finally, instead of the last method, we may, if we please. Change the vulgar fraction into a decimal, and continue the division until the number of decimal places is double the number of places required in the root. Then, extract the root of the decimal by the last rule, Ex. 1. Extract the square root of —- to within ,001. This number, reduced to decimals, is 0.785714 to within 0,000001 ; but the root of 0,785714 to the nearest unit, is .886 ; hence 0,886 is the root of — to within ,001. 14 2. Find the \/2— to within 0,0001. V 15 Ans. 1,6931 + . 3. What is the square root of — ? Ans. 0,24253 + . 7 4. What is the square root of -— ? Ans. 0,93541 + . 8 5 5. What is the square root of — - ? Ans, 1,29099 + . o Quest. — 102. By what other method may the root be foimdl 144 FIRST LESSONS IN ALGEBRA. Extraction of the Square Root of Monomials. 103. In order to discover the process for extracting the square root, we must see how the square of the monomial is formed. By the rule for the multiplication of monomials (Art. 35), we have that is, in order to square a monomial, it is necessary to square its coefficient, and double each of the exponents of the different letters. Hence, to fmd the root of the square of a monomial, we have the following RULE. 1. Extract the square root of the coefficient. II. Divide the exponent of each letter by 2. Thus, V64a66*= 8^352 for Sa'^b'^xSaW=zQAa%^. 2. Find the square root of Q2ba%^c^. Ans. 25a¥c^. 3. Find the square root of 576a*5^c^. Ans. 24a^^c^. 4. Find the square root of 1960:^^2^^. ^1??^. l^x^yz"^, 5. Find the square root of 441 a^b^c^^d^^. Ans. 21 a-b^c^d^. 6. Find the square root of 7S4a^'^b^^c^hP. Ans. 2Sa%'c^d, 7. Find the square root of 81a^5^c^. Ans. da'^b^c^. QuAsi*. — 103. How do you extract the square root of a monomial 1 EXTRACTION OF THE SQUARE ROOT. 145 104. From the preceding rule it follows, that when a monomial is a perfect square, its numerical coefficient is a perfect square, and all its exponents even numbers. Thus, 25f<*52 is a perfect square, but 98a5* is not a perfect square, because 98 is not a perfect square, and a is affected with an uneven exponent. In the latter case, the quantity is introduced into the cal- culus by affecting it with the sign -y/ , and it is written thus : ^/9Sab\ Quantities of this kind are called radical quantities, or irra- tional quantities, or simply radicals of the second degree. They are also, sometimes called Surds, These expressions may often be simplified, upon the prin- ciple that, the square root of the product ofttoo or more factors is equal to the product of the square roots of these factors ; or, in algebraic language, •yjahcd . . . =zy/ a , ^h . -^ c . ^d . . . This being the case, the above expression, VoSo^^, can be put under the form ■y/\9¥x2a=: ^/A9¥x ^/2a, Now ^/A^h^ may be reduced to 7Z>2 ; hence, -v/98^=: 7*2^2^ In like manner, ^/AbaWM:=z y/^Wc^x^bd^: Zahc ^553", y/B6W^c^=z ^/XiA^b^c^^xUcz=zl2ah'^c^ ^/Wc. 13 146 FIRST LESSONS IN ALGEBRA. The quantity which stands without the radical sign is called the coejfficient of the radical. Thus, in the expressions the quantities 75^, ^ahc, I2ab'^c^, are called coefficients of the radicals. Hence, to simplify a radical expression of the second degree, we have the following RULE. I. Separate the expression into two parts, of which one shall contain all the factors that are perfect squares^ and the other the remaining ones. II. Take the roots of the perfect squares a?id place them before the radical sign, under which leave those factors which are not perfect squares. O 105. Remark. — To determine if a given number has any factor which is a perfect square, we examine and see if it is divisible by either of the perfect squares 4, 9, 16, 25, 36, 49, 64, 81, &c ; and if it is not, we conclude that it does not contain a fac- tor which is a perfect square. Quest. — 104. When is a monomial a perfect square 1 When it is not a perfect square, how is it introduced into the calculus 1 What are quantities of this kind called 1 May they be simplified \ Upon what principle '? What is a coefficient of a radical 1 Give the rule for reducing radicals. — 105. How do you determine whether a given number has a factor which is a perfect square 1 EXTRACTION OF THE SQUARE ROOT. 147 EXAMPLES. 1. Reduce -y/lWiFbc to its simplest form. Ans. 5a^^3aEc. 2. Reduce ■\/l28¥a^ to its simplest form. Ans, Sh'^a?d^/2f. 3. Reduce -\/32«^Z>^c to its simplest form. ' ""- Ans. Aa'^¥'y/2ac, 4. Reduce -y/'IsSo^p^ ^q j^g simplest form. Ans. IGai^c*. 5. Reduce '\/\02Aa^^ to it's simplest form. w Ans. 22a'^b^c^ -y/Wc. 6. Reduce 'y/l29d^h^c^d to its simplest form. Ans. 2'7aWc'^^/^I, 7. Reduce ^/Qlba'h^c^d to its simplest form. Ans. l5a^Pc^3abd. 8. Reduce ^/]A45a^c^ to its simplest form. Ans. 17 ac^d^-x/da. 9. Reduce ^/l008a^d''m^ to its simplest form. Ans. Ua'^d^m'^ 'x/TaJ, 10. Reduce ■\/2l 56a^%^'c^ to its simplest form. Ans. Ua^b^c^^TT. 11. Reduce y/~i05a^b^d^ to its simplest form. Ans. da^Pd'^^^^. 148 FIRST LESSONS IN ALGEBRA. 106. Since like signs in both the factors give a plus sign in the product, the square of — «, as well as that of + «, will be «2 J hence the root of a" is either -\-a or —a. Also, the square root of 25a^^ is either -]-5ab'^ or —5ab^. Whence we may conclude, that if a monomial is positive, its square root may be affected either with the sign -|- or — ; thus, V^a^^dtSa^ ; for, -j-Sa^ or —Sa^, squared, ffives 9a^. The double sioni ± with which the root is affected is read plus ar minus. Ift-the proposed monomial were iiegative, it would be im* possible to extract its root, since it has just been shown that the square of every quantity, whether positive or negative, is essentially positive. Therefore, are algebraic symbols which indicate operations that cannot be performed. They are called imaginary quantities^ or rather imaginary expressions, and are frequently met with in the resolution of equations of the second degree. These symbols can, however, by extending the rules, be simplified in the same manner as those irrational expressions which indicate operations that cannot be performed. Thus, V— 9 may be reduced by (Art. 104). Thus, and ^/'^^^Aa^=: ^/a^ X ^/^-i =2a ^/^^iTi also, V — 8a2^,z= 'y/Aa^X—2:h=:2a ^/^^2b = 2a ^/Wx ■y/'^^^. Quest. — 106. What sign is placed before the square root of a monor- mian Why may you place the sign plus or minus 1 What is an ima- ginary quantity *? Why is it called imaginary 1 RADICALS OF THE SECOND DEGREE 149 Of the Calculus of Radicals of the Second Degree. 107. A radical quantity is the indicated root of an imperfect power. The extraction of the square root gives rise to such ex- pressions as -y/a^ 3 V^, 7 ^2, which are called irra-^ tional quantities, or radicals of the second degree. We will now establish rules for performing the four fundamental operations on these expressions. 108. Two radicals of the second degree are similar, when the quantities under the radical sign are the same in both. Thus, 3 -y/^ and 5c ^/b are similar radicals ; and so also are 9 ^2 and 7 ^/2. Addition. 109. Radicals of the second degree may be added together by the following RULE. I. If the radicals are similar add their coefficients, and to the sum annex the common radical, II. If the radicals are not similar, connect them together with their proper signs. Thus, ^a^/h+^c^/h = {^a+bc)^/h. Quest. — 107. What is a radical quantity 1 What are such quantities called 1 — 108. When are radicals of the second degree similar 1 — 109. How do you add similar radicals of the second degree 1 How do you add radicals which are not similar 1 13* 150 FIRST LESSONS IN ALGEBRA. In like manner. Two radicals, which do not appear to be similar at first sight, may become so by simplification (Art. 104). For example, ^4Sab^-^b '\/75a=4b -|/3«+56 ^/3a — 9h ^/3a; and 2^/45'+3^=6^/b + 3^/'^=z9^/5. When the radicals are not similar, the addition or sub- traction can only be indicated. Thus, in order to add 3 '\/b to 5 "v/a, we write EXAMPLES. 1. What is the sum of •y/WcT' and V^SoM Ans, Ta-y/S. 2. What is the sum of ^/ma^b'^ and y/'72a^b'^ 1 Ans. l\a%'y/2'. 3. What is the sum of W and \/- ■ ? Ans. 4a \, 4. What is the sum of Vl25 and V^OOa^ ? Ans. (5 + 10a)V57 RADICALS OP THE SECOND DEGREE. 151 „r, . , ^ / 50 , /Too , 5. What IS the sum of \/ ,^ and \/ ^^^ ? 147 V 294 6. What is the sum of VMS* and V36x2--36a2? j4w5. '1a^J^Zx-\-^^JW—a^. 7. What is the sum of V^^^^^ 3,nd V^SSo*^? ^W5. (7a+12«2a:2)V2Z 8 Required the sum of ^^72" and -v/^^^- Ans. 14 V^ 9. Required the sum of -y^ and -v^l47. ^;i;y. 10 -y/'S"^ /2^ /~27" 10. Required the sum of -i/—- and 3 V 50 Ans. ^V~6*. 1 1 . Required the sum of 2 'y/o^J and 3 ^64^0?^. Ans. (2a+24a:2)y^T: 12. Required the sum of ^243 and lO-y/sesT u4n5. 119V^ 13. What is the sum of -y/S^Oo^P and VSTS^sp ? ^n^. (8a5+7«^i3)y^, 14. What is the sum of VtSo^F and -x/SOOa^^ ? 152 FIRST LESSONS IN ALGEBRA, Subtraction. 1 1 0. To subtract one radical expression from another, we have the following RULE. I. If the radicals are similar, subtract their coejfficients^ and to the difference annex the common radical. II. If the radicals are not similar, their difference can only he indicated hy the minus sign, EXAMPLES. 1. What is the difference between ^a^fb and a^l ? Here ^a'}/~b—a'\/~bz=z2a^\fb7Ans. 2. From27«V27F subtract Qa^/~27¥7 First, 27av57P~=27a5VX and 6a V^TP^lSa^^V^ and 'Z^ah^J~^ — \Qah^/~^z=z9ah^/~^ Ans. 3. What is the difference of ^/~Tb and V^ • A71S. -y/ 3 . 4. What is the difference of ^~2i^¥ and V^^F"? Ans. (2a5— 362)ynr Quest. — 110. How do you subtract similar radicals 1 How do you subtract radicals which are not similar ? RADICALS OF THE SECOND DEGREE. 163 ip: 5. Required the difference of \ / — and \ /— -. ^ ^ V 5 V 27 4 . Ans. -— vl5. 45 6. What is the difference of vn[28a352 and '}/32a^ 1 Ans. {8ab—4:a'^)^/2a. 7. What is the difference of ^/4:8a^b^ and -x/dab ? Ans. 4ab\^3ab — 3 -^ ab. 8. What is the difference of 'v/24~2«^^ and '\/2a^Wl Ans. {na^^-ab)^^/2^. >f^9. What is the difference of \/- — and \/—r- ? Ans. -^V~^' 10. What is the difference of ^320^2 and -x/SOa^ ? Ans. 4a^ 5 , 1 1 . What is the difference between 'v/720a3p and y^5ab^'^ ? Ans. {I2ab—7cd)^5ab, 12. What is the difference between V968^2j2 and '}/200a'^b^ ? Ans. 12ab\/2, 13. What is the difference between 'v/ll2a8Z>6 and -y/gSa^ ? Ans, 2a*J3'v/7r 154 FIRST LESSONS IN ALGEBRA. Multiplication. 111. For the multiplication of radicals, we have the following RULE. I. Multiply the quantities under the radical signs together, and place the common radical over the product. II. If the radicals have coefficients, ice multiply them to- gether, and place the product before the common radical. Thus, -y/lt X V'^= V^ 5 This is the principle of Art. 104, taken in the inverse order. EXAMPLES. 1 . What is the product of 3 ^/bah and 4 -y/^Oa'l Ans. I20a^/T, 2. What is the product of la^hc and '^a^/~hc'\ Ans. 6a%c. 3. What is the product of 2a ^/c^-fW' and — 3a V^^+F ? Ans. —Qa^a^+I)^), Quest. — 111. How do you multiply quantities which are under radi- cal signs 1 "When the radicals have coefficients, how do you multiply them % RADICALS OF THE SECOND DEGREE. 155 4. What is the product of 3 V"2~ and 2 -y/WJ A71S. 24. 5. What is the product of f ^/^a^b and ^^ -y/^c'^b 1 Ans. ^^ahc yTS^ 6. What is the product of 2a? + -\/T and 2x— ^/l^ 1 Ans. Ax^ — b. 7. What is the product of -y/a^^^/b and ^/ a—2^/b\ Ans. -y/W—Ab. 8. What is the product of '^a-^fa^ by -y/^ ? Ans. 9a^y^ Division, 112. To divide one radical by another, we have the following RULE. I. Divide one of the quantities under the radical sign by the other, and place the common radical over the quotient. II. If the radicals have coefficients, divide the coejfficieiit of the dividend by the coefficient of the divisor, and place the quotient before the common radical. Quest. — 112. How do you divide quantities which are under the radical sign 1 When the radicals have coefficients, how do you divide theml 156 FIRST LESSONS IN ALGEBRA. Thus, — =rzi\/-— ; for the squares of these two expressions are equal to the same quantity — ; hence the expressions themselves must be equal. EXAMPLES. 1. Divide ba^Jh by ^h-yfc. Arts, -^x/ — • 2. Divide \2ac-\/Wc by 4c ^2?^ Ans. '^ayJYc. 3. Divide G^V^GF by SV^F. Ans. \ahyj^. 4. Divide 4a2y^5oF by 2a^ ^/W, Jlus. 2Z>2yTo; 5. Divide "Z^a^hyJ^^^ by 13a ^9^?^ Ans. Qa^b y/ah. 6. Divide %\a%^yJWac by \1ab yj^a, Ans. ^d^y^ ^/c, 7. Divide y/^cP- by -ypZ. Ans. \a. 8. Divide ^aWypZ^'^^ by 12^5^. Ans. o?}P: 9. Divide ^ayflW by 3'v/"57 .4^^. 2ab^/2: 10. Divide iSb^^W by 2b'^^^'. Ans. 360b^. 11. Divide Sa^^^c^ ^73^ ^y 2a'x/2M: Ans. 2a¥c^d, 12. Divide ^Qa^c^y/dW by 48aZ»cV'2?T RADICALS OF THE SECOND DEGREE. 157 13. Divide 21 a^h^ -y/^lU? by ^J'U. Arts, 27^656 -y/3. 14. Divide ISaW^gJ? by ^ah^fcF: Ans. ^a%^ ^J2. To Extract the Square Root of a Polynomial, 113. Before explaining the rule for the extraction of the square root of a polynomial, let us first examine the squares of several polynomials : we have {a+hf — a?+2al-\-h'^, {a+h+cY=a?+2ah+P-^2{a+h)c+c\ {a+h-\-c-\-df—a?+2ah + h'^+2{a+h)c+c^ ^2{a^h + c)d+d?. The law by which these squares are formed can be enun- ciated thus : The square of any polynomial contains the square of the first term, plus twice the product of the first term hy the second^ plus the square of the second ; plus twice the first two terms multiplied hy the third, plus the square of the third ; plus twice the first three terms multiplied hy the fourth, plus the square of the fourth ; and so on» Quest. — 113. What is the square of a binomial equal to] What is the square of a trinomial equal to 1 What is the square of any polynomial equal to 1 14 158 FIRST LESSONS IN ALGEBRA. 114. Hence, to extract the square root of a polynomial we have the following RULE. I. Arrange the 'polynomial with reference to one of its letters and extract the square root of the first term : this will give the first term of the root. II. Divide the second term of the polynomial hy double the first term of the root, and the quotient will be the second term of the root. III. Then form the square of the two terms of the root found, and subtract it from the first polynomial, and then divide the first term of the remainder by double the first term of the root, and the quotient will be the third term. IV. Form the double products of the first and second terms, by the third, plus the square of the third ; then subtract all these products from the last remainder, and divide the first term of the result by double the first term of the root, and the quotient will be the fourth term. Then proceed in the same manner to find the other terms. EXAMPLES. 1 . Extract the square root of the polynomial 49«2^2_24aZ/3 4.25a4_30«35-]_1654. First arra,nge it with reference to the letter a. 25«^ — 30^3^ + A9d^b'^—2Aab^ -f 1 65-^ 25a4_30a3Z>+ 9a%'^ 10^2 40a2^2_24a53_|_i6Z,4 1st Rem. 40a252_24a53_|_i6Z>^ . . . 2d Rem, ^ RADICALS OF THE SECOND DEGREE 159 After having arranged the polynomial with reference to a, extract the square root of 25a*, this gives Sa^, which is placed at the right of the polynomial ; then divide the second term, — 30a^Z>, by the double of ba^, or lOa^ ; the quotient is — 3ah, and is placed at the right of ba?. Hence, the first two terms of the root are baF'—'iah. Squaring this binomial, it becomes 25a*— SOa^^+^a^^^, which, subtracted from the proposed polynomial, gives a remainder, of which the first term is 4:00?})^. Dividing this first term by \Oa?j (the double of ba?'), the quotient is +Ah'^ ; this is the third term of the root, and is written on the right of the first two terms. By forming the double product of ba? — 3ah by 4:h'^, and at the same time squaring 4Z>2, we find the polynomial A0a^h'^—2Aah^-\-\Qh^, which, subtracted from the first re- mainder, gives 0. Therefore 5a? — ^ah-\-Ah'^ is the required root. 2. Find the square root of a'^-{-Aa'^x+Qa"x^-\-Aax'^-\-x^, Ans. aF'-{-2ax-{-x'^. 3. Find the square root of a^—Aa^x+Qa?x'^~Aax'^+x^. Ans, aF- — 2ax-\-x'^, 4. Find the square root of Ax^-\-\2x^+bx^—2x'^+lfx'^—2x+l. Ans, 2x^+^x'^—x+l. 5. Find the square root of * 9a*-12a35+28«2^2_i6«^>3+16K Ans, 3a^—2ab+4b^. Quest. — 114. Give the rule for extractxig the square root of a poly- nomial 1 What is the first step 'l What the second 1 What the third '^ What the fourths 160 FIRST LESSONS IN ALGEBRA. 6. What is the square toot of Ans. a?2— 2«a?— 2. 7. What is the square root of 9^2 — 12a;+6a?y + y^ — 4y + 4. Ans. Sic+y— 2. 8. What is the square root of y^—2y'^x'^-^2x'^—2i/-{-l ' ^ • Ans. y'^—x'^ — l, 9. What is the square root of 9a^6^—30«3Z>3+ 25^2^2 ? Ans. Sa^b^—dab, 10. Find the square root of 2ba^b^—40a^^c-{-76a'^b^c^—48ab^c^-{-36b^c'^—30a^bc + 24a^bc^ — 36a2&c3 + ^a^c^. Ans. ba?b — 32-j- 64^254 is a, perfect square, since -^/d^^Sa^, and ^/MM^=—8ab\ and also 2 X 3a^ X — 8ab^= —4Sa^b^= the middle term. But 4a^-\-l4ab-\-9b^ is not a perfect square : for although 4a" and 4-9b^ are the squares of 2a and 3b, yet 2 X2ax 36 is not equal to I4ab. 3rd. In the series of operations required in a general ex- ample, when the first term of one of the remainders is not exactly divisible by twice the first term of the root, we may conclude that the proposed polynomial is not a perfect square. This is an evident consequence of the course of reasoning, by which we have arrived at the general rule for extracting the square root. 4th. When the polynomial is not a perfect square, it may be simplified (See Art. 104.) Take, for example, the expression '\/a^-\-4a'^b^-{-4ab^. The quantity under the radical is not a perfect square ; but it can be put under the form ab{a'^-\-4ab-\-4b'^). Now, the factor between the parenthesis is evidently the square of a -\- 2b, whence we may conclude that, /' ^a^ + 4a^b'^+4ab^ = {a-\-2b) VoF. 2. Reduce" 2a2Z>— 4a J2_j_ 253 to its simple form. Ans. {a — b)^2F. Quest. — 115. Can a binomial ever be a perfect power'? "WTiy not 1 When is a trinomial a perfect power 1 When, in extracting the square root we find that the first term of the remainder is divisible by twice the root, is the polynomial a perfect power or not 1 14* 162 FIRST LESSONS IN ALGEBRA. CHAPTER VL Equations of the Second Degree, 116. An equation of the second degree is an equation involving the second power of the unknown quantity, or the product of two unknown quantities. Thus, aj^zzrcr, ax^-\-hxzz::c, and xyz^zd?, are equations of the second degree. 117. Equations of the second degree are of two kinds, viz : equations involving two terms, which are called incom- plete equations ; and equations involving three terms, which are called complete equations. Thus, x^=:a and ax^ =: h, are incomplete equations ; and x^-\-2axz=zb, and ax'^-\-bx=zdf are complete equations. Quest. — 116. What is an equation of the second degree 1 — 117. How many kinds aie there 1 What is an incomplete equation 1 What is a complete equation 1 EQUATIONS OF THE SECOND DEGREE 163 118. When we speak of an equation involving tvv^o terms, and of an equation involving three terms, we under- stand that the equation has been reduced to its simplest form. Thus, if we have the equation although in its present form there are four terms, yet it may- be reduced to an equation containing but two. For, by- adding 3x^ to 4a;2 and transposing —4, we have 7ic2:=10. Also, if we have 3x'^-i-5x-{-7x+5 — 9, we get by reducing 3x'^-^l2x=z4, an equation containing but three terms. Again, if we take the equation ax^+bx^+d=:f we have (a-{-b)x^=f-d and a;2r=£-^, an equation of two terms. Quest. — 118. When you speak of an equation involving two terms, do you speak of the equation after it has been reduced, or before 1 When you speak of an equation of three terms, is it the reduced equation to which you refer '? To what forms, then, may every equation of the second degree be reduced 1 164 FIRST LESSONS IN ALGEBRA. Also, if we have ax^ + dx"" ■\-fx + 5 = c we obtain {a-\-d)x^-\-fx=.c—'b, and consequently a-\-d a-\-d^ an equation of three terms. Hence we may conclude : That every equation of the second degree may he reduced to an incomplete equation involv' ing two terms, or to a complete equation involving three terms. Of Inco7nplete Equations, 1 . What number is that which being multiplied by itself the product will be 144. Let X— the number : then a:x xzzzx'^zizlii. It is plain that the value of x will be found by extracting the square root of both members of the equation : that is '^x'^zzi -1/144 : that is, a: =12. 2. A person being asked how much money he had, said if the number of dollars be squared and 6 be added, the sum will be 42 : How much had he ? Jjct x=z the number of dollars. Then by the conditions hence, x'^=z42 — 6 = 36 and xz=e. Ans. $6. EQUATIONS OF THE SECOND DEGREE. 166 3. A person being asked his age said, if from the square of my age you take 192, the remainder will be the square of half my age : what was his age 1 Denote his age by x. Then, by the conditions of the question -192: 4. and by clearing the fractions 4x^—768=x^; hence, 4a;2— a;2=:768, and 3x^=768 0:2=256 X = 16. Ans. 16. 119. There is no difficulty in the resolution of an equa- tion of the form ax'^=b. We deduce from it x^= — , a whence ^=Vt- When — is a particular number, either entire or frac- a tional, we can obtain the square root of it exactly, or by ap- proximation. If — is algebraic, we apply the rules estab- lished for algebraic quantities. QuBst. — 119. How do you resolve an incomplete equation 1 166 FIRST LESSONS IN ALGEBRA. Hence, to find the value of x we have the following RULE. I. Find the value of'^. II. Then extract the square root of both members of the equation. 4. What is the value of x in the equation 3a:H-8=:5a?2— 10. By transposition 3a;2 — 5a?2 — — 1 — 8, by reducing —2x'^z=:i — \B, by dividing by 2 and changing the signs by extracting the square root a;=:3. We should, however, remark that the square root of 9, is either +3, or —3. For, + 3x+3 = 9 and — 3x— 3=9. Hence, when we have the equation we have £c=: + 3 and a?=:~3. 1 20. A root of an equation is such a number as bemg substituted for the unknown quantity, will satisfy the equa- tion, that is, render the two members equal to each other. Thus, in the equation there are two roots, +3 and — 3 ; for either of these numbers being substituted for x will satisfy the equation. EQUATIONS OF THE SECOND DEGREE. 5. Again, if we take the equation we shall have a;= + \/ — and x^=. — \/ — . ^ a y a 167 For, =:h, or ax — = b, a and ax = £>, or aX — =6. a Hence we may conclude, 1st. That every incomplete equation of the second degree has two roots. 2nd. That those roots are numerically equal but have con- trary signs. 6. What are the roots of the equation 3a:2+6 — 4a:2_10. Ans. x=:-\-4 and x=z—4, 7. What are the roots of the equation 1 x^ 0:2-8=:: — +10. 3 9 Ans. a?™ -{-9 and x^z—d. Quest. — 120. What is the root of an equation^ What are the roots of the equation a:2 = 9 1 Of the equation ax2=b 1 How many roots has every incomplete eqvation 1 How do those roots compare with each other 1 168 FIRST LESSONS IN ALGEBRA. 8. What are the roots of the equation 6ic2— 7 = 3a;2+5. Ans. x=+2, a:=~2. 9. What are the roots of the equation. 8+5x^ = ^ + 4x^+28, Ans. x=:+5j x='—5, 10. Find a number such that one-third of it multiplied by one-fourth shall be equal to 108 ? Ans. 36. 1 1 . What number is that whose sixth part multiplied by its fifth part and product divided by ten, shall give a quo- tient equal to 3 ? Ans. 30. 12. What number is that w^hose square, plus 18, shall be equal to half its square plus 30^. Ans. 5. 13. What numbers are those v^hich are to each other as 1 to 2 and the difference of v^hose squares is equal to 75. Let x=: the less number. Then 2x=z the greater. Then by the conditions of the question 4a;2— a;2=75, hence, 3x'^=75 ; and by dividing by 3, x'^=:25 and x=5, and 2a?=10. Ans. 5 and 10. EQUATIONS OF THE SECOND DEGREE. 169 ;•- 14. What two numbers are those which are to each other as 5 to 6, and the difference of whose squares is 44. Let x=: the greatest number. 5 Then -^x= the least. 6 By the conditions of the question .25 x^- ~— -a? =44. 36 by clearing fractions, 36a?2- -25a?2=1584; hence, lla;2=1584. and a?2=144. hence, X =12, and 4. =10. Ans. 10 and 12, 15. What two numbers are those which are to each other as 3 to 4, and the difference of whose squares is 28 ? Ans. 6 and 8. 16. What two numbers are those which are to each other as 5 to 1 1 , and the sum of whose square is 584 ? Ans. 10 and 22. 17. A says to B, my son's age is one quarter of yours, and the difference between the squares of the numbers re- presenting their ages is 240 : what were their ages ? . (Eldest 16. Ans. < ^^ ( Younger 4. 15 170 FIRST LESSONS IN ALGEBRA. When there are two unknown quantities, 121. When there are two or more unknown quantities, eliminate one of them by the rule of Article 77 : there will thus arise a new equation with but a single unknown quantity, the value of which may be found by the rule already given. 1. There is a room of such dimensions, that the differ- rence of the sides multiplied by the less is equal to 36, and the product of the sides is equal to 360 : what are the sides 1 Let x=z the less side ; y=z the greater. Then, by the 1st condition, (y— a?)a; = 36 ; and by the 2nd, ' 0:3/ =360. From the first equation, we have icy— a;2=36 ; and by subtraction, a:2=r324. Hence, cc=^d2i'—lS\ ''' .20. 18 Ans. a?=rl8,y=20. Quest. — 121. How do you resoh-e the equation when there are two or more unknown quantities 1 EQUATIONS OF THE SECOND DEGREE. 171 2. A merchant sells two pieces of muslin, which together measure 12 yards. He received for each piece just so many dollars per yard as the piece contained yards. Now, he gets four times as much for one piece as for the other : how many yards in each piece 1 Let x=z the number in the larger piece ; y= the number in the shorter piece. Then, by the conditions of the question, 'fiitetf xXx=zx'^= what he got for the larger piece ; yXy=y'^= what he got for the shorter. And x^zzz^y^, by the 2nd condition. X =^y, by extracting the square root Substituting this value of x in the first equation, we have iy+y-:13; and consequently, y= 8, and a?= 4. Ans. 8 and 4. 3. What two numbers are those whose product is 30, and quotient 3 J ? Ans. 10 and 3. 4. The product of two numbers is a, and their quotient b : what are the numbers ? A71S. -yj ah and \ / -7- . V 5. The sum of the squares of two numbers is 117, and the difference of their squares 45 : what are the numbers ? Alls, 9 and 6. 172 FIRST LESSONS IN ALGEBRA. 6. The sum of the squares of two numbers is «, and the difference of their squares is h : what are the numbers ? Ans. x= y^a-i-b, y= -y/a — b. 7. What two numbers are those which are to each other as 3 to 4, and the sum of whose squares is 225 ? Ans. 9 and 12. 8. What two numbers are those which are to each other as m to n, and the sum of whose squares is equal to a^ ? ma na Ans. , — — , , =r. V?^^+^^ ym^-^-n^ 9. What two numbers are those which are to each other as 1 to 2, and the difference of whose squares is 75 ? Ans. 5 and 10. 10. What two numbers are those which are to each other as m to n, and the difference of whose squares is equal to ^2? mb nb Ans. yrn^ — n"^ ' yw ' m'' — n^ 11. A certain sum of money is placed at interest for six months, at 8 per cent, per annum. Now, if the amount be multiplied by the number expressing the interest, the pro- duct will be 562500 : what is the amount at interest ? Ans. $3750. 12. A person distributes a sum of money between a num- ber of women and boys. The number of women is to the number of boys as 3 to 4. Now, the boys receive one- half as many dollars as there are persons, and the women twice as many dollars as there are boys, and together they receive 138 dollars : how many women were there, and how many boys ? ( 36 women. i 48 boys. EQUATIONS OF THE SECOND DEGREE. 173 Of co7nplete Equations. 1 22. We have already seen (Art. 117), that a complete equation of the second degree, after it has been reduced, contains three terms, viz : the square of the unknown quantity in the first term ; the first power of the unknown quantity in the second term ; and a known quantity, in a third term. Thus, if we have the equation 5a?2— 2a;24-8 = 9a;4-32, we have, by transposing and reducing, 3x'^—9x—24, and by dividing by 3, x'^—3x=S, which contains but three terms. 2. If we have the equation a^x"^ -f 3abx -\-x'^:=:zcx+d, by collecting the coefficients of x"^ and x, we have (^2+ l)x^+{3ab — c)x = d ; and dividing by the coefficient of x^, we have 3ab—c d X^-\ —-—X=:- a?-\-l d^+l Quest. — 122. How many terms does a complete equation of the second degree contain 1 Of what is the first term composed 1 The second 1 The thirds 15* 174 FIRST LESSONS IN ALGEBRA. If we represent the coefficient of x by 2p, and the known term by q, we have an equation containing but three terms ; and we see, from the above examples, that every complete equation of the second degree may be reduced to this form. 1!33. We wish now to show that there are four forms under which this equation will be expressed, each depend- ing on the signs of 2p and q. 1st. Let us for the sake of illustration, make 2p=r+4, and ^=+5: We shall then have x'^-\-4x=5. 2nd. Let us now suppose 2j9 = — 4, and q=z-{-5 : we shall then have x^--4x=5. 3rd. If we make 25'= +4, and $'=—5, we have x^+4xz=: — 5. 4th. If we make 2pz=z—4, and q= — 5, we have x'^—bx=z—5. Quest. — 123. Under how many forms may every equation of the second degree be expressed 1 On what will these forms depend 1 What are the signs of the coefficient of z and the known term, in the first form 1 What in the second 1 What in the third 1 What in the fourth I Repeat the four forms. EQUATIONS OF THE SECOND DEGREE. 175 We therefore conclude that every complete equation of the second degree may be reduced to one of these forms : a?2 + 2^a? =^+q, 1st form. a;2 — 2px = + ^, 2nd form. x^ + 2px ——q^ 3rd form. x^ — 2px = — 5', 4th form. 124. Remark. — If, in reducing an equation to either of these forms, the second power of the unknown quantity should have a negative sign, it must be rendered positive by changing the sign of every term of the equation. 125. We are next to show the manner in which the *-alue of the unknown quantity may be found. We have ^en (Art. 38), that {x-\-pY=:x'^+2px+p'^ ; and comparing this square with the first and third forms, we see that the first member in each contains two terms of the square of a binomial, viz : the square of the first term plus twice the product of the 2nd term by the first. If, then, we take half the coefiicient of x, viz : p, and square it, and add to both members, the equations take the form x'^+2px-{-p'^=q -\-p\ x^+2px -}-p^ zn-^q'^ +p^i in which the first members are perfect squares. This is Quest. — 124. If in reducing an equation to either of these forms the coefficient of x^ is negative, what do you do ?— 125. What is the square of a binomial equal to 1 What does the first member in each form con- tain 1 How do you render the first member a perfect square 1 What is this called 1 176 FIRST LESSONS IN ALGEBRA. called completing the square. Then, by extracting the square root of both members of the equation, we have and X -\-p = ± 'v/— 5' +i^^> which gives, by transposing p, x=—p±'\/q-{-p'^, x= — p i -y/ — q-\-p'^. 126. If we compare the second and fourth forms with the square {x — pY =: a;2 — 2px +/>^, we also see that half the coefficient of x being squared and added to both members, will make the first members perfect squares. Having made the additions, we have ic^ — 2px -j-p"^ = q -\-p'^, x"^ — 2px -\-p'^ =~q +i?^- Then, by extracting the square root of both members, we have and x—pz=z ± ^—q+p^; and by transposing —p^ we find x=pdz^/^'p, and X =:p dz 'y/—q-{-p'i. Quest. — 126. In the second form, how do you make the first mem- ber a perfect square 1 EQUATIONS OF THE SECOND DEGRlEE. 17T 127, Hence, for the resolution of every equation of the econd degree, we have the following RULE, I. Reduce the equation to one of the known forms. II. Take half the coefficient of the second term, square it^ 4€nd add the result to both members of the equation. III. Then extract the square root of both members of the equation ; cfter which, transpose the known term to the second member. Remark, — The square root of the first member is always equal to the square root of the first term, plus or minus half the coefHcient of x, EXAMPLES IN THE FIRST FORM, 1 . What are the values of x in the equation 2a:24-8ir— 64 ? If we first divide by the coefficient 2, we obtain a:2+4a:=32. Then, completing the square, a;2+4^-f4=32 + 4=36. Extracting the root, x+2 = ±:^/ZQz=z + Q or —6. Hence, a;=— 2 + 6=r-f 4 ; or, x=—2~Qz=z—S. Quest. — 127. Give the general rule for resolving an equation of the second degree. What is the first step 1 What the second \ What the third 1 What is the square root of the first member always eqnal to 1 178 FIRST LESSONS IN ALGEBRA. That is, in this form the smaller root is positive, and the larger negative. Verification. If we take the positive value, viz : a;= + 4, the equation a:2+4a;=32 gives 42 -{-4 X 4=:32 : and if we take the negative value of a?, viz : xz=. — 8, the equation x^+Ax=z^2 gives (-8)2+4(-8)=:64-32=:32. From which we see that either of the values of x, viz : a:=+4 or x=: — 8, will satisfy the equation. 2. What are the values of x in the equation 3a;2+I2aj — 19=— aj2_x2a;+89 ? By transposing the terms, we have 3a;2+a;2+12a: + 12a; = 89 + 19 ; and by reducing, 4a;2+ 2407 =108; and dividing by the coefficient of a?^, a;2^6a;=27. Now, by completing the square, a;2+6a;2+9 = 36; extracting the square root, a;+3 = ± V36'= + 6 or —6: hence, a;= + 6— 3 = + 3; or, «=— 6—3=— 9. EQUATIONS OF THE SECOND DEGREE. 179 Verification. If we take the plus root, the equation gives (3)2 4-6(3)=27; and for the negative root, a?2+6a?=27 gives (-~9)2+6(~9)=:81 -54=27. 4. What are the values of x in the equation a?2— 10a? 4-15=^ 34a;+155. 5 By clearing the fractions, we have 5a;2— 50a;+75 = a?2— 170a;+775 : by transposing and reducing, we obtain 4a;2 4-120a?=700; then, dividing by the coefficient of a;^, we have a;2+30a?=175; and by completing the square, ^2+30^+225=400; and by extracting the square root, a?+15 = ih V400 = +20 or —20. Hence, a:= + 5 or —35. Verification, For the plus value of x, the equation a:2+30a;=175 gives (5)2+30x5=25+150 = 175. 180 FIRST LESSONS IN ALGEBRA. And for the negative value of x, we have (-.35)2+30(-35) = 1225-1050=175. 5. What are the values of x in the equation _^2 ^ _,^_.=s--x-x^+-j^ ? Clearing the. fractions, we have 10a:2-6a:+9=96-8a;-12x2+273 ; transposing and reducing, 22a;2+2ir=360; dividing both members by 22, 2 360 ^22 22 / 1 \2 Add ( — ] to both members, and the equation becomes ,2 . / 1 V 360 / 1 \2 ^ +22^+122) =-22-+V22) ' whence, by extracting the square root, . 1 /360 , / r^ ^+22 = -V-22-+V22> therefore, 1 , /360-- / 1 x2 1\^ «,d ,= _^^_y/^^Q' EQUATIONS OP THE SECOND DEGREE. 181 It remains to perform the numerical opwations. In the first place, •^+(^) ^^^st be reduced to a single num- ber, having (22 )'^ for its denominator. N — ( ^ Y_ 360x224-l _7921 ^^' 22"''\22/"" (22)2 -(22)2' extracting the square root of 7921, we find it to be 89; therefore, ' V 22 "^V22/ - 89 ^22- Consequently, the plus value of x is '^""~22'^22~22"~ ' and the negative value is ___1 89____45 ^~ 22 22"" ""IT' that is, one of the two values of x which will satisfy the proposed equation is a positive whole number, and the other a negative fraction. 6. What are the values of x in the equation 3a?2+2a?— 9=76. 7. What are the values of x in the equation 2ir2-f8a;+7=^~^ + 197. 4 o 16 182 FIRST LESSONS IN ALGEBRA. 8. What are the values of x in the equation Ans. i x=9 -64^. 9. What are the values of x in the equation ~-— -8-— -7^4-61 14^-2 ^''+^2- c x=z2 (00= -^7 1 10. What are the values of x in the equation Ans, x^ X x^ a? 1 3 Ans, \ i x=.-- X=:\ 2*. EXAMPLES IN THE SECOND FORM. 1. What are the values of x in the equation a?2 — 8a;+10=:19. By transposing, a;2-8a:=19-10=9, then by completing the square a;2>-.8ir+ 16 = 9 + 16=25, and by extracting the root a;— 4=zfc V25'= + 5 or —5. Hence, a?=4+5=9 or a;=4— 5 = — 1. That is, in this form, the largest root is positive and the smaller negative. EQUATIONS OF THE SECOND DEGREE. 183 Verification, If we take the positive value of x, the equation x^—Qx—^ gives (9)2—8x9=81 — 72=9; and if we take the negative vakie, the equation a;2_-8a;=9 gives {-l)2-8(-l) = l+8=9 ; from which we see that both values alike satisfy the equa- tion. 2. What are the values of x in the equation By clearing the fractions, we have 6a;2+4a;— 180 = 3a;2+12a?— 177 and by transposing and reducing 3a;2_8a;=3, and dividing by the co-efRcient of a;^, we obtain X^ —X=z\, Then, by completing the square, we have ^3'^"^9~'^9 9' and by extracting the square root, _±____^ /25 ^_5_ ___5_ '^'~T~~\/ 9""'^T ^^ ""3' Hence, 4,5 . ^ 451 3 3 * 3 3 3 184 FIRST LESSONS IN ALGEBRA. Verification, For the positive value of x, the equation x^ — —x=\ 8 gives 32__x 3=9-8 = 1: and for the negative value, the equation 2 Q 1 x^ — —x=l o V 3 7 3 ^■"3~"9""^"9— ^• 1 \2 8 gives 3. What are the values of a? in the equation Clearing the fractions, and dividing by the coefficient of «2, we have Completing the square, we have ''3''^ 9 -^^+9 -36' then, by extracting the square root, we have 1/^7 7 hence, *-T==^V36=+-6- ""' --e- 1,79,, 17 5 EQUATIONS OF THE SECOND DEGREE. 185 Verification, If we take the positive value of x, the equation 2 2 11 gives (ii)2_|.Xli=2l-l = lJ: and for the negative value, the equation ac2 — jx=^H / 5\2 2 5 25 ,10 45 ,, 4. What are the values of x in the equation 4a2_2a;2+2aa;=18a5 — 18J2 ? By transposing, changing the signs, and dividing by 2, it becomes x^—ax=2a^—^ab+W^ ; whence, completing the square, x'^—ax+—=^—^ 9a^+962 ; 4 4 extracting the square root. x=-^±: ^^~9a5-W Now, the square root of -- — Qdb+W, is evidently ^—35. Therefore, 2 \ 2 / ( «j=r — a+3J. 16* 186 FIRST LESSONS IN ALGEBRA. What will be the numerical values of oc, if we suppose fl=6 and 5==! ? t 5. What are the values of x in the equation 1 4 3 5 —aj— 4— a;2+2a; — —x'^=:45 — 3x^-^4x 1 , ( a;= 7,12 ) to within ix=z-5,73y 0,01. 6. What are the values of x in the equation 8a;2-.14a:+10=:2a?+34 ? . ( x= 3. Ans, < p^, the conditions of the question will be incompatible with each other. 134. Before showing this it will be necessary to estab- lish a proposition on which it depends : viz. If a given number he decomposed into two parts and those parts multiplied together, the product will he the greatest pos- sible when the parts are equal. Let 2p be the number to be decomposed, and d the differ- ence of the parts. Then p -\ — the greater part (page 80, Ex. 7.) and p = the less part. and p2 — F~^' their product (Art. 40.) Now it is plain that P will increase as d diminishes, and that it will be the greatest possible when d—0 : that is, py^pz^p^ is the greatest product. Quest. — 133. In which forms do the values of x become imaginary 1 When will the values of x be imaginary 1 Why will the values of x be then imaginary T or THE \ UNIVERSITY 1 1} EQUATIONS OF THE SECOND DEGREE. 201 Now, since in the equation Xp is the sum of the roots, and q their product, it follows that q can never be greater than 'p^. The conditions of the equation, therefore, fix a limit to the value of §', and if we make ^>p^, we express by the equation a condition which cannot be fulfilled, and, this contradiction is made apparent by the values of x becoming imaginary. Hence we may conclude that. When the values of the unknown quantity are imaginary, the conditions of the question are incompatible with each other, EXAMPLES. 1. Find two numbers whose sum shall be 12 and pro- duct 46. Let X and y be the numbers. By the 1st condition, £c+y=12 ; and by the 2d, xy=:AQ. The first equation gives x—\2—y. Substituting this value for x in the second, we have 12y— y2 — 46; and changing the signs of the terms, we have /— 12y— .— 46. Quest. — 134. What is the proposition demonstrated in Article 1341 If the conditions of the question are incompatible, how will the values of the unknown quantity be 1 202 FIRST LESSONS IN ALGEBRA. Then by completing the square y2_l2y+36 = -46 + 36= -10 which gives y = 6 + -y/— 10, and y = 6— V — 10; both of which values are imaginary, as indeed they should be, since the conditions are incompatible. 2. The sum of two numbers is 8, and their product 20 : what are the numbers ? Denote the numbers by x and y. By the first condition, x+y=S\ and by the second, xy=:20. The first equation gives a;=8 — y Substituting this value of x in the second, we have 8y-y2=20 ; changing the signs, and completing the square, we have y2-8y+16 = — 4 ; and by extracting the root, y__.4_|_yCr4 and y=:4 — -y/ — 4. These values of y maybe put under the forms (Art. 106), y=4+2y — 1 and y=A:—2^/^^. 3. What are the values of x in the equation a?2+2a?= — 10. Ans. (— l + 3/^i: (a?=: — 1-3V^^1. EQUATIONS OF THE SECOND DEGREE. 203 Examples with more than one unknown quantity. c X 4-2/ — 14 ) 1. Given \ ^ \ .^^ I" to find x and y. By transposing y in the first equation, we have a;==:14 — y ; and by squaring both members, Substituting this value for x^ in the 2nd equation, we have 196-28y+y2+y2_ioO; from which we have y2_l4y=— 48; and by completing the square, y2_14^_[_49_l ; and by extracting the square root, y-7=:d=VT=r+l or -1: hence, y=:7+l=:8, or y=:7 — 1=6. If we take the larger value, we find xz=zQ \ and if we take the smaller, we find a? = 8. Verification. For the largest value, y=8, the equation a:-f-y=:14 gives 6 + 8r=14; and a:2-fy2_ioo gives 36 + 64 = 100. For the value y = 6, the equation a;4-y=14 gives 8 + 6 = 14 ; and a;2+y2=ioo gives 64 + 36 = 100. Hence, both sets of values will satisfy the given equation. 204 FIRST LESSONS IN ALGEBRA. 2. Given { ' \,~~_ > to find x and y. ( a;2— y2__45 ) ^ Transposing y in the first equation, we have a?=3+y ; and then squaring both members, x"z=:z9-{-6i/-\-y'^. Substituting this value for x"^ in the second equation, we have 9 + 6y+y2 — y2__45 . whence we have 6yz=:36 and yr=6. Substituting this value of y in the first equation, we hav^ a?— 6 = 3, and consequently x=3 + 6=z9. Verification. x—y=z3 gives 9 — 6 = 3 ; and ir2— 3/2 — 45 gives 81—36=45. ,. ^. ^ a;24-3^y =22 ) 3. Given { ^ , ^ , ^ n .r^\ to find x and y, I a;2-f-3a:y+ 21/2 — 40 ) ^ Subtracting the first equation from the second, we have 2y2 = 18, which gives y^ = 9, and y=-}-3 or —3. Substituting the plus value in the first equation, we have a:2+9a;=22; EQUATIONS OF THE SECOND DEGREE. 205 from wh^ch we find x—-{-2 and ic— — 11. If we take the negative value, y=r:— 3, we have from the first equation, 0^2—9^—22 ; from which we find iczz: + ll and x=—2. Verification. For the values y=r:+3 and x=:+2, the equation a:2+3a:y=22 gives 22+3x2x3 = 4+18=22: and for the second value, a: =—11, the same equation a:2 + 3a:y=22 gives ( — ll)2 + 3x -11X3 = 121— 99=22. If now we take the second value of y, that is, y— — 3, and the corresponding values of x, viz, a:= + ll, and a:=— 2; for a:= + ll, the equation ir2+3a:y=22 gives 112+3x11 X-3 = 121-99=22 ; and for x=z—2, the same equation 5c2 + 3a:y=22 gives (-2)2+3x -2 X -3=4+18=22. r xz=f (l)x Jiven < X +y +z =7 (2) > to find x,- ( a;2 I ^2_j_^2_2i (3) ) xz=zy'^ (1) 4. Given ^ x +y +z =7 (2) ^ to find x,i/, and a;2+y2_j_^2_2i (3) 18 I 206 FIRST LESSONS IN ALGEBRA. Transposing y in the second equation, we have x+z^l-y (4) ; then squaring the members, we have a:2+2a:;^+;^2_49_i4y_|_y2. If now we substitute for 2xz its value taken from the first equation, we have a;2+2y24-;^23^49„14y^y2 . and cancelKng y^ ij^ each member, there results a:2+y2+;^2_49_i4y. But, from the third equation we see that each member of the last equation is equal to 21 : hence 49-14y=21, and 14y=i49— 21rz:28. 28 ^ hence, y=:— -=:3. ^ 14 Placing this value for y in equation (1) gives xzz=L^ ; and placing it in equation (4) gives x-\-z:=zb, and a:z=5 — 2-. Substituting this value of x in the previous equation, we obtain 52^— >^2_4 Qj. 2:2_52:— _4j and by completing the square, we have ^2_5^_|.6^25=2,5, and £r— 2,5r=±V2^=-{-l>5 or —1,5; hence, ^=2,5 + 1,5=4 or 2^=4-2,5 — 1,5=1. EQUATIONS OF THE SECOND DEGREE. 207 If we take the value z=:4, we find a?=l : if we take the less value z=:}, we find ocz=4. t 3. Given x +^/ocy+y = 19 > ^ r i i ' y y ' ^ > to find X and y. and x^+ x7/-\-7/=:l33 ) ^ Dividing the second equation by the first, we have x—'\/^-\- y= 7 but a:+'\/^+ y=19 hence, by addition, 2x +2y=26 or a? + y=:13 and substituting in 1 st equa. '\/Ty+ 13 = 19 or -y/xy^zz 6 and by squaring a?y=36 From 2d equation, x'^+xy +y'^ = l33 and from the last 3xy =108 Subtracting x'^—2xy-\-y'^z:= 25 hence, x—y:=ziz 5 but a?+y=: 13 hence a?=9 or 4 ; and y=4 or 9. 6. Given the s«um of two numbers equal to a, and the sum of their cubes equal to c, to find the numbers By the conditions < „ ^, 208 FIRST LESSONS IN ALGEBRA. Putting x=:s-{-z, and y—s—z, we have a=2s, or 5=—; \ 7/-s^ — 3s^z+3sz^—z^, hence, by addition, x^-\-y''^=2s^ -\-6sz'^z=zc, whence z^=z — - — and z=dz\/ — ; OS \ OS r^zz2s'3 7 a?=:5±Y/— ^^; and y=s^i^^ c—2s^ 6s or by putting for s its vakie, a3 '=-2-V V-37-)=T-V-l27-' and y==-J=Fy/(i^)=.|-^y/i^ 2a ' Note. — What are the numbers when a — 5 and c=r35. What are the numbers when az=i9 and c = 243. QUESTIONS. 1. Find a number such, that twice its square, added to three times the number, shall give 65. Let X denote the unknown number. Then the equation of the problem will be 2x'^+3x.=z65 , whence EQUATIONS OF THE SECOND DEGREE. 209 Therefore, 3 , 23 ^ ^ 3 23 13 Both these values satisfy the question in its algebraic sense. For, 2x(5)2+3x5=2x25 + 15r=:65 ; 13\2 ^ 13 169 39 130 and / 13V . o 13 169 39 13U ^^ Remark. — If we wish to restrict the enunciation to its arithmetical sense, we will first observe, that when x is replaced by —x, in the equation 2x'^-\-3x=z65j the sign of the second term 3x only, is changed, because (—xY=x^. 3 23 Therefore, instead of obtaining a:= — — ±— , we should find xz= — dt — , or xz= — , and a: =—5, values which only 4 4 2 differ from the preceding by their signs. Hence, we may 13 say tliat the negative solution , considered indepen- dently of its sign, satisfies this new enunciation, viz : To find a numher such, that twice its square, diminished hy three times the numher, shall give 65. In fact, we have « /13\2 ^ 13 169 39 ^ Remark. — The root which results from giving the plus sign to the radical, generally resolves the question both in its arithmetical and algebraic sense, while the second root resolves it in its algebraic sense only. 18* 210 FIRST LESSONS IN ALGEBRA. Thus, in the example, it was required to find a number, of which twice the square added to three times the mmiber shall give Qb. Now, in the arithmetical sense, added means increased ; but in the algebraic sense it implies diminution, when the quantity added is negative. In this sense, the second root satisfies the enunciation. 2. k certain person purchased a number of yards of cloth for 240 cents. If he had received 3 yards less of the same cloth for the same sum, it would have cost him 4 cents more per yard. How many yards did he purchase ? Let 0?= the number of yards purchased. Then will express the price per yard. If, for 240 cents, he had received 3 yards less, that is a?— 3 yards, the price per yard, under this hypothesis, would this last cost would exceed the first by 4 cents. Therefore, we have the equation 240 240 , =4; X — 6 X whence, by reducing a?^ — 3a; =180, 3 /W ,^^ 3=t27 2 ' therefore x=l6 and x=: — \2. The value a:=15 satisfies the enunciation; for, 15 yards 24-0 for 240 cents gives , or 16 cents for the price of 1 o one yard, and 12 yards for 240 cents, gives 20 cents forth© price of one yard, which exceeds 16 by 4. EQUATIONS OF THE SECOND DEGREE. 211 As to the second solution, we can form a new enuncia- tion, with which it will agree. For, going back to the equation, and changing x into —a?, it becomes 240 240 , 240 240 , -=4, or r^=4, — X — 3 — X X jr+3 an equation which may be considered the algebraic transla- tion of this problem, viz : A certain person purchased a num- ber of yards of cloth for 240 cents : if he had paid the same sum for 3 yards more, it would have cost him 4 cents less per yard. How many yards did he purchase ? Ans. a; = 12, and a?= — 15. 3. A man bought a horse, which he sold after some time for 24 dollars. At this sale, he loses as much per cent, upon the price of his purchase as the horse cost him. What did he pay for the horse ? Let X denote the number of dollars that he paid for the horse, a;— 24 will express the loss he sustained. But as he lost X per cent, by the sale, he must have lost ^ -^ ' 100 upon each dollar, and upon x dollars he loses a sum de- noted by -jTTfr » ^'^ h^B.YQ then the equation z=zx—2A, whence a?2 — 100^?=:— 2400. 100 and a?=50zbv^2500— 2400 = 50dzl0. Therefore, a? =60 and a:=:40. Both of these values satisfy the question. For, in the first place, suppose the man gave %Q0 for the horse and sold him for 24, he loses 36. Again, from the enunciation, he should lose 60 per cent, of 60, that is, 212 FIRST LESSONS IN ALGEBRA* of 60, or — , which reduces to 36 ; there- 100 ' 100 fore, 60 satisfies the enunciation. Had he paid $40, he would have lost $16 by the sale ; 40 for, he should lose 40 per cent, of 40, or 40 x — -f-^, which reduces to 16 ; therefore, 40 verifies the enunciation. 4. A man being asked his age, said the square root of my own age is half the age of my son, and the sum of our ages is 80 years : what was the age of each 1 Let a?= the age of the father. y =r that of the son. Then by the first condition and by the second condition a:+y=80. If we take the first equation and square both members, we have If we transpose y in the second, we have a:=80— y: from which we find y=:— 2zfc'v/324'=:16; by taking the plus root, which answers to the question in its arithmetical sense. Substituting this value, we find a? =64. ^ ^ Father's age 64 ( Son's 16. ^' EQUATIONS OF THE SECOND DEGJIEE. 213 5. Find two numbers, such that the sum of their pro- ducts by the respective numbers a and J, may be equal to Si", and that their product may be equal to p. Let X and y be the required numbers, we have the equa- tions ax-\-hyz=.%s, and xyz=i'p, 2s — ax From the first y= — ; whence, by substituting in the second, and reducing, ax? — 2sx =z — bp. 1 Therefore, x— — ± — a/ s^—abp, a a and consequently, s 1 y=—::pj-^S^-abp. This problem is susceptible of two direct solutions, be- cause s is evidently > ^s^—abp ; but in order that they may be real, it is necessary that s^^ or =zabp. Let a = b—i ; the values of x and y reduce to x=:sdz^s^ — p and y — ^^p-y/^^ — p. Whence we see, that the two values of x are equal to those of y, taken in an inverse order ; which shows, that if s-{- ^/ s^ — p represents the value of a?, s — aJs^ — p will re- present the corresponding value of y, and reciprocally. This circumstance is accounted for, by observing that in this particular case the equations reduce to \ xy—p ; S 214 FIRST LESSONS IN ALGEBRA. and then the question is reduced to, finding two numbers of lohich the sum is 2s, and their product p, or in other words, to divide a number 2s, into two such parts, that theit product may be equal to a given number p. Let us now suppose 2sz=l4: and jt> — 48: what will then be the values of x and y ? Ans. \ ^ :8 or 6. ' y=6 or 8. 6. A grazier bought as many sheep as cost him £Q0, and after reserving fifteen out of the number, he sold thtj re- mainder for jC54, and gained 2s. a head on those he sold : how many did he buy I Ans. 75. 7. A merchant bought cloth for which he paid .£"33 156\, which he sold again at £2 Qs. per piece, and gained by the bargain as much as one piece cost him : how many pieces did he buy ? Ans. 15. 8. What number is that, which, being divided by the pro- duct of its digits, the quotient is 3 ; and if 18 be added to it, the digits will be inverted ? Ans. 24. 9. To find a number, such that if you subtract it from 1 0, and multiply the remainder by the nmnber itself, the product shall be 21. Ans. 7 or 3. 10. Two persons, A and B, departed from different places at the same time, and travelled towards each other. On meeting, it appeared that A had travelled 18 miles more than B ; and that A could have gone B's journey in 15|^ days, but B would have been 28 days in performing A's journey. How far did each travel ? ^ ( A 72 miles, Ans. \ B 54 milea. EQUATIONS OF THE SECOND DEGREE. 215 11. There are two numbers whose difference is 15, and half their product is equal to the cube of the lesser num- ber. What are those numbers ? Ans. 3 and 18. 12. What two numbers are those whose sum, multiplied by the greater, is equal to 77 ; and whose difference, multi- plied by the lesser, is equal to 12 ? Ans. 4 and 7, or | -y/T and y ^/^, 13. To divide 100 into two such parts, that the sum of their square roots may be 14. Ans. 64 and 36. 14. It is required to divide the number 24 into two such parts, that their product may be equal to 35 times their dif- ference. Ans, 10 and 14. 15. The sum of two numbers is 8, and the sum of their cubes 152. What are the numbers ? Ans, 3 and 5. 16. Two merchants each sold the same kind of stuff; the second sold 3 yards more of it than the first, and to- gether they receivne 35 dollars. The first said to the second, *' 1 would have received 24 dollars for your stuff;" the other replied, " And I should have received 12^ dollars for yours." How many yards did each of them sell ? . C 1st merchant a? =15 oc=b. Ans. I , or ;i2nd „ y=\Q "^ y=8. 17. A widow possessed 13,000 dollars, which she divided into two parts, and placed them at interest, in such a man- ner, that the incomes from them were equal. If she had put out the first portion at the same rate as the second, she would have drawn for this part 360 dollars interest ; and if she had placed the second out at the same rate as the first, she would have drawn for it 490 dollars interest. What were the two rates of interest ? Ans. 7 and 6 per cent. 216 FIRST LESSONS IN ALGEBRA. CHAPTER VII. Of Proportions and Progressions. 135. Two quantities of the same kind may be compared together in two ways : — 1st. By considering how much one is greater or less than the other, which is shown by their difference ; and, 2nd. B;y considering how many times one is greater or less than the other, which is shown by their quotient. Thus, in comparing the numbers 3 and 12 together with respect to their difference, we find that ] 2 exceeds 3 by 9 ; and in comparing them together with respect to their quo- tient, we find that 12 contains 3 four times, or that 12 is 4 times as great as 3. The first of these methods of comparison is called Arith- metical Proportion, and the second Geometrical Proportion. Hence, Arithmetical Proportion considers the relation of quantities with respect to their difference, and Geometrical Proportion the relation of quantities with respect to their quotient. Quest. — 135. In how many ways may two quantities be compared together 1 What does the first method consider 1 What the second 1 What is the first of these methods called 1 What is the second called ? How then do you define the two proportions ^ ARITHMETICAL PROPORTION. 217 Of Arithmetical Proportion and Progression, 136. If we have four numbers, 2, 4, 8, and 10, of which the difference between the first and second is equal to the difference between the third and fourth, these num- bers are said to be in arithmetical proportion. The first term 2 is called an antecedent^ and the second term 4, with which it is compared, a consequent. The number 8 is also called an antecedent, and the number 10, with which it is compared, a consequent. When the difference between the first and second is equal to the difference between the third and fourth, the four num- bers are said to be in proportion. Thus, the numbers 2, 4, 8, 10, are in arithmetical proportion. 137. When the difference between the first antecedent and consequent is the same as between any two adjacent terms of the proportion, the proportion is called an arith- metical progression. Hence, a progression by differences, or an arithmetical progression, is a series in which the succes- sive terms continually increase or decrease by a constant number, which is called the common difference of the progression. Thus, in the two series 1, 4, 7, 10, 13, 16, 19, 22, 25, . . . 60, 56, 52, 48, 44, 40, 36, 32, 28, . . . Quest. — 136. When are four numbers in arithmetical proportion 1 What is the first called 1 What is the second called 1 What is th© third called 1 What is the fourth called "? 19 218 FIRST LESSONS IN ALGEBRA. the first is called an increasing progression, of which, the common difference is 3, and the second a decreasing pro- gression, of which the common difference is 4. In general, let a, b, c, d, e,f, . . . designate the terms of a progression by differences ; it has been agreed to write them thus : a,h.c.d,e.f.g.h.i.h... This series is read, a is to h, as h is to c, as c is to d, as d is to e, Slc. This is a series of continued equi-differences, in which each term is at the same time a consequent and antecedent, with the exception of the first term, which is only an antecedent, and the last, which is only a consequent. 138. Let T represent the common difference of the progression a * h . c , d . e . f . g . h, &c, which we will consider increasing. From the definition of the progression, it evidently follows that hz:za-\-r, c=:5 + rr=a+2r, d—c-\-rz=:a+Zr', and, in general, any term of the series is equal to the first term plus as many times the common difference as there are preceding terms. Thus, let I be any term, and n the number which marks the place of it : the expression for this general term is l=a-{-{n — \)r. Quest. — 137. What is an arithmetical progression 1 What is the number called by which the terms are increased or diminished 1 What is an increasing progression 'I What is a decreasing progression '^ Which term is only an antecedent % "WTiich only a consequent 1 f ARITHMETICAL PROGRESSION. 219 Hence, for finding the last term, we have the following RULE. I. Multiply the common difference by one less than the numher of terms. II. To the product add the first term: the sum will he the .last term. EXAMPLES. The formula Z~a+(» — 1)?' serves to find any term whatever, without our being obliged to determine all those which precede it. 1. If we make n=i, we have /=«; that is, the series will have but one term. 2. If we make ni=2, we have l^^a-\-r ; that is, the series will have two terms, and the second term is equal to the first plus the common diflference. 3. If a — 2 and r—2, what is the 3rd term? Ans, 7. 4. If «=5 and r—A, what is the 6th term? Ans. 25. 5. If az=7 and r=:5, what is the 9th term? Ans. 47. 6. If a=z8 and r=5, what is the 10th term ? Ans. 53. 7. If a— 20 and r—^, what is the 12th term? Ans. 64. 8. If a—AO and r=z20, what is the 50th term ? Ans. 1020. Quest. — 138. Give the rule for finding the last term of a series when the progression is increasing. 22© FIRST LESSONS IN ALGEBRA. 9. If a=45 and r=30, what is the 40th tenii ? Ans. 1215. la. If a=z30 and r=20, what is the 60th term? Ans. 1210. 11. If a = 50 and rz=:10, what is the 100th term? Ans. 1040. 12. To find the 50th term of the progression 1 . 4 . 7 . 10 . 13 . 16 . 19 . . ., we have /r=l 4-49x3 = 148. 13. To find the 60th term of the progression 1 . 5 . 9 . 13 . 17 . 21 . 25 . . ., we have Z= 1+59x4=237. \ 139. If the progression were a decreasing one, we should have l=a — {n — l)r. Hence, to find the last term of a decreasing progression^ we have the following RULE. I. Multiply the common difference hy one less than the num-- her of terms. II. Subtract the product from the first term : the remainder will be the last term. Quest. — 139. Give the rule for finding the last term of a series* when the progression is decreasing. ARITHMETICAL PROGRESSION. 221 EXAMPLES. 1 . The first term of a decreasing progression is 60, the number of terms 20, and the common difference 3 : what is tne last term ? l~a — {n — l)r gives Zr:::60— (20 — 1)3 = 60 — 57=3. 2. The first term is 90, the common difference 4, and the number of terms 15 : what is the last term ? Ans, 34. 3. The first term is 100, the number of terms 40, and the common difference 2 : what is the last term ? Ans. 22. 4. The first term is 80, the number of terms 10, and the common difference 4 : what is the last term ? Ans, 44. 5. The first term is 600, the number of terms 100, and the common difference 5 : what is the last term ? Ans. 105. 6. The last term is 8^00, the number of terms 200, and the common difference 2 : what is the last term ? Ans. 402. 140. A progression by differences being given, it is proposed to prove that, the sum of any two terms, taken at equal distances from the two extremes, is equal to the sum of the two extremes. That is, if we have the progression 2 . 4 . 6 . 8 . 10 . 12, we wish to prove that 4+10 or 64-8 is equal to the sum of the two extremes 2 and 12. 19* 222 FIRST LESSONS IN ALGEBRA. Let a,h,c.d,e.f,...i.k.l be the pro- posed progression, and n the number of terms. We will first observe that, if x denotes a term which has p terms before it, and y a term which has p terms after it, we have, from what has been said, x=a-\-pXr, and 2/=l—px T ; whence, by addition, x-\-yz=.a-\-l. Which demonstrates the proposition. Referring this proof to the previous example, if we sup- pose, in the first place, x to denote the second term 4, then y will denote the term 10, next to the last. If x denotes the 3rd term 6, then y will denote 8, the third term from the last. Having proved the first part of the proposition, write the progression below itself, but in an inverse order, viz : a,h.c.a.e.f.,.i.li.l. I , k . i c . b . a. Calling S the sum of the terms of the first progression, 2S will be the sum of the terms in both progressions, and we shall have 2S=:(a+l) + {b+k) + {c+i) . . . -^{i-^c) + {k-\-b)+{l+a). Now, since all the parts a+I, b-{-k, c+i . . . are equal to each other, and their number equal to n, 2S=(a+l)n, or S=(^)n. ARITHMETICAL PROGRESSION. 223 Hence, for finding the sum of an arithmetical series, we have the following RULE. I. Add the two extremes together, and take half their sum. II. Multiply the half-sum hy the number of terms ; the product will he the sum of the series. EXAMPLES. 1. The extremes are 2 and 16, and the number of terms 8 : what is the sum of the series ? S-- =:(-^jXn, gives Srr:— ^ x8 = 72. 2. The extremes are 3 and 27, and the number of terms 12 : what is the sum of the series 1 Ans. 180. 3. The extremes are 4 and 20, and the number of terms 10 : what is the sum of the series ? Ans. 120. 4. The extremes are 100 and 200, and the number of terms 80 : what is the sum of the series ? Ans. 12000. 5. The extremes are 500 and 60, and the number of terms 20 : what is the sum of the series ? Ans. 5600. 6. The extremes are 800 and 1200, and the number of terms 50 : what is the sum of the series ? Ans. 50000. Quest. — 140. In every progression, what is the sum of the two ex- tremes equal to 1 What is the rule for finding the sum of an arithmeti cal series 1 224 FIRST LESSONS IN ALGEBRA. > '^ 1 4 1 . In arithmetical proportion there are five numbers to be considered : — 1st. The first ^rm, a. 2nd. The common difference, r. 3rd. The number of terms, n. 4th. The last term, Z. 5th. The sum, S. The formulas l—a-{-[n — \)r and Sz=i( )xn contain five quantities, a, r, w, Z, and S, and consequently give rise to the following general problem, viz : Any three of these five quantities being given, to determine the other two. We already know the value of S in terms of a, n, and r. From the formula l=:a-\-[n—\)r, we find a—l—{n—\)r. That is : The first term of an increasing arithmetical pro- gression is equal to the last term, minus the product of the common difference hy the number of terms less one. From the same formula, we also find I — a That is : In any arithmetical progression, the common differ- ence is equal to the difference between the two extremes divided by the number of terms less one. Quest. — 141. How many numbers are considered in arithmetical proportion 1 What are they 1 In every arithmetical progression, what is the common diiference equal to 1 ARITHMETICAL PROGRESSION. 225 The last term is 16, the first term 4, and the number of terms 5 : what is the common difference ? The formula rz=z n—\ 16—4 ^ gives rz± —3. 2. The last term is 22, the first term 4, and the number of terms 10 : what is the common difierence ? Ans. 2. 142. The last principle affords a solution to the follow- ing question : To find a number m of arithmetical means between two given numbers a and b. To resolve this question, it is first necessary to find the common difference. Now, we may regard a as the first term of an arithmetical progression, h as the last term, and the required means as intermediate terms. The number of terms of this progression will be expressed by m4-2. Now, by substituting in the above formula, h for Z, and w-}-2 for n^ it becomes h — a h — a m-}-2 — 1 m-\-\ that is, the common difference of the required progression is obtained by dividing the difference between the given num- bers a and 5, by one more than the required number of means. Quest. — 142. How do you find any number of arithmetical means between two given numbers 1 226 FIRST LESSONS IN ALGEBRA. Having obtained the common difference, form the second term of the progression, or the first arithmetical mean^ by adding r. or , to the first term a. The second mean "^ ' m-\-\ is obtained by augmenting the first by r, &c. 1. Find three arithmetical means between the extremes 2 and 18. The formula m+1 18-2 , gives r= — - — =4 ; hence, the progression is 2 . 6 . 10 . 14 . 18. 2. Find twelve arithmetical means between 12 and 77. h-a The formula m-\-l 77-12 ^ gives r:=z — — — —5, Hence the progression is 12 . 17 . 22 . 27 .... 77 143. Remark. If the same number of arithmetical means are inserted between all the terms, taken two and two, these terms, and the arithmetical means united, will form but one and the same progression. For, let a.h.c,d.e.f... be the proposed progression, and m the number of means to be inserted between a and h, h and c, c and d . , , ARITHMETICAL PROGRESSION. 227 From what has just been said, the common difference of each partial progression will be expressed by h — a c — h d—e m+l ' m+1 ' m+1 which are equal to each other, since a, b, c . . . are in progression : therefore, the common difference is the same in each of the partial progressions ; and since the last term of the first, forms the Jlrst term of the second, &c, we may conclude that all of these partial progressions form a single progression. EXAMPLES. 1 . Find the sum of the first fifty terms of the progression 2 . 9 . 16 . 23 . . . For the 50th term we have /z=2 + 49x7.~345. 50 Hence, ^ = (2 + 345) x— =347x25 = 8675. 2. Find the 100th term of the series 2 . 9 . 16 . 23 . . . Ans. 695. 3. P^ind the sum of 100 terms of the series 1.3.5. 7.9 ... Ans. 10000. 4. The greatest term is 70, the common difference 3, and the number of terms 21 : what is the least term and the sum of the series ? Ans. Least term 10 ; sum of series 840. 228 FIRST LESSONS IN ALGEBRA. 5. The first term is 4, the common difference 8, and the number of terms 8 : what is the last term, and the smn of the series ? . C Last term 60. Ans. < _ < Sum irz 256. 6. The first term is 2, the last term 20, and the number of terms 10 : what is the common difference ? A71S. 2. 7. Insert four means between the two numbers 4 and 19 : what is the series 1 Ans. 4 . 7 . 10 : 13 . 16 . 19. 8. The first term of a decreasing arithmetical progres- sion is 10, the common difference — , and the number of o terms 21 : required the sum of the series. Afis. 140. 9. In a progression by differences, having given the common difference 6, the last term 185, and the sum of the terms 2945 ; find the first term, and the number of terms. Ans. First term ==5 ; number of terms 31. 10. Find nine arithmetical means between each antece- dent and consequent of the progression 2.5.8.11.14 ... Ans. Common dif, or r=:0,3. 1 1 . Find the number of men contained in a triangular bat- talion, the first rank containing one man, the second 2, the third 3, and so on to the n^, which contains n. In other words, find the expression for the sum of the natural num- bers 1, 2, 3 . . ., from I to n inclusively. Ans. «=^i). GEOMETRICAL PROPORTION. 229 9. Find the sum of the n first terms of the progression of uneven numbers 1, 3, 5, 7, 9 . . . Arts* S = /i2. 10. One hundred stones being placed on the ground in a straight line, at the distance of 2 yards from each other, how far will a person travel who shall bring them one by one to a basket, placed at 2 yards from the first stone ? Ans. 11 miles, 840 yards. Geometrical Proportion and Progression. 144. Ratio is the quotient arising from dividing one quantity by another quantity of the same kind. Thus, if the numbers 3 and 6 have the same unit, the ratio of 3 to 6 will be expressed by And in general, if A and B represent quantities of the same kind, the ratio of A to i? will be expressed by A' 145. If there be four numbers 2, 4, 8, 16, having such values that the second divided by the first is equal to the fourth divided by the third, the numbers are QUEST.—144. What is ratio ] What is the ratio of 3 to 6 1 Of 4 to 121 20 230 FIRST LESSONS IN ALGEBRA. said to be in proportion. And in general, if there be four quantities, A, B, C, and D, having such values that A- C then A is said to have the same ratio to B that C has to D ; or, the ratio of A to jB is equal to the ratio of C to D, When four quantities have this relation to each other, they are said to be in proportion. H-ence, proportion is an equality of ratios. To express that the ratio of ^ to 5 is equal to the ratio of C to Z), we write the quantities thus ; A : B : : C : D] and read, A is to 5 as C to D. The quantities which are compared together are called the terms of the proportion. The first and last terms are called the two extremes, and the second and third terms, the two means. Thus, A and D are the extremes, and B and C the means. 146. Of four proportional quantities, the first and third are called the antecedents, and the second and fourth the consequents ; and the last is said to be a fourth proportional to the other three taken in order. Thus, in the last pro- portion A and C are the antecedents, and B and D the consequents. Quest. — 145. What is proportion 1 How do you express that four numbers are in proportion 1 What are the numbers called 1 What are the first and fourth called'? What the second and third] — 146. In four proportional quantities, what are the first and third called ] What the second and fourth 1 GEOMETRICAL PROPORTION. 231 1 47 . Three quantities are in proportion when the first has the same ratio to the second that the second has to the third ; and then the middle term is said to be a mean pro- portional between the other two. For example, 3 : 6 : : 6 : 12; and 6 is a mean proportional between 3 and 12. 148. Quantities are said to be in proportion by muer- sion, or inversely, when the consequents are made the ante- cedents and the antecedents the consequents. Thus, if we have the proportion 3 : 6 : : 8 : 16, the inverse proportion would be 6 : 3 : : 16 : 8. 1 49. Quantities are said to be in proportion by alterna- tion, or alternately, when antecedent is compared with ante- cedent and consequent with consequent. Thus, if we have the proportion 3 : 6 : : 8 : 16, the alternate proportion would be 3 : 8 : : 6 : 16. Quest. — 147. When are three quantities proportionaU What is the middle one called 1 — 148. When are quantities said to be in proportion by inversion, or inversely \ — 149. When are quantities in proportion by aHemation 1 232 FIRST LESSONS IN ALGEBRA. 150. Quantities are said to be in proportion by compo- sition, when the sum of the antecedent and consequent is compared either with antecedent or consequent. Thus,, if we have the proportion 2 : 4 : : 8 ; 16, the proportion by composition would be 2 + 4 : 4 : : 8+16 : 16; that is, 6:4:: 24 : 16. 151. Quantities are said to be in proportion by division, when the difference of the antecedent and consequent is compared either with antecedent or consequent. Thus, if we have the proportion 3 : 9 : : 12 : 36, the proportion by division will be 9-3 : 9 : : 36-12 : 36; that is, 6 : 9 : : 24 : 36. 152. Equi-multiples of two or more quantities are the products which arise from multiplying the quantities by the same number. Thus, if we have any two numbers, as 6 and 5, and mul- tiply them both by any number, as 9, the equi-multiples will be 54 and 45 ; for 6x9=54, and 5x9=45. Quest. — 150. When are quantities in proportion by composition'? — 151. When are quantities in proportion by division? — 153. What are equi-multiples of two or more quantities 1 GEOMETRICAL PROPORTION. 233 Also, mxA and mx B are equi-multiples of A and D, the common multiplier being m. 153. Two quantities, A and B, are said to be recipro- colly proportional, or inversely proportional, when one in- creases in the same ratio as the other diminishes. When this relation exists, either of them is equal to a constant quantity divided by the other. Thus, if we had any two numbers, as 2 and 4, so related to each other that if we divided one by any number we must multiply the other by the same number, one would increase just as fast as the other would diminish, and their product would be constant. 154. If we have the proportion A : B : : C : D, we have —=z—, (Art. 145); and by clearing the equation of fractions, we have BC^AD, That is. Of four proportional quantities, the product of the two extremes is equal to the product of the two means. This general principle is apparent in the proportion be- tween the numbers 2 : 10 : : 12 : 60, which gives 2 x 60zizl0x 12 = 120. Quest. — 153. When are two quantities said to be reciprocally pro- portional 1 — 154. If four quantities are proportional, what is the product of the two means equal to 1 20* 234 FIRST LESSONS IN ALGEBRA. 155. If four quantities, A, B, C, D, are so related to each other that AxD=:BxC, we shall also have -—=:•—-; A G and hence, A : B : : C : D. That is ; If the product of two quantities is equal to the prO' duct of two other quantities, two of them may he made the extremes, and the other two the means of a proportion. Thus, if we have 2x8r=4x4, we also have 2 : 4 : : 4 : 8. 1 56. If we have three proportional quantities A : B : : B : C, we have — rr=^t- ; A B hence, B^=:AC. That is : The square of the middle term is equal to the product of the two extremes. Thus, if we have the proportion 3 : 6 : : 6 : 12, we shall also have 6x6=z62=:3xl2 = 36. Quest. — 155. If the product of two quantities is equal to the product of two other quantities, may the four be placed in a proportion 1 How 1 — 156. If three quantities are proportional, what is the product of the extremes equal to 1 GEOMETRICAL PROPORTION. 2B5 157. If we liave A : B : : C : D, and consequently -r-r=-T7, C multiply both members of the last equation by -^, we then obtain, A~B' and hence, A : C : : B : D. That is : If four quantities are proportional, they will he in proportion by alternation. Let us take, as an example, 10 : 15 : : 20 : 30. We shall have, by alternating the terms, 10 : 20 : : 15 : 30. 158. If we have A : B :: C : D and A : B : : E : F, we shall also have B D :, B F B F hence, —^ =-^= and C : D : : E : F. L> XL That is : If there are two sets of proportions having an Quest. — 157. If four quantities are proportional, will they be in pro- portion by alternation 1 236 FIRST LESSONS IN ALGEBRA. antecedent and consequent in the one equal to an antecedent and consequent of the other, the remaining terms will he pro- portional. If we have the two proportions 2 : 6 : : 8 : 2*4 and 2 : 6 : : 10 : 30, we shall also have 8 : 24 : : 10 : 30. 150. If we have B D A : B : : C : D, and consequently —=:—-, we have, by dividing 1 by each member of the equation, A C -—=-—, and consequently B : A : : D : C. That is : Four proportional quantities will be in proportion^ lohen taken inversely. To give an example in numbers, take the proportion 7 : 14 : : 8 : 16 ; then, the inverse proportion will be 14 : 7 : : 16 : 8, in which the ratio is one-half. 160. The proportion A : B :: C : D gives AxB=BxC. Quest. — 158. If you have two sets of proportions having an ante- cedent and consequent in each, equal; what will follow 1 — 159. If four quantities are in proportion, will they be in proportion when taken in- versely 1 GEOMETRICAL PROPORTION. 237 To each member of the last equation add BxD. We shall tlien have {A+B)xD={C+D)xB; and by separating the factors, we obtain A+B : B : : C+D : D. If, instead of adding, we subtract BxD from both mem- bers, we have {A-B)xD={C^D)xB; which gives A-^B : B : : C-D : D, That is : If four quantities are proportional, they will he in proportion hy composition or division. Thus, if we have the proportion 9 : 27 : : 16 : 48, we shall have, by composition, 9+27 : 27 : : 16 + 48 : 48; that is, 36 : 27 : : 64 : 48, in which the ratio is three-fourths. The proportion gives us, by division, 27-9 : 27 : : 48-16 : 48; that is, 18 : 27 : : 32 : 48, in which the ratio is one and one -half. Quest. — 160. If four quantities are in proportion, will they be in pro- portion by composition 1 Will they be in proportion by division 1 What is the difference between composition and division 1 288 FIRST LESSONS IN ALGEBRA. ^ 1 6 1 . If we have B__D_ and multiply the numerator and denominator of the first member by any number m, we obtain -=— - and mA : mB : : C : D. mA That is : Equal multiples of two quantities have the same ratio as the quantities themselves. For example, if we have the proportion 5 : 10 : : 12 : 24, and multiply the first antecedent and consequent by 6, we have 30 : 60 : : 12 : 24, in which the ratio is still 2. 162. The proportions A : B :: C : D and A : B \ : E : F, give AxD:=iBxC and AxF=iBxE\ adding and subtracting these equations, we obtain A(P±F):=zB{C±E), or A . B : : C±E : D±.F. That is : If and D, the antecedent and consequent, he aug- mented or diminished hy quantities E and F, which have the same ratio as C to D, the resulting quantities will also have the same ratio. Quest. — 161. Have equal multiples of two quantities the same ratio as the quantities'! — 162. Suppose the antecedent and consequent be augmented or diminished by quantities having the same ratio '^ GEOMETRICAL PROPORTION. 239 Let us take, as an example, the proportion 9 : 18 : : 20 : 40, in which the ratio is 2. If we augment the antecedent and consequent by 15 and 30, which have the same ratio, we shall have 9+15 : 18 + 30 : : 20 : 40 ; that is, 24 : 48 : : 20 : 40, in which the ratio is still 2. If we diminish the second antecedent and consequent by the same numbers, we have 9 : 18 : : 20-15 : 40-30; that is, 9 : 18 : : 5 : 10, in which the ratio is still 2. 163. If we have several proportions A : B : : C : D, which gives AxD = BxC, A : B : : E : F, „ „ AxF=:BxE, A : B : : G : H, „ „ AxH=BxG. &c, &:c, we shall have, by addition, A{D-]-F-]-H)=B{C-hE-hG); and by separating the factors, A : B : : C-\-E-j-G : D+F+H. That is : In any number of 'proportions having the same ratio^ any antecedent will he to its consequent, as the sum of the antecedents to the sum of the consequents. 240 FIRST LESSONS IN ALGEBRA. Let us take, for example, 2 : 4 : : 6 : 12 and 1 : 2 : : 3 : 6, SlQ Then, 2:4:: 6 + 3 : 12 + 6; that is, 2 : 4 : : 9 : 18, in which the ratio is still 2. 164. If we have four proportional quantities A : B : : C : D, we have —=-— ; A. G and raising both members to any power, as n, we have and consequently A" : 5" : : C" : D\ That is : If four quantities are proportional , any like powers or roots will he proportional. If we have, for example, 2 : 4 : : 3 : 6, we shall have 2^ : 42 : : 3^ : 6^ ; that is, 4 : 16 : : 9 : 36, in which the terms are proportional, the ratio being 4. 165. Let there be two sets of proportions, A : B : : C : D, which gives -—=::-—-, A O F H E : F : : G : H, „ „ ~e~'g' Quest. — 163. In any number of proportions having the same ratio, how will any one antecedent be to its consequent 1 — 164. In four pro- portional quantities, how are like powers or roots 1 GEOMETRICAL PROGRESSION. 241 Multiply them together, member by member, we have -4^=-^ which gives AE : BF i: CG : BH. That is : In two sets of proportional quantities, the products of the corresponding terms will he proportional. Thus, if we have the two proportions 8 16 : : 10 : 20 ^nd 3 : 4 : : 6 : 8 we shall have 24 , 64 : : 60 : 160 Geometrical Progression, 166. We have thus far only required that the ratio of the first term to the second should be the same as that of the third to the fourth. If we impose the farther condition, that the ratio of the second term to the third shall also be the same as that of the first to the second, or of the third to the fourth, we shall have a series of numbers, each one of which, divided by the preceding one, will give the same ratio. Hence, if any term be multiplied by this quotient, the product will be the succeeding term. A series of numbers so formed is called a geometrical progression. Hence, A Geometrical Progression, or progression hy quotients, is a series of terms, each of which is equal to the product of Quest. — 165. In two sets of proportions, how are the products of the corresponding terms 1 21 242 FIRST LESSONS IN ALGBBRA. that which precedes it by a constant number, which number is called the ratio of the progression. Thus, 1 : 3 : 9 : 27 : 81 : 243, (fee, is a geometrical progression, which is written by merely placing two dots between each two of the terms. Also, 64 : 32 : 16 : 8 : 4 : 2 : 1 is a geometrical progression, in which the ratio is one-half. In the first progression each term is contained three times in the one that follows, and hence the ratio is 3. In the second, each term is contained one-half times in the one which follows, and hence the ratio is one-half. The first is called an increasing progression, and the second a decreasing progression. Let «, h, c, d, e, f, . . . be numbers in a progression by quotients ; they are written thus : a X h I c I d ', e I f I g . . . and it is enunciated in the same manner as a progression by differences. It is necessary, however, to make the distinc- tion, that one is a series of equal differences, and the other a series of equal quotients or ratios. It should be remarked that each term is at the same time an antecedent and a con- sequent, except the first, which is only an antecedent, and the last, which is only a consequent. Quest. — 166. What is a geometrical progression'? What is the ratio of the progression 1 If any term of a progression be multiphed by the ratio, what will the product be 1 If any term be divided by the ratio, what will the quotient be 1 How is a progression by quotients written 1 Which of the terms is only an antecedent 1 Which only a consequent ^ How may each of the others be considered 1 GEOMETRICAL PROGRESSION. 24S 167. Let q denote the ratio of the progression a \ h \ c \ d , . .\ q being >1 when the progression is increasing, and 5'<1 when it is decreasing. Then, since b c d e ^ T=^' T=?' T=^' T=^' '^^' we have hz=^aq, c=zhqz=zaq^, d=icq=zaq^, e=^dq^:zaq^, ' f—eq — aq^ . . .; that is, the second term is equal to aq, the third to aq^, the fourth to acf', the fifth to aq^, &c ; and in general, any term w, that is, one which has n—1 terms before it, is expressed by fl5^"-i. Let I be this term ; we then have the formula by means of which we can obtain any term without being obliged to find all the terms which precede it. Hence, to find the last term of a progression, we have the following RULii L Raise the ratio to a power whose exponent is one less titan the number of terms. II. Multiply the power thus found by the first term : the product will be the required term. Quest. — 167. By what letter do we denote the ratio of the progres- sion 1 In an increasing progression is q greater or less than 11 In a decreasing progression is q greater or less than 11 If a is the first term and q the ratio, what is the second term equal to 1 What the third 1 What the fourth 1 What is the last term equal to 1 Give the rule for finding the last term. 244 FIRST LESSONS IN ALGEBRA. EXAMPLES. 1 . Find the 5th term of the progression 2 : 4 : 8 : 16 . . in which the first term is 2 and the common ratio 2. 5th term=2 X2^=2 X 16 = 32 Atis, 2. Find the 8th term of the progression 2 : 6 : 18 : -54 . . . 8th term=2 X 3*^=2 X 2187=4374 Ans. 3. Find the 6th term of the progression 2 : 8 : 32 : 128 . . . 6th term =2 x 4^=2 X 2048=4096 Aris. 4. Find the 7th term of the progression 3 : 9 : 27 : 81 . . . 7th term=3 x 3^=3 X 729=2187 Ans. 5. Find the 6th term of the progression 4 : 12 : 36 : 108 . . , 6th term =4 x 3^=4 X 243 = 972 A7is. 6. A person agreed to pay his servant 1 cent for the first day, two for the second, and four for the third, doubling every day for ten days : how much did he receive on the tenth day? Ans. $5,12 GEOMETRICAL PROGRESSION. 245 1. What is the 8th term of the progression 9 : 36 : 144 : 576 . . . 8th term=r9x4^ = 9x 16384 = 147456 Ans. 8. Find the 12th term of the progression 64 : 16 : 4 : 1 : 4- • • • 4 / 1 \ii 43 1 1 12th term=:64( — ) z=z — — = — Ans. \4J 411 48 65536 1G8. We will now proceed to determine the sum of n terms of the progression a : b : c : d : e : f : . . . : i : k : I; I denoting the nth. term. We have the equations (Art. 167), hzzzaq, c=zhq, d^=:cq, e:=idq, . . . k^ziq, lz=zkq\ and by adding them all together, member to member, we deduce Sum of \st members. Sum of 2nd members. h+c+d+e+ . . . +k+l=z{a+h + c+d-{- . . . -\-i+k)q\ in which we see that the first member wants the first term a, and the polynomial within the parenthesis in the second member wants the last term I. Hence, if we call the sum of the terms S, we have S— a=:(S — Z)9'=S$'— Zg, or ^q—^^lq—a\ whence S = — — — -. q-l 246 FIRST LESSONS IN ALGEBRA. Therefore, to obtain the sum of the terms of a geometrical progression, we have the following RULE. I. Multiply the last term hy the ratio. II. Subtract the first term from the product. III. Divide the remainder hy the ratio diminished hy unity ^ and the quotient will he the sum of the series. 1 . Find the sum of eight terms of the progression 2 : 6 : 18 : 54 : 162 . . . 2x3'7rrr4374. q — l 2 2. Find the sum of the progression 2 : 4 : 8 : 16 : 32. q-l 1 3. Find the sum of ten terms of the progression 2 : 6 : 18 : 54 : 162 . . . 2x39=.39366. Ans. 59048. 4. What debt may be discharged in a year, or twelve months, by paying $1 the first month, $2 the second month, Quest. — 168. Give the rule for finding the sum of the series. What is the first step 1 What the second 1 What the third 1 GEOMETRICAL PROGHESSION. 247 $4 the third month, and so on, each succeeding payment being double the last ; and what will be the last payment 1 ^^^5 Debt, . . $4095. ' < Last payment, $2048. 5. A gentleman married his daughter on New Year's day, and gave her husband 1^. towards her portion, and was to double it on the first day of every month during the year : what was her portion ? Ans. jE^204 1 5^. 6. A man bought 10 bushels of wheat on the condition that he should pay 1 cent for the 1st bushel, 3 for the second, 9 for the third, and so on to the last : what did he pay for th^ last bushel and for the ten bushels ? A ^ ^ Last bushel, $196,83. ^^' i Total cost, $295,24. 7. A man plants 4 bushels of barley, which, at the first harvest, produced 32 bushels ; these he also plants, which, in like manner, produce 8 fold ; he again plants all his crop, and again gets 8 fold, and so on for 16 years : what is his last crop, and what the sum of the series ? j^^ i Last, 140737488355328SW. ^** i Sum, 160842843834660. 169. When the progression is decreasing, we have 5'<1 and Z