. *. J. DABBY V INSTITUTE DAUBY WISKE. University of California Berkeley THE THEODORE P. HILL COLLECTION of EARLY AMERICAN MATHEMATICS BOOKS I TREATISE ON ALGEBRA. BY ELIAS LOOMIS, A.M., FBOFESSOB OP MATHEMATICS AND NATURAL PHILOSOPHY IN THE UNIVERSITY OF THE CITY OF NEW YORK, MEMBER OF THE AMERICAN PHILOSOPHICAL SOCIETY, OF THE AMERICAN ACADEMY OF ARTS AND SCIENCES, AND AUTHOR OF "ELEMENTS OF GEOMETRY." THIRD EDITION. NEW YORK: HARPER & BROTHERS, PUBLISHERS, 82 CLIFF STREET. 1850, Entered, according to Act of Congress, in the year one thousand eight hundred and forty-seven, by HARPER & BROTHERS, in the Clerk's Office of the District Court of the Southern District of New York. TO THE COUNCIL OF THE UNIVERSITY OP THE CITY OP NEW YORK, THIS TREATISE respectfully THEIR OBEDIENT SERVANT, THE AUTHOR. PR E F ACE. THE first edition of my Algebra was received with unex- pected favor. Almost immediately after its publication, it was adopted as a text-book in half a dozen colleges, besides nu- merous academies and schools ; and the most flattering testi- monials were received from every part of the country. I have thus been stimulated to increased exertions to render it less unworthy of public favor. Every line of it has been sub- jected to a thorough revision. The work has been read by two successive classes in the University, and wherever im- provement seemed practicable, alterations have been freely made. I have also availed myself of the suggestions of sev- eral professors in other colleges. This edition will accord- ingly be found to differ considerably from the preceding. Al- terations, more or less important, have been made on nearly every page. Among these may be mentioned the addition of Continued Fractions, the Extraction of the Roots of Numbers, Elimination by means of the Greatest Common Divisor, and a large collection of Miscellaneous Examples. It is believed that this treatise contains as much of Algebra as can be profitably read in the time allotted to this study in most of our colleges, and that those subjects have been se- lected which are most important in a course of mathematical study. These materials I have endeavored to combine, so as to form a consistent treatise. I have aimed to cultivate in the mind of the student a habit of generalization, and to lead him to IV PREFACE. reduce every principle to its most general form. At the same time, I have been solicitous not to discourage the young begin- ner, who frequently finds it much more difficult to comprehend a general than a particular proposition. Accordingly, many of the Problems have been twice stated. I first give a simple numerical problem, and then repeat the same problem in a more general form. I have labored to develop, in a clear and intelligible manner, the most important properties of equa- tions, and have bestowed great pains upon the selection of examples to illustrate these properties. Throughout the work I have endeavored to render the most important principles so prominent as to arrest attention ; and I have reduced them, as far as practicable, to the form of concise and simple rules. It is believed that, in respect of difficulty, this treatise need not discourage any youth of fifteen years of age who possesses average abilities, while it is designed to form close habits of reasoning, and cultivate a truly philosophical spirit in more mature minds. CONTENTS. SECTION I. DEFINITIONS AND NOTATION. DEFINITIONS. Difference between Algebra and Arithmetic 1 Signs of Addition and Subtraction 4 Signs of Multiplication and Division 5 Coefficient. Exponent. Power. Boot 6 Algebraic Quantity. Monomial. Polynomial 8 Degree of a Term. Homogeneous Polynomials 9 SECTION II. ADDITION . . 12 SECTION III. SUBTRACTION 16 SECTION IV. MULTIPLICATION. Case of Monomials. Rule for the Exponents 21 Case of Polynomials. Rule for the Signs 23 Degree of a Product. Number of Terms in a Product 26 Theorems proved. Quantities resolved into Factors 27 Multiplication by detached Coefficients 29 SECTION V. DIVISION. Case of Monomials. Rule of Exponents 32 Negative Exponents. Symbol aP 33 Case of Polynomials 36 a" _ Jn Divisible by a b 39 Quantities resolved into Factors 40 Division by detached Coefficients 41 SECTION VI. FRACTIONS. Fundamental Principles. Signs of the Terms 43 To reduce a Fraction to lower Terms 45 To reduce a Fraction to an entire or mixed Quantity 4G v j CONTENTS. Page To reduce a mixed auantity to the Form of a Fraction ' To reduce Fractions to a common Denominator < Addition and Subtraction of Fractions Multiplication and Division of Fractions ! SECTION VII. SIMPLE EQUATIONS. Definitions. Axioms employed ! Equations solved by Subtraction and Addition ( Equations solved by Division and Multiplication < Equations cleared of Fractions ( Solution of Problems '" 65 SECTION VIII. EQUATIONS WITH TWO OR MORE UNKNOWN QUANTITIES. Elimination by Substitution 78 By Comparison. By Addition and Subtraction 79 Equations containing three unknown Quantities - 86 Equations containing m unknown Quantities 88 SECTION IX. DISCUSSION OF EQUATIONS OF THE FIRST DEGREE. Positive Values of a;. Negative Values - 92 Infinite Values. Indeterminate Values 96 Zero and Infinity Inequalities 10 SECTION X. INVOLUTION AND POWERS. Involution of Monomials. Sign of the Result 105 Involution of Polynomials 108 t SECTION XI. EVOLUTION AND RADIPAL QUANTITIES. To extract a Hoot of a Monomial 110 Sign of the Root. Square Root of a Trinomial 112 Irrational Quantities. Fractional Exponents 115 To reduce Surds to their most simple Forma 117 To reduce a Rational Quantity to the Form of a Surd 120 To reduce Surds to a common Index 121 Addition and Subtraction of Surd Quantities 122 Multiplication and Division of Surd Quantities 124 Involution and Evolution of Surd Quantities 127 To find Multipliers which shall render Surds Rational 129 To reduce a Surd Fraction to a Rational Numerator or Denominator 131 To free an Equation from Radical Quantities 133 Calculus of Imaginary Expressions ,,., 135 CONTENTS. Vll SECTION XII. EQUATIONS OF THE SECOND DEGREE. Page Solution of Pure Quadratics 137 Solution of Complete Quadratics 142 Solution of the Equation aft" -\-px n = g 148 Quadratic Equations containing two unknown Quantities 156 Discussion of the general Equation of the Second Degree 161 Discussion of particular Problems 168 Double Values of a Imaginary Values 169 SECTION XIII. RATIO AND PROPORTION. Definitions. Ratios compared with each other 171 Proportion. Product of Extremes and Means I 75 Equal Ratios. Alternation. Inversion - 178 Composition. Division. Conversion I 79 Like Powers of Proportionals. Continued Proportion 180 Harmonical Proportion. Variation 182 SECTION XIV. PROGRESSIONS. Arithmetical Progression. L ast Term. Sum of the Terms 188 Geometrical Progression. L ast Term. Sum of the Terms 194 Progressions having an infinite Number of Terms 198 Harmonical Progres sion 201 SECTION XV. GREATEST COMMON DIVISOR. CONTINUED FRACTIONS. PERMUTATIONS AND COM- BINATIONS. Greatest common Divisor. How found 203 Rule applied to Polynomials 205 Continued Fractions 207 Permutations and Combinations 210 SECTION XVI. INVOLUTION OF BINOMIALS. Powers of a Binomial 21 5 Law of the Exponents. Coefficients 216 Binomial Theorem 219 Theorem applied to any Polynomial 221 When the Exponent is Negative 222 When the Exponent is a Fraction 224 When the Exponent is a Negative Fraction 225 Theorem applied to find the Roots of Numbers 226 SECTION XVII. EVOLUTION OF POLYNOMIALS. Method of extracting the Square Root of a Polynomial 228 Method of extracting the Square Root of Numbers 229 Method of extracting the Cube Root of a Polynomial 232 1* Vlii CONTENTS. Page Method of extracting the Cuhe Root of Numbers 234 Method of extracting any Boot of a Polynomial 236 Square Root of a Jb. Formula 239 SECTION xvm. INFINITE SERIES. Definitions. Orders of Differences 243 To find the nth Term of a Series 244 To find the Sum of n Terms of a Series 246 To find any Term of a Series by Interpolation 247 Fractions expanded into Infinite Series. 249 Infinite Series obtained by extracting the Square Root 250 Method of unknown Coefficients .Rule 251 SECTION XIX. GENERAL THEORY OF EQUATIONS. Definitions. General Form of E quations 255 An Equation whose Root is a is divisible by x a 256 An Equation of the mth Degree has m Roots 257 Law of the Coefficients of every Equation 259 What Equations have no Fractional Roots 261 How the Signs of all the Roots may be changed 262 The Number of imaginary Roots must be even 263 Number of Positive and Negative Roots 264 The Limits of a Root determined 266 The Roots may be increased or diminished by any Quantity 267 The second Term of an Equation taken away 268 Derived Polynomials. Equal Roots 269 Theorem of Sturm 272 Elimination by means of the Greatest Common Divisor 279 Ctauul wuli -w : G iuvuuij 1 * J* ^Jort> SECTION XX. JOiil'*i JjCBfliJiKjO SOLUTION OF NUMERICAL EQUATIONS. To find the Integral Roots of an Equation 281 Homer's Method for Incommensurable Roots 284 Equations of the fourth and higher Degrees 293 Newton's Method of Approximation 297 Method of Double Position 299 The different Roots of Unity 301 SECTION XXI. LOGARITHMS. Logarithms. Rule for the Characteristic 304 Multiplication and Division. Involution and Evolution 306 Computation of Logarithms 311 Logarithmic Series 313 Naperian Logarithms. Common Logarithms 316 Exponential Equations solved 3.30 Compound Interest. Increase of Population 321 MISCELLANEOUS EXAMPLES 305 ALGEBRA. SECTION I. PRELIMINARY DEFINITIONS AND NOTATION. (Article 1.) WHATEVER is capable of increase or diminution, or will admit of mensuration, is called magnitude or quantity. A sum of money, therefore, is a quantity, since we may in- crease it or diminish it. A line, a surface, a weight, and other things of this nature, are quantities ; but an idea is not a quantity. (2.) Mathematics is the science of quantity, or the science which investigates the means of measuring quantity. The operations of the mind, therefore, such as memory, imagina- tion, judgment, &c., are not subjects of mathematical investi- gation, since they are not quantities. (3.) Mathematics is divided into pure and mixed. Pure mathematics comprehends all inquiries into the relations of magnitude in the abstract, and without reference to material bodies. It embraces numerous subdivisions, such as Arith- metic, Algebra, Geometry, &c. In the mixed mathematics these abstract principles are ap- plied to various questions which occur in nature. Thus, in Surveying, the abstract principles of Geometry are applied to the measurement of land ; in Navigation, the same principles are applied to the determination of a ship's place at sea ; in Optics, they are employed to investigate the properties of light ; and in Astronomy, to determine the distances of the heavenly bodies. (4.) Algebra is that branch of mathematics which enables us, by means of letters and other symbols, to abridge and generalize the reasoning employed in the solution of all questions relating to numbers. A. 2 PRELIMINARY DEFINITIONS AND NOTATION. Arithmetic is the art or science of numbering. It treats of the nature and properties of numbers, but it is limited to cer- tain methods of calculation which occur in common practice. Algebra is more comprehensive, and has been called by New- ton, Universal Arithmetic. (5.) The following are the main points of difference between Arithmetic and Algebra. First, the operations of Algebra are more general than those of Arithmetic. In Arithmetic we represent quantities by par- ticular numbers, as 2, 5, 7, &c., which numbers always retain the same value. The results obtained, therefore, are applicable only to the particular question proposed. Thus, if it is re- quired to find the interest of a thousand dollars for three months at six per cent., the question may be solved by Arith- metic, and we obtain an answer, which is applicable only to this problem. But in the solution of a general Algebraic problem we em- ploy letters, to which any value may be attributed at pleasure. The results obtained, therefore, are equally applicable to all questions of a particular class. Thus, if we have given the sum and difference of two quantities, we may obtain by means of Algebra a general expression for the quantities themselves. This result will always be found true, whatever may be the magnitude of the quantities. Hence Algebra is adapted to the investigation of general principles, while Arithmetic is confined to operations upon particular numbers. Secondly, Algebra enables us to solve a vast number of problems, which are too difficult for common Arithmetic. Some of the problems in Sections VII. and VIII. may be solved by Arithmetical methods ; but others can not thus be resolved, particularly such problems as are given in Sections XII., XIV., &c. Thirdly, in Arithmetic all the different quantities which en- ter into a problem are blended together in the result, so as to leave no trace of the operations to which they have been sub- jected. From a simple inspection of the result, we can not tell whether it was derived by multiplication or division, invo- lution or evolution, or what connection it has with the given quantities of the problem. But in a general Algebraic solu- tion, all the different quantities are preserved distinct from each PRELIMINARY DEFINITIONS AND NOTATION. 3 other, and we see at a glance how all the data of the problem are combined in the result. Illustrations of this remark will be found in Section VII., &c. Fourthly, the operations of Algebra are often far more con- cise than those of Arithmetic. Thus, although some of the problems in Sections VII. and VIII. may be solved Arith- metically, these solutions are generally much more tedious than the Algebraic. This advantage which is possessed by Algebra is partly due to the representation of the unknown quantities by letters, and their introduction into the operations as if they were already known, and partly to the fact that the operations of multiplication, division, &c., are at first merely indicated, and are not actually performed until an Algebraic expression has been reduced to its simplest form. Finally, perhaps the most striking difference between Arith- metic and Algebra springs from the use of negative quantities, which give rise to many peculiar results. The full purport of these remarks will be best apprehended after the student has made some progress in the study of Al- gebra. (6.) A definition is the explanation of any term or word. It is essential to a perfect definition that it distinguish the thing defined from every thing else. Thus, if we say that man is a biped, it is an imperfect definition of man, because there are many other bipeds. (7.) A theorem is the statement of some property, the truth of which is required to be proved. Thus the principle that the sum of the three angles of any triangle is equal to two right angles, is a theorem, the truth of which is demonstrated by Geometry. (8.) A problem is a question requiring something to be done. Thus, to draw one line perpendicular to another is a problem. Theorems and problems are both known by the general term of propositions. (9.) A determinate problem is one which admits of a certain or definite answer. An indeterminate problem commonly ad- mits of an indefinite number of solutions ; although when the answers are required in positive whole numbers, they are in some cases confined within certain limits, and in others the problem may be impossible. 4 PRELIMINARY DEFINITIONS AND NOTATION. (10.) The solution of a problem is the process by which we obtain the answer to it. A numerical solution is the obtaining an answer in numbers. A geometrical solution is the obtaining an answer by the principles of geometry. A mechanical so- lution is one which is gained by trials. (11.) The principal symbols employed in Algebra are the following : The sign + (an erect cross) is named plus, and is employed to denote the addition of two or more numbers. Thus, 5+3 signifies that we must add 3 to the number 5, in which case the result is 8. In the same manner, 11+6 is equal to 17; 14+10 is equal to 24, &c. We also make use of the same sign to connect several num- bers together. Thus, 7+5+9 signifies that to the number 7 we must add 5 and also 9, which make 21. So, also, the sum of 8+5+13+11 + 1+3+10 is equal to 51. (12.) In order to generalize numbers we represent them by letters, as a, 6, c, d, &c. Thus the expression a -\-b signifies the sum of two numbers, which we represent by a and b, and these may be any numbers whatever. In the same manner, m+n+p+x signifies the sum of the numbers represented by these four letters. If we knew, therefore, the numbers repre- sented by the letters, we could easily find by arithmetic the value of such expressions. The first letters of the alphabet are commonly used to rep- resent known quantities, and the last letters those which are unknown. (13.) The sign (a horizontal line) is called minus, and in- dicates that one quantity is to be subtracted from another. Thus, 85 signifies that the number 5 is to be taken from the number 8, which leaves a . remainder of 3. In like manner, 127 is equal to 5, and 2014 is equal to 6, &c. Sometimes we may have several numbers to subtract from a single one. Thus, 1654 signifies that 5 is to be subtract- ed from 16, and this remainder is to be further diminished by 4, leaving 7 for the result. In the same manner, 501 35 7 9 is equal to 25. So, also, a b signifies that the number designated by a is to be diminished by the number designated by b. Quantities preceded by the sign + are called positive quan PRELIMINARY DEFINITIONS AND NOTATION. 5 titles; those preceded by the sign , negative quantities. When no sign is prefixed to a quantity, + is to be understood. Thus, a+bc is the same as +a+b c. (14.) The sign X (an inclined cross) is employed to denote the multiplication of two or more numbers. Thus, 3X5 signi- fies that 3 is to be multiplied by 5, making 15. In like man- ner, axb signifies a multiplied by b ; and aXbXc signifies the continued product of the numbers designated by , b, and c, and so on for any number of quantities. Multiplication is also frequently indicated by placing a point between the successive letters. Thus, a . b . c . d signifies the same thing as aXbXcXd. Generally, however, when numbers are represented by let- ters, their multiplication is indicated by writing them in suc- cession without the interposition of any sign. Thus, a b sig- nifies the same thing as a . b or a X b ; and a b c d is equivalent to a. b. c . d, or aXbXcXd. It must be remarked that the notation a . b or a b is seldom employed except when the numbers are designated by letters. If, for example, we attempt to represent the product of the numbers 5 and 6 in this manner, 5 . 6 might be confounded with an integer followed by a decimal fraction ; and 56 would be read fifty-six, according to the common system of nota- tion. The multiplication of numbers may, however, be expressed by placing a point between them, in cases where no ambiguity can arise from the use of this symbol. Thus, 1.2.3.4 is sometimes used to represent the continued product of the num- bers 1, 2, 3, 4. (15.) When two or more quantities are multiplied together, each of them is called a factor. Thus, in the expression 7X5, 7 is a factor, and so is 5. In the product abc there are three factors, a, fc, c. When a quantity is represented by a letter, it is called a literal factor, to distinguish it from a numerical factor, which is represented by an Arabic numeral. Thus, in the expression Sab, 5 is a numerical factor, while a and b are literal factors. (16.) The character -r (a horizontal line with a point above and below) shows that the quantity which precedes it is to be divided by that which follows. 6 PRELIMINARY DEFINITIONS AND NOTATION. Thus, 24-T-6 signifies that 24 is to be divided by 6, making 4. So, also, a-^-b is a divided by b. Generally, however, the division of two numbers is indi- cated by writing the dividend above the divisor, and drawing a line between them. Thus, 24-7-6 and a-^b are usually written ? and I. (17.) The sign = (two horizontal lines) when placed be- tween two quantities, denotes that they are equal to each other. Thus, 7+6=13 signifies that the sum of 7 and 6 is equal to 13. So, also, $1 = 100 cents, is read one dollar equals one hundred cents; 3 shillings =36 pence, is read three shillings are equal to thirty-six pence. In like manner, a=b signifies that a is equal to b; and a+bcd signifies that the sum of the numbers designated by a and b is equal to the difference of the numbers designated by c and d. (18.) The symbol > is called the sign of inequality ', and when placed between two numbers, denotes that one of them is greater than the other, the opening of the sign being turned toward the greater number. Thus, 3<5 signifies that 3 is less than 5, and 11>6 denotes that 11 is greater than 6. So, also, a>6 shows that a is greater than b, and c=-ab ^ therefore ^ -ab-+b=-a. +aXb=ab j j ab b=+a. aX-b=+abJ [+ab b= a. Hence we have the following RULE FOB THE SIGNS. When both the dividend and divisor have the same sign, the quotient will have the sign + ; when they have different signs, the quotient will have the sign . 36 DIVISION. X EXAMPLES. Ex. 1. Divide 15ay 8 by Say. Ex. 2. Divide 18ax*y by 9ax. Ex. S. Divide I5Qa*bc by 5ac. Ex. 4. Divide 40a 3 6V by abc. CASE II. (72.) When the divisor is a monomial, and the dividend a polynomial. We have seen, Art. 51, that when a single term is multi- plied into a polynomial, the former enters into every term of the latter. Thus, a(a+b)=a*+ab. Hence (a*+ab)-t-a=a+b. Whence we deduce the following RULE. Divide each term of the dividend by the divisor, as in the for- mer case. \ EXAMPLES. Ex. 1. Divide 3x*+6x*+3ax-I5x by Sx. Ans. x*+2x+a 5. Ex. 2. Divide 3abc+12abx-9a' i b by Sab. Ex. S. Divide 400 8 & 3 +60a 2 & 2 -17a& by -ab. Ex. 4. Divide I5a*bc 10flcx 2 +5ac 2 ^ 2 by 5# 2 c. Ex. 5. Divide 6aVy 6 -12 8 ;ry+15ffV?/ 3 by 3V?/ a . Ex. 6. Divide x n+i x n+z +x n -^-x n + 4 by x". Ex. 7. Divide 12y 16y+20y-28ay by 4y. CASE III. (73.) When the divisor and dividend are both polynomials. Let it be required to divide 2afe+ 2 +6 2 by a+b. The object of this operation is to find a third polynomial which, multiplied by the second, will reproduce the first. It is evident that the dividend is composed of all the partial products arising from the multiplication of each term of the divisor by each term of the quotient, these products being add- ed together and reduced. Hence, if we can discover a term DIVISION. 37 of the dividend which is derived without reduction from the multiplication of a term of the divisor by a term of the quo- tient, then dividing this term by the corresponding term of the divisor, we shall be sure to obtain a term of the quotient. But from Art. 58, it appears that the term a 2 , which contains the highest exponent of the letter #, is derived, without reduc- tion, from the multiplication of the two terms of the divisor and quotient which are affected with the* highest exponent of the same letter. Dividing then the term a 2 by the term a of the divisor, we obtain a, which we are certain must be one term of the quotient sought. Multiplying each term of the di- visor by a, and subtracting this product from the proposed dividend, the remainder may be regarded as the product of the divisor by the remaining terms of the quotient. We shall then obtain another term of the quotient by dividing that term of the remainder affected with the highest exponent of a. by the term a of the divisor, and so on. Thus we perceive that at each step we are obliged to search for that term of the dividend which is affected with the high- est exponent of one of the letters, and divide it by that term of the divisor which is affected with the highest exponent of the same letter. We may avoid the necessity of searching for this term by arranging the terms of the divisor and dividend in the order of the powers of one of the letters. The operation will then proceed as follows : The arranged dividend = Q ab a+b= the divisor. a+b= the quotient. 2 = first remainder. ab+b* It is generally convenient in Algebra to place the divisor on the right of the dividend, and the quotient directly under the divisor. (74.) From this investigation we deduce the following RULE FOE THE DIVISION OF POLYNOMIALS. 1 . Arrange the dividend and divisor according to the powers of the same letter. 38 DIVISION. 2. Divide the first term of the dividend by the first term of the divisor, the result will be the first term of the quotient. 3. Multiply the divisor by this term, and subtract the product rom the dividend. 4. Divide the first term of the remainder by the first term of he divisor, the result will be the second term of the quotient. 5. Multiply the divisor by this term, and subtract the product from the last remainder. Continue the same operation till all the terms of the dividend are exhausted. If the divisor is not exactly contained in the dividend, the quantity which remains after the division is finished must be placed over the divisor in the form of a fraction, and annexed to the quotient. EXAMPLES. 1. Divide 2a*b+b 3 +2ab*+a 3 by a*+b*+ab. Ans. a+b. 2. Divide x 3 a*+3a*x Sax* by xa. Ans. x*2ax+a 9 . 3. Divide 6 +z e +2aV by a'-az+z*. Ans. a 4 +a 3 z+az 5 +z*. 4. Divide a*-I6a 3 x*+G4x by a*-4ax+4x\ 5. Divide a*+6d 1 x*4a 3 x+x*4ax 3 by a* 2ax+x\ Ans. a?2ax+x*. 6. Divide x*+x*y* +?/ 4 by x*+xy+y*. 7. Divide I2x*-192 by 3x-6. Ans. 4x 3 +8x*+16x+32. 8. Divide 6x 6 -6y* by 2x*-2y*. 9. Divide Ans. ~ 46 FRACTIONS. 14 4. Reduce ^ ~ to its lowest terms. la Ans. . 5c Ans. a . 5 - Reduce to its lowest terms - 6. Reduce ft , , , a to its lowest terms. a a 2ab+b* a*x* 7. Reduce -5 ; 5 to its lowest terms. a* 2ax+x* PROBLEM II. (84.) 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. 27 Thus, =27-r-5=5f. Also, =(ax+a > )-^-x=a-{ . X X EXAMPLES. 1. Reduce to an entire quantity. K FRACTIONS. 47 2. Reduce r to a mixed quantity. fl a +a; a 3. Reduce to a mixed quantity. ax Ans . 4. Reduce to an entire quantity. xy 5. Reduce to a mixed quantity. DX 6. Reduce rr to a mixed quantity. PROBLEM III. (85.) To reduce a mixed quantity to the form of a fraction RULE. Multiply the entire part by the denominator of the fraction ; to the product add the numerator with its proper sign, and place the result over the denominator. 3X5+2 15+2 17 Thus, S|= = T - This result may be proved by the preceding Rule. For _ c c c EXAMPLES. a a # 2 1. Reduce x-\ to the form of a fraction. x 2. Reduce x-\ - to the form of a fraction. 2a Ans. . x 3ax+x* 48 FRACTIONS. _ 3. Reduce 5H - to the form of a fraction. 4. Reduce 1 H -- to the form of a fraction. _ 5. Reduce 1 +2x-\ - to the form of a fraction. . 6. Reduce 7H r~n~ to the form of a fraction. PROBLEM IV. (86.) To reduce fractions to a common denominator. RULE. Multiply each numerator into all the denominators, except its own, for a new numerator, and all the denominators together for a common denominator. EXAMPLES. a c 1. Reduce 7 and -3 to a common denominator. b a ad be bd ; bd" Here it will be seen that the numerator and denominator of the first fraction are both multiplied by d, and in the second fraction they are both multiplied by b. The value of the frac- tions, therefore, is not changed by this operation. 2. Reduce j- and to equivalent fractions having a com- * mon denominator. ac ab+b* Ans. be' be ' 3x 2& 3. Reduce , , and d to fractions having a common de- nominator. O 4. Reduce -, , and a+-j- to fractions having a common denominator. FRACTIONS. 49 5. Reduce -, , and - to fractions having a common 2 7 a x denominator. np '>* ^| _. 1 1 i Q* 6. Reduce -, -, and to fractions having a common O O 1 ~T~X denominator. 7. Reduce - and - to fractions having a common de- 3x 3 4z nominator. Following the Rule, we obtain Sax which fractions have a common denominator, and are equiva- lent to those originally proposed. Nevertheless, it may be ob- served, that these fractions are not reduced to their least com- mon denominator, for every term is divisible by x. The least common denominator is the least common multiple of the de- nominators of the proposed fractions. A common multiple of two or more numbers is any number which they will divide without a remainder ; and the least com- mon multiple is the least number which they will so divide. Thus, 12x* is the least common multiple of 3x* and 4x ; and the above fractions reduced to their least common denomina- tor are 8a , 15 and The least common multiple of two numbers is their product divided by their greatest common divisor. o c 8. Reduce and to equivalent fractions having the least common denominator. The product of the denominators is 294, which, divided by 7 (their greatest common divisor), gives 42, the least common denominator, and the required fractions are 9 10 9. Reduce the fractions and to others which have the 10 15 least common denominator. D 50 FRACTIONS. 10. Reduce ^- and ^T-J to equivalent fractions having the 4ac , d Ans. TTT-T and least common denominator. 11. Reduce r and a , a to equivalent fractions having the least common denominator. (a+b)* . c+d Ans . __ and __ PROBLEM V. (87.) To add fractional quantities together. RULE. Reduce the fractions to a common denominator ; add the nu- merators together, and place their sum over the common denom- inator. The fractions must first be reduced to a common denomina- tor to render them like parts of unity. Before this reduction, they must be considered as unlike quantities. EXAMPLES. 1. What is the sum of - and -? * > Reducing to a common denominator, the fractions become 3x 2x 6 /} * o 5x Adding the numerators, we obtain . It is plain that three sixths of x and two sixths of x make five sixths of x. ft f* f> 2. Required the sum of ^, ^, and > adf+bcf+bde bdf ' FRACTIONS. 51 3. Required the sum of r and r. a+b ab 2a . a+2x 4. Required the sum of 5x, ^-5, and - . OX 437 5. Required the sum of 2a 9 3a+, and #+-g- 6. Required 7. Required 8. Required 9. Required Ans. 58o: 45 ' x* the sum of and . Ans. a. a a 2m a+2m the sum ot , . , and -. mab na+b the sum of ; and ; . m+n m-\-n PROBLEM VI. (88.) To subtract one fractional quantity from another. RULE. Reduce the fractions to a common denominator, subtract one numerator from the other, and place their difference over the common denominator. EXAMPLES. 23; 3x 1. From subtract . o o Reducing to a common denominator, the fractions become 103; , 9a; 103J 9x x Hence TTT T^=T^; 15 15 15 and it is plain that ten fifteenths of x, diminished by nine fit- teenths of x, equals one fifteenth of x. 52 FRACTIONS. I2x 3x 2. From -=- subtract -. 7 & 9x 4y 5x3y 3. From subtract -- ^-A 7 o It must be remembered, that the minus sign before the di- viding line of a fraction affects the quotient (Art. 82) ; and since a quantity is subtracted by changing its sign, the result of the subtraction in this case is which fractions may be reduced to a common denominator, and the like terms united, as in addition. 4. From ^- subtract ^-. bc b+c 2acx Ans ' " 5. From 2x-] subtract x ft 91 355Z-6 Ans. 168 _ x . xa 6. From 3x+^-i subtract x . 2b c a+b ab 7. From subtract ^ . & iii 13 3 , x b 3cx 5. Divide -r - by -7-7. Scd J 4d Ans. ^ b >~- ^^Jj G^^.6 TV -i 6. Btvide c ' Tk- -j a b a b 7. Divide -H r by r -7. a+b ab J ab a+b Ans. Unity. (92.) Ex. 1. Divide i by t . According to the Rule of the preceding Article, we have 56 FRACTIONS. But ~i may be written or 6 ; may be written a 3 ; and is equal to or 3 . Hence a- 5 +a-^=a-\ That is, the Rule of Art. 66 is general, and applies to nega- tive as well as positive exponents. Ex. 2. Divide ZH 5 by b~\ Ans. ft- 3 . 3. Divide a 8 by or*. 4. Divide 1 by a 4 . 5. Divide 6a n by Sa- 3 . 6. Divide #"-* by b m . 7. Divide I2x~ 3 y 4 by 4xy*. 8. Divide (x y)~* by (x y)- 6 . (93.) According to the definition, Art. 33, the reciprocal of a quantity is the quotient arising from dividing a unit by that quantity. Hence the reciprocal of j- a b b is l-f~=lx-=-. b a a That is, the reciprocal of a fraction is the fraction inverted. a Thus the reciprocal of ?. is - -. b+x a The reciprocal of ^ is b+c. Hence, to divide by any quantity is the same as to multiply by its reciprocal, and to multiply by any quantity is the same as to divide by its reciprocal. (94.) The numerator or denominator of a fraction may be itself a fraction ; . f As b or - c c d Such expressions are easily reduced by applying the pre- ceding principles. FRACTIONS. 57 (-} Thus, \bJ means r -=-, which, according to Remark second, Art. 81, equals . oc b Again, /b\ means a-:--, I - I c \c/ 2 SIMPLE EQUATIONS. the equation is solved by multiplying both members by this known quantity. (103.) V. Several terms of an equation may be fractional. Let the equation be ? ? 4 2~3 + 5* Multiplying each of these equals by 2, the result is 48 * = 3 + 5- Multiplying each of these last equals by 3, we obtain and multiplying again by 5, we obtain 15*=20+24, an equation free from fractions. We might have obtained the same result by multiplying the original equation at once by the product of all the denom- inators. Thus, multiplying by 2X3X5, we have 30z_60 120 ~2~-~3~ + ~5~' or reducing, we have 15z=20+24, as before. So, also, in the equation x_b d a^c + e f multiplying successively by all the denominators, or by a c e at once, we obtain acex_dbce acde a c e ' Canceling from each term the letter which is common to its numerator and denominator, we have cex=abe+acd, an equation clear of fractions. Hence it appears that An equation may be cleared of fractions by multiplying each member into all the denominators. (104.) From the preceding remarks, we deduce the fol- lowing SIMPLE EQUATIONS. (53 RULE FOB THE SOLUTION OF A SIMPLE EQUATION CONTAINING ONE UNKNOWN QUANTITY. 1. Clear the equation of fractions, and perform in both mem- bers all the algebraic operations indicated. 2. Transpose all the terms containing the unknown quantity to one side, and all the remaining terms to the other side of the equation, and reduce each member to its most simple form. 3. Divide each member by the coefficient of the unknown quantity. EXAMPLES. 1. Given 5x+ 8=4^ + 10, to find the value of x. Transposing 4x to the first member of the equation, and 8 to the second member, taking care to change their signs (Art. 100), we have 5^-4^=10-8. Uniting similar terms, x=2. In order to verify this result, put 2 in the place of x wher- ever it occurs in the original equation, and we shall obtain 5X2+8=4X2 + 10. That is, 10+8=8 + 10, or 18=18, an identical equation, which proves that we have found the correct value of x. IT 3T 2. Given x 7= - + , to find the value of x. O o Multiplying every term of the equation by 5 and also by 3, in order to clear it of fractions (Art. 103), we obtain Hence, by transposition, or 7z=105, and therefore x=- = 15. To verify this result, put 15 in the place of x in the original equation, and we have _15 15 n ~T + y 4 64 SIMPLE EQUATIONS. That is, 15-7=3+5, or 8=8, an identical equation. 3. Given Sax 4ab=2ax 6ac, to find the value of x in terms of b and c. Dividing every term by a, we have 3x-4b=2x-6c. By transposition, or x=4b 6c. This result may be verified in the same manner as the ceding. 4. Given 3# a 10x=8x+x*, to find the value of x. Ans. x=9. a(d*+x*) , ax 5. Given -^ = L =ac+-j, to find x. dx d 6. Given j +6#= ~ , to find 7. Given =bc+d+-, to find a;. a: x . #=-. c Ans. x=9. , ^ t* -W7 , ' 6 , U*-37 4 fi , 8. Given 3a;H =5H , to find x. Oi i{ n*4g / if. ~? 9. Given 5ax2b+4bx=2x+5c, to find #. J.WS. .T = 5a+4b-2' Qg. _ g g^ _ 10. Given #H -- - =12 -- - , to find the value of#. Ans. .T=5. Sz-ll 5x 5 97 7x 11. Given 21 +^=5+ - , to find x. ID o Z (105.) An equation may always be cleared of fractions by multiplying each member into all the denominators according SIMPLE EaUATIONS. 65 10 Art. 103. But sometimes the same object may be attained by a less amount of multiplication. Thus, in the preceding example, the equation maybe cleared of fractions by multiplying each term by 16, instead of 16X8 X2, and it is important to avoid all useless multiplication. In general, it is sufficient to multiply by the least common multiple of all the denominators. See Art. 86. 12. Given 3z-^- 4= y^, to find x. 13. Given 3x a+cx= , to find x. o (I Ans. x 8a+3ac-& Oj. y> Oj* 14. Given c-f r=4oH T, to find x. a b a abed Ans. x=- 3bd+ ad 4abd 2ab' afc 15. Given (a+x) (b+x)-a(b+c)=-^-+x\ to find x. ac Ans. x=. o ll-3x 4x+2 7+14 16. Given =5 6oH , to find x. DO O 3x-3 20-x Gx-S 4x-4 17. Given x \-4= 1 , to find x. o fi 7 o 7a;+16 x+S x 18. dven -^ 5 fTg. ^ find x. ftc+7 7Z-13 2x+4 19. G,ven -+--= -jj-, to find x. 20. Given -ab+-ac -cx=-ac-{-2ab 6cx, to find the value o 5 o 4 of x. Ans. x= 320c SOLUTION OF PROBLEMS. (106.) The solution of a Problem by Algebra consists of two distinct parts : 66 SIMPLE EQUATIONS. 1. To express the conditions of the problem algebraically ; that is, to form the equation. 2. To solve the equation. The second operation has already been explained, but the first is often more embarrassing to beginners than the second. Sometimes the statement of a problem furnishes the equation directly; and sometimes it is necessary to deduce from the statement new conditions, which are to be expressed alge- braically. The former are called explicit conditions ; and those which are deduced from them, implicit conditions. It is impossible to give a general rule which will enable us to translate every problem into algebraic language. The power of doing this with facility can only be acquired by re- flection and practice. The following directions may be found of some service. Denote one of the required quantities by x ; then, by means of this letter, with the algebraic signs, perform the same opera- tions which would be necessary to verify its value if it was al- ready known. Problem 1. What number is that, to the double of which if 16 be added, the sum is equal to four times the required num- ber ? Let x represent the number required. The double of this will be 2x. This increased by 16 should equal 4x. Hence, by the conditions, 2x-\-lQ=4x. The problem is now translated into algebraic language, and it only remains to solve the equation in the usual way. Transposing, we obtain I6=4x 2x=2x, and 8= x, or x=S. To verify this number, we have but to double 8, and add 16 to the result ; the sum is 32, which is equal to four times 8, according to the conditions of the problem. Prob. 2. What number is that, the double of which exceeds its half by 6? Let x = the number required. Then, by the conditions, SIMPLE EdUATIONS. 67 Clearing of fractions, 4x-x=I2, or 3x=12. Hence x=4. To verify this result, double 4, which makes 8, and dimmish ,t by the half of 4, or 2 ; the result is 6, according to the con- ditions of the problem. Prob. 3. The sum of two numbers is 8, and their difference 2. What are those numbers ? Let x = the least number. Then x+2 will be the greater number. The sum of these is 2z+2, which is required to equal 8. Hence we have 2x+2=8. By transposition, 2x=S 2=6, and x=3, the least number. Also, x+2 =5, the greater number. Verification. 5+3=8 ) 5_o_o ( according to the conditions. The following is a generalization of the preceding Problem. Prob. 4. The sum of two numbers is , and their difference b. What are those numbers ? Let x represent the least number. Then x +b will represent the greater number. The sum of these is 2x+b, which is required to equal a. Hence we have 2x+b=a. By transposition, 2x=ab, a b a b or x= ~~7r~ = o~o tne * ess num ker. a b a b Hence x+b=-~+b= x+-, the greater number. / <* Z As these results are independent of any particular value at- tributed to the letters a and b, it. follows that Half the difference of two quantities, added to half their sum, is equal to the greater ; and 68 SIMPLE EQ,UATIONS. Half the difference subtracted from half the sum is equal to the less. The expressions -+- and - - are called formulas, because & A they may be regarded as comprehending the solution of all questions of the same kind; that is, of all problems in which we have given the sum and difference of two quantities. Thus, let a=S j ag ^ the preceding pro blem. o S ** . R I O<~ 5 2 10 " i " Prob. 5. From two towns which are 54 miles distant, two travelers set out at the same time with an intention of meet- ing. One of them goes 4 miles and the other 5 miles per hour. In how many hours will they meet ? Let x represent the required number of hours. Then 4x will represent the number of miles one traveled, and 5x the number the other traveled ; and since they meet, they must together have traveled the whole distance. Consequently, 4#+5#=54. Hence 9#=54, or x=6. Proof. In 6 hours, at 4 miles an hour, one would travel 24 miles ; the other, at 5 miles an hour, would travel 30 miles. The sum of 24 and 30 is 54 miles, which is the whole distance. This Problem may be generalized as follows : Prob. 6. From two points which are a miles apart, two bodies move toward each other, the one at the rate of m miles SIMPLE EQUATIONS. 09 per hour, the other at the rate of n miles per hour. In how many hours will they meet? Let x represent the required number of hours. Then mx will represent the number of miles one body moves, and nx the miles the other body moves, and we shall obviously have Hence x= - . m-\-n This is a general formula, comprehending the solution of all problems of this kind. Thus, c =150; ^ 6; 4 ^ ^g 90 || 8 -I 7 Si -> > O J3 135 | 15 9 12 'g n 210 20 -5 15 Required the time of meeting. We see that an infinite number of problems may be pro- posed, all similar to Prob. 5 ; but they are all solved by the formula of Prob. 6. We also see what is necessary in order that the answers may be obtained in whole numbers. The given distance (a) must be exactly divisible by m+n. Prob. 7. A gentleman meeting three poor persons, divided 60 cents among them ; to the second he gave twice, and to the third three times as much as to the first. What did he give to each? Let x = the sum given to the first. Then 2x = the sum given to the second, and 3x = the sum given to the third. Then, by the conditions, That is, 6z=60, or #=10. Therefore he gave 10, 20, and 30 cents to them respectively. The learner should verify this, and all the subsequent results. The same problem generalized. Prob. 8. Divide the number a into three such parts, that the 70 SIMPLE EQUATIONS. second may be m times, and the third n times as great as the first. a ma na 1+m+n' 1+m+n' l+m+n What is necessary in order that the preceding values may be expressed nn whole numbers? Prob. 9. A bookseller sold 10 books at a certain price, and afterward 15 more at the same rate. Now at the last sale he received 25 dollars more than at the first. What did he re- ceive for each book ? Ans. Five dollars. The same Problem generalized. Prob. 10. Find a number such that when multiplied success- ively by m and by n, the difference of the products shall be a. Ans. . m n Prob. 11. A gentleman dying, bequeathed 1000 dollars to three servants. A was to have twice as much as B, and B three times as much as C. What were their respective shares ? Ans. A received $600, B $300, and C $100. Prob. 12. Divide the number a into three such parts that the second may be m times as great as the first, and the third n times as great as the second. a . ma mna A ft O - - _._ r l+m+mn 9 l+m+mn j l+m+mn' Prob. 13. A hogshead which held 120 gallons was filled with a mixture of brandy, wine, and water. There were 10 gallons of wine more than there were of brandy, and as much water as both wine and brandy. What quantity was there of each ? Ans. Brandy 25 gallons, wine 35, and water 60 gallons. Prob. 14. Divide the number a into three such parts, that the second shall exceed the first by m, and the third shall be equal to the sum of the first and second. a 2m a+2m a ns ' ~T~' ~T~ ; 2* Prob. 15. A person employed four workmen, to the first of whom he gave 2 shillings more than to the second ; to the SIMPLE EQUATIONS. 71 second 3 shillings more than to the third ; and to the third 4 shillings more than to the fourth. Their wages amount to 32 shillings. What did each receive ? Ans. They received 12, 10, 7, and 3 shillings respectively. Prob. 16. Divide the number a into four such parts, that the second shall exceed the first by m, the third shall exceed the second by n, and the fourth shall exceed the third by p. a3m2np a+m2np Ans . . __ P. . __L, a+m+2np a+m+2n+3p ~4~ ~~4~ (107.) Problems which involve several unknown quantities may often be solved by the use of a single unknown letter. Most of the preceding examples are of this kind. In general, when we have given the sum or difference of two quantities, both of them may be expressed by means of the same letter. For the difference of two quantities added to the less must be equal to the greater ; and if one of two quantities be sub- tracted from their sum, the remainder will be equal to the other. Prob. 17. At a certain election 36000 votes were polled ; and the candidate chosen wanted but 3000 of having twice as many votes as his opponent. How many voted for each ? Let x = the number of votes for the unsuccessful candidate Then 36000 x = the number the successful one had, And 36000-o;+3000=2;r. Ans. 13000 and 23000. Prob. 18. Divide the number a into two such parts, that one part increased by b shall be equal to m times the other part. mab a-\-b m+l Prob. 19. A train of cars moving at the rate of 20 miles per hour, had been gone three hours, when a second train followed at the rate of 25 miles per hour. In what time will the second train overtake the first ? Let x = the number of hours the second train is in motion, x-}-3 = the time of the first train. Then 25x the number of miles traveled by the second train, 2Q(x-\-3) = the miles traveled by the first train. 72 SIMPLE EQUATIONS. But at the time of meeting they must both have traveled the same distance. Therefore 25z=20z+60. By transposition, 5#=60, and x=l2. Proof. In 12 hours, at 25 miles per hour, the second train goes 300 miles ; and in 15 hours, at 20 miles per hour, the first train also goes 300 miles ; that is, it is overtaken by the sec- ond train. Prob. 20. Two bodies move in the same direction from two places at a distance of a miles apart ; the one at the rate of n miles per hour, the other pursuing at the rate of m miles per hour. When will they meet ? Ans. In - hours. w n This Problem, it will be seen, is essentially the same as Prob. 10. Prob. 21. Divide the number 197 into two such parts, that four times the greater may exceed five times the less by 50. Ans. 82 and 115. Prob. 22. Divide the number a into two such parts, that m times the greater may exceed n times the less by b. mab na+b Ans - When 7i=l, this Problem reduces to Problem 18. When b=Q, this Problem reduces to Problem 24. Prob. 23. A prize of 2329 dollars was divided between two persons, A and B, whose shares were in the ratio of 5 to 12. What was the share of each ? Beginners almost invariably put x to represent one of the quantities sought in a problem ; but a solution may often be very much simplified by pursuing a different method. Thus, in the preceding problem, we may put x to represent one fifth of A's share. Then 5x will be A's share, and I2x will be B's, and we shall have the equation and x =137, consequently their shares were 685 and 1644 dollars. SIMPLE EQUATIONS. Prob. 24. Divide the number a into two such parts, that the first part may be to the second as m to n. ma na Ans. ; ; . m+n m-\-n Prob. 25. What number is that whose third part exceeds its fourth part by 16? Let I2x = the number. Then 4z-3x=16, or x=lQ. Therefore the number = 12X 16=192. Prob. 26. Find a number such that when it is divided suc- cessively by m and by n, the difference of the quotients shali be a. . mna Ans. . n m Prob. 27. What two numbers are as 2 to 3, to each of which, if four be added, the sums will be as 5 to 7 ? A strict adherence to system would have required this ex- ample to be placed after the subject of Proportion, which is treated of in Section XIII. It is, however, only necessary to assume one simple principle which is employed in Arithmetic, viz., If four quantities are proportional, the product of the ex- tremes is equal to the product of the means. Thus, if a:b::c:d. Then ad=bc. In the preceding Problem, let 2x and 3x be the numbers. Then 2x+4 : Sx+4 : : 5 : 7, and by the last principle, 1 4^+28 = 15+20. Prob. 28. What two numbers are as m to n, to each of which, if a be added, the sums shall be as p to q ? ma(p-q) _ na(p-q) J\.'HS. mqnp mqnp Prob. 29. A gentleman divides a dollar among 12 children, giving to some 9 cents each, and to the rest 7 cents. Ho\v many were there of each class ? Prob. 30. Divide the number a into two such parts, that if 74 SIMPLE EaUATIONS. the first is multiplied by m and the second by n, the sum of the products shall be b. b na ma b Ans. - ; - . m n mn Prob. 31. If the sun moves every day one degree, and the moon thirteen, and the sun is now 60 degrees in advance of the moon, when will they be in conjunction for the first time, second time, and so on ? Prob. 32. If two bodies move in the same direction upon the circumference of a circle which measures a miles, the one at the rate of n miles per day, the other pursuing at the rate of m miles per day, when will they meet for the first time, second time, &c., supposing them to be b miles apart at starting ? b a+b 2a+b Ans. In -- ; - ; -- , &c., days. mn mn mn It will be seen that this Problem includes Prob. 20. Prob. 33. Divide the number 12 into two such parts, that the difference of their squares may be 48. Prob. 34. Divide the number a into two such parts, that the difference of their squares may be b. 2a 2a' Prob. 35. The estate of a bankrupt, valued at 21000 dollars, is to be divided among three creditors according to their re- spective claims. The debts due to A and B are as 2 to 3, while B's claims and C's are in the ratio of 4 to 5. What sum must each receive ? Prob. 36. Divide the number a into three parts, which shall be to each other as m : n : p. ma na pa \ M O .______, - _ ** - m-\-n-\-p ' m+n+p ' m-\-n+p' When p=l 9 Prob. 36 leduces to the same form as Prob. 8. Prob. 37. A grocer has two kinds of tea, one worth 72 cents per pound, the other 40 cents. How many pounds of each must be taken to form a chest of 80 pounds, which shall be worth 60 cents ? Ans. 50 pounds at 72 cents, and 30 pounds at 40 cents. Prob. 38. A grocer has two kinds of tea, one worth a cents per pound, the other b cents. How many pounds of each must SIMPLE EaUATIUNS. 75 be taken to form a mixture of n pounds, which shall be worth c cents ? n(c-b) Ans. - j pounds at a cents, a b , n(ac) and - T pounds at b cents. a b r Prob. 39. A can perform a piece of work in 6 days ; B can perform the same work in 8 days ; and C can perform the same work in 24 days. In what time will they finish it if all work together? Prob. 40. A can perform a piece of work in a days, B in b days, and C in c days. In what time will they perform it if all work together ? abc Ans. -= -- r- days. ab+ac+bc Prob. 41. There are three workmen, A, B, and C. A and B together can perform a piece of work in 27 days ; A and C together in 36 days ; and B and C together in 54 days. In what time could they finish it if all worked together ? A and B together can perform ^ T of the work in one day. A and C " -'_ one " B and C " _'_ one " Therefore, adding these three results, 2A+2B+2C can perform ^ T + --+^_ in one day. = ^5- in one day. Therefore, A, B, and C together can perform ^ of the work in one day ; that is, they can finish it in 24 days. If we put x to represent the time in which they would all finish it, then they would together perform ~ part of the work in one day, and we should have *V+*V+ik=l. Prob. 42. A and B can perform a piece of labor in a days ; A and C together in b days ; and B and C together in c days. In what time could they finish it if all work together ? 2abc Ans. r- -- r- days. ab+ac+bc J This result, it will be seen, is of the same form as that of Problem 40. 76 SIMPLE EQUATIONS. Prob. 43. A broker has two kinds of change. It takes 20 pieces of the first to make a dollar, and 4 pieces of the second to make the same. Now a person wishes to have 8 pieces for a dollar. How many of each kind must the broker give him? Prob. 44. A has two kinds of change; there must be a pieces of the first to make a dollar, and b pieces of the second to make the same. Now B wishes to have c pieces for a dol- lar. How many pieces of each kind must A give him ? Ans. a ( c ~~V of the firgt kind . b(a-c) secon d. ab ab Prob. 45. Divide the number 45 into four such parts, that the first increased by 2, the second diminished by 2, the third multiplied by 2, and the fourth divided by 2, shall all be equal. In solving examples of this kind, several unknown quantities are usually introduced, but this practice is worse than super- fluous. The four parts into which 45 is to be divided, may be represented thus : The first =x 2, second =x+2, third =|, fourth 2x ; for if the first expression be increased by 2, the second dimin- ished by 2, the third multiplied by 2, and the fourth divided by 2, the result in each case will be x. The sum of the four parts is 4fc which must equal 45. Hence x=W. Therefore the parts are 8, 12, 5, and 20. Prob. 46. Divide the number a into four such parts, that the first increased by m, the second diminished by m, the third multiplied by m, and the fourth divided by m, shall all be equal. ma ma a m~a ' (m+l) a ~ '* (m+iy +m; (m+iy' (m+1) 2 ' Prob. 47. A merchant maintained himself for three years at an expense of $500 a year; and each year augmented that part of his stock which was not thus expended by one third SIMPLE EQUATIONS. 77 thereof. At the end of the third year his original stock was doubled. What was that stock ? Prob. 48. A merchant supported himself for three years at an expense of a dollars per year ; and each year augmented that part of his stock which was not thus expended by one third thereof. At the end of the third year his original stock was doubled. What was that stock ? 148 Substituting this value of y in the expression for the value of x given above, it becomes 67-7X7_67-49_18_ ~3~ ~3~ ~T Thus we have y =I 7, and x=6. By the second method. From the first equation we find 5x=584y, 58 -4y whence x r . D 67- 7y From the second equation, x TT-^-* o 58-4v 67 7y Therefore ^-= --^. 5 o Clearing of fractions, 174 12?/=335-35y. By transposition, 35y 12^=335 174, or 23y=161. Therefore y =r l> whence, as before, .77=6. CONTAINING TWO OR MORE UNKNOWN QUANTITIES. 83 By the third method. Multiplying the second equation by 5 and the first by 3, we obtain and 15z+12?/=174. By subtraction, 23?/=161, or y= 7. Whence, from equation first, 5z=58-4y=58 28=30, and therefore x =6. Thus the same example may be solved by either of the three methods, and each method has its advantages in particular cases. Generally, however, the first two methods give* rise to fractional expressions which occasion inconvenience in prac- tice, while the third method is not liable to this objection. When the coefficient of one of the unknown quantities in one of the equations is equal to unity, this inconvenience does not occur, and the method by substitution may be preferable ; the third will, however, commonly be found most convenient. Ex. 2. Given llx+3y=WO \ . . _ 4 1 to find the values of x and y. Multiplying the first equation by 7 and the second by 3, we obtain 77z+21y=700, I2x-2ly= 12. Therefore, by addition, 89^=712, or x= 8. From equation first, 3y= 1001 Ix, = 100-88=12, and 2/=4. These values of x and y may be easily verified by substitu- tion in the original equations. Thus, 11X8+3X4=100; or 88 + 12=100. And 4X87X4= 4; or 32-28= 4. Ex. 3. Given 2+|= 7 I ^ to find the values of x and y. ZC *J i o o .Arcs. x=6, v=12. 84 SIMPLE EQUATIONS x+2 4- wy=ai | to find the values of x and y. Ex. 4. Given ^+ 8y=31 1 ^+10*=192J x+3 Ex. 5. Given 2y- =7 find . 6. Given -+-=m ^ ^ to find the values of # and y. d y c be ad be ad Ans. x=r - -3; y= -- . no ma mcna (113.) When a problem involves a large number of quanti- ties, it is common to designate a part of them by different let ters, and for the remaining quantities to employ the same let ters accented or numbered. Thus, a, a', a", a'", a"" . %. . f,f , ' :V) v '. a<"> (1) , a (3) , a (3) , (4) .' r-, (>n are used to denote different quantities, though they generally imply some connection between the quantities which they rep- resent. a 1 is read a prime; a", a second; a" 1 , a third, &c. We must carefully distinguish between a. 2 and a 2 ; -between a 4 and a\ &c. In the one case, the numerals are exponents, an ' denote powers of a ; while in the other case, the numerals are only used for the sake of convenience to denote distinct quan- tities. Examples showing the convenience of this notation will be found in Sections XIX. and XX. Ex. 7. Given ax +by =c ( , , 7 , _ / C to find the values of x and y. b'cbc' ac'a'c Ans. XT-. - 77,* < y = T~, - n> ab'a'b y ab'a'b The symmetry of these expressions is well calculated to fix them in the memory. Ex. 8. What fraction is that, to the numerator of which, if 4 CONTAINING TWO OR MORE UNKNOWN QUANTITIES. 85 be added, the value is one half; but if 7 be added to the de- nominator, its value is one fifth ? Let - represent the fraction required. Then, by the first condition, z+4 1 - =-; whence 2x+S=y. By the second condition, =- ; whence &c=y+7. Subtracting the first equation from the second, we have 3z-8=7, whence 3#=15, or x=5. Therefore, y=2x+8=W+8=lQ, and the fraction is T 5 . P f 5+41 Pro f- -w=* 5 1 18+7=5' Ex. 9. A certain sum of money, put out at simple interest, amounts in 8 months to 81488, and in 15 months it amounts to 81530. What is the sum and rate per cent. ? Ex. 10. A sum of money put out at simple interest amounts in m months to a dollars, and in n months to b dollars. Required the sum and rate per cent. ? , namb ba Ans. The sum is --- ; the rate is 1200X n m namb' Ex. 11. There is a number consisting of two digits, the second of which is greater than the first; and if the number be divided by the sum of its digits, the quotient is 4 ; but if the digits be inverted, and that number be divided by a num- ber greater by two than the difference of the digits, the quo- tient is 14. Required the number. Let x represent the left hand digit, and y " right hand digit. Then, since x stands in the place of tens, the number will be represented by IQx+y. fiii ....; SIMPLE EdUATIONS Hence, by the first condition, Wx+y_ x+y = By the second condition, Wy+x__ y- X +2~ Whence x=4, y=8, and the required number is 48. Ex. 12. A boy expends thirty pence in apples and pears, buying his apples at 4 and his pears at 5 for a penny, and afterward accommodates his friend with half his apples and one third of his pears for 13 pence. How many did he buy of each ? Ex. 13. A father leaves a sum of money to be divided among his children, as follows : the first is to receive $300 and the sixth part of the remainder ; the second 8600 and the sixth part of the remainder ; and, generally, each succeeding one receives 8300 more than the one immediately preceding, to- gether with the sixth part of what remains. At last it is found that all the children receive the same sum. What was the fortune left and the number of children ? Ans. The fortune was $7500, the number of children 5. Ex. 14. A sum of money is to be divided among several persons, as follows : the first receives a dollars together with the rath part of the remainder ; the second 2a together with the nth part of the remainder; and each succeeding one a dol- lars more than the preceding, together with the nth part of the remainder ; and it is found, at last, that all have received the same sum. What was the amount divided, and the num- ber of persons ? Ans. The amount =a(n I) 2 , the number of persons n I. EdUATIpNS WHICH CONTAIN THREE OB MORE UNKNOWN dUANTITIES. _,. r , (114.) Let us now consider the case of three equations in- volving three unknown quantities. Take the system of equations, 3x+2y+ z=I6, (1.) 2x+2y+2z=l8, (2.) 2x+3y+ z = l7. (3.) CONTAINING TWO OR MORE UNKNOWN QUANTITIES. 87 In order to eliminate z between equations (1.) and (2.), we will divide both members of the second equation by two ; we thus obtain Subtracting this from the first equation, we find a new equa- tion containing but two unknown quantities, 2x+y=1. (a.) In order to eliminate z between equations (1.) and (3.), sub- tract the former from the latter, which gives -x+y=l. ' ((3.) From the two equations (a.) and ((3.), one may be deduced containing only one unknown quantity. For, by subtracting the one from the other, we have 3x=6, or x=2. Substituting this value of x in equation (0.), we obtain y=3. Substituting these values of x and y in equation (1.), we ob- tain 3X2+2X3+z=16. Hence z 4. These values of x, y, and z may be verified by substitution in the original equations. We have effected the elimination in this case by method third, Art. Ill; but either of the other methods might have been employed. Hence, to solve three equations containing three unknown quantities, we have the following RULE. (115.) From the three equations, deduce two containing only two unknown quantities ; then from these two deduce one con- taining only one unknown quantity. Ex. 15. Given x+ y+ z=29 (1.) > x+2y+3z=62 (2.) > to find x, y, and z. i*+iy+i*=io (3.) ) Subtract equation (1.) from (2.), and we obtain y+2z=33; (a.) clearing equation (3.) of fractions, we have 6x+4y+3z=120. (4.) 5 88 SIMPLE EQUATIONS Multiplying equation (1.) by 6, 6z+6y+6z=174. (5.) Subtracting (4.) from (5.), 2y+3z=54. (0.) We have thus obtained two equations, (a.) and (0.), contain- ing two unknown quantities. Multiplying (a.) by 2, we have 2y+4z=66, (6.) Subtracting (|3.) from (6.), z=l2. Substituting this value of % in (0.), we obtain 2y+36=54. Whence y=$> Substituting these values of y and z in equation (1.), Whence x=S. These values may be verified as in former examples. Ex. 16. Given 2x+4y-3z=22 \ 4x2y+5z=l8 > to find x, y, and z. Ans. # . 17. Given x+ya \ x+ z=b > to find x, y, and z. 18. Given =15 > to find x, y, and z. (116.) If we had /owr equations involving four unknown quantities, we might, by the methods already explained, elim- inate one of the unknown quantities. We should thus obtain three equations between three unknown quantities, which might be solved according to Art. 114. So, also, if we had Jive equations involving Jive unknown quantities, we might, by the same process, reduce them to four equations involving four unknown quantities ; then to three, and so on. By following the same method, we might resolve a system of any number of equations of the first degree. Hence, if we have m equa- tions involving m unknown quantities, we proceed by the fol- lowing RULE. 1. Combine successively any one of the equations with each of the others, so as to eliminate the same unknown quantity ; we CONTAINING TWO OR MORE UNKNOWN QUANTITIES. 69' thus obtain ml new equations containing m 1 unknown 2. Eliminate another unknown quantity by combining any one of these new equations with the others; there will result m 2 equations containing m 2 unknown quantities. 3. Continue this series of operations until there results a single equation containing but one unknown quantity, from which the value of this unknown quantity is easily deduced. Then by going back, step by step, to one of the original equa- tions, the values of the other unknown quantities may be suc- cessively determined. Ex. 19. Given 7x-2z +3^= 4y 2z + t=ll 5y3x2u= 8 )> to find x, y, z, u, and t. 4y3u+2 t= 9 3z+8u=33j Ans. x=2, y=4, z=3, u=3, t=l. Either of the unknown quantities may be selected as the one to be first exterminated. It is, however, generally best to begin with that which has the smallest coefficients ; and if each of the unknown quantities is not contained in all the proposed equations, it is generally best to begin with that which is found in the least number of equations. Ex. 20. A person owes a certain sum to two creditors. A* one time he pays them 8530, giving to one four elevenths of the sum which is due, and to the other $ 30 more than one sixth of his debt to him. At a second time he pays them $420, giving to the first three sevenths of what remains due to him, and to the other one third of what remains due to him. What were the debts ? Ex. 21. If A and B together can perform a piece of work in 12 days, A and C together in 15 days, and B and C in 20 days, how many days will it take each person to perform the same work alone ? This Problem is readily solved by first finding in what time they could finish it if all worked together. Ex. 22. If A and B together can perform a piece of work in a days, A and C together in b days, and B and C in c days, 90 SIMPLE EQUATIONS how many days will it take each person to perform the same work alone ? 2abc Ans. A requires -- days, 2abc B " , , , - days, ab+bcac J days . ab+acbc J (11*7.) Hitherto we have supposed the number of equations equal to the number of symbols employed to denote the un- known quantities. This must be the case with every problem, in order that it may be determinate ; that is, that it may not admit of an indefinite number of solutions. Suppose, for example, that a problem involving two un- known quantities (x and y) leads to the single equation x y=3. Now if we make y=l, then x=4 ; y=2, then #=5; y=3, then x=Q ; y=4, then #=7, &c., &c. ; and each of these systems of values, 1 and 4, 2 and 5, 3 and 6, &c., substituted for x and y in the original equation, will sat- isfy it equally well. Hence the problem is indeterminate ; that is, admits of an indefinite number of solutions. (118.) If we had two equations involving three unknown quantities, we could, in the first place, eliminate one of the un- known quantities by means of the proposed equations, and thus obtain one equation containing two unknown quantities, which would be satisfied by an infinite number of systems of values. Therefore, in order that a problem may be determ- inate, its enunciation must contain as many different condi- tions as there are unknown quantities, and each of these con- ditions must be expressed by an independent equation. Equations are said to be independent when they express conditions essentially different ; and dependent when they ex- press the same conditions under different forms. Thus, x+y= 7 ) 2x+v=lO \ are independent equations. CONTAINING TWO OB MORE UNKNOWN QUANTITIES. 91 But x+ y= 7 = 7 ) = 14 J are not Dependent, because the one may be deduced from the other. (119.) If, on the contrary, the number of independent equa- tions exceeds the number of unknown quantities, these equa- tions will be contradictory. For example, let it be required to find two numbers such that their .sum shall be 7, their difference 1, and their product 100. From these conditions we derive the following equations : x+y= 7, x-y= 1, xy=WO. From the first two equations we easily find x=4, and y=S. Hence the third condition, which requires that their product ihall be equal to 100, can not be fulfilled. SECTION IX. DISCUSSION OF EQUATIONS OF THE FIRST DEGREE. INEQUALITIES. (120.) To discuss a problem or an equation is to determine the values which the unknown quantities assume for particular hypotheses made upon the values of the given quantities, and to interpret the peculiar results obtained. The term, there- fore, is not strictly applicable, except to problems which are stated in the most general form, like some of those in Arts. 106 and 107. If the sum of two numbers is represented by a and their difference by &, the greater number will be expressed by a b . . a b -+-, and the less by 5 - -"-ere a and b may have any values whatever, and still these formula will always hold true. It frequently happens that, by .attributing different values to the letters which represent known quantities, the values of the un- known quantities assume peculiar forms which deserve con- sideration. (121.) We may obtain five species of values for the unknown qi lantity in a problem of the first degree. I. Positive values. II. Negative values. III. Values of the form of zero, or -r. A IV. Values of the form of . V. Values of the form of -. We will consider these five cases in succession. DISCUSSION OF EQUATIONS, ETC. 93 I. Positive values are generally answers to problems in the sense in which they are proposed. Nevertheless, all positive values will not always satisfy the enunciation of a problem. If, for example, a problem requires an answer in whole num- bers, and we obtain a fractional value, the problem is impossi- ble. Thus, in Problem 17, page 71, it is implied that the value of x must be a whole number, although this condition is not expressed in the equations. It would be easy to change the data of the problem so as to obtain a fractional value of x, which would indicate an impossibility in the problem pro- posed. Problem 43, page 76, is of the same kind ; also Ex. 11, page 85. If the value obtained for the unknown quantity, even when positive, does not satisfy all the conditions of the problem, the problem is impossible in the form proposed. (122.) II. Negative values. Let it be proposed to find a number, which, added to the number 6, gives for a sum the number a. Let x = the required number. Then, by the terms of the problem, x+b=a, whence x=ab. This formula will give the value of x for every case of the proposed problem. For example, let =7, and b=4. Then x=l 4=3. Again, let a = 5, and b=S. Then x =5_8=-3. We thus obtain for x a negative value. How is it to be in- terpreted ? By referring to the problem, we see that it is proposed to find a number which, added to 8, shall make it equal to 5. Considered arithmetically, the problem is plainly impossible. Nevertheless, if in the equation 8+x=5, we substitute for +z its value 3, it becomes 8-3=5, an identical equation ; that is, 8 diminished by 3 is equal to 5. The negative solution x= 3, shows, therefore, the impossi- bility of satisfying the enunciation of the problem as above stated ; but, taking this value of a: with a contrary sign, we see that it satisfies the enunciation when modified as follows : 94 DISCUSSION or EQUATIONS To find a number which, subtracted from 8, gives a differ- ence of 5 ; an enunciation which differs from the former only in this, that we put subtract for add, and difference for sum. If we wish to solve this new question directly, we shall have 8 x5. Whence x =8 5, or x=3. (123.) For another example, take Problem 50, page 77. The age of the father being represented by a, and that of the son by b ; then y- will represent the number of years be- fore the age of the father will be n times that of the son. Thus, suppose #=54, &=9, and ?i=4. 54-36 18 Then ,=___=_=, That is, the father having lived 54 years and the son 9, in 6 years more the father will be 60 years old and the son 15. But 60 is 4 times 15 ; hence this value, x=6j satisfies the enun- ciation of the problem. Again, suppose #=45, &=15, and n=4. 45-60 -15 Then x= - =-= 5. Here again we obtain a negative solution. How are we to interpret it ? By referring to the problem, we see that the age of the son is already more than one fourth that of the father, so that the time required is already past by five years. The value of x just obtained, taken with a contrary sign, satisfies the following enunciation : A father is 45 years old, his son 15 ; how many years since the age of the father was four times that of his son ? The equation corresponding to this new enunciation is __ _ 45 ~ x Whence 60 4x=45 x; and x=5. (124.) Reasoning from analogy, we deduce the following general principles : 1. Every negative value found for the unknown quantity in a OF THE FIRST DEGREE. 95 problem of the first degree, indicates an absurdity in the condi- tions of the problem, or at least in its algebraic statement. 2. This value, taken with a contrary sign, may be regarded as the answer to a problem', whose enunciation only differs from that of the proposed problem in this, that certain quantities which were ADDED should have been SUBTRACTED, and recipro- cally. (125.) In what case would the value of the unknown quan- tity in Prob. 20, page 72, be negative ? Ans. When Thus, let m=20, n=25, and a=6Q miles. Then *= = - 18 - To interpret this result, observe that it is impossible that the second train, which moves the slowest, should overtake the first. At the time of starting, the distance between them was 60 miles, and every subsequent hour the distance increases. If, however, we suppose the two trains to have been moving uniformly along an endless road, it is obvious that at some former time they must have been together. This negative solution then shows an absurdity in the con- ditions of the problem. The problem should have been stated thus: Two trains of cars, 60 miles apart, are moving in the same direction, the forward one 25 miles per hour, the other 20. How long since they were together ? To solve this problem, let x = the required number of hours. Then 25x the distance traveled by the first train, 20;e = " " second train. And since they are now 60 miles apart, Hence 5z=60, and 2;= + 12. We thus obtain a positive value of x. In order to include both of these cases in the same enuncia- tion, the question should have been asked, Required the time of 'heir being together, leaving it uncertain whether the time was vast or future. In what case would the value of one of the unknown qnan 96 DISCUSSION OF EQUATIONS titles in Problem 34, page 74, be negative ? Why should it be negative ? and how could the enunciation be corrected for this case? In what case would the value of one of the unknown quan- tities in Problem 4, page 67, be negative ? (126.) III. Values of the form of zero, or -T. In what case would the value of the unknown quantity in Problem 20, page 72, become zero, and what would this value signify ? Ans. This value becomes zero when a=0, which signifies that the two trains are together at the outset. In what case would the value of the unknown quantity in Problem 50, page 77, become zero, and what would this value signify ? Ans. When a=nb, which signifies that the age of the fa- ther is now n times that of the son. In what case would the values of the unknown quantities in Problem 38, page 75, become zero, and what would this sig- nify? When a problem gives zero for the value of the unknown quantity, this value is sometimes applicable to the problem, and sometimes it indicates an impossibility in the proposed question. (127.) IV. Values of the form of-. In what case does the value of the unknown quantity in oblem 20, pa pret this result ? Ans. When m=n. On referring to the enunciation of the problem, we see that it is absolutely impossible to satisfy it ; that is, there can be no point of meeting, for the two trains being separated by the distance a, and moving equally fast, will always continue at the same distance from each other. The result - may then be regarded as indicating an impossibility. Problem 20, page 72, reduce to ? and how shall we inter OF THE FIRST DEGREE. 97 The symbol is sometimes employed to represent infinity ; and for the following reason : When the difference m n, without being absolutely nothing, is very small, the quotient _ is very large. For example, let m n=0.01. Then x= =-^-= mn .01 Let m n=0.0001, a a mn .0001 Hence, if the difference in the rates of motion is not zero, the two trains must meet, and the time will become greater and greater as this difference is diminished. If, then, we sup- pose this difference less than any. assignable quantity, the time represented by - - will be greater than any assignable quan- tity, or infinite. j^ Hence we infer, that every expression of the form , found for the unknown quantity, indicates the impossibility of satis- fying the problem, at least in finite numbers. In what case would the value of the unknown quantity m ^ Problem 10, page 70, reduce to the form ? and how shall we interpret this result ? (128.) V. Values of the form of-. In what case does the value of the unknown quantity in Problem 20, page 72, reduce to - ? and how shall we interpret this result? Ans. When a=0, and m=n. To interpret this result, let us recur to the enunciation, and observe that, since a is zero, both trains start from the same point ; and since they both travel at the same rate, they will always remain together, and therefore the required point of meeting will be any where in the road traveled over. Th G 98 OF ZERO AND INFINITY. problem, then, is entirely indeterminate, or admits of an infinite number of solutions, and the expression - may represent any finite quantity. We infer, therefore, that an expression of the form - found for the unknown quantity, generally indicates that it may have any value whatever. In some cases, however, this value is subject to limitations. In what case would the values of the unknown quantities in Problem 44, page 76, reduce to - ? and how would they satisfy the conditions of the problem? Ans. When a=b=c, which indicates that the coins are all of the same value. B might therefore be paid in either kind of coin ; but there is a limitation, viz., that the value of the coins must be one dollar. In what case do the values of the unknown quantities in Problem 38, page 75, reduce to - ? and how shall we interpret this result ? OF ZERO AND INFINITY. (129.) From Art. 127, it is seen that in Algebra we some- times have occasion to consider infinite quantities. It is nec- essary, therefore, to establish some general principles respect- ing them. An infinite quantity is one which exceeds any assignable limit. It is often expressed by the character ce . Thus, a line pro- duced beyond any assignable limit is said to be of infinite length. A surface indefinitely extended, and also a solid of indefinite extent in any one of its three dimensions, are ex- amples of infinity. An infinite quantity does not mean an infinite number of terms. Thus, the fraction | reduced to a decimal, is .333333, &c., without end, but the value of this series is less than unity. Infinite quantities are not all equal among themselves. OF ZEIiO AND INFINITY. 99 Thus the series 1 + 1 + 1 + 1 + 1 +, &c., 2+2+2+2+2+, &c., 3+3+3+3+3+, &c., continued to an infinite number of terms, will each be infinite, although the second series will be double, and the third treble the first. So, also, a line may be infinitely extended both ways ; or it may be infinitely extended in one direction, and limited in the other. In either case, the line is said to be infinite. A quantity less than any assignable quantity is called an in- finitesimal, and is sometimes represented by 0. Thus, take the series of fractions T V, T ^ , r ^> Toio . So, also, a finite quantity is not altered by the addition or subtraction of zero ; that is, a0=7and7>6. As also 5<8 and 3<4, are inequalities which subsist in the same sense ; but the ine- qualities 10>6 and 3<7, subsist in a contrary sense. (132.) I. If we add the same quantity to both members of an inequality, or subtract the same quantity from both members, the resulting inequality will always subsist in the same sense. Thus, 8>3. Adding 5 to each member, 8+5>3+5; and subtracting 5 from each member, 8-5>3 5. Again, take the inequality 3< 2. OF INEQUALITIES. 101 Adding 6 to each member, we have -3+6<-2+6, or 3<4; and subtracting 6 from each member, -3-6<-2-6, or -9<-8. The student must here bear in mind what was stated in Art. 47, of two negative quantities, that is the least whose numer- ical value is the greatest. This principle enables us to transpose any term from one member of an inequality to the other by changing its sign. Thus, a 2 +& 2 >36 2 -2a a . Adding 2 2 to each member of the inequality, it becomes Subtracting 6 2 from each member, or 3 2 >2& a . (133.) II. If we add together the corresponding members of two or more inequalities which subsist in the same sense, the re- sulting inequality will always subsist in the same sense. Thus, 5>4 4>2 Adding, we obtain 16>9. III. But if we subtract the corresponding members of two or more inequalities which subsist in the same sense, the resulting inequality will NOT ALWAYS subsist in the same sense. Take the inequalities 4<7 2<3 Subtracting, we have 4 2<7 3, or 2<4, where the resulting inequality subsists in the same sense. But take 9<10 and 6< 8. Subtracting, the result is 9 6> (not <) 108, or 3>2, where the resulting inequality subsists in the contrary sense. We should therefore avoid as much as possible the use of this transformation, or when we employ it, determine in what sense the resulting inequality subsists. (134.) IV. If we multiply or divide the two members of an in- equality by a positive number, the resulting inequality will sub- sist in the same sense. 102 OF INECIUALITIES. Thus, if a < b. Then ma&. Then na>nb. And - a ->- b -. n n This principle will enable us to clear an inequality of frac- tions. Thus, suppose we have a 2 -fr a a - 3a ' Multiplying both members by Gad, it becomes V. If we multiply or divide the two members of an inequality by a negative number, the resulting inequality will subsist in a contrary sense. Take, for example, 8>7. Multiplying both members by 3, we have the opposite in- equality, -24<-21. So, also, 15>12. Dividing each member by 3, we have -5<-4. Therefore, if we multiply or divide the two members of an inequality by an algebraic quantity, it is necessary to ascer- tain whether the multiplier or divisor is negative, for in this case the inequality subsists in a contrary sense. VI. If we change the signs of both members of an inequality, we must reverse the sense of the inequality, for this transforma- tion is evidently the same as multiplying both members by -1. (135.) VII. If both members of an inequality are positive numbers, we can raise them to any power without changing the sense of the inequality. Thus, 5>3, so also, 5 a >3 2 , or 25 > 9. OF INEQUALITIES. 103 And if a >&, then will a n >b n . VIII. If both members of an inequality are not positive num- bers, and they be raised to any power, the resulting inequality will not always subsist in the same sense. Thus, 2<+3, gives (-2) a <3 2 , or 4<9, where the resulting inequality subsists in the same sense. But -3>-5, gives (-3) a <(-5) 2 , or 9<25, where the resulting inequality subsists in a contrary sense. IX. In extracting the root of both members of an inequali- ty^ it is sometimes necessary to reverse the sense of the ine- quality. Thus, from 9<25, by extracting the square root, we obtain either 3<5, or -3>-5. EXAMPLES. 1. Given 7x 3<25, to find the limit of x. 2. Given 2#+-8<6, to find the limit of a:. o 3. Given |+|+5+|+-j|- 7 > 9 to find the limit of x - bx ab^\ 4. Given -\-cx- c<- r 2 " dxbd X to find the limits of a;. Ans. Ans. #6. 5. A man being asked how many dollars he gave for his watch, replied, If you multiply the price by 4, and to the product add 60, the sum will exceed 256 ; but if you multiply the price by 3, and from the product subtract 40, the re- 104 OF INEQUALITIES. mainder will be less than 113. Required the price of the watch. 6. What number is that whose half and third part added together are less than 105, but its half diminished by its fifth part is greater than 33 ? 7. The double of a number diminished by 6 is greater than 24, and triple the number diminished by 6 is less than double the number increased by 10. Required the number. SECTION X. INVOLUTION AND POWERS. (136.) According to Art. 20, the products formed by the suc- cessive multiplication of the same number by itself are called the powers of that number. Thus, the first power of 3 is 3. The second power of 3 is 9, or 3X3. The fourth power of 3 is 81, or 3X3X3X3, &c., &c., &c. According to Art. 21, the exponent is a number or letter writ- ten a little above a quantity to the right, and denotes the number of times that quantity enters as a factor into a product. Thus, the first power of a is a 1 , where the exponent is 1, which, however, is commonly omitted. The second power of a is a X a, or a 3 , where the exponent 2 denotes that a is taken twice as a factor to produce the pow- er aa. The third power of a is aXaXa, or a 3 , where the exponent 3 denotes that a is taken three times as a factor to produce the power aaa. The fourth power of a is aXaXaXa, or a 4 . Also, the nth power of a is aXaXaXa . . . repeated as a factor n times, and is written a n . Exponents may be applied to polynomials as well as to -mo- nomials. Thus (a+b+c)* is the same as (a+b+c) x (a + b+c) x (a+b+c), or the third power of the entire expression a+b + c. (137.) According to the rule for the multiplication of mono- mials, Arts. 49 and 50. 106 INVOLUTION AND POWERS. (3a6 2 ) a =3a& 2 X 3ab*=9a*b\ So, also, (4a*bc*)*=4a*bc 3 X 4a*bc 3 = 16aW. Hence it appears that, in order to square a monomial, we must square its coefficient, and multiply the exponent of each of the letters by 2. EXAMPLES. 1. Required the square of7axy. Ans. 49aVy a . 2. Required the square of lla*bcd*. 3. Required the square of 12d*xy. 4. Required the square of I5ab*cx*. 5. Required the square of I8x*yz*. ' According to Art. 53, + multiplied by -f, and multiplied by , give +. Now the square of any quantity being the product of that quantity by itself, it necessarily follows that whatever may be the sign of a monomial, its square must be affected with the sign +. Thus the square of +3ax or of Sax is +9aV. (138.) The method of involving a quantity to any power, is easily derived from the preceding principles. Let it be required to form the fifth power of 2a a b z . According to the rules for multiplication, 2 8 & 2 *= 2ab* X 2a*b* Where we perceive 1. That the coefficient has been raised to the fifth power. 2. That the exponent of each of the letters has been multi plied by 5. In like manner, =27a 6 b'c*. Hence, to raise a monomial to any power, we have the fol- lowing RULE. Raise the numerical coefficient to the given power, and multi- ply the exponent of each of the letters by the exponent of the power required. INVOLUTION AND POWERS. 107 EXAMPLES. 1. Required the fourth power of 4 &c., &c., &c. The product of several factors which are all positive, is in- variably positive. Hence, Every EVEN power is positive, but an ODD power has the same sign as its root. EXAMPLES. 1. Required the square of 2x\ Ans. +4z 10 . 2. Required the square of 3x n . 3. Required the cube of 3a 3 . 4. Required the fourth power of 3a?Uh. 5. Required the fifth power of 2a 3 X3x*y. (140.) A fraction is involved by involving both the numerator and denominator. 1. Thus, the square of ^ is ^X^ ; which, by Art. 89, is equal to 77, which, by Art. 68, may be written a 2 &- a . 108 INVOLUTION AND POWERS. A Sa * b ' 9 . * ' r 2. Required the cube of . 3. Required the nth power of -. (141.) Hence, expressions with negative exponents are in- volved by the same rule as those with positive exponents. Thus, let it be required to find the square of a~ a . This expression may be written , which, raised to the second power, becomes or a~\ the same result as would be obtained by multiplying the exponent 3 by 2. Ex. 1. Required the square of 3a?b *. Ex. 2. Required the square of 7a-*b*cr-*dx-\ Ex. 3. Required the cube of 6ab~*dy-\ Ex. 4. Required the fourth power of 3a~ n 6. Ex. 5. Required the fifth power of 2ab~ V. (142.) A polynomial is involved by multiplying it into itself as many times less one as is denoted by the exponent of the power. Ex. 1. Required the fourth power of a+b. a +b a+b a*+ab > a , the second power of a+b. a+b _ a*+2a*b+ab* + a*b+2ab*+b* (a+by=a a +3a*b+3ab*+b a , the third power. a+b _ + ab* (a+by=a'+4a*b+6a*b*+4ab a +b*, the fourth power. Ex. 2. Required the fourth power of a b. Ans. a t -4 Ex. 3. Required the cube of 2a 1. INVOLUTION AND POWERS. 109 Ex. 4. Required the fourth power of Sa h. Ex. 5. Required the square of a+b+c. Hence it appears that the square of a trinomial is composed of the sum of the squares of all the terms, together with twice the sum of the products of all the terms multiplied together two and two. Ex. 6. Required the cube of 2ab+cd. Ex. 7. Required the fourth power of a?+b 3 . Ex. S. Required the cube of a+-. Ex. 9. Required the cube . Z Ex. 10. Required the square of a+b+c+d+e. From this example we infer that the square of any polynomial is composed of the sum of the squares of all the terms, together with twice the sum of the products of all the terms multiplied to- gether two and two, and this proposition may be rigorously demonstrated. It is obvious that this rule for a polynomial includes the pre- ceding rule for a trinomial, and that in Art. 60 for a binomial. SECTION XL EVOLUTION AND RADICAL QUANTITIES. (143.) The square root of a quantity is a factor which, multi- plied by itself once, will produce that quantity. Thus, the square root of a 2 is a, because a when multiplied by itself produces a 2 . The square root of 144 is 12 for the same reason. According to Art. 22, the square root is indicated by the sign V . Thus, Va*=a, and Vl44a*=12a. (144.) According to Art. 137, in order to square a monomial, we must square its coefficient, and multiply the exponent of each of its letters by 2. Therefore, in order to derive the square root of a monomial from its square, we must I. Extract the square root of its coefficient. II. Divide each of the exponents by 2. Thus we shall have This is manifestly the true result, for (8a 3 6 2 ) 2 == 8a 8 6 2 X 8a 3 6 2 = So, also, For, (25ab*c*Y=25ab t c*X25ab i c a , =625aW. 1. Required the square root of 196a a 6Vd 8 . 2. Required the square root of 225a* m b 10 x*. (145.) According to Art. 140, a fraction is involved by in- volving both the numerator and denominator ; hence it is ob- EVOLUTION AND RADICAL QUANTITIES. HI vious that the square root of a fraction is equal to the root of the numerator divided by the root of the denominator. Thus the square root of rj is r. 4a a 1. Find the square root of--:. ** c a 2. rind the square root of e . ia . (146.) It appears, from Art. 144, that a monomial can not be a perfect square unless its coefficient be a square number, and the exponents of its letters all even numbers. Thus, 7ab* is not a perfect square, for 7 is not a square num- ber, and the exponent of a is not an even number. Its square root may be indicated by the usual sign, thus, Vlab*. Ex- pressions of this nature are called surds, or radicals of the sec- ond degree. (147.) We have seen, Art. 137, that whatever may be the sign of a monomial, its square must be affected with the sign +. Hence we conclude that If a monomial be positive, its square root may be either posi- tive or negative. Thus, V9a l =+3a\ or 3 3 , for either of these quantities, when multiplied by itself, pro- duces 9a*. We therefore always affect the square root of a quantity with the double sign , which is read plus or minus. Thus, (148.) If a monomial be affected with a negative sign, the extraction of its square root is impossible, since we have just seen that the square of every quantity, whether positive or negative, is necessarily positive. Thus, v/^4, x/^9, V-5a, are algebraic symbols representing operations which it is im- possible to execute. Quantities of this nature are called im- aginary or impossible quantities, and are symbols of absurdity which we frequently meet with in resolving quadratic equa tions. 6 EVOLUTION AND RADICAL QUANTITIES. Such quantities may be represented by the form V a, which equals VaX 1= VaV~l. So tnat VaV l is a general form for all imaginary quan- tities of the second degree. Thus, -l= 2 V^l That is, the square root of a negative quantity may always be represented by the square root of a positive quantity multiplied by the square root of I. (149.) According to Art. 138, in order to raise a monomial to any power, we raise the numerical coefficient to the given power, and multiply the exponent of each of the letters by the exponent of the power required. Hence, reciprocally, to ex- tract any root of a monomial, we obtain the following RULE. I. Extract the root of the numerical coefficient. II. Divide the exponent of each letter by the index of the re- quired root. Thus, ^/64a e & 3 =4a*b. From Art. 145, it is obvious that to extract ANY root of a fraction, we must divide the root of the numerator by the root of the denominator. , ,270 6 6 3 . 3a*b Thus the cube root of - is - - ; 8x s y 9 %xy Q which may be written -a?bx-*y-\ IB (150.) Let us now consider the sign with which the root should be affected. We have seen, Art. 139, that every even power is positive/but an odd power has the same sign as its root. Thus a, when raised to different powers in succession, will give a, +a a , a 8 , +a 4 , , +a 6 , +a 7 ,&c. Since every even number may be expressed by 2n, every even power may be considered as the square of the nth power, or 2n =(a n ) 2 , and must, therefore, be positive ; and, in like manner, since an odd number may be expressed by 2n+l, every power of an uneven degree may be considered as the product of the 2nth power by the original quantity, ancf must, therefore, have the same sign with the monomial. Hence it appears, I. An odd root of any quantity must have the same sign as the quantity itself Thus, II. An even root of a negative quantity is ambiguous. Thus, V8la*b l *=3ab*. III. An even root of a negative quantity is impossible. For no quantity can be found which, when raised to an even power, can give a negative result. Thus, \/ a, %/b, are symbols of operations which can not be performed, and they are therefore called impossible or imaginary quantities, as V , in Art. 148. EXAMPLES. 1. Find the fourth root of 8 la 8 . Ans. 2. Find the fifth root of 243a 10 & 6 c- 16 . 3. Find the cube root of I25a a x'y 9 . 4. Find the square root of . 5. Find the fifth root of (151.) According to the rule of Art. 149, we perceive that, in order that a monomial may be a perfect power of any degree, its coefficient must be a perfect power of that degree, and the H 114 EVOLUTION AND RADICAL QUANTITIES. exponent of each letter must be divisible by the index of the root. When the quantity whose root is required is not a perfect power of the given degree, we can only indicate the operation to be performed. Thus, if it be required to extract the cube root of 4a 2 6 5 , the operation may be indicated by writing the expression thus, Expressions of this nature are called surds, or irrational quantities, or radicals of the second, *lhird, or nth degree, ac- cording to the index of the root required. (152.) The method of extracting the roots of polynomials will be considered in Section XVII. There is, however, one class so simple and of so frequent occurrence that it may prop- erly be introduced here. In Arts. 60 and 61 we have seen that the square of a+b is a*+2ab-\-b 9 , and the square of a b is a 2 2ab-\-b*. Therefore, the square root of a?2ab+b* is ab. Hence a trinomial is a perfect square when two of its terms are squares, and the third is the double product of the roots of these squares. Whenever, therefore, we meet with a quantity of this de- scription, we may know that its square root is a binomial ; and the root may be found by extracting the roots of the two terms which are complete squares, and connecting them by the sign of the other term. Ex. 1. Find the square root of a a +4a&+46 a . The two terms, a 2 and 4& 2 are complete squares, and the third term 4ab is twice the product of the roots a and 2b ; hence /2 ra- tional. Given surd, v/5+ v/3 v/2 First multiplier, v/5+x/3+v/2 3 x/6 +N/10+X/6-2 _ First product, 2 x/ 1 5 + 6 Second multiplier, 2 -/1 5 6 60+12x/15 -12V/15-36 Second product, 6036=24, a rational quantity. !#. 2. Find multipliers that shall make ^/a+^/b+^/c ra- tional, and determine the product. PROBLEM XL (174.) To reduce a fraction containing surds to another hav- ing a rational numerator or denominator. RULE. Multiply both numerator and denominator by a factor which will render either of them rational, as the case may require. Ex. 1. If both terms of the fraction - be multiplied by v/a, it will become - r, in which the numerator is rational. Or if both terms be multiplied by -/&, it will become r , in which the denominator is rational. 2 Ex. 2. Reduce the fraction to one that shall have a ra- v*> tional denominator. 2v/3 Ans. . 132 IRRATIONAL ClUANTITIES. Ex. 3. Reduce - - - to a fraction having a rational de- \/5 to find the values of x and y. 9y-9x=l8 Ans. x=2, y=4. (188.) When the same algebraic expression is involved to different powers, it is sometimes best to regard this expression as the unknown quantity. Ex. 4. Given x*+2xy+y* +2x=I202y ) to find the val- xy y* =8 S uesofa;andy. Here the first equation may be put under the form where x+y may be regarded as a single quantity, and, by completing the square, we shall find its value to be either 10, or 12. Proceeding now as in the last Article, we shall find x=6, or 9, or 9qpv/5, y=4, or 1, or 3v/5. Ex. 5. Given 4xy=96x*y* ) y I to find the values of x and y. x=Q Here we may regard xy as the unknown quantity, and we shall find its value from the first equation to be either 8, or 12. Proceeding again as in the former Article, we shall find 158 QUADRATIC EQUATIONS z=2, or 4, or 3 to find the values of # and y. x Here - may be treated as the unknown quantity, and we shall find its value to be 5 17 either g, or . From which we find ,_.. ,^^. (189.) When the sum of the dimensions of the unknown quantities is the same in every term of the two equations, it is sometimes best to substitute for one of the unknown quantities the product of the other by a third unknown quantity. Ex. 7. Given x*+xy =56 xy t0 find the ValueS f x and Here, if we assume x=vy, we shall have From the first of these equations, Jfc 1) -f-V 60 and from the second, y 8= ~~T7; v "r~2 . 60 56 therefore, = -r . v+2 ir+t> From which, after completing the square, we obtain 4 7 v=-, or --. o O ; -at ,}- Substituting either of these values in one of the preceding expressions for y a , we shall obtain the values of y ; and since x=vy, we may easily obtain the values of #. .^=14, or An*. \ -, 10> or CONTAINING TWO UNKNOWN QUANTITIES. 159 Ex. 8. Given 2x*+3xy+ y*=20 ) to find the values of x 5x* +4y 2 =41 ) and y. If we assume x=vy, we shall find _1 13 whence, as before, we shall obtain #=1, or > 2 '=3, or V. Ex. 9. Given _ i to find the values of x and v. xyy 1 =12 If we assume xvy 9 we shall find _7 U whence, as before, 11 : fc7, or ^^ Ans. . g ^ 4 ' r 72 (190.) When the unknown quantities in each equation are similarly involved, it is sometimes best to substitute for the unknown quantities the sum and difference of two other quan- tities, or the sum and product of two other quantities. Ex. 10. Given +^-=18 ) y x > to find the values of x and y. x+y=12 ) Here let us assume x=z+v, y=zv. Then, by adding these two equations together, we shall have x+y=2z=I2, or %=6; that is, x=6+v, and y=Q v. But, from the first equation, we find 8 160 UUADRATIC EQUATIONS Substituting the preceding values of x and y in this equa- tion, and reducing, we obtain 432+36u 2 =648-18v a . Whence t>=2. Therefore, #=4, or 8, and y=S, or 4. Ex. 11. Given # 6 +V 6 =3368 y to find the values of x and y. x +y= 8 ) ( x=3, or 5, Ans. ] ( y 5, or 3. - . .r=5, or 0, Ex. 12. Given x 3 +y 3 =341 ),<.,,. , - . a a _ | to find the values of x and x=5, or y=6, or 5. PROBLEMS. 1. Divide the number 100 into two such parts, that the sum of their square roots may be 14. Ans. 64 and 36. 2. Divide the number a into two such parts, that the sum of their square roots may be b. 3. The sum of two numbers is 8, and the sum of their fourth powers is 706. What are the numbers ? Ans. 3 and 5. 4. The sum of two numbers is 2#, and the sum of their fourth powers is 2b. What are the numbers ? Ans. 5. The sum of two numbers is 6, and the sum of their fifth powers is 1056. What are the numbers ? Ans. 2 and 4. 6. The sum of two numbers is 2a, and the sum of their fifth powers is b. What are the numbers ? Ans. ayV-+-^~a'. CONTAINING TWO UNKNOWN QUANTITIES. 16f 7. What two numbers are those whose product is 120 ; and if the greater be increased by 8 and the less by 5, the product of the two numbers thus obtained shall be 300 ? Ans. 12 and 10, or 16 and 7.5. 8. What two numbers are those whose product is a ; and if the greater be increased by b and the less by c, the product of the two numbers thus obtained shall be d? m /m* ab m /m? ab dabc where m=. . c 9. Find two numbers such that their sum, their product, and the difference of their squares may be all equal to one an- other. 3 5,1 5 Ans. 2+\/4 and 2 +v/ 4' that is, 2.618, and 1.618, nearly. 10. Divide the number 100 into two such parts, that their product may be equal to the difference of their squares. Ans. 38.197, and 61.803. 11. Divide the number a into two such parts, that their prod- uct may ftfe equal to the difference of their squares. Ans. and DISCUSSION OP THE GENERAL EQUATION OP THE SECOND DEGREE. (191.) We have seen, Art. 181, that every equation of the second degree may be reduced to the form a; 3 +p x =q, where p and q represent known quantities, either positive OP negative, integral or fractional. The value of x in this equation is either x= -l or 162 DISCUSSION OF THE And, since these values necessarily result from the general equation, we infer, astfi) ^nsJiiturj ow) srh V> PROPERTY I. Every equation of the second degree has two roots, and only two. A root of an equation is such a number as, being substituted for the unknown quantity, will satisfy the equation. This principle has been often exemplified in the preceding pages. Two values have uniformly been found for x, although both values may not be applicable to the problem which fur- nishes the equation. This property will be found demon- strated in a general manner in Art. 294. (192.) If we multiply Id ,OE we shall obtain x*+pxq=Q, which was the equation originally proposed. Tj ' . PROPERTY II. Every equation of the second degree, whose roots are a and b, may be resolved into the two factors x a and x b. Ex. 1. Thus the equation may be resolved into the factors j X J^ where 8 and 2 are the roots of the given equation. It is also obvious that if a is a root of an equation of the sec- ond degree, this equation must be divisible by x a. Thus the preceding equation is divisible by x8, giving the quotient Ex. 2. The roots of the equation GENERAL EdUATION OF THE SECOND DEGEEE. 163 are 2 and 4. Resolve it into its factors. Ex. 3. The roots of the equation are +3 and 9. Resolve it into its factors. Ex. 4. The roots of the equation x* 2x 24=0, are +6 and 4. Resolve it into its factors. (193.) If we add together the two values of x in the gen- eral equation of the second degree, the radical parts having opposite signs disappear, and we obtain "~2~2 = ~^' Hence, PROPERTY III. The algebraic sum of the two roots is equal to the coefficient of the second term of the equation, taken with a contrary sign. Thus, in Ex. 1, page 145, the two roots are 8 and 2, whose sum is +10, the coefficient of x taken with a contrary sign. In the equation the two roots are 2 and 4. In the equation the two roots are 6 and 10. If the two roots are equal numerically, but have opposite signs, their sum is zero, and the second term of the equation vanishes. Thus the two roots of the equation x 2 =16, are +4 and 4, whose sum is zero. This equation may be written (194.) If we multiply together the two values of # (observ- ing that the product of the sum and difference of two quan- tities is equal to the difference of their squares), we obtain 164 DISCUSSION OF THE Hence, PROPERTY IV. The product of the two roots is equal to the second member of the equation, taken with a contrary sign. Thus, in the equation the product of the two roots 8 and 2 is +16, which is equal to the second member of the equation taken with a contrary sign. So, also, in the equation whose two roots are +3 and 9, their product is 27. The two last properties enable us readily to form an equa- tion when its roots are known. Ex. 1. Let it be required to form the equation whose roots are 2 and 8. According to Property III., the coefficient of the second term of the equation must be 10; and, from Property IV., the second member of the equation must be 16. Hence the equation is Ex. 2. Form the equation whose roots are 3 and 5. Ex. 3. Form the equation whose roots are 4 and 7. Ex. 4. Form the equation whose roots are 5 and 9. Ex. 5. Form the equation whose roots are 6 and +11. REAL AND IMAGINARY VALUES OF THE UNKNOWN QUANTITY. (195.) The values of a: in the general equation of the second degree are Values of the unknown quantity which are not imaginary are, for the sake of distinction, called real. GENERAL EUUATION OP THE SECOND DEGREE. 165 Since ^-, being a square, is positive for all real values of p, P* it follows that the expression q+^ can only be rendered neg- ative by the sign of q. When q is positive, or when q is negative and numerically 2? 59 less than , then will <7+^r be positive, and, consequently, \/q+ ~r wnl l be real. This happens in nearly all the preced- ing examples. When q is negative, and numerically greater than -, then q+*-r will be negative, and, consequently, y<7+^r will be im- aginary. This happens in Ex. 5, page 146. CASE I. When + ~r i 1. When, in the equation x*+px=q, p is negative, and is numerically greater than y^+x* both values of x w;i7/ &e raz/ and positive. This happens in the equation x* 6x=-8, whose two roots are 4 and 2. Also in the equation whose two roots are 8 and 2. f) 2. When p is positive, and is numerically greater than / P" Y/ ?+ 4 > ^ ;0 ^ values of x wu7/ 6e real and negative. 4 This happens in the equation 166 DISCUSSION OF THE whose two roots are 2 and 4. Also in the equation whose two roots are 6 and 10. 3. When |- is numerically less than y <7+ ~T both values of x will be real, the one positive and the other negative. This happens in the equation whose roots are +3 and 9. Also in the equation ' whose roots are +6 and 4. CASE II. (196.) When y q+^r is imaginary. In this case, both values of x are imaginary. This happens in the equation x*-8x=-18, whose roots are 4V 2. We will now prove that in this case the conditions of the question are incompatible with each other, and therefore the values of x ought to be imaginary. The demonstration de- pends upon the following principle : The greatest product which can be obtained by dividing a number into two parts and multiplying them together, is the square of half that number. Let p = the given number, and d = the difference of the parts. Then, from page 67, ^-f-g = the greater part, p d ~~- - the less part, GENERAL EQUATION OF THE SECOND DEGREE. 167 8 d 3 and ^ = the product of the parts. Now, since p is a given quantity, it is plain that this expres- 2 sion will be the greatest possible when d=0 ; that is, ^- is the greatest product, which is the square of ^, half the given number. For example, let 12 be the number to be divided. We have 12=1 + 11; and 11X1 = 11. 12=2 + 10; and 10X2=20. 12=3+ 9; and 9X3=27. 12=4+ 8; and 8X4=32. 12=5+ 7; and 7X5=35. 12=6+ 6; and 6x6=36. We here see that the smaller the difference of the two parts, the greater is their product ; and this product is greatest when the two parts are equal. Now, in the equation x*px=-q, p is the sum of the two roots, and q is their product. There- P* fore q can never be greater than ~. If, then, any problem furnishes an equation in which q is negative, and greater than ^-, we infer that the conditions of the question are incompatible with each other. Thus, in the example =^-=9, which is numerically less than q. The equation re- quires us to divide the number 6 into two parts whose product shall be 10, which is an impossibility; and, accordingly, in solving the equation, we obtain imaginary values for x. Hence an imaginary root indicates an absurdity in the pro- posed question which furnished the equation. Suppose it is required to divide 8 into two such parts that their product shall be 18. 168 DISCUSSION OF PARTICULAR PROBLEMS. Let x = one of "the parts, and S-x = the other. Then, by the conditions, ^v ~#)~- Whence x*-8x=-l8. This equation, solved by the usual method, gives x=4V 2, an imaginary expression. Hence we infer that it is impossible to find two numbers whose sum is 8, and product 18. This is obvious from the Proposition above demonstrated, from which it appears that 16 is the greatest product which can be obtained by dividing 8 into two parts, and multiplying them together. 3 (197.) When q is negative, and numerically equal to ^-, the radical part of both values of x becomes zero, and both values of a; reduce to ~. The two roots are then said to be equal. Thus, in the equation the two roots are 3 and 3. We say that in this case the equation has two roots, because it is the product of the two factors, a; 3=0, and a; 3=0. DISCUSSION OF PARTICULAR PROBLEMS. (198.) In discussing particular problems which involve equa- tions of the second degree, we meet with all the different cases which are presented by equations of the first degree, and some peculiarities besides. We may therefore have, 1. Positive values of x. 2. Negative values. 3. Values of the form of -r. ^ 4. Values of the form of . 5. Values of the form of -. All these different cases are presented by Problem 10, DISCUSSION OF PARTICULAR PROBLEMS, IG'J page 155, when we make different suppositions upon the values of a, m, and n ; but we need not dwell upon them here. The peculiarities exhibited by equations of the second de- gree are, 6. Double values of x. 7. Imaginary values. We will consider the last two cases. (199.) Double values of the unknown quantity. We have seen that every equation of the second degree has two roots. Sometimes both of these values are applicable to the problem which furnishes the equation. Thus, in Problem 20, page 155, we obtain either 100 or 180 miles for the dis- tance between the places C and D. C E D I I I Let E represent the situation of A when B sets out on his journey. Then, if we suppose CD equals 100 miles, ED will equal 55 miles, of which A will travel 30 miles (being 6 miles an hour for 5 hours), and B will travel 25 miles (being 5 miles an hour for 5 hours). If we suppose CD equals 180 miles, ED will equal 135 miles, of which A will travel 54 miles (being 6 miles an hour for 9 hours), and B will travel 81 miles (being 9 miles an hour for 9 hours). This problem, therefore, admits of two positive answers, both equally applicable to the question. Problem 22, page 156, is of the same kind; and another will be found on page 193. In Problem 18, page 155, one of the values of x is positive, and the other negative. C' A C B I I I Let the weakest magnet be placed at A, and the strongest at B ; then C will represent the situation of a needle equally attracted by both magnets. According to the first value, the distance AC=8 inches, and CB=12. Now at the distance of R inches, the attraction of the weakest magnet will be repre- 4 sented by ; and at the distance of 12 inches, the attraction o 170 DISCUSSION OP PARTICULAR PROBLEMS. g of the other magnet will be represented by an d tnese two powers are equal ; foi |;a\ What is the absurdity involved in this supposition ? Ans. It is absurd to suppose that the product of two num- bers can be greater than the square of half their sum. When will the values of x in Problem 11, page 154, be imag- inary ? Ans. When a a >6; or (24&. What is the absurdity of this supposition ? Ans. The square of the sum of two numbers can not be greater than twice the sum of their squares. When will the values of x in Problem 17, page 155, be im- aginary ? Ans. When a*>b; or (2a)*>Qb. What is the absurdity of this supposition ? Ans. The cube of the sum of two numbers can not be greater than four times the sum of their cubes. When will the values of x in Problem 4, page 140, be im- aginary, and what is the absurdity of this supposition ? SECTION XIII. RATIO AND PROPORTION. Numbers may be compared in two ways : either by means of their difference, or by their quotient. We may in- quire how much one quantity is greater than another ; or, how many times the one contains the other. One is called Arith- metical, and the other Geometrical Ratio. The difference between two numbers is called their Arith- metical Ratio. Thus, the arithmetical ratio of 9 to 7 is 9 7, or 2 ; and if a and b designate two numbers, their arithmetical ratio is represented by a b. Numbers are more generally compared by means of quo- tients ; that is, by inquiring how many times one number con- tains another. The quotient of one number divided by another is called their Geometrical Ratio. The term Ratio, when used without any qualification, is always understood to signify a geometrical ratio, and we shall confine our attention to ratios of this description. (202.) By the ratio. of two numbers, then, we mean the quo- tient which arises from dividing one of these numbers by the other. 12 Thus, the ratio of 12 to 4 is represented by , or 3. The ratio of 5 to 2 is -, or 2.5. The ratio of 1 to 3 is -, or .333, &c. o We here perceive that the value of a ratio can not always be expressed exactly in decimals ; but, by taking a sufficient 173 RATIO AND PROPORTION. number of terms, we can approach as nearly as we please to the true value. If a and b designate two numbers, the ratio of a to b is the quotient arising from dividing a by 6, and may be represented by writing them a : b, or 7. The first term, , is called the antecedent of the ratio ; the last term, b, is called the consequent of the ratio. Hence it appears that the theory of ratios is included in the theory of fractions, and a ratio may be considered as a fraction whose numerator is the antecedent, and whose denominator is the consequent. (203.) When the antecedent of a ratio is greater than the consequent, the ratio is called a ratio of greater inequality ; as, 5 12 -, . When the antecedent is less than the consequent, it is 2 5 called a ratio of less inequality ; as, -, -. When the antece- O y dent and consequent are equal, it is called a ratio of equality ; Q fi as, o o' I* i s pl a i n that a ratio of equality may always be o o represented by unity. (204.) When the corresponding terms of two or more sim- ple ratios are multiplied together, the ratios are said to be compounded. Thus, the ratio of r, compounded with the ratio c ac of -, becomes 7-3. a oa When a ratio is compounded with itself, the result is called 2 4 a duplicate ratio. Thus, the duplicate ratio of - is - ; and the o y duplicate ratio of 7 is -75. b b A ratio compounded of three equal ratios is called a tripli- 2 8 cate ratio. Thus, the triplicate ratio of - is ; and the tripli- c a. a 3 cate ratio of 7 is 7^. o o The ratio of the square roots of two quantities is called a EATIO AND PROPORTION. 173 4 2 subduplicate ratio. Thus, the subduplicate ratio of - is - ; and 9 3 the subduplicate ratio of =- is j-. The ratio of the cube roots of two quantities is called a sub- 8 2 triplicate ratio. Thus, the subtriplicate ratio of is - ; and the subtriplicate ratio of 7 is -TTT- (205.) If the terms of a ratio are both multiplied, or both di- vided by the same quantity, the value of the ratio remains un- changed. The ratio of a to b is represented by the fraction T, and the value of a fraction is not changed if we multiply or divide both numerator and denominator by the same quantity. Thus, a n a b or a : o=ma : mo=- : -. n n (206.) Ratios are compared with each other by reducing the fractions which represent them to a common denominator. In order to ascertain whether the ratio of 2 to 7 is greater or less than that of 3 to 8, we represent these ratios by the 2 3 fractions - and -, and reduce them to a common denominator. 7 o They thus become 16 , 21 and, since the latter of these is the greatest, we infer that the ratio of 2 to 7 is less than the ratio of 3 to 8. (207.) A ratio of greater inequality is diminished, and a ratio of less inequality is increased, by adding the same quantity to both terms. 174 RATIO AND PROPORTION. 22+1 3 To prove the proposition generally, let T represent any ra- tio, and let x be added to each of its terms. The two ratios will then be a a+x which, reduced to a common denominator, become , . , , , ao-\-ax ab-\-bx &(6+a5' b(b+x)' Now if >&, that is, if =- is a ratio of greater inequality, then, since ax is greater than bx, the first of these fractions is great- er than the second, and therefore r is diminished by the addi- tion of the same quantity to each of its terms. But if a<&, that is, if =- is a ratio of less inequality, then, since ax is less than bx, the first of the above fractions is less than the second, and therefore 7 is increased by the addition of the same quantity to each of its terms. (208.) If, in a series of ratios, the consequent of each is the antecedent of the following ratio, then the ratio of the first an- tecedent to the last consequent is equal to that which is corn,' pounded of all the intervening ratios. Let the proposed ratios be a b c d e b' c' d' e' f Compounding them by Art. 204, we obtain abcde which, being divided by be de, reduces to i RATIO AND PROPORTION. 175 PROPORTION. (209.) Proportion is an equality of ratios. Thus, if a, b, c, d are four quantities, such that , when di- vided by b, gives the same quotient as c when divided by d, then a, b, c, d are called proportionals, and we say that a is to b as c is to d; and this is expressed by writing them thus : a : b : : c : d, or a : b=c : d, a c or i=d- So, also, 3, 4, 9, 12 are proportionals ; that is, 3:4::9:12 3 9 4 = l2* In ordinary language, the terms ratio and proportion are confounded with each other. Thus, two quantities are said to be in the proportion of 3 to 5, instead of the ratio of 3 to 5. A ratio subsists between two quantities, a proportion only be- tween four. Ratio is the quotient arising from dividing one quantity by another ; two equal ratios form a proportion. (210.) In the proportion a : b : : c : d, a,b,c,d are called the terms of the proportion. The first and last terms are called the extremes, the second jand third the means. The first term is called the first antecedent, the second term the first consequent, the third term the second antecedent, and the fourth term the second consequent. The word term, when applied to a proportion, is used in a slightly different sense from that v explained in Art. 27. The terms of a proportion may be polynomials. Thus, a+b:c+d::e+f:g+h. (211.) When the second and third terms of a proportion are identical, this quantity is called a mean proportional between the other two. Thus, if we have three quantities, a, b, c, such that a : b : : b : c, 176 RATIO AND PROPORTION. then b is called a mean proportional between a and c, and c is called a third proportional to a and b. If, in a series of proportional magnitudes, each consequent is identical with the next antecedent, these quantities are said to be in continued proportion. Thus, if we have a, ft, c, d, e, f such that a : b :: b : c :: c : d :: d : e :: e :/, a b c d e or -=-=-=-=, the quantities a, b, c, d, e, f are in continued proportion. (212.) If four quantities are proportional, the product of the extremes is equal to the product of the means. Let a : b : : c : d. Then will ad=bc. For, since the four quantities are proportional, tt^'i' a c b d' Multiplying each of these equals by bd, the expression be- comes abd_bcd or ad=bc. Thus, if 3 : 4 : : 9 : 12, then 3X12=4X9. (213.) Conversely, if the product of two quantities is equal to the produQt of two others, the first two quantities may be made the extremes, and the other two the means of a proportion. Let ad=bc. Then will a : b : : c : d. For, since ad=bc, dividing each of these equals by bd, the expression becomes a_c c _a that is, a : b : : c : d, or c : d : : a : b. Thus, if 3X12=4X9, then 3 : 4 : : 9 : 12, or 9 : 12 : : 3 : 4. RATIO AND PROPORTION. 177 (214.) The preceding proposition is called the test of propor- tions, and any change may be made in the form of a propor- tion which is consistent with the application of this test. In order, then, to decide whether four quantities are proportional, we must compare the product of the extremes with the product of the means. Thus, to determine whether 5, 6, 7, 8 are proportional, we multiply 5 by 8, and obtain 40. Multiplying 6 by 7, we ob- tain 42. As these two products are not equal, we conclude that the numbers 5, 6, 7, 8 are not proportional. Again, take the numbers 5, 6, 10, 12. The product of 5 by 12 is 60, and the product of 6 by 10 is also 60. Hence these numbers are proportional ; that is, 5:6:: 10: 12. (215.) If three quantities are in continued proportion, the product of the extremes is equal to the square of the mean. If a:b::b:c. Then, by Art. 212, ac=bb, which is equal to b'\ Conversely, if the product of two quantities is equal to the square of a third, the last quantity is a mean proportional be- tween the other two. Thus, let ac=b\ Dividing these equals by be, we obtain a_b ft"? or a :b : : b : c. Thus, if 4 : 6 : : 6 : 9, then 4X9:=6 2 . And conversely, if 4X9=6 2 , then 6 is a mean proportional between 4 and 9. EXAMPLES. 1. Given the first three terms of a proportion, 24, 15, and 40, to find the fourth term. 2. Given the first three terms of a proportion, 3fe 3 , 4a 2 6% and 9a s b, to find the fourth term. 3. Given the last three terms of a proportion, 4 b a Subtract unity from each of these equals, and we have a c ab cd --l = -_l,or =^-; that is, ab :b :: cd : d. (221.) If four quantities are proportional, they will be pro- portional by conversion ; that is, the first will have to the dif- ference of the first and second the same ratio that the third has to the difference of the third and fourth. Let a:b::c:d; then will a : ab : :c : cd. For since a : b : : c : d 9 by inversion, b : a : : d : c ; b d whence -=-. a c Subtract each of these equals from unity, and we have 180 RATIO AND PROPORTION'. b d ab cd 1 =1 , or - = - ; a c a c that is, ab : a :: cd : c, or inversely, a : ab :: c : cd. (222.) If four quantities are proportional, the sum of the first and second will have to their difference the same ratio that the sum of the third and fourth has to their difference. hh Let a:b: c-.d; then will a+b : ab : c+d : c-d. For since a:b: c:d, by composition, a+b : b : c+d : d, and by alternation, a+b : c+ i::b:d. Also, since a:b: c:d, by division, ab : b : cd : d. and by alternation, ab : cd :: b : d. Hence, by equality of ratios, a+b : ab : c+d : cd. (223.) If four quantities are proportional, like powers or roots of these quantities will also be proportional. Let a : b : : c : d ; then will a n : b n : : c n : d n . For since a : b : : c : d, a c we have T= > o a Raising each of these equals to the rath power, we obtain lr~I ; that is, a n :b n ::c n : d\ where n may be either a whole number or a fraction. (224.) If there is any number of proportional quantities all having the same ratio, the first will have to the second the same, ratio that the sum of all the antecedents has to the sum of all the consequents. Let a, &, c, d, e, f be any number of proportional quantities such that a : b :: c : d :: e :/, then will a : b : : a+c+e : b+d+f. RATIO AND PROPORTION. 181 For since a : b : : c : d, we have ad=bc; and since a : b : : e :/, we have af=be. To these equals add ab=ba, and we obtain a(b+d+f)=b(a+c+e). Hence, by Art. 213, a : b : : a*\-c+e : b+d+f. (225.) If three quantities are in continued proportion, the first will have to the third the duplicate ratio of that which it has to the second. Let a : b : : b : c. Then a : c : : a a : b*. For since a : b : : b : c, by Art. 212, ac=b*. Multiplying each of these equals by a, we obtain a a c=a& 3 ; that is, a*Xc=aXb*. Resolving this equation into a proportion by Art. 213, we have a : c : : a* : b\ (226.) If four quantities are in continued proportion, the first will have to the fourth the triplicate ratio of that which it has to the second. Let a, bj c, d be four quantities in continued proportion, so that a :b :: b : c :: c : d, then will a : d : : a 9 : b 9 . For since a : b : : c : d, we have ad=bc; and since a : b : : b : c, we have ac=b*. Multiplying these equals by ab, we obtain a\bdc)=b\abc\ or a*Xd=b*Xa. Hence, by Art. 213, a : d - : a 9 : b 3 . 182 RATIO AND PROPORTION. (227.) If there are two sets of proportional quantities, the products of the corresponding terms will be proportional. Let a : b : : c : d, and e:fi:g:h. Then will ae:bf::cg: dh. For, since a : b : : c : d, by Art. 212, ad=bc. And since e :f :: g : h, by Art. 212, eh=fg. Multiplying these equals together, we have . aeXdh=bfXcg. Hence, by Art. 213, ae : bf : : eg : dh. (228.) Three quantities are said to be in harmonical propor- tion when the first is to the third as the difference between the first and second is to the difference between the second and third. Thus, 2, 3, 6 are in harmonical proportion ; for 2: 6:: 3-2 : 6-3. Let a, 6, c be in harmonical proportion ; then *v i. i. a : c : : ab : bc. Multiplying the extremes and means, and reducing, we have c"z. r- 2a b Hence, to find a third harmonical proportional to two quan- tities, divide the product of the first and second by twice the first diminished by the second. Ex. 1. Find a third harmonical proportional to 3 and 5. Ex. 2. Find a third harmonical proportional to 5 and 8. (229.) Four quantities are said to be in harmonical propor- tion when the first is to the fourth as the difference between the first and second is to the difference between the third and fourth. Thus, 2, 3, 4, 8 are in harmonical proportion ; for 2:8:: 3-2 : 8-4. Let , &, c, d be in harmonical proportion ; then aid:' ab : cd. Multiplying the extremes and means, and reducing, we have RATIO AND PROPORTION. 183 > .'. ,.i cJl. Hence, to find a fourth harmonical proportional to three quantities, divide the product of the first and third by twice the first diminished by the second. Ex. 1. Find a fourth harmonical proportional to 4, 5, and 6. Ex. 2. Find a fourth harmonical proportional to 5, 8, and 10. (230.) Proportions are often expressed in an abridged form. Thus, if A and B represent two sums of money put out for one year at the same rate of interest, then A : B : : interest of A : interest of B. This is briefly expressed by saying that the interest varies as the principal. A peculiar character QD is used to denote this relation. Thus, we write .,. , v the interest QD the principal. One quantity varies directly as another, when both increase or diminish together in the same ratio. Thus, in the above example, A varies directly as the interest of A. In such a case either quantity is equal to the other multiplied by some con- stant number. Thus, if the interest varies as the principal, then the interest equals the principal multiplied by a constant quantity, which is the rate of interest. If A QD B, then A=wB. If the space (S) described by a falling body varies as the square of the time (T), then m representing some constant quantity. (231.) One quantity may vary directly as the product of several others. Thus, if a body moves with uniform velocity, the space described is measured by the product of the time by the velocity. If we put S to represent the space described, T the time of motion, and V the uniform velocity, then we shall have SooTxV. Also the area of a rectangle varies as the product of its length and breadth. The weight of a stick of timber varies as its length X its breadth X its depth X its density. 9 184 RATIO AND PROPORTION. If the density is given, then the weight varies as the length X the breadth X the depth. If the depth also is given, then the weight varies as the length X the breadth. If the breadth is given, then the weight varies as the length. Finally, if the length also is given, then the weight is equal to a constant quantity. (232.) One quantity varies inversely as another when one increases in the same ratio that the other diminishes. Thus, the altitude of a triangle whose area is given, varies inversely as its base. If the product of two quantities is constant, then one varies inversely. as the other. j In uniform motion, the space is measured by the product of the time by the velocity ; that is, Whence T=. If the space be supposed to remain constant, then that is, the time required to travel a given distance varies in- versely as the velocity. Suppose the distance is 360 miles : then, if the velocity is 12 miles per hour, the time will be 30 hours ; " 20 " " 18 " " 24 " " 15 " that is, if the velocity is doubled, the time is halved. The one varies inversely as the other. Conversely, if one quantity varies inversely as another, the product of the two quantities is constant. n . 3fltrr 9*i; ', Thus, if T OD i then the space (S) is a constant quantity. (233.) One quantity may vary directly as a second, and in- versely as a third. Thus, according to the Newtonian law of gravitation, the attraction (G) of any heavenly body varies ' RATIO AND PROPORTION. 185 directly as the quantity of matter (Q), and inversely as the square of the distance (D). That is, G oo ^. (234.) Application of the preceding principles. Ex. 1. Given x+yix:-. 5:3, j | to find the values of a: and y. Since x+y : x : : 5 : 3. By division, Art. 220, y : x : : 2 : 3. Q/w Therefore, 3y=2#, and y=-^-. o Substituting this value of y in the second equation, we obtain |U Therefore, z and y jBa:. 2. Given x+y : x y : : 3 : 1, / to find the values of x x 9 y 3 =56, S and y. From the first equation, by Art. 222, we obtain 2x : 2y : : 4 : 2 ; whence, # : y : : 2 : 1, and x=2y. Substituting this value of x in the second equation, we ob- tain Ex. 3. Given x+y : x y : : 64 : 1, j to find the values of a: xy= 63, ) and y. By Art. 223, x+y : xy : : 8 : 1. By Art. 222, 2x : 2y : : 9 : 7 ; whence a; : y : : 9 : 7. Therefore, *=T. Substituting this value of x in the second equation, we ob- tain 186 RATIO AND PROPORTION. Ex. 4. Given x* y* : x y : : 61 : 1, | to find the values of xy=320, ) x and y. \J?. f^ A Ar4 r jT' Since a; 3 -v 3 : x*3x*v+3xy* y 9 : : 61 : 1. By division, Art. 220, 3xyx(xy) : xy : : 60 : 1. 2 60: 1, 1:1. Hence 960 : xy : ? and 16 : xy : Therefore, xy =4. Also, since And By addition, ; Extracting the root, x+y=36. Hence x=20, or 16, y=16, or 20. Ex. 5. Given x 3 y 3 : cc a y xy a : : 7 : 2, > to find the values x+y 6, y ofxandy. Ans. x=4, or 2 ; y=2, or 4. Ex. 6. Given Vy Vax= Vyx, ) to find the Vyx+ Vax : Vax : : 5 : 2, ) values of x and y. 4# 5a 4 S . X =-;y=-. Ex. 7. Given x+ tfhQ| .';i?k 9tiiL saotiw daod^irur tiaiiffrwa owj Ifiii W . U .^ SECTION XIV. ikda lioidw lo i9la'j aiUsxadaiuu 1/- . . J . Ex. 12. Find m arithmetical means between two given num- bers. In order to solve this problem, we must first find the com- mon difference. The whole number of terms consists of the two extremes and all the intermediate terms. If, then, m rep- resent the number of means, m+2 will be the whole number of terms,,,;, w 3r Substituting m+2 for n, in Formula 9, page 190, we have 2*11' .?!- I a d= r-r= the common difference, m+1 whence .the required means are easily obtained by addition. Ex. 13. Find 6 arithmetical means between 1 and 50. PROGRESSIONS. 193 Ex. 14. Find three numbers in arithmetical progression, the sum of whose squares shall be 1232, and the square of the mean greater than the product of the two extremes by 16. Ans. 16, 20, and 24. In examples of this kind, it is generally best to represent the series in such a manner that the common difference may dis- appear in taking the sum of the terms. Thus a progression of three terms may be represented by a d, a, a-\-d; one of four terms by a 3d, ad, a+d, a+3d, &c. Ex. 15. Find three numbers in arithmetical progression, the sum of whose squares shall be , and the square of the mean greater than the product of the two extremes by b. /a-2b /a-2b /a-2b . y 3 Vb; \/~3~ ; and V "3"+ Vb ' Ex. 16. Find four numbers in arithmetical progression whose sum is 28, and continued product 585. Ans. 1, 5, 9, 13. Ex. 17. A sets out for a certain place, and travels 1 mile the first day, 2 the second, 3 the third, and so on. In five days afterward B sets out, and travels 12 miles a day. How long will A travel before he is overtaken by B ? Ans. 8 or 15 days. This is another example of an equation of the second de- gree, in which the two roots are both positive. The following diagram exhibits the daily progress of each traveler. The di- visions above the horizontal line represent the distances trav- eled each day by A ; those below the line the distances trav- eled by B. A. B. 1234 II 1 1 5 1 6 7 1 1 8 1 9 1 10 1 11 1 12 1 13 1 14 1 1 1 1 1 2 3 4 1 5 6 1 7 -8 1 9 1 It is readily seen from the figure that A is in advance of B until the end of his 8th day, when B overtakes and passes him. After the 12th day, A gains upon B, and passes him on the 15th day, after which he is continually gaining upon B, and could not be again overtaken. Ex. 18. A goes 1 mile the first day, 2 the second, and so on. N 194 PROGRESSIONS. B starts a days later, and travels b miles per day. How long will A travel before he is overtaken by B ? 2b-IV(2b-iy-8ab , Ans. *~ days. In what case would B never overtake A ? Ans. When >M-J ' 9c "" h- For instance, in the preceding example, if B had started one day later, he could never have overtaken A. Ex. 19. A traveler set out from a certain place and went 1 mile the first day, 3 the second, 5 the third, and so on. After he had been gone three days,- a second traveler sets out, and goes 12 miles the first day, 13 the second, and so on. In how many days will the second overtake the first ? Ans. In 2 or 9 days. Let the student illustrate this example by a diagram like the preceding. GEOMETRICAL PROGRESSION. (240.) A Geometrical Progression is a series of quantities, each of which is equal to the product of that which precedes it by a constant number. > yrj < Thus, the series 2, 4, 8, 16, 32, &c., and 81, 27, 9, 3, &c., are geometrical progressions. In the former, each number is derived from the preceding by multiplying it by 2, and the series forms an increasing geometrical progression. In the latter, each number is derived from the preceding by multiply- ing it by J, and the series forms a decreasing geometrical pro- gression. In each of these cases, the common multiplier is called the common ratio. (241.) To find the last term of a geometrical progression. Let a represent the first term of the progression, and r the common ratio ; then the successive terms of the series will be a, ar, ar a , r 8 , ar 4 , &c. The exponent of r in the second term is 1, in the third term PROGRESSIONS. 195 is 2, in the fourth term 3, and so on ; hence the rath term of the series will be ar n ~\ If, therefore, we put I for the last term, and n the number of terms of the series, we shall have That is, The last term of a geometrical progression is equal to the product of the first term by that power of the ratio whose expo- nent i$., Qjte less than the number of terms. (242.) To find the sum of all the terms of a geometrical pro- gression. If we take any geometrical series, and multiply each of its terms by the ratio, a new series will be formed, of which ev- ery term except the last will have its corresponding term in the first series. Thus, take the series 1, 2, 4, 8, 16, 32, the sum of which we will represent by S, so that 8=1+2+4+8 + 16+32. Multiplying each term by 2, we obtain 28=2+4+8+16+32+64. The terms of the two series are identical, except the first term of the first series and the last term of the second series. If, then, we subtract one of these equations from the other, all the remaining terms will disappear, and we shall have 28-8=64-1. In order to generalize this method, let a, ar, ar 2 , &c., rep- resent any geometrical series, and 8 its sum ; then S =a +ar+ar*+ar a + ...... +ar n Multiplying this equation by r, we have Subtracting the first equation from the second, we obtain rS-S=ar"-. Hence 8= -y- ; or, substituting the value of I already found, we shall have 196 PROGRESSIONS. Hence, to find the sum of the terms of a geometrical pro- gression, Multiply the last term by the ratio, subtract the first term, and divide the remainder by the ratio less one. If the series is a decreasing one, and r consequently repre- sents a fraction, it is convenient to change the signs of both numerator and denominator in this expression, which then be- comes a ar alr =7=r=T^r- (243.) In the two fundamental equations s= Ir-a r-1' \+r. 3 - it-i 1 J. * 'i t* there are five variable quantities, a, /, r, 7i, S, of which any three being given, the other two may be found. Accordingly, as in arithmetical progression, 20 different cases may arise, all of which are readily solved, with the exception of those in which n is the quantity sought. The value of n can only be found by the solution of an exponential equation. See Art. 352. These different cases are all exhibited in the following table for convenient reference. No. Given. | Required. Formulae. 1 2 3 4 a, r, n a, r, S #, n, S r, 7i, S / l=ar n ~\ a+(r-l)S r ' /(S-/r>=a(S-arV (r-l)Sr"- 1 * r"-l ' 5 6 7 8 a, r, n a, r, I a, n, / r, n, / S o ar"-a = r-l' Ir-a -$rp *-yf_->fif. "-i/l-"-!/a ' If -I ^-^-r-'- PROGRESSIONS. 197 No. Given. Required. Formulae. 9 a, n, I r= v/!' 10 a, n, S r ar"-rS=a-S, 11 a,/, S S-a s-r 12 n,/, S (S-O^-Sr"-^-/. 13 r, n, I =: j, 14 r, n, S a (r-l)S 15 r, /, S a=/r-(r-l)S, 16 n, /, S a(S~a) ^/(S-O 1 . 17 a, r, / log. I- log. a log. r 18 a, r, S log.[a+(r 1)S] log. a n , log.r 19 0, /, S log. I- log. a log.(S-a)-log.(S-l) ] 20 r, /, S log.llog.\lr(i 1)S] t ^ log.r n - EXAMPLES. Ex. 1. Required the sum of the series 1, 3, 9, 27, &c., continued to 12 terms. Ans. 265720. This example is solved by Formula 5. Ex. 2. Required the sum of the series 1, 2, 4, 8, 16, &c., continued to 14 terms. Ans. 16383. Ex. 3. Given the first term 2, the ratio 3, and the number of terms 10, to find the last term. Ans. 39366. Ex. 4. Given the first term l,the last term 512, and the sum of the terms 1023, to find' the ratio. 198 PROGRESSIONS. Ex. 5. Given the last term 2048, the number of terms 12, and the ratio 2, to find the first term. Ex. 6. A person being asked to dispose of his horse, said he would sell him on condition of receiving one cent for the first nail in his shoes, two cents for the second, and so on, doubling the price of every nail to 32, the number of nails in his four shoes. What would the horse cost at that rate ? Ans. 842,949,672.95. (244.) To find any number of geometrical means between two given numbers. In order to solve this problem, it is necessary to know the ratio. If m represent the number of means, m+2 will be the whole number of terms. Substituting m+2 for n in Formula 9 Art. 243, we obtain That is, to find the ratio, divide the last term by the first term, and extract the root denoted by the number of means plus one. When the ratio is known, the required means are obtained by continued multiplication. Ex. 1. Find three geometrical means between 2 and 162. Ex. 2. Find two geometrical means between 4 and 256. (245.) Of decreasing progressions having an infinite number of terms. The formula a ai* = T-^T' which represents the sum of n terms of a decreasing series, may be put under the form a ar" -1^~I=? In a decreasing progression, since r is a proper fraction, r" is less than unity, and the larger the number n, the smaller will be the quantity r". If, therefore, we take a very large num- ber of terms of the series, the quantity r n , and, consequently, the term , will be very small ; and if we take n greater PROGRESSIONS. 199 ar n than any assignable number, then will be less than any assignable number. We shall therefore have Hence the sum of an infinite series decreasing in geometrical progression is found by the following RULE. Divide the first term by unity diminished by the ratio. Ex. 1. Find the sum of the infinite series Here a =l,r=$. Therefore, S=-?-=-l=2. Ex. 2. Find the sum of the infinite series Ans. f . Ex. 3. Find the sum of the infinite series Ex. 4. Find the ratio of an infinite progression, whose first term is 1, and the sum of the series f. Ans. }. Ex. 5. Find the first term of an infinite progression, whose ratio is ~, and the sum f . Ans. | Ex. 6. Find the first term of an infinite progression, of which the ratio is -, and the sum - -. n nl PROBLEMS. (246.) Prob. 1. Of four numbers in geometrical progression, the sum of the first and second is 15, and the sum of the third and fourth is 60. Required the numbers. Let #, xy, xy*, xy*, be the numbers. Therefore, x+xy =15, and xy*+xy*=QO. JJOO PROGRESSIONS. Multiplying the first equation by y*, Therefore, y fl =4, W and y = 2. Also, x2x=l5. Therefore, *=5,or-15. Taking the first value of x, and the corresponding value of y, we obtain the series K 5>1 1 ' 20 ' 4 fi ; which numbers may be easily verified. Taking the second value of x, and the corresponding value of y, we obtain the series -15, +30, -60, +120; which numbers also perfectly satisfy the problem understood algebraically. If, however, it is required that the terms of the progression be positive, the last value of x would be inapplica- ble to the problem, though satisfying the algebraic equation. Several of the following problems also have two solutions, if we admit negative values. Prob. 2. There are three numbers in geometrical progres- sion whose sum is 210, and the last exceeds the first by 90 What are the numbers ? Ans. 30, 60, and 120. Prob. 3. There are three numbers in geometrical progres- sion whose continued product is 64, and the sum of their cubes is 584. Required the numbers. t abi889igof} gjtftot i mtf fcnii^ 715 - 2 ' 4 > and 8 - Prob. 4. There are four numbers in geometrical progres- sion, the second of which is less than the fourth by 24 ; and the sum of the extremes is to the sum of the means as 7 to 3. Re- quired the numbers. Ans. 1, 3, 9, and 27. Prob. 5. Of four numbers in geometrical progression, the difference between the first and second is 4, and the difference between the third and fourth is 36. What are the numbers ? Ans. 2, 6, 18, and 54. Prob. 6. Of four numbers in geometrical progression, the PROGRESSIONS. 201 sum of the first and third is , the sum of the second and fourth is b. What are the numbers ? a 3 a*b ab* b 9 /i 77 o -' HARMONICA! PROGRESSION. (247.) .A series of quantities is said to be in harmonical pro- gression when, of any three consecutive terms, the first is to the third as the difference of the first and second is to the difference of the second and third. Thus the numbers 60, 30,20, 15, 12, 10, are in harmonical progression ; for 60 : 20 :: 60-30 : 30-20 30 : 15 :: 30-20 : 20-15 20 : 12 :: 20-15 : 15-12 15: 10 :: 15-12: 12-10. So, also, the numbers 1 J i i i J &c., form an harmonical progression. (248.) The reciprocals of a series of terms in harmonical pro- gression form an arithmetical progression. Thus, the reciprocals of 60, 30, 20, &c., are sV sV aV 1*5 rV rV which are respectively equal to eV /or* ^o"> e*o> /o"> ro being an arithmetical progression whose common difference is V- If six musical strings of equal weight and tension have their lengths in the ratio of the numbers 1 ! i> i> y J> .the second will sound the octave of the first ; the third will sound the twelfth ; the fourth will sound the double octave ; the fifth will sound the eighteenth ; and the sixth will sound the third octave of the first. Hence the origin of the term harmonical or musical proportion. 202 PROGRESSIONS. Let im;^ 7 7 bx & 3 Whence x-t-b is the greatest common divisor. Ex. 3. Required the greatest common divisor of 4a 3 -2a a -3a+l and 3a a -2a-l. Ans. a I. Ex. 4. Find the greatest common divisor of x*a* and x*a*. Ans. xa Ex. 5. Find the greatest common divisor of j3*fi -:f?fi7 ^AfEfhroittcri) bn5 /i 3 The permutations of the same letters taken two and two, are ...... % ., ..- ulu 1 .. >>). ; t-- ii; i-'SBp^^lDKQ I''.' ! .U^IJ'il 2> *(.* oiil.s^ J '!' '' The permutations of the same letters taken singly, or \ , ' one by one, are f . ? ' f IX ^ ^. ^ . . . ) ' c. (258.) To find the number of permutations of n toers, taken m flTid m together. Let a,b,c,d A, be the n letters. The number of permutations of n letters taken singly, or one by one, is evidently equal to the number of letters, or to n. The number of permutations of n letters taken two and two is n(n 1). For if we reserve one of the letters, as a, there will remain nl letters, PERMUTATIONS AND COMBINATIONS. 211 b, C, d k. Writing a before each of these letters, we shall have ab, ac, ad dk ; that is, we obtain n 1 permutations of the n letters taken two and two, in which a stands first. Proceeding in the same manner with 6, we shall find n 1 permutations of the n letters taken two and two, in which b stands first ; and so for each of the n letters. Hence the whole number of permutations will be n(n-l). The number of permutations of n letters taken three and three together is n(n-l) (n-2). For if we reserve one of the letters, as a, there will remain ,il letters. Now we have found the number of permuta- tions of n letters taken two and two to be 71(71!). Hence the permutations of n 1 letters taken two and two must be (n-l) (n-2). Writing a before each of these permutations, we shall have (n 1) (n 2) permutations of the n letters taken three and three, in which a stands first. Proceeding in the same manner with bj we shall find (nl) (n2) permutations of the n let- ters taken three and three, in which b stands first ; and so for each of the n letters. Hence the whole number of permuta- tions will be n(n-l) (n-2). In like manner, we can prove that the number of permuta- tions of n letters taken four and four is n(n-l) (n-2) (n-3). When the letters are taken two and two, the last factor in the formula representing the number of permutations is n 1. When the letters are taken three and three, the last factor is n2. When the letters are taken four and four, the last factor is n3. Hence, when the letters are taken m and m together, the last factor will be n (m 1) or n m+1; and the number of per- mutations of n letters taken m and m together will according- ly be n(n-l) (n-2) (n-3) (n-m+l). 212 PERMUTATIONS AND COMBINATIONS. EXAMPLES. Ex. 1. Required the number of permutations of the 8 letters a, 6, c, d, e,f, g, h, taken 5 and 5 together. Here w=8, m=5, nm+l=4, and the above formula becomes 8.7.6.5.4=6720, Ans. Ex. 2. Required the number of permutations of the 26 let- ters of the alphabet, taken 4 and 4 together. Ans. 358800. Ex. 3. Required the number of permutations of 12 letters, taken 6 and 6 together. Ans. 665280. (259.) If we suppose that each permutation comprehends all the n letters ; that is, if m=n, the preceding formula becomes w(rc-l) (n-2) 2X1; or, inverting the order of the factors, 1.2.3.4 (n-l)n; which expresses the number of permutations of n letters taken all together. Ex. 1. Required the number of changes which can be rung upon 8 bells. According to the preceding formula, we have 1.2.3.4.5.6.7.8=40320, Ans. Ex. 2. How many permutations may be formed from the letters of the word Roma ? Ex. 3. What is the number of permutations which may be formed from the letters composing the word " virtue ?" Ex. 4. What is the number of different arrangements which can be made of 12 persons at a dinner-table ? Ans. 479001600. (260.) The combinations of any number of quantities signify the different collections which may be formed of these quanti- ties, without regard to the order of their arrangement. Thus, the three letters a, 6, c, taken all together, form but one combination, abc. Taken two and two, they form three combinations, ab, ac, be. PERMUTATIONS AND COMBINATIONS. 213 (261.) To find the number of combinations of n letters, taken m and m together. The number of combinations of n letters taken separately, or one by one, is evidently n. The number of combinations of n letters taken two and two, (n-l) 1.2 For the number of permutations of n letters taken two and two is n(n 1) ; and there are two permutations (ab, ba) cor- responding to one combination of two letters. Therefore the number of combinations will be found by dividing the number of permutations by 2. The number of combinations of n letters taken three and n(n-l) (7i-2) three together, is - - --5 - -. For the number of permutations of n letters taken three and three, is n(n 1) (n 2) ; and there are 1.2.3 permutations for one combination of three letters. Therefore the number of combinations will be found by dividing the number of permu- tations by 1.2.3. In the same manner, we shall find the number of combina- tions of n letters, taken m and m together, to be n(n-l) (n-2) ..... (n-m+l) 1.2.3 ..... m Ex i. Required the number of combinations of six letters taken three and three together. Here 71=6, m 3, n wz+l=4, and the formula becomes 6.5.4 u^T 20 - Ex. 2. Required the number of combinations of 8 letters taken 4 and 4. Ans. 70. Ex. 3. Required the number of combinations of 10 letters taken 6 and 6. Ans. 210. The following table, which is computed by the preceding for- mula, shows the number of combinations of 1, 2, 3, 4, &c., let- 214 PERMUTATIONS AND COMBINATIONS. ters taken singly, or two and two, three and three, &c. An important application of these principles will be seen in the next Section. Letters. Singly. 2 and 2. 3 and 3. 4 and 4. 5 and 5. 6 and 6. 7 and 7. 8 and 8. 9 and 9. [10 and 10. 1 1 2 3 2 3 1 3 1 Number of combinations. 4 4 6 4 1 5 5 10 10 5 1 6 6 15 20 15 6 1 7 7 21 35 35 21 7 1 8 8 28 56 70 56 28 8 1 9 9 36 84 126 126 84 36 9 1 10 10 45 120 210 252 210 120 45 10 1 ia. ' : ; SECTION XVI. INVOLUTION OF BINOMIALS. (262.) We have shown, in Art. 142, how to obtain any power of a binomial by actual multiplication. We now pro- pose to develop a theorem by which this labor may be greatly abridged. Taking the binomial a+b, its successive powers found by actual multiplication are as follows : (a+b) l =a +6, (a+by=a i +2ab +b\ The powers of a b, found in the same manner, are as fol- lows : (a-b) l =a -b, (a-by=a*-2ab -f& 2 , On comparing the powers of a+b with those of a b, we perceive that they only differ in the signs of certain terms. In the powers of a+b, all the terms are positive. In the powers of a 5, the terms containing the odd powers of b have the sign, while the even powers retain the sign +. The reason of this is obvious ; for, since b is the only negative term of the root, the terms of the power can only be rendered nega- 10* 216 INVOLUTION OF BINOMIALS. tive by b. A term which contains the factor b an even number of times, will therefore be positive ; if it contain it an odd number of times, it must be negative. Hence it appears that it is only necessary to seek for a method of obtaining the powers of a+b ; for these will become the powers of a b by simply changing the signs of the alternate terms. (263.) If we consider the exponents of the preceding pow- ers, we shall find that they follow a very simple law. Thus, of a are 2, 1,0, of b are 0, 1,2. In the square, the exponents . . j ( of a are 3,2,1,0, In the cube, the exponents . . ( of b are 0, 1, 2, 3. ( of a are 4, 3,2, 1, 0, In the fourth power, the exponents ] c . . , _ ( of b are 0, 1, 2, 3, 4. &c., &c., &c. In the first term of each power, a is raised to the required power of the binomial ; and in the following terms, the expo- nents of a continually decrease by unity to ; while the ex- ponents of b increase by unity from up to the required power of the binomial. It is obvious that this will always be the case, to whatever extent the involution may be carried. Also, the sum of the exponents of a and b in any term is equal to the ex- ponent of the power required. Thus, in the second power, the sum of the exponents of a and b in each term is 2 ; in the third power it is 3 ; in the fourth power, 4, &c. We hence infer, that for the seventh power the terms, with- out the coefficients, must be a\ a*b, a*V, ctb\ "&*, a*b\ ab\ V; and for the nth power, a", a n ~ l b, a n ~*b\ a n ~ 3 b* a'b"-*, ab n ~ l , b\ (264.) It remains to determine the coefficients which belong to these terms ; and in order to discover the law of their forma- tion, let us take the coefficients already found by themselves. The coefficients of the 1st power are 1 1 " 2d " 121 " 3d " 1331 " 4th " 14641 " 5th 1 5 10 10 5 1 " 6th " 1 6 15 20 15 6 1 INVOLUTION OF BINOMIALS. 217 The numbers in this table are identical with those in the ta- ble of combinations on page 214. For example, the coefficients of the fifth power denote the number of combinations of five letters taken one and one, two and two, &c. ; the coefficients of the sixth power denote the number of combinations of six letters taken one and one, two and two, &c. The reason of this will appear if we observe the law of the product of several binomial factors, x+a, x+b, x+c, x+d, &c. Multiplying x + a by x + b, we obtain x*+(a+b)x+ab=lst product. Multiplying by x + c, we obtain x*+ (a+b + c)x* + (ab + ac + bc)x + abc = 2d product. Multiplying by x + d, we obtain x* + (a+b+c+d)x* + (ab+ac+ad+bc + bd+ cd)x* + (abc + abd+ acd+ bcd)x + abcd= 3d product. We observe that in each of these products the coefficient of x in the first term is unity ; the coefficient of the second term is the sum of the second terms of the binomial factors ; the coefficient of the third term is the sum of all their products taken two and two ; the coefficient of the fourth term is the sum of all their products taken three and three, &c. It is easily seen that if we multiply the last product by a new factor, x+e, the same law of the coefficients will be pre- served. Hence the law is general. If now, in the preceding binomial factors, we suppose , b, c. d, &c., to be all equal to each other, the product (x+a) (x+b) (x+c) (x+d) becomes (x+a) n . The coefficient of the second term of the product, or a+b-\- c+d , becomes a+a+a-\-a ; that is, a taken as many times as there are letters a, b, c, d, and is, consequently, equal to na. The coefficient of the third term, or ab+ac, &c., reduces to a a_l_ a a _l_ a 9 9 or a 2 repeated as many times as there are different combinations of n letters taken two and two ; that is, n(nl) , hv Art. 261, to v ' a\ 218 INVOLUTION OF BINOMIALS. The coefficient of the fourth term reduces to a 3 repeated as many times as there are different combinations of n letters n(n-l) (n2) , taken three and three ; that is, - j-^-~ - a 3 , and so on. Thus we find that the nth power of x+a may be expressed as follows : which is called the BINOMIAL FORMULA, and is generally as- cribed to Sir Isaac Newton. So important was it regarded, that it was engraved on his monument in Westminster Abbey as one of his greatest discoveries. On comparing the different terms of this development, we perceive that any coefficient may be derived from the preced- ing one by the following rule : If the coefficient of any term be multiplied by the exponent of x in that term, and divided by the exponent of a increased by one, it will give the coefficient of the succeeding term. Thus, the fifth power of x+a is If the coefficient 5 of the second term be multiplied by 4, the exponent of x in that term, and divided by 2, which is the exponent of a increased by one, we obtain 10, the coefficient of the third term. So, also, if 10, the coefficient of the fourth term, be multi- plied by 2, the exponent of x, and divided by 4, the exponent of a increased by one, we obtain 5, the coefficient of the fifth term ; and so of the others. The coefficients of the sixth power will also be found as fol- lows: 6X5 15X4 20X3 15X2 6X1 ' ' 2 ' 3~~' 4 '~5~~' ~6~ ; that is, 1,6, 15, 20, 15, 6, 1. The coefficients of the seventh power will be 7 7x6 21x5 35X4 35X3 21X2 7X1 2 ' 3 ' 4 ' ~~5~' ~~6~' ~T~ 5 that is, 1,7, 21, 35, 35, 21, 7, 1. INVOLUTION OF BINOMIALS. 219 Therefore, the seventh power of x +a is It is sometimes preferable to retain the factors of the coeffi- cients distinct from each other, as follows : 7.6.5.4.3 a 7.6.5.4.3.2 6 7.6.5.4.3.2.1 7 1.2.3.4.5^ X + 1.2.3.4.5.6 ff ; + 1.2.3.4.5.6/7* ' The factor 1 is retained for the sake of symmetry, and to exhibit more clearly the law of the coefficients. (265.) The following, therefore, is the BINOMIAL THEOREM. In any power of a binomial x+a, the exponent of x begins in the first term with the exponent of the power, and in the follow- ing terms continually decreases by one. The exponent of a com- mences with one in the second term of the power, and continually increases by one. The coefficient of the first term is one ; that of the second is the exponent of the power ; and if the coefficient of any term be multiplied by the exponent of x in that term, and divided by the exponent of a increased by one, it will give the coefficient of the succeeding term. (266.) The number of terms in the power is always greater by unity than the exponent of the power. Thus, the number of terms in (a+b)* is 4+1, or 5 ; in (a+b) 6 is 6 + 1, or 7. Also, if we examine the table in Art. 264, it will be per- ceived that, after we pass the middle term, the same coeffi- cients are repeated in the inverse order. Thus, the coeffi- cients of (a+b)* are 1, 5, 10, 10, 5, 1 ; of (a+b) are 1, 6, 15, 20, 15, 6, 1. Hence it is only necessary to compute the ^coefficients for half the terms ; we then repeat the same numbers in the in- verse order. (267.) The sum of the coefficients for each power is equal to the number 2 raised to the same power. For, let x=l and a=l, then each term without the coefficients reduces to unity, and 220 INVOLUTION OF BINOMIALS. the value of the power is simply the sum of the coefficients. .Also, in this case, (x+a) n becomes (l + l) w , or 2 n . Thus the coefficients of the first power are 1 + 1 =2=2* ; second " l+2+l=4=2 a ; v% third " 1+3+3+1 =8=2"; fourth " 1 +4+6+4+ 1 = 16=2 4 , &c., &c., &c. EXAMPLES. Ex. 1. Raise x+a to the 9th power. The terms without the coefficients are x g , ax 6 , a?x\ a*x e , a*x* 9 cfx*, a'x*, cfx*, a*x, a 9 . And the coefficients are 9X8 36X7 84X6 126X5 126X4 84X3 36X2 9X1 lf ' ~1T' "3"' "IT"' ~~5~' ~~6~~' ~T~' ~~8~' ~9~' that is, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, Prefixing the coefficients, we obtain It should be remembered that, according to Art. 266, it is only necessary to compute the coefficients of half the terms in- dependently. Ex. 2. What is the 6th power ofxa? (268.) If the terms of the given binomial are affected with coefficients or exponents, they must be raised to the required powers, according to the principles already established for the involution of monomials. Ex. 3. Raise 2#+5a 3 to the fourth power. For convenience, let us substitute b for 2x, and c for 5a\ Then (6+c) 4 =6 4 +46 8 c+66V+4&c s +c 4 . Restoring the values of b and c, The first term will be (2x)* =I6x*. The second term " 4(2x) 3 X 5 3 +5a- 6 6 4 , &c. Here the coefficients increase regularly by 1, and the signs are alternately positive and negative. We might have ob- tained the same result by division, as in the former example. Ex. 3. Expand into a series r or (a 6)" 1 . Here the coefficients furnished by the Rule are + 1, -1, +1, -1, &c. But the factor b being negative, all its odd powers are nega- tive. Hence the second term contains two negative factors, so that its resulting sign is +. The same remark applies to 224 INVOLUTION OF BINOMIALS. the fourth and sixth terms, &c., making the terms of the series all positive. Ex. 4. Expand into a series , ,. a or (ab)~*. Ex. 5. Expand into a series (a+b)~*. Ex. 6. Expand into a series (a b)~*. (272.) We have now considered the powers of a binomial when the exponent is an integer, either positive or negative. It remains to consider the case when the exponent is a. fraction. EXAMPLES. Ex. I. Expand Va+b or (a+b)* into an infinite series. The terms without the coefficients are The exponents of a decrease by unity, while those of b in- crease by unity. The coefficient of the first term is 1. " second " -J . 2 ~2A' fourth - -^= + S " fifth 4 2.4.6.8* The series, therefore, is The factors which form the coefficients are kept distinct, in order to show more clearly the law of the series. The numer- ators of the coefficients contain the series of odd numbers, 1,3, 5, 7, &c., while the denominators contain the even numbers, 2, 4, 6, 8, 10, &c, The above series expresses the square root of a+b. We shall obtain the same result if we extract the square root by the usual method. See Art. 299. INVOLUTION OF BINOMIALS. 225 Ex. 2. It is required to convert (a*+x) 2 into an infinite series. 3.5.QTV 3.5.7a~V 2.4.6.8 + 2.4.6.8.10 ' x_ x* 3x* 3.5x* 3.5.7x 5 + 2^ ~25^ + 2Z6^~2Z6^ + 2.4.6.8.10a 9 ' & where the law of the series is evident. i Ex. 3. It is required to convert (ax) 2 into an infinite series. Ex. 4. It is required to convert (a+b)^ into an infinite series. b 26 3 2.56 3 2.5.S6 4 Ex. 5. Expand (a b)* into an infinite series. T Ql.2 Q fy7i3 Q T 1 1 A* * \ , 00 d.70 6. 7.110 . Ji.715. a 1 1 2 o 1 ; &C. t . 4 +2ab+b*\a+b = the root. 2ab+b' 2a+b = the divisor. 2db+V Hence we derive the following EVOLUTION OF POLYNOMIALS. RULE FOB, EXTRACTING THE SdUARE ROOT OF A POLYNOMIAL. Arrange the terms according to the powers of some one letter ; take the square root of the first term for the first term of the re- quired root, and subtract its square from the given polynomial Divide the first term of the remainder by double the root al- ready found, and annex the result both to the root and the divi- sor. Multiply the divisor thus increased by the last term of the root, and subtract the product from the last remainder. Pro- ceed in the same manner to find the additional terms of the root. Ex. 1. Required the square root of a 4 2a a x+3d l x*2az*+x'. a*2a*x+3a*x'*2ax 3 +x* I a 2 ax+x* = the root. -2a 3 x+3a*x* -2a*x+ a 2a?ax = the first divisor. 2a\x*-2ax 3 +x* 2d*2ax+x* = the second divisor. For verification, multiply the root 2 ax+x* by itself, and we shall obtain the original polynomial. Ex. 2. Required the square root of a 2 +2a&+2ac+6 a +2&c+c a . Ex. 3. Required the square root of Wx* Wx 9 I2x*+5x*+ Ex. 4. Required the square root Ans. 2x*+2ax+4b\ Ex. 5. Extract the square root of 15a 4 6 a +a 6 6a*b 2Qa*b* Ex. 6. Extract the square root of 8'"=d-3c+3b-a; " d=a+3V'+3V"+ D"', V""=e-4d+6c-4b+a; " e= otc. a a a a Ex. 4. Resolve ^ into an infinite series. b+x Ex. 5. Resolve into an infinite series. lx We may proceed in the same manner when there are more than two terms in the divisor. Ex. 6. Resolve ; 5 into an infinite series. Ans. I+aa* a 4 + 6 +, &c. Ex. 7. Resolve -: rr into an infinite series. (299.) Infinite series obtained by extracting the square root. In Art. 272, Va+b has been expanded into an infinite series by the Binomial Theorem. It was also remarked that the same result might have been obtained by extracting the square root according to the usual rule, Art. 275. The operation will proceed as follows : a+b a*-\ r jH 5, &c., = the square root of a+b. 2a * 2a 2 H r, first divisor. 2 * b b* r H Y 3 , second divisor. a 2 Sa 2 ___ 4a 8a a 64a* , j^_ JL 8a 9 64a 8 * INFINITE SERIES. 251 This result is the same as that obtained in Art. 272. Ex. 2. Extract the square root of 1 #. x x* x 3 5x* Ans . !_________, &c . Ex. 3. Extract the square root of cf+b. Ex. 4. Extract the square root of a* b. METHOD OF UNKNOWN COEFFICIENTS. (300.) The method of unknown coefficients is a method of developing algebraic expressions into series, by assuming a series having unknown coefficients, and afterward finding the value of these coefficients. This method is founded on the fol- lowing THEOREM. If an equation of the form A+Bz+Oc 2 +Da; 3 +, &c., =A'+R'x+C'x*+V'x 3 +, &c., must be verified by any value given to x, the terms involving the same powers in the two members are respectively equal. For, since this equation must be verified for every value of x, it must be verified when x=0. But, upon this supposition, all the terms vanish except two, and we have A=A'. Suppressing these two equal terms, we have Bz+Oc 2 +D.r'+, & c ., -=B^+CV+DV+, &c. Dividing every term by x, we obtain B+Cx+Vx>+, &c., =B / +C'z+D'or 2 +, &c. Since this equation must be verified for every value of #, it must be verified when x=0. But, upon this supposition, B=B'. In the same manner, we can prove that C=C', D=D', &c. (301.) Let it be proposed to develop the expression into a series arranged according to the powers of x. It is plain that this development is possible, for we may divide the numerator by the denominator, as explained in Art. 298. 252 INFINITE SERIES. Let us, then, assume , &c., where the coefficients A, B, C, D are supposed to be independ- ent of x t but dependent on the known terms of the fraction. In order to obtain the values of these coefficients, let us mul- tiply both members of the above equation by the denominator 1 +x, and we shall have l-a;=A+(A+B>+(B+CX+(C+D)a; 8 +(D+EX+, &c. But, according to the preceding Theorem, the terms involving the same powers of a; in the two members of the equation must be equal to each other. Therefore, A= 1, A+B = -1; hence B = -2. = 0; C = +2. 0; " D=-2. D+E= 0; " E=+2. &c., &c. Substituting these values of the coefficients in the assumed series, we obtain <-, &c. 1+x (302.) The method thus exemplified is expressed in the fol- lowing RULE. Assume a series with unknown coefficients as equal to the pro- posed expression ; then, having cleared the equation of fractions, or raised it to its proper power, find the value of each of these coefficients by equating the corresponding terms of the two ex- pressions, or putting such of them as have no corresponding terms, equal to zero. Ex. 2. Expand the fraction - - ; r into an infinite series. 1 2x+x* Assume X 9 =A+Ex+Cx*+Dx*+ftx'+, &c. 1 ~~~ x~\~x Multiplying by 1 2x+x*, we have INFINITE SERIES. C)x 4 +, &c. Hence we must have A=l B-2A=0 v B=2A =2, C-2B+ A=OvC=2B-A=3, D-2C+ B=0 v D=2C-B=4, E-2D+ C=OvE=2D-C=5 ? &c., &c. Therefore, ^-7-^= l+2x+3x* +4x 3 +5x 4 + , &c. 1 I %x Ex. 3. Expand the fraction - -- ; into an infinite series. 1 -~ic x , &c., where the coefficient of each term is equal to the sum of the coefficients of the two preceding terms. 1 _ , Ex. 4. Expand - - - ^ into an infinite series. 1 <&x ox Ans. l+^4-5^ 2 +13a; 3 +4l2: 4 +121x 6 +, &c What is the law of the coefficients in this series ? 1 I O-y Ex. 5. Expand - - into an infinite series. 1 ox Ans. l + 5x + 15x a +45x 3 +135o; 4 +, &c. What is the law of the coefficients in this series 1 Ex. 6. Expand Vlx into an infinite series. x x* 3^ 3 3.5z 4 3.5.7x s ~2~2A~2A^~2A&8~2A.6.8.1Q~' C ' (303.) The method of unknown coefficients requires that we should know beforehand the form of the development with re- spect to the powers of x. Generally, we suppose the develop- ment to proceed according to the ascending powers of x, com- mencing with x ; but sometimes this form is inapplicable, in which case the result of the operation is sure to indicate it. Let it be required, for example, to develop the expression into a series. 3x x 2 Assume -=A+Bx+Cx*+T>x>+, &c. 254 INFINITE SERIES. Clearing of fractions, we have , &c. ; whence, according to Art. 300, we conclude 1=0, 3A=0, &c. Now the first equation, 1=0, is absurd, and shows that the assumed form is not applicable in the present case. But if we put the fraction under the form X - , and suppose that x 3 x X^-=-(A.+Kx+Cx>+Vx*+, &c.), X O X X it will become, after the reductions are made, l=3A+(3B-A)z+(3C-BX+(3D-C)z 3 -f, &c., which gives the equations 3A=1 ; whence A= J. 3B-A=0; " B=J. 3C-B=0; " C=sV 3D-C=0; " D= T V Therefore, _l- J =i(I + | + g + g + , &c .) x- 1 x x , x' , = T + 9-+27 + 81 + ' &C - ; that is, the development contains a term affected with a nega- tive exponent. We ought, then, to have assumed at the outset , &c. The particular series which should be adopted in each case may be determined by putting x=Q, and observing the nature of the result. If, in this case, the proposed expression becomes equal to a finite quantity, the first term of the series will not contain x. If the expression reduces to zero, the first term ^ will contain x ; and if the expression reduces to the form -, then the first term of the development must contain x with a negative exponent. SECTION XIX. GENERAL THEORY OF EQUATIONS. (304.) It is proposed in this Section to exhibit the most im- portant propositions relating to the theory of equations, to- gether with the Theorem of Sturm, by which we are enabled to determine the number of real roots of an equation. A function of a quantity is any expression involving that quantity. Thus, ax*+b is a function of x. ay*+cy+d is a function of y. ax* by* is a function of a; and y. In a series of terms, two successive signs constitute a per- manence when the signs are alike, and a variation when they are unlike. Thus, in the polynomial a+b c+d, the signs of the first two terms constitute a permanence ; the signs of the second and third constitute a variation : and those of the third and fourth also a variation. (305.) A cubic equation is one in which the highest power of the unknown quantity is of the third degree ; as, for ex- ample, All equations of the third degree, with one unknown quan- tity, may be reduced to the form x*+ax*+bx+c=0. A biquadratic equation is one in which the highest power of the unknown quantity is of the fourth degree ; as, for example, 12 256 GENERAL THEORY OF EQUATIONS. Every equation of the fourth degree, with one unknown quantity, may be reduced to the form x*+ax*+bx*+cx+d=Q. The general form of an equation of the fifth degree is and the general form of an equation of the mth degree, with one unknown quantity, is x m +Ax m - 1 +Bx m -*+Cx m - 3 + ..... +Tx+V=0 (m). This equation will be frequently referred to hereafter by the name of the general equation of the mth degree, or simply by the letter (m). It is obvious, that if we could solve this equation, we should have the solution of every equation which could be proposed. Unfortunately, no general solution has ever been discovered ; yet many important properties are known, which enable us to solve any numerical equation which can ever occur. PROPOSITION I. (306.) If a is a root of the general equation of the mth degree, the equation will be exactly divisible by x a. For if a is one value of x, the equation must be verified when we substitute a in the place of x. Hence we must have a m +Aa m - 1 +Ba m - 2 +Ca m - 3 + ..... +T+V=0 (1). Subtracting equation (1) from equation (m), we obtain (x m -a m )+A(x m - l -a m - l )+B(x m -*-a m -*) + . . . + T(x-a)=0 (2). But, by Art. 76, each of the expressions (x m a m ), (x m ~ 1 a m ~ l ), &c., is divisible by xa, and therefore equation (2) is also di- visible by xa. Now equation (m) is but another form for equation (2) ; for if we take the value of V, as found from equation (1), and substitute it for V in equation (m), it will give us equation (2) ; therefore, equation (m) is divisible by x a. Conversely, if equation (m) is divisible by xa, then a is a root of the equation. It will be noticed that this property is but a generalization of what has been proved of equations of the second degree, in Art. 192. Ex. 1. Prove that 1 is a root of the equation GENERAL THEORY OF EQUATIONS. 257 This equation is divisible by x 1, and gives #* 5#+6=0. Ex. 2. Prove that 2 is a root of the equation x *- x -6=0. This equation is divisible by x 2, and gives x*+2x+3=Q. Ex. 3. Prove that 2 is a root of the equation Ex. 4. Prove that 4 is a root of the equation x*+x*34x+56=0. Ex. 5. Prove that 1 is a root of the equation Ex. 6. Prove that 5 is a root of the equation x 5 +6^ 4 - Wx*- 1 12x*-207x- 110=0. Ex. 7. Prove that 3 is a root of the equation PROPOSITION II. (307.) Every equation of the mth degree containing but one unknown quantity, has m roots and no more. For, suppose a to be a root of the general equation of the mth degree. By the last Proposition, this equation is divisible by xa; and if we actually perform the division, the equation will be reduced to one of the next inferior degree. If we represent the coefficients of the different powers of x by A', B', &c., the quotient will be This equation must also have a root, which we will repre- sent by b ; and dividing by x fc, the equation will be reduced to one of the next inferior degree, and so on. We may continue this series of operations (ml) times, when we shall arrive at a simple equation which has only one root. Hence the proposed equation will have m roots, a,b,c,d, ..... /; and its successive divisors, or the factors of which it is com- posed, will be xa, xb t x c, xd f ..... x /, being equal in number to the units contained in m, the highest exponent of the equation. R 258 GENERAL THEORY OF EQUATIONS. We have seen that when one root of an equation is known, the equation is readily reduced to one of the next inferior de- gree ; and if we can depress any equation to a quadratic, its roots can be determined by methods already explained. Ex. 1. One root of the equation is 1. Find the remaining roots. Ex. 2. Two roots of the equation a: 4 - 10z 3 -f35:e'-502;+24=0 are 1 and 3. Find the remaining roots. Ex. 3. Two roots of the equation x* 12a; 3 +48:c a -68:c-j-15=0 are 3 and 5. Find the remaining roots. Ans. Ex. 4. Two roots of the equation are 2 and 3. Find the remaining roots. Ans. Ex. 5. Two roots of the equation are 2 and 2. Find the remaining roots. Ans. (308.) It should be observed that this Proposition only proves that an equation of the mth degree may be continually depress- ed by division, and finally exhausted after m operations. The divisors are not necessarily unequal. Any number, and indeed all of them, may be equal. When we say that an equation of the mth degree has m roots, we mean that the polynomial can be decomposed into m binomial factors, equal or unequal, each containing one root. Thus, the equation x 3 -6a; a +12a;-8=0 can be resolved into the factors (x-2) (x-2) (x-2)=0; or (x-2) s =0; whence it appears that the three roots of this equation are 2, 2, 2. But, in general, the several roots of an equation differ from each other numerically. GENERAL THEORY OF EQUATIONS. 259 The equation x*=8 has apparently but one root, viz., 2; but by the method of the preceding article we can discover two other roots. Dividing a; 8 8 by x2, we obtain Solving this equation, we find Thus, the three roots of the equation x*=8 are 2; -l+V^S; -l-V^3. These last two values may be verified by multiplication as follows : -1+ V^3 -I- V]3 -1+ V-3 -1- V-3 1- V~^3 1 + V~-3 - V^-3-3 + V 33 22 V 3=the square. 2+2 ^^-3= the square. -1+ V^3 -1- V^S 3 2-2V-3 -2V-3+6 +2^-3+6 8 the cube. 8= the cube. If the last term of an equation vanishes, as in the example the equation is divisible by #0, and, consequently, is one of its roots. If the last two terms vanish, then two of its roots are equal toO. PROPOSITION III. To discover the law of the coefficients of every equation. (309.) In order to discover the law of the coefficients, let us form the equation whose roots are , &, c, d, ..... /. This equation will contain the factors (x a), (x 6), (x c) t &c. ; that is, we shall have '<:-) (x-b) (x-c) (x-d) ..... (z-/) = 0. 260 GENERAL THEORY OF EdUATIONS. x m a x^+ab x m ~* abc -b + ac -abd c +ad acd -d + bc -bed &c. +bd &c. + cd If we perform the multiplication as in Art. 264, we shall have + ..... -(abc ..... /)=0. i> ow , -& yti a- - &c. Hence we perceive, 1. The coefficient of the second term of any equation is equal to the sum of all the roots with their signs changed. 2. The coefficient of the third term is equal to the sum of the products of all the roots taken two and two. 3. The coefficient of the fourth term is equal to the sum of the products of all the roots taken three and three, with their signs changed. 4. The last term is the product of all the roots with their signs changed. It will be perceived that these properties include those of quadratic equations mentioned on pages 163 and 164. If the roots are all negative, the signs of all the terms of the equation will be positive, because the factors of which the equation is composed are all positive. If the roots are all positive, the signs of the terms will be al- ternately + and -. Ex. 1. Form the equation whose roots are 1, 2, and 3. For this purpose, we must multiply together the factors a; 1, a; 2, x3, and we obtain This example conforms to the rules above given for the co- efficients. Thus, the coefficient of the second term is equal to the sum of all the roots (1+2+3) with their signs changed. The coefficient of the third term is the sum of the products of the roots taken two and two ; thus, 1X2+1X3+2X3=11. The last term is the product of all the roots (1X2X3) with their signs changed. GENERAL THEORY OF EQUATIONS. 261 Ex. 2. Form the equation whose roots are 2, 3, 5, and 6. Ans. x* 4x*-29x*+ 156^-1 80=0. Show how these coefficients conform to the laws above given. Ex. 3. Form the equation whose roots are 1, 3, 5, 2, 4, -6. Ans. x 6 +3x 6 -41x*-87x*+4QOx*-4443;-1l20=0. (310.) Every rational root of an equation is a divisor of the last term ; for, since this term is the product of all the roots, it must be divisible by each of them. If, then, we wish to find a root by trial, we know at once what numbers we must employ. For example, take the equation X *- X -G=0. If this equation has a rational root, it must be a divisor of the last term, 6 ; hence we must try the numbers 1, 2, 3, 6, either positive or negative. If x=I, we have 1 16= 6, x=2, " 8-2-6= 0, z=3, " 27-3-6= 18, #=6, " 216-6-6=204, Hence we see that 2 is one of the roots of the given equa- tion, and by the method of Art. 307, we shall find the remain- ing roots to be PROPOSITION IV. (311.) No equation whose coefficients are all integers, and that of the first term unity, can have a root equal to a rational frac- tion. For, take the general equation of the third degree, and suppose, if possible, that the fraction r is one value of x, this fraction being reduced to its lowest terms. If we substi- tute this value for x in the given equation, we shall have 262 GENERAL THEORY OF EaUATIONS. Multiplying each term by 6 a , and transposing, we obtain Now, by supposition, A, B, C, a and b are whole numbers. Hence the entire right-hand member of the equation is a whole number. But by hypothesis, is an irreducible fraction ; that is, a and b contain no common factor. Consequently, a 9 and b will contain no common factor, that is, T is a fraction in its lowest terms. Hence, if 7- were a root of the proposed equation, we should have a fraction in its lowest terms equal to a whole number, which is absurd. The same mode of demonstration is applicable to the general equation of the rath degree. This proposition only asserts that in an equation such as is here described, the real roots must be integers, or they can not be exactly expressed in numbers. They may often be express- ed approximately by fractions, as is seen in the examples on pages 288-301. A real root which can not be exactly ex- pressed in numbers is called incommensurable. PROPOSITION V. (312.) If the signs of the alternate terms in an equation are changed, the signs of all the roots will be changed. If we take the general equation of the mth degree, and change the signs of the alternate terms, we shall have x m -Ax m - 1 +Rx m -'-Cx n -*+ ..... =0 (1) : or, changing the sign of every term of the last equation, -x^+Ax^-Bx^+Cx"-*- ..... =0 (2). Now, substituting +a for x in equation (m) will give the same result as substituting a in equation (1), if m be an even number; or, substituting am equation (2), if m be an odd number. If, then, a is a root of equation (m), a will be a root of equation (1), and, of course, a root of equation (2), which is identical with it. GENERAL THEORY OF EQUATIONS. 263 Hence we see that the positive roots may be changed into negative roots, and the reverse, by simply changing the signs of the alternate terms ; so that the finding the real roots of any equation is reduced to finding positive roots only. Ex. 1. The roots of the equation are 1, 3, and 2. What are the roots of the equation x*+2x*-5x-6=0? Ex. 2. The roots of the equation x*-6x*+llx-6=0 are 1, 2, and 3. What are the roots of the equation PROPOSITION VI. (313.) If an equation whose coefficients are all real, contains imaginary roots, the number of these roots must be even. If an equation whose coefficients are all real, has a root of the form then will a 6V 1 be also a root of the equation. For, let a-\-bV 1 be substituted for x in the equation, the result will consist of a series of terms, of which those involving only the powers of a, and the even powers of bV 1 will be real, and those which involve the odd powers of bV 1 will be imaginary. If we denote the sum of the real terms by P, and the sum of the imaginary terms by Qv 7 1, then we must have which relation can only exist when P=0 and Q=0. Again, let abV l be substituted for x in the proposed equation, the only difference in the result will be in the signs of the odd powers of bV 1, so that the result will be P Q, V 1. But we have found that P=0 and Q=0 ; hence 12 264 GENERAL THEORY OF EQUATIONS. And, since a b V 1 substituted for x gives a result equal to zero, it must be a root of the equation. Ex. 1. Find the roots of the equation Ans. 2, and 1-/ 1. Ex. 2. Find the roots of the equation Ans. 3, and 2 V 1. Hfc. 3. Find the roots of the equation 5x*+2x 44=0. ' A -T -' M ~ 1,1 / - - .Arcs. 2, and -1V 3.4. Hence every equation of the third degree whose coefficients are all real, must have one real root. The same is true of every equation of an odd degree. i U sm> i^$H!ft*o'i oio&B i'.omws !} PROPOSITION VII. (314.) Every equation must have as many variations of sign as it has positive roots, and as many permanences of sign as there are negative roots. To prove this Proposition, it is only necessary to show that the multiplication of an equation by a new factor, x a, cor- responding to a positive root, will introduce at least one varia- tion t and that the multiplication by a factor x+a will intro- duce at least one permanence. T-* , , For an example, take the equation in which the signs are +H , giving one variation. Multiply this equation by a; 2=0, as follows : x*+3x*-10x -24 x-2 il38t ! <|OU[ 3i. z*+ x'-16x*- 4^+48=0. In this last product the signs are +H h, giving two va- riations ; that is, the introduction of a positive root has intro- duced one new variation in the signs of the terms. GENERAL THEORY OF EaUATIONS. 265 To generalize this reasoning, we perceive that the signs in the upper line of the partial products are the same as in the given equation ; but those in the lower line are all contrary to those of the given equation, and advanced one term toward the right. Now, if each coefficient of the upper line is greater than the corresponding one in the lower, the signs of the upper line will be the same as in the total product, with the exception of the last term. But the last term introduces a new variation, since its sign is contrary to that which immediately precedes it; that is, the product contains one more variation than the original equation. When a term in the lower line is larger than the correspond- ing one in the upper line, and has the contrary sign, there is a change from a permanence to a variation ; for the lower sign is always contrary to the preceding upper sign. Hence, when- ever we are obliged to descend from the upper to the lower line in order to determine the sign of the product, there is a variation which is not found in the proposed equation ; and as all the remaining signs of the lower line are contrary to those of the proposed equation, there must be the same changes of sign in this line as in the given equation. If we are obliged to reascend to the upper line, the result may be either a variation or a permanence. But even if it were a permanence, since the last sign of the product is in the lower line, it is necessary to go once more from the upper line to the lower, than from the lower to the upper. Hence each factor, corresponding to a positive root, must introduce at least one new variation ; so that there must be as many variations as there are positive roots. In the same manner, we may prove that the multiplication by a factor x-\-a, corresponding to a negative root, must intro- duce at least one new permanence ; so that there must be as many permanences as there are negative roots. Ex. 1. The roots of the equation are 1, 2, 3, 1, and 2. There are also three variations of sign, and two permanences, as there should be, according to the Proposition. 266 GENERAL THEORY OF EQUATIONS. Ex. 2. The equation has four real roots. How many of these are negative ? Ex. 3. The equation has six real roots. How many of these are positive ? If all the roots of an equation are real, the number of posi- tive roots must be the same as the number of variations, and the number of negative roots must be the same as the number of permanences. If any term of an equation is wanting, we must supply its place with before applying the preceding Rule. PROPOSITION VIII. (315.) If two numbers, when substituted for the unknown quantity in an equation, give results with, contrary signs, there is at least one root comprised between those numbers. Take, for example, the equation x*2x*+3x 44=0. If we substitute 3 for x in this equation, we obtain 26 ; and if we substitute 5 for x, we obtain +46. There must, therefore, be a real root between 3 and 5 ; for, when we sup- pose x=3, we have But when we suppose x=5, we have Now both the quantities x*+3x and 2# a +44 increase while x increases. And since the first of these quan- tities, which was originally less than the second, has become the greater, it must increase more rapidly than the second. There must, therefore, be a point at which the two magnitudes are equal, and that value of x which renders these two magni- tudes equal must be a root of the proposed equation. In general, if two numbers, p and q, substituted for x in an equation, give results with contrary signs, we may suppose the less of the two numbers to increase by imperceptible degrees GENERAL THEORY OF EQUATIONS. 26"? until it becomes equal to the greater number. The results of these successive substitutions must also change by impercepti- ble degrees, and must pass through all the intermediate values between the two extremes. But the two extreme values are affected with opposite signs ; there must, therefore, be some number between p and q which reduces the given equation to zero, and this number will be a root of the equation. In the same manner, it may be proved that if any quantity p, and every quantity greater than p, substituted in an equation, renders the result positive, then p is greater than the greatest root. Hence, also, if the signs of the alternate terms are changed, and if q, and every quantity greater than q, renders the result positive, then q is less than the least root. If the two numbers, which give results with contrary signs, differ from each other only by unity, it is plain that we have found the integral part of a root. Ex. 1. Find the integral part of one of the roots of the equa- tion When x =2, the equation reduces to 12; and when x =3, it reduces to +71. Hence there must be a root between 2 and 3 ; that is, 2 is the first figure of one of the roots. Ex. 2. Find the first figure of one of the roots of the equa- tion x*+x*+x- 100=0. Ans. 4. Ex. 3. Find the first figure of each of the roots of the equa- tion PROPOSITION IX. (316.) Every equation may be transformed into another, whose roots are greater or less than those of the former by any given quantity. Let it be required to transform the general equation of the mth degree into another whose roots are greater by r than those of the given equation. Take y=x+r, or x=yr t 268 GENERAL THEORY OF EQUATIONS. and substitute y r for x in the proposed equation; we shall then have i-l)(w-2)r 8 +A -(w-l)Ar +B 2T- 2 - 2.3 (m-l)(m-2)Ar* -(m-2)Br +C which equation evidently fulfills the required conditions, smce y is greater than x by r. If we take y=xr, or xy-\-r, we shall obtain in the same way an equation whose roots are less than those of the given equation by r. Ex. 1. Find the equation whose roots are greater by 1 than those of the equation We must here substitute y I in place of x. Ans. y- Ex. 2. Find the equation whose roots are less by 1 than those of the equation x*-2x*+Zx-4=0. Ans. y-fy 9 +2y-2=0. Ex. 3. Find the equation whose roots are greater by 3 than those of the equation x'+9x*+I2x*-14x=0. Ans. y 4 3?/ 8 15y 3 +49y 12=0. Ex. 4. Find the equation whose roots are less by 2 than those of the equation Ans. 5 Ex. 5. Find the equation whose roots are greater by 2 than those of the equation Ans. PROPOSITION X. (317.) Any complete equation may be transformed into an- other whose second term is wanting. GENERAL THEORY OF EQUATIONS. 269 Since r in the preceding Proposition is indeterminate, we may put mr+A equal to zero, which will cause the second ^ term of the general development to disappear. Hence r= , A and x=y -- . y m Hence, to remove the second term of an equation, substitute for the unknown quantity a new unknown quantity, together with such a part of the coefficient of the second term, taken with a contrary sign, as is denoted by the degree of the equation. Ex. 1. Transform the equation into another whose second term is wanting. Here we take a new unknown quantity, and annex to it a third part of the coefficient of the second term of the equation with its sign changed ; that is, we put x=y+2. Making this substitution, we obtain y s 4y2=0. Ans. Ex. 2. Transform the equation into another whose second term is wanting. Here we put x=y+4. Ans. y*96y*518y 777=0. Ex. 3. Transform the equation into another whose second term is wanting. Ans. 5 - Since the coefficient of the second term is equal to the sum of the roots with their signs changed, it is obvious that when the second term of an equation is wanting, the sum of the posi- tive roots must be equal to the sum of the negative roots. PROPOSITION XL (318.) To discover the law of Derived Polynomials. When we substitute y+r for x in the general equation of the with degree, the coefficients of r follow a remarkable law. The equation, before it is developed, is 270 GENERAL THEORY OF EQUATIONS. ' n - a + ..... + T(y+r)+V=0. If we actually involve the several terms (y+r) m , (y+r)'*" 1 , &c., as was done in Art. 316, we obtain certain terms inde- pendent of r, others which contain the first power of r, others the second power of r, and so on ; and the development is of the following form : where the values of X, X lf X 2 , &c., are X =y m +Ay m - l +By m -*+Cy m -*+ ..... + Ty+V. Each of these polynomials may be derived from that imme- diately preceding it, by multiplying each term by the exponent of y in that term, and diminishing the exponent by unity. The expressions X lf X 2 , &c., are called derived polyno- mials of X. Xj is called the first derived polynomial, X 2 the second derived polynomial, X 3 the third, and so on. Ex. 1. Find the equation whose roots are less by r than those of the equation Here we shall have X = y a -5y+6, But we have seen that when y+r is substituted for x, the equation reduces to the form Substituting the values of X, X,, X 2 , &c., above found, we obtain which is the development of Ex. 2. Find the equation whose roots are less by / than those of the equation GENERAL THEORY OF EQUATIONS. 271 Here we shall have X = y 8 - 7y a +8y-3, X ,=3^-1% +8, X 4 =0; and, substituting these values in the same formula as above, we obtain which is the development of Ex. 3. Find the successive derived polynomials of the equa- tion Ex. 4. Find the successive derived polynomials of the equa- tion x 6 +3x 4 +2x s 3x*-2x2=0. PROPOSITION XII. (319.) To find the equal roots of an equation. We have seen, in Art. 308, that an equation may have two or more equal roots. Thus, the equation or (x-2) 3 =0, has the three equal roots 2, 2, 2. Such an equation and its first derived polynomial always contain a common divisor ; for the first derived polynomial of the above equation is or 3(#-2) a , where it is evident that (x 2) a is a common divisor of both equations. In general, let a be one of the equal roots which occurs n times as a root of the given equation ; the first member will therefore contain the factors (xa), (xa), (xa), - ; that is, (x a)". The first derived polynomial will contain the factor n(x a)" ' ; that is, x a occurs (n 1) times as a factor 272 GENERAL THEORY OF EQUATIONS. in the first derived polynomial. The greatest common divisor of the given equation and its first derived polynomial must therefore contain the factor (x a) repeated once less than in the given equation. To determine, therefore, whether an equation has equal roots, fin d the greatest common divisor between the equation and its first derived polynomial. If there is no common divisor, the equation has no equal roots. If there is a common divisor, solve the equation obtained by putting this divisor equal to zero. Ex. 1. Find the equal roots of the equation The first derived polynomial of this equation is The greatest common divisor between this and the given equation is x-3. Hence the equation has two roots, each equal to 3. Ex. 2. Find the equal roots of the equation Ans. Two roots equal to 5. Ex. 3. Find the equal roots of the equation Ans. Two roots equal to 2. Ex. 4. Find the equal roots of the equation x'-6x*-8x-3=0. Ans. Three roots equal to 1 PROPOSITION XIII. (320.) To find the number of real and imaginary roots of an equation. In 1829, M. Sturm discovered a theorem which determines the precise number of real roots, and of course the number of imaginary ones, since the real and imaginary roots are to- gether equal in number to the degree of the equation. We propose now to develop this theorem. Let X represent the first member of the general equation of the mth degree, which we suppose to have no equal roots, and GENERAL THEORY OF EQUATIONS. 273 let X, be its first derived polynomial, found by the method of Art. 318. Divide X by X, until the remainder is of a lower degree than the divisor, and call this remainder X 77 ; that is, let X /7 designate the remainder with a contrary sign. Divide X 7 by X 77 in the same manner, and so on, designating the successive remainders with contrary signs by X //7 , X /7// , &c., until the di- vision terminates by leaving a numerical remainder independ- ent of x ; which must always be the case, according to the pre- ceding Proposition, since the equation having no equal roots, there can be no factor, which is a function of x, common to the equation and its first derived polynomial. Let this re- mainder, having its signs changed, be called X. The operation thus described will stand as follows : X |X, X, X,Q,Q, X,, x// x /7 X /7/ Q,,, X X 7 Q 7 X 77 ; X, X 77 Q /7 X //7 ; X /7 X 7// Q, 777 X 7777 . We thus obtain the series of quantities X, X 7 , X 7/ , X /7/ , X ///7 , X m , each of which is of a lower degree with respect to x than the preceding, and the last is altogether independent of x, that is, does not contain x. We now substitute for x in the above functions any two numbers p and q, of which p is less than q. The substitution of p will give results either positive or negative. If we only take account of the signs of the results, we shall obtain a certain number of variations and a certain number of permanences. The substitution of q for x will give a second series of signs, presenting a certain number of variations and permanences. The following, then, is THE THEOREM OF STURM. The difference between the number of variations of the first row of signs and that of the second, is equal to the number of real roots of the given equation comprised between p and q. (321.) In order to simplify the demonstration of this theorem, we shall premise three Lemmas ; and, for convenience, we shall call X the primitive function, and X 7 , X 77 , X /7/ , &c., auxiliary functions. S GENERAL THEORY OF EQUATIONS. LEMMA I. If we substitute any number for x in the series of functions X, X,, X //5 &c., two consecutive functions can not both reduce to zero at the same time. For, from the method in which X/, X /; , &c., are obtained, we have the following equations : X =X, Q, -X,, (1). X, =X // Q / , -X,,, (2). X// : =X /// Q, /// X,,,, (3). =X m _ 1 Q OT _ 1 X M (w 1). Now, if possible, suppose X/=0, and X//=0 ; then, by equa- tion (2), we shall have X/,^0. Also, since X/^0, and X 7// = ; therefore, by equation (3), we must have X/,,^0 ; and, pro- ceeding in the same manner, we shall find that X m =0, which is absurd, since it was shown, Art. 320, that this final remainder must be independent of x, and must therefore remain un- changed for every value of x. LEMMA II. 'When one of the auxiliary functions vanishes for a particular value of x, the two adjacent functions must have contrary signs. For, by equation (3), we have X// = A. /// (qJ,/ // -A-//// and if X //7 reduces to zero, then X,,^ X /7// ; that is, X /7 and X 7/// have contrary signs. LEMMA III. If & is a root of the equation X=0, the signs of X and X/ will constitute a variation for a value of x which is a little less than a, and a permanence for a value ofx which is a little greater than a. For if we substitute a+r for x in the equation X=0, the de- velopment of the function X, according to Art. 318, will be of the form A+AV+ other terms involving higher powers of r. Now if a is a root of the equation X=0, the first term of the development becomes zero, and there remains AV+ other terms involving higher powers of r. Also, if we substitute a+r for x in the first derived polyno- mial, the development will contain A' 4- other terms involving r. GENERAL THEORY OP EQUATIONS. 275 Now we may take r so small that each of these develop- ments shall have the same sign as its first term, A'r and A'. Hence they must both have the same sign when a is positive, and contrary signs when r is negative. That is, the signs of the two functions X and X, constitute a variation for x=a r, and a permanence for x=a+r. DEMONSTRATION OF THE THEOREM. (322.) Suppose all the real roots of the equations X=0, X 7 =0, X,,=:0, X ///= 0, &c., to be arranged in a series in the order of magnitude, beginning with the least. Let p be less than the least of these roots, and let it increase continually until it becomes equal to q, which we suppose to be greater than the greatest of these roots. Now so long as p is less than any of the roots, no change of signs will occur from the substitution of p for x in any of these functions, Art. 315 ; but when p arrives at a root of any of the auxiliary equations, its substitution for x reduces that polyno- mial to zero, and neither the preceding nor succeeding func- tion can vanish for the same value of a: (Lemma I.), and these two adjacent functions have contrary signs (Lemma II.). Hence the entire number of variations of sign is not affected by the vanishing of any of the auxiliary functions ; for the three adjacent functions must reduce to +, 0, -, or -, 0, +. Here is one variation, and there will also be one variation if we supply the place of the with either + or ; thus, +, +, , or , +, +, +, -, -, or , -, +. Suppose, now, p to pass from a number very little smaller, to a number very little greater than a root of the primitive equation X=0, the sign of X will be changed from + to , or from to +, Art. 315. The signs of X and X/ constitute a variation before the change, and a permanence after the change (Lemma III.} 27ft GENERAL THEORY OF EQUATIONS. Hence the change of sign of the function X occasions a loss of one variation of sign. Again, while p increases from a number very little smaller to a number very little greater than another root of the equa- tion X=0, a second variation will be changed into a perma- nence, and so on for the other roots of the primitive equation. Now, since all the real roots must be comprised within the limits oo and + 00 , if we substitute these values for x in the series of functions X, X /5 &c., the number of variations lost will indicate the whole number of real roots. A third suppo- sition, that x=0, will show how many of these roots are posi- tive and how many negative ; and if we wish to determine smaller limits of the roots, we must try other numbers. It is generally best to make trial in the first instance of such num- bers as are most convenient in computation, as, 1, 2, 10, &c. EXAMPLES. (323.) Ex. 1. How many real roots has the equation Here we have X.,=3x* Dividing x a 6x*+llx 6 by 3x* 12z-hll, as in the meth- od for finding the greatest common divisor, Art. 251, we have for a remainder 2#+4. Hence, rejecting the factor 2, X// =x 2. Dividing X, by X,,, we have for a remainder 1. Therefore, X /;/ = + l. Hence we have X = x a - 6a: a +llz-6. X, =3x*-l2x +11. A-// == X ~~ < x /// =+i. If we substitute GO for # in the first polynomial x 9 6x*+ llx 6, the sign of the result is ; substituting oo for x in the second polynomial 3x 2 12#+11, the sign of the result is -j- ; substituting the same in x2, the sign of the result is ; and X //7 , being independent of x, will remain + for every value of x, so that by supposing x= oo , we obtain the series of signs - + - +. Proceeding in the same manner for other assumed values of x, we shall obtain the following results : GENERAL THEORY OF EQUATIONS. 277 Assumed Values of x. Resulting Signs. Variations. oo -1 h giving 3 variations. 1 1- 3 + .9 h + "3 " + 1 + - + " 2 + 1.1 + + - + "2 + 1.9 H h "2 " +2 - + " 1 " +2.1 h + "1 " +2.9 - + + + "1 + 3 + + + " + 3.1 + + + + "0 -{- 00 _|_ _j_ _|- _{- "0 " Here the three roots of this equation are seen to be 1, 2, 3, and no change of sign in either function occurs by the substi- tution for x of any number less than 1 ; but when p exceeds 1, there is a change of sign in the original equation from to +, by which one variation is lost. When p=2, two of the functions disappear simultaneously, showing that 2 is a root of the second derived function as well as of the original equation, and a second variation of sign is lost. Also, when p becomes equal to 3, a third variation is lost ; and there are no further changes of sign arising from the substitution of any numbers between 3 and +00 . There are three changes of sign of the primitive function, two of the first auxiliary function, and one of the second auxiliary function ; but no variation is lost by the change of sign of any of the auxiliary functions ; while every change of sign of the primitive function occasions a loss of one variation. Ex. 2. How many real roots has the equation Here we find X = x 3 - X, =3x*-Wx +8. X,,=2x-3l. X^-2295. When x 00, the signs are -- 1 --- , giving 2 variations; Hence this equation has but one real root, and, consequent 278 GENERAL THEORY OP EQUATIONS. ly, must have two imaginary roots. Moreover , % it is easilv proved that the real root lies between and +1. Ex. 3. How many real roots has the equation Here we have X = x*- 2x 3 - 7 X, = 4x*- 6x 3 -14a:+10; or 2x*-3x*- f tx+5. X,, = 17x*-23x-45 X,,, =152x -305. X ///y = +524785. When x= oo, the signs are + H --- h, giving 4 variations ; x= + ct>, " + + + + +, " " Hence the four roots of this equation are real. If we try different values for x, we shall find that When #=3, the signs are H --- 1 --- h, giving 4 variations; x=-2, " - + + -+, " 3 x=-l, " - + + -+, " 3 x= 0, " + + -- +, " 2 a?= + l, " + --- +, " 2 x=+2, " + --- +, " 2 x=+3, " + + +++, " Hence this equation has one negative root between 2 and 3 ; one negative root between and 1 ; and two positive roots between 2 and 3. Ex. 4. How many real roots has the equation Ans. Three : viz., two between 1 and 2, and one between 3 and 4. Ex. 5. How many real roots has the equation 2x*-20x + l9=Ql Ans. Two. Ex. G. How many real roots has the equation Ans. One between 1 and 2. Ex. 7. How many real roots has the equation x'+3x*+5x- 178=0? Ans. One between 4 and 5. GENERAL THEORY OF EQUATIONS. 279 Ex. 8. How many real roots has the equation x t -l2x*-\-l2x-3=<) f ! Ans. Four. Ex. 9. How many real roots has the equation x*-8x 3 +I4x*+4x+-8=0 ? Ans. Four. PROPOSITION XIV. (324.) To discover a method of elimination for equations of any degree. The principle of the greatest common divisor affords one of the most general methods for the elimination of unknown quantities from a system of equations. Suppose we have two equations involving x and y reduced to the form of If we proceed to find the greatest common divisor of A and B, we shall have, according to Art. 249, A=QB+R. But since A and B are each equal to zero, it follows that R must equal zero. Hence we see that, if we divide one of the polynomials by the other, as in the method of finding the greatest common divisor, each successive remainder may be put equal to zero. If we arrange the polynomials before di- vision with reference to the letter x, we shall at last obtain a remainder which does not contain x ; which remainder, being put equal to zero, is the equation from which x has been elim- inated. Ex. 1 . Eliminate x from the equations a: 2 +y-13=0, x +y - 5=0. Divide the first polynomial by the second, as follows : x'+y*-13 \x+y-5 x*+(y-5)x\x-y+5 -(y-5)x+ y'-13 -(y-5)x- 2y a 10y+12 = remainder. 13 280 OUNERAL THEORY OP EQUATIONS. This remainder we have already proved must be equal to zero ; that is, an equation from which x has been eliminated. Ex. 2. Eliminate x from the equations x*+xy -56=0, xy +2y* 60=0. Ans. y 4 -118y 2 -f 1800=0. Ex. 3. Eliminate x from the equations .x*+y*x y-78=0, xy+x+y 39=0. Ans. y 4 +2/ 3 -77t/ 2 -273s/+ 1404=0. Ex. 4. Eliminate x from the equations x 2 3xy+y*+y=0, x*xy+l =0. .Arcs. 2/ 4 -5y a +2y-l=0. If we have three equations containing three unknown quan- tities, we must first eliminate one of the unknown quantities by combining either of the equations with each of the others. We thus obtain two new equations involving but two unknown quantities, from which we may obtain a final equation involv- ing but one unknown quantity. Ex. 5. Eliminate x and y from the equations xyz c=0, xz+xy+yz 6=0, x + y+ z a=0. Ans. z* #z a +6% c=0. Ex. 6. Eliminate x and y from the equations x*+= 7, Ans. z 8 - SECTION XX. SOLUTION OF NUMERICAL EQUATIONS. (325.) We* will first consider the method of finding the in- tegral roots of an equation, and will begin with forming the equation whose roots are 2, 3, 4, and 5. This equation must be composed of the factors. (x-2) (x-S) (x-4) (x-5)=0. If we perform the multiplication (which is most expeditious- ly done by the method of detached coefficients shown in Art. 64), we obtain the equation We know that this equation is divisible by x5. Let us perform the division by the method of detached coefficients shown in Art. 80. A B C D V a 1 14+71-154+12011 5= divisor. 1-5 |l 9+26-24= quotient. " 9+71 - 9+45 +26-154 +26-130 - 24+120 24+120. Supplying the powers of x, we obtain for a quotient This operation may be still further abridged, as follows : Represent the root 5 by , and the coefficients of the given equation by A, B, C, D, ..... V, 282 SOLUTION OF NUMERICAL EQUATIONS. We first multiply a by A, and subtract the product from B; the remainder, 9, we multiply by a, and subtract the product from C ; the remainder, +26, we multiply again by a, and subtract from D ; the remainder, 24, we multiply by a, and, subtracting from V, nothing remains. If we take the root a with a positive sign, we may substitute addition for subtraction in the above statement ; and if we set down only the successive remainders, the work will be as follows : ABC D V a I 14+71 -154+120[5 1- 9+26- 24, and the rule will be, Multiply A by a, and add the product to B ; set down the sum, multiply it by a, and add the product to C ; set down the sum, multiply it by a, and add the product to D, and so on. The final product should be equal to the last term V, taken with a contrary sign. The coefficients above obtained are the coefficients of a cubic equation whose roots are 2, 3, 4. The equation may therefore be divided by #4, and the operation will be as fol- lows : l-9+26-24|4 1-5+6. These, again, are the coefficients of a quadratic equation whose roots are 2 and 3. Dividing again by #3, we have l-5+6|3 1 2 which are the coefficients of the binomial factor x2. These three operations of division may be exhibited together as follows : 1-14+71-154+120 1- 9+26- 24 5, first divisor. 4, second divisor. 3, third divisor. 1- 5+ 6 1- 2. (326.) The method here explained will enable us to find all the integral roots of an equation. For this purpose, we make trial of different numbers in succession, all of which must be divisors of the last term of the equation. If any division leaves SOLUTION OF NUMERICAL EQUATIONS. 283 a remainder, we reject this divisor ; if the division leaves no remainder, the divisor employed is a root of the equation. Thus, by a few trials, all the integral roots may be easily found. Ex. 2. Find the seven roots of the equation x^+x* 14# 5 14# 4 +49# 3 +49:2r 2 3Qx 36=0. We take the coefficients separately, as in the last example, and try in succession all the divisors of 36, both positive and negative, rejecting such as leave a remainder. The operation is as follows : 1, first divisor. 2, second divisor. 3, third divisor. 1, fourth divisor. I, fifth divisor 2, sixth divisor. 3, seventh divisor. -14-14+49+49-36-36 1+2-12-26+23+72+36 1+4 4344518 1+7+17+17+ 6 1+6+11+ 6 1+5+ 6 1+3 Hence the seven roots are, 1,2,3, -1, -1, -2, -3. Ex. 3. Find the six roots of the equation 1584=0. 1. 4. 6. - 2. - 3. -11. 1+ 581 85+964+ 7801584 1+ 6-75-160+804+1584 1 + 10-35300-396 1 + 16+61+ 66 1 + 14+33 1 + 11 The six roots, therefore, are 1,4, 6, -2, -3, -11. Ex. 4. Find the five roots of the equation x*+6x*10x*112x*2Q7x 110=0. 1+6-10-112-207-110 -1. 1+515 97110 2. 1+3-21- 55 -5. 1-2-11 Three of the roots, therefore, are -1, -2, -5. 284 SOLUTION OF NUMERICAL EdUATIONS. The two remaining roots may be found by the ordinary method of quadratic equations. Supplying the letters to the last coefficients, we have x*-2x-ll=0. Hence x=l^12. Ex. 5. Find the four roots of the equation x*+2x*-1x*-8x+l2=0. Ex. 6. Find the four roots of the equation 4 Ex. 7. Find all the roots of the equation Ex. 8. Find all the roots of the equation x 6 +5x*+x* I6x* 20x 16=0. Ex. 9. Find all the roots of the equation A -, ^ J Ans. 1, 2, 3, and 6. HORNER'S METHOD. (327.) The preceding method furnishes the roots of an equa- tion only when they are expressed by whole numbers. When the roots are incommensurable, we employ the following meth- od, which is substantially the same as published by Homer in 1819. The Theorem of Sturm, together with Art. 315, enables us to find the integral part of any real root of the equation pro- posed. We then transform the equation into another having its roots less than those of the preceding by the number just found, Art. 316. We discover again, by Art. 315, the first figure of the root of this equation, which will be the first deci- mal figure of the root of the original equation. Again, we transform the last equation into another having its roots less than those of the preceding by this decimal figure. W e thus discover the second decimal figure of the root ; and proceed- ing in this manner from one transformation to another, we are enabled to discover the successive figures of the root, and may carry the approximation to any degree of accuracy re- quired. SOLUTION OF NUMERICAL EaUATION8. 285 Ex. 1. Find a root of the cubic equation We have found, page 278, that this equation has but one real root, and that it lies between 4 and 5. The first figure of the root, therefore, is 4. To ascertain the second figure, we transform the given equation into another in which the value of x is diminished by 4, which is done by substituting for x. We thus obtain The first figure of the root of this equation, according to Art. 315, is .5. Now transform the last equation into another in which the value of y is diminished by .5, which is done by substituting for ?/, .54-%. We thus obtain The first figure of the root of this equation is .03. We must now transform this equation into another in which the value of z is diminished by .03, which is done by substituting for z, 03+u. We thus obtain u 3 +16.59u 2 +93.7427v=.827623. The first figure of the root of this equation is .008. In order to find the next figure, we must transform the last equation into another in which the value of v is diminished by .008, and so on. (328.) This method would be very laborious if we were obliged to deduce the successive equations from each other by the ordinary method of substitution ; but they may all be de- rived from each other by a very simple law. Thus, let Ax*+Bx*+Cx=D (1) be any cubic equation ; and let the first figure of its root be denoted by , the second by a', the third by a", and so on. If we substitute a for x in equation (1), we shall have Aa 3 +Ba 2 +Ca=D, nearly. Whence a If we put y for the sum of all the figures of the root except the first, we shall have x=a+y ; and substituting this value for x in equation (1), we obtain 286 SOLUTION OF NUMERICAL EQUATIONS. or, arranging according to the powers of y, we have Ay+(B+3A=D-Ca-Ba 2 -Aa 8 (3) Let us put B' for the coefficient of y 2 , C' for the coefficient of y, and D' for the right member of the equation, and we have Ay 3 +B'y 2 +C'y=D' (4). This equation is of the same form as equation (1) ; and, pro- ceeding in the same manner, we shall find D' a ' = C'+B'a'+Aa (5) ' where a 1 is the first figure of the root of equation (4), or the second figure of the root of equation (1). Putting % for the sum of all the remaining figures, we have y=a'+z; and substituting this value in equation (4), we shall obtain a new equation of the same form, which may be written Az 3 +B"z 2 +C"z=D" (6) ; and in the same manner we might proceed with the remaining figures. Equation (2) furnishes the value of the first figure of the root; equation (5) the second figure, and similar equations would furnish the remaining figures. Each of these expres- sions involves the unknown quantity which is sought, and might therefore appear to be useless in practice. When, however, the root has been already found to several decimal places, the value of the terms Ba and Aa* will be very small compared with C, and a will be very nearly equal to -~. We may there- fore employ C as an approximate divisor, which will probably furnish a new figure of the root. Thus, in the above example, all the figures of the root after the first are found by division. 46 -7-77 =.5. 3.625-7- 92.75=.03. .827-7-93.74=.008. Tf we multiply the first coefficient A by , the first figure of SOLUTION OF NUMERICAL EQUATIONS. 287 the root, and add the product to the second coefficient, we shall have B+Aa (7). If we multiply this expression by a, and add the product to the third coefficient, we shall have C+B co to t ^ to g co to o S K o en o O5 co en O tfk. H- CO t ^ 1-1 O ifk. H- -3 O O O gggi rf^ ^ co co H- ~r co to oo en en en CO co -^ en o en "- 1 t-i O CO O5 tO 00 *5 oa to >-* q> ^ en -* ^- oo ^ to to || P- CA? >^ Cb f^ to -5 OD S to <. o en o 05 & w to to co ^. if ( " * 1 & o >- co en en o to to 00 H- 00 H CO en to to 00 Cn QD to ISs-i c Oi O5 CO 01 'P ^ a. to 1 en 3 CO 1 14 1 S co 0- g- cp S ^ en 00 to & g -^ 1 en II CO to en < r- 1 ^ -* pi zr ct> 296 SOLUTION OF NUMERICAL EQUATIONS. Dividing the fourth dividend by the fourth divisor, we ob- tain the figures 789. When we wish to obtain a root correct to a limited numbei of places, we may save much of the labor of the operation by cutting off all figures beyond a certain decimal. Thus if, in the example above, we cut off all beyond five decimal places in the successive dividends, and all beyond four decimal places in the divisors, it will not affect the first six decimal places in the root. Ex. 3. Find the roots of the equation C- 3.907378, j + .443277, Ans ' | + .606018, i. +2.858083. Ex. 4. Find the roots of the equation x 4 - 16z 3 +79.r 2 - 140a-= -58. r +0.58579, I +3.35425, Ans.< + 3 >41421) t + 8.64575. Ex. 5. Find the roots of the equation r +0.934685, +3.308424, Ans. J +3.824325, | +4.879508, 1^+7.053058. Ex. 6. Required the fourth root of 18339659776. Ans. 368. Ex. 7. Required the fifth root of 26286674882643. Ans. 483. Ex. 8. There is a number consisting of four digits such that the sum of the first and second is 9 ; the sum of the first and third is 10 ; the sum of the first and fourth is 11 ; and if the product of the four digits be increased by 36 times the product SOLUTION OF NUMERICAL EdUATIONS. 297 of the first and third, the sum will be equal to 3024 diminished by 300 times the first digit. Required the number. 6345, or 7234, 9012. RESOLUTION OF EQUATIONS BY APPROXIMATION. (331.) The method of Horner for finding the incommensura- ble roots of a numerical equation is generally better than any other ; nevertheless, the method by approximation may some- times be preferred. We shall explain the method of Newton, and that of Double Position. METHOD OF NEWTON. This method supposes that we have already determined nearly the value of one root ; that we know, for example, that such a value exceeds , and that it is less than a+l. In this case, if we suppose the exact value =a+y, we are certain that y expresses a proper fraction. Now, as y is less than unity, the square of ?/, its cube, and, in general, all its higher powers, will be much less with respect to unity ; and for this reason, since we only require an approximation, they may be neglect- ed in the calculation. When we have nearly determined the fraction ?/, we shall know more exactly the root a-\-v ; from which we proceed to determine a new value still more exact, and we may continue the approximation as far as we please. We will illustrate this method by an easy example, requiring by approximation the root of the equation x*=20. Here we perceive that x is greater than 4, and less than 5. If we suppose x=4+y, we shall have But, as y a must be quite small, we shall neglect it, and we have 16+8y=20, or 8^=4. Whence y=.5, and x=4.5, which already approaches near the true root. If we now suppose x=4.5+z, we are sure thm 298 SOLUTION OF NUMERICAL EQUATIONS. z expresses a fraction much smaller than y, and that we may neglect z a with greater propriety. We have, therefore, a; a =20.25-|-9z=20, or 9z= .25. Consequently z= .0278. Therefore, z=4.5-.0278=4.4722. If we wish to approximate still nearer to the true value, we must make x = 4.4722 +v, and we should have z a =20.00057284+8.9444u=20. So that 8.9444t>= .00057284. Whence v= - .0000640. Therefore, z=4.4722-.0000640=4.4721360, a value which is correct to the last decimal place. (332.) The preceding method is expressed in the following RULE. Find by trial a number (a) nearly equal to the root sought, and represent the true root by a+y. Substitute a+y/or x in the given equation, and there will re- sult a new equation containing only y and known quantities. Reject all the terms of this equation which contain the second or higher powers of y, and the approximate value of y will then be given by a simple equation. Havin^ applied this correction to the assumed root, the op- eration must be repeated with the corrected value of a, when a second correction will be obtained which will give a nearer value of the root, and the process may be repeated as often as is thought necessary. EXAMPLES. Ex. 1. Find a root of the equation If we substitute a +y for x in this equation, and reject all the terms containing the higher powers of y, we shall have Whence y= - SOLUTION OF NUMERICAL EaUATIONS. 299 We find by trial that x is nearly equal to 3. If we substi- tute 3 for , we shall have 2 y = -5r Whence x =2. 9 nearly. And if we substitute this new value instead of a, we shall find another still more exact. Ex. 2. Find a root of the equation x*-6x=IO. If we make x=a+y, we shall have Therefore, y Assume /5i or z=--iv/5. The values of #, deduced from the equation or x* zx= 1, are z /?^4 z /z a -4 =g+ v~T~' and ^"s" V ~~T"' from which, by substituting the value of z, we obtain or =-iIV5+l=Fv'-10+2V5]. Hence the five fifth roots of unity are 1 SOLUTION OF NUMERICAL EQUATIONS. 303 Ex. 5. Find the six roots of the equation # 6 =1. These are found by taking the square roots of the cube roots. Hence we have, + 1, -1, Thus we see that unity has two square roots, three cube roots, four fourth roots, jive fifth roots, six sixth roots, and, generally, the nth root of unity admits of n different algebraic values. As, however, most of these roots are imaginary, they can not be found by Horner's Method. 14 .SftdtYA npg adt m ;>mfc &s SECTION XXI. ;> {^i o tear d* edi >nu : a HBO 1 LOGARITHMS. (335.) In a system of logarithms, all numbers are considered as the powers of some one number, arbitrarily assumed, which is called the base of the system ; and the exponent of that power of the base which is equal to any given number is called the log- arithm of that number. Thus, if a be the base of a system of logarithms, and (2). Multiplying together equations (1) and (2), we obtain o m- -NN'= r ar B/ ^ , PMSfifi Therefore, according to the definition of logarithms, a: -fa;' is the logarithm of NN ; , since x+x' is the exponent of that power of the base a which is equal to NN' ; hence .ic 2 a.i 100. bns fO. nawJIsd rad'mTm v*isr? *lo mHHT PROPERTY I The logarithm of the product of two or more factors is equal to the sum of the logarithms of those factors. Hence we see that if it is required to multiply two or more numbers by each other, we have only to add their logarithms ; the sum will be the logarithm of their product. We must then look in the Table for the number answering to that logarithm, in order to obtain the required product. EXAMPLES. Ex. 1. Find the product of 8 and 9 by means of logarithms. On page 318, the logarithm of 8 is given 0.903090 " 9 " 0.954243 The sum of these two logarithms is 1.857333, which, according to the same Table, is seen to be the loga- rithm of 72. ^pla -ai 900, lo mrhn'rap! erL f gr/riT Ex. 2. Find the continued product of 2, 5, and 14 by means of logarithms. Ex. 3. Find the continued product of 1, 2, 3, 4, and 5 by means of logarithms. (339.) If, instead of multiplying, we divide equation (1) by equation (2), we shall obtain LOGARITHMS. 307 Therefore, according to the definition, x x' is the logarithm N of ^7, since x x' is the exponent of that power of the base a N which is equal to ^ ; hence, PROPERTY II. The logarithm of a fraction, or of the quotient of one number divided by another, is equal to the logarithm of the numerator, minus the logarithm of the denominator. Hence we see that if we wish to divide one number by an- other, we have only to subtract the logarithm of the divisor from that of the dividend ; the difference will be the logarithm of their quotient. EXAMPLES. Ex. 1. It is required to divide 108 by 12 by means of loga- rithms. The logarithm of 108 is 2.033424 12 1.079181 The difference is 0.954243, which is the logarithm corresponding to the number 9. Ex. 2. Divide 133 by 7 by means of logarithms. Ex. 3. Divide 136 by 17 by means of logarithms. Ex. 4. Divide 135 by 15 by means of logarithms. The preceding examples are designed to illustrate the prop- erties of logarithms. In order to exhibit fully their utility in computation, it would be necessary to employ larger numbers ; but that would require a more extensive Table than the one given on page 318. (340.) Logarithms are attended with still greater advantages in the involution of powers and in the extraction of roots. For if we raise both members of equation (1) to the mth power, we obtain 308 LOGARITHMS. Therefore, according to the definition, mx is the logarithm of N m , since mx is the exponent of that power of the base which is equal to N M ; hence PROPERTY III. The logarithm of any power of a number is equal to the loga- rithm of that number multiplied by the exponent of the power. EXAMPLES. Ex. 1 . Find the third power of 4 by means of logarithms. The logarithm of 4 is 0.602060 Multiply by 3 The product is 1.806180, which is the logarithm of 64. Ex. 2. Find the fourth power of 3 by means of logarithms. Ex. 3. Find the seventh power of 2 by means of loga- rithms. Ex. 4. Find the third power of 5 by means of logarithms. (341.) Also, if we extract the mth root of both members of equation (1), we shall obtain x therefore, according to the definition, is the logarithm of N w ; hence PROPERTY IV. The logarithm of any root of a number is equal to the loga- rithm of that number divided by the index of the root. EXAMPLES. Ex. 1. Find the square root of 81 by means of logarithms. The logarithm of 81 is 1.908485 Divided by 2 The quotient is .954243, which is the logarithm of 9. LOGARITHMS. 309 Ex. 2. Find the square root of 121 by means of loga- rithms. Ex. 3. Find the sixth root of 64 by means of logarithms. Ex. 4. Find the third root of 125 by means of logarithms. The preceding examples will suffice to show, that if we had tables which gave the logarithms of all numbers, they would prove highly useful when we have occasion to perform fre- quent multiplications, divisions, involutions, and extraction of roots. (342.) The following examples will show the application of some of the preceding principles. Ex. I. log. (abcd)= log. a+ log. b+ log. c-f log. d. Ex. 2. log. f j = log. a+ log. b+ log. c log. d log. e. Ex. 3. log. (a m b n c p )=m log. a+n log. b+p log. c. Ex. 4. log. ( \=m log. a+n log. b p log. c. Ex. 5. log. (a a -# a )=log. [(<*+z) (a-x)]= log. (a+x) + log. (a-x). Ex. 6. log. Vrf r= ?=i log. (a+x)+% log. (a-x}. Ex. 7. log. ( an( j v< a;t)=a, y=6a?J \/ a Ex. 5. Given xyz = 105, > j xyv=I35, I to find the values of #, y, 2, and xzv=189, f v. y2v=315,J jins. #=3, y=5, 2=7, v9. 326 MISCELLANEOUS EXAMPLES. X* Ex. 6. Given x *+-+y'=84, to find the values of x and . . #=4, y=2 or\8. 7. Given - -- =b } to find the values of x. a+ Ans. x Ex. 9. Given v/y Jax Vyx, ) to find the values 2 Vy-a;+2 Va-a;=5 Va-x, ) of and y- 4 5 -a, y~: D Tt . 10. Given -- 1 -- - = \/ r , to find the values of x. = \/ r , v b Ans. x= Ex. 11. Given re 8 +xy*=ay, y =ay, ) e . , [ to find the values of ^ and y. =bx, ) , . _ ., JEa:. 12. Given - - -- --- =- - 77-7, to find a+x ax 3a4b+x the values of #. 2a> Ans. x=3a, or 3a r-. ax bx a+b f Ex. 13. Given -r-, --- ; = - r> to find the values of #. o+x a+x ab , .? Ex. 14. Given MISCELLANEOUS EXAMPLES. Vl _ X 1+Vl+x 1Vlx 327 to find the values ofx. Ex. 15. Given a+x Ans. #= c =6, to find the values of x. . X ~~ = 10, tO find th _ '+ Vy> values of =275, ) x and y . Ans. x=9, y=4. f to find the values x+y x+y ofa;and =41, J ^^= ( Ex. 16. Given Ex. 17. Given ^x. 18. Given (a;+y) 8 +x+y=30, ) to find the values of x xy= 1, ) and y. Ans. x=2, y=l. Ex. 19. Given x* 4x*+7x* 6z=18, to find the values of a; by a quadratic equation. Ans. x=3, or 1. Ex. 20. Given (x +y ) (x y +1)= 18x y, ) to find the values +l)=2Q8 Ans. #= (x +y ) (x y +1)= 18x y, ) (x*+y*) (xy+l)=2Q8xy, \ of x and y. E&. 21. Given (x a +y 2 )x?/= 13090, ) to find the values of x x+y =18, j and y. Ans. x=7, or 11, y=ll, or 7. JBa;. 22. Given 5(z 2 +?/ 2 )+4:ry=356, ) to find the values of x*+y*+x+y=62, } x and y. Ans. x=4, y=Q. Ex. 23. Given (x'+y*)xy=300, ) to find the values of a; and x'+y* =337, \ y. Ans. #=4, y~3. 15 328 MISCELLANEOUS EXAMPLES. Ex. 24. Given (x*+y*) (aj'+y')=455, ) to find the values of x+y =5, j x and y. Ans. x=3, y=2. Ex. 25. Given , y +14, ] x+y I to find the values of a; and and x-y #=12, y=Q. to find the Ex. 26. Given (:r a xy+y*) ( a +y a ) =91, . (^-ay+y-) " fc 5 values of Ans. x2, or 3; y=3, or 2. fin and 27. Given (x+y)xy =30, ) to find the values of x :y =468, J Sr. 28. The sum of two numbers is a, and the sum of their reciprocals is b. Required the numbers. a /a* a Ans - *v 4- jE;r. 29. In the composition of a certain quantity of gunpow- der, the nitre was ten pounds more than two thirds of the whole ; the sulphur was four and a half pounds less than one sixth of the whole ; and the charcoal was two pounds less than one seventh of the nitre. How many pounds of gunpowder were there ? Ans. 69 pounds. Ex. 30. Find three numbers such that if six be subtracted from the first and second, the remainders will be in the ratio of 2 : 8 ; if thirty be added to the first and third, the sums will be in the ratio of 3 : 4 ; but if ten be subtracted from the sec- ond and third, the remainders will be as 4 : 5. Ans. 30, 42, 50. Ex. 31. Divide the number 165 into five such parts that the first increased by one, the second increased by two, the third diminished by three, the fourth multiplied by 4, and the fifth divided by five, may all be equal. Ans 19, 18, 23, 5, and 100. MISCELLANEOUS EXAMPLES. Ex. 32. A criminal having escaped from prison, traveled ten hours before his escape was known. He was then pur- sued, so as to be gained upon three miles an hour. After his pursuers had traveled eight hours, they met an express going at the same rate as themselves, who met the criminal two hours and twenty-four minutes before. In what time from the commencement of the pursuit will they overtake him ? Ans. 20 hours. Ex. 33. A and B engage to reap a field of wheat in twelve days. The times in which they could severally reap an acre are as 2 : 3. After some days, finding themselves unable to finish it in the stipulated time, they call in C to help them, whose rate of working was such that, if he had wrought with them from the beginning, it would have been finished in nine days. Also, the times in which he could have reaped the field with A alone, and with B alone, are in the ratio of 7 : 8. When was C called in ? Ans. After six days. Ex. 34. A laborer is engaged for n days, on condition that he receives p pence for every day he works, and pays q pence for every day he is idle. At the end of the time he receives a pence. How many days did he work, and how many was he idle? Ans. He worked -^- , and was idle -~ days. p+q p+q Ex. 35. The fore wheel of a carriage makes three revolu- tions more than the hind wheel in going sixty yards ; but, if the circumference of each wheel be increased one yard, it will make only two revolutions more than the hind wheel in the same space. Required the circumference of each. Ans. 4 and 5 yards. Ex. 36. There is a wagon with a mechanical contrivance by which the difference of the number of revolutions of the wheels on a journey is noted. The circumference of the fore wheel is a feet, and of the hind wheel b feet. What is the dis- tance gone over when the fore wheel has made n revolutions more than the hind wheel ? abn Ans. r - feet. 6 a 330 MISCELLANEOUS EXAMPLE*. Ex. 37. A merchant has two casks, each containing a cer- tain quantity of wine. In order to have an equal quantity in each, he pours out of the first cask into the second as much as the second contained at first ; then he pours from the second into the first as much as was left in the first ; and then again. from the first into the second as much as was left in the sec- ond, when there are found to be a gallons in each cask. How many gallons did each cask contain at first ? lla . 5 Ans. and . Ex. 38. A and B engage to reap a field for $24 ; and as A alone could reap it in nine days, they promise to complete it in five days. They found, however, that they were obliged to call in C to assist them for the last two days, in consequence of which B received one dollar less than he otherwise would have done. In what time could B or C alone reap the field ? Ans. B in 15, and C in 18 days. Ex. 39. A cistern can be filled by four pipes ; by the first in a hours, by the second in b hours, by the third in c hours, and by the fourth in d hours. In what time will the cistern be filled when the four pipes are opened at once ? abed /\ 77 *? n ~~ i ^ r - abc+abd+acd+bcd' Ex. 40. The sum of the cubes of two numbers is 35, and the sum of their ninth powers is 20195. Required the num- bers. Ans. 2 and 3. Ex. 41. A number consisting of three digits, which are in Arithmetical Progression, being divided by the sum of its dig- its, gives a quotient 26; and if 198 be added to it, the digits will be inverted. Required the number. Ans. 234. Ex. 42. There are three numbers in Geometrical Progres- sion, the difference of whose differences is six, and their sum is forty-two. Required the numbers. Ans. 6, 12, and 24. Ex. 43. There are three numbers in harmonical proportion ; MISCELLANEOUS EXAMPLES. 331 the sum of the first and third is 18, and the product of the three numbers is 576. Required the numbers. Ans. 12, 8, and 6. Ex. 44. There are three numbers in harmonica! proportion, the difference of whose differences is 2, and four times the product of the first and third is 960. Required the numbers. Ans. 20, 15, 12. Ex. 45. There are two numbers whose product is 300 ; and the difference of their cubes is thirty-seven times the cube of their difference. What are the numbers ? Ans. 20 and 15. Ex. 46. There are three numbers in geometrical progres- sion, the greatest of which exceeds the least by 24 ; and the difference of the squares of the greatest and the least is to the sum of the squares of all the three numbers as 5 : 7. What are the numbers ? Ans. 8, 16, and 32. Ex. 47. A merchant had 826,000, which he divided into two parts, and placed them at interest in such a manner that the incomes from them were equal. If he had put out the first portion at the same rate as the second, he would have drawn for this part $720 interest ; and if he had placed the second out at the same rate as the first, he would have drawn for it $980 interest. What were the two rates of interest ? Ans. 6 per cent, for the larger sum, and 7 for the smaller. Ex. 48. A grocer has a cask containing 20 gallons of bran- dy, from which he draws off a certain quantity into another cask of equal size, and, having filled the last with water, the first cask was filled with the mixture. It now appears that if 6| gallons of the mixture are drawn off from the first into the second cask, there will be equal quantities of brandy in each. Required the quantity of brandy first drawn off. Ans. 10 gallons. Ex. 49. A miner bought two cubical masses of ore for $820. Each of them cost as many dollars per cubic foot as there were feet in a side of the other ; and the base of the greater con- tained a square yard more than the base of the less. What was the price of each ? Ans. 500 and 320 dollars. 15* MISCELLANEOUS EXAMPLES. Ex. 50. A and B traveled on the same road, and at the same rate, from Cumberland to Baltimore. At the 50th mile- stone from Baltimore, A overtook a drove of geese, which were proceeding at the rate of three miles in two hours ; and two hours afterward met a wagon, which was moving at the rate of nine miles in four hours. B overtook the same drove of geese at the 45th milestone, and met the same wagon 40 min- utes before he came to the 31st milestone. Where was B when A reached Baltimore ? Ans. 25 miles from Baltimore. Ex. 51. A gentleman bought a rectangular lot of land at the rate of ten dollars for every foot in the perimeter. If the same quantity had been in a square form, and he had bought it at the same rate, it would have cost him $330 less ; but if he had bought a square piece of the same perimeter, he would have had 12 J rods more. What were the dimensions of the lot? Ans. 9 by 16 rods. Ex. 52. A and B put out at interest sums amounting to $2400. A's rate of interest was one per cent, more than B's ; his yearly interest was five sixths of B's ; and at the end of ten years his principal and simple interest amounted to five sevenths of B's. What sum was put at interest by each, and at what rate ? Ans. A $960, at 5 per cent. B $1440, at 4 " Ex. 53. Two merchants sold the same kind of cloth. The second sold three yards more of it than the first, and together they received $35. The first said to the second, I should have received $24 for your cloth ; the other replied, I should have received $12j for yours. How many yards did each of them sell? Ans. The first merchant, 5 or 15 yards. The second " 8 or 18 " Ex. 54. A person bought a quantity of cloth of two sorts for $63. For every yard of the best piece he gave as many dol- lars as he had yards in all ; and for every yard of the poorer, as many dollars as there were yards of the better piece more than of the poorer. Also, the whole cost of the best piece was MISCELLANEOUS EXAMPLES. 333 six times that of the poorer. How many yards had he of each? Ans. 6 yards of the better, and 3 of the poorer. Ex. 55. A and B, 165 miles distant from each other, set out with a design to meet. A travels 1 mile the first day, 2 the second, 3 the third, and so on. B travels 20 miles the first day, 18 the second, 16 the third, and so on. In how many days will they meet ? Ans. 10 or 33 days. Ex. 56. There are three numbers in Geometrical Progres- sion whose continued product is 216, and the sum of their cubes is 1971. Required the numbers. Ans. 3, 6, and 12. Ex. 57. There are four numbers in Geometrical Progression whose sum is 350 ; and the difference between the extremes is to the difference of the means as 37 : 12. What are the num- bers? Ans. 54, 72, 96, 128. Ex. 58. A commences a piece of work alone, and labors for two thirds of the time that B would have required to perform the entire work. B then completes the job. Had both labor- ed together, it would have been completed two days sooner ; and A would have performed only half what he left for B. Required the time in which they would have performed the work separately. Ans. A in 6 days, and B in 3 days. Ex. 59. A ship, with a crew of 175 men, set sail with a sup- ply of water sufficient to last to the end of the voyage ; but in 30 days the scurvy made its appearance, and carried off three men every day ; and at the same time a storm arose which protracted the voyage three weeks. They were, however, just enabled to arrive in port without any diminution in each man's daily allowance of water. Required the time of the passage, and the number of men alive when the vessel reached the harbor. Ans. The voyage lasted 79 days, and the number of men alive was 28. Ex. 60. The number of deaths in a besieged garrison 334 MISCELLANEOUS EXAMPLES. amounted to 6 daily ; and, allowing for this diminution, their stock of provisions was sufficient to last 8 days. But on the evening of the sixth day 100 men were killed in a sally, and afterward the mortality increased to 10 daily. Supposing the stock of provisions unconsumed at the end of the sixth day to support 6 men for 61 days, it is required to find how long it would support the garrison, arid the number of men alive when the provisions were exhausted. Ans. The provisions last 6 days, and 26 men survive. THE END. C0omi0 1 bourse ot 4iilcttl)ematu$, PUBLISHED BT HARPER & BROTHERS, NEW YORK, THE Publishers of the course of Mathematics by Prof. Loomis invite the at- tention of professors of colleges and teachers generally to an examination ot these works. They are the fruits of a long series of years devoted to collegiate instruction, and it is believed that they possess in an eminent degree the quali- ties of simplicity, conciseness, and lucid arrangement, and are adapted to the wants of students generally in our colleges and academies. LOOMIS' TREATISE ON ALGEBRA. 8vo, p. 334, Sheep, $1 00. Third Edition. I hare carefully examined the work of Prof. Loomis on Algebra, and am much pleased with it. The arrangement is sufficiently scientific, yet the order of the topics is obviously, and, I think, ju- diciously made with reference to the development of the powers of the pupil. The most rigorous modes of reasoning are designedly avoided in the earlier portions of the work, and deferred till the student is better fitted to appreciate them. All the principles are, however, established with suffi- cient rigor to give satisfaction. Much care seems to have been taken, by generalizing particular ex- amples and other means, to develop the faculty and induce the habit of generalizing, a point which- I think, has not received sufficient attention hitherto. On the whole, therefore, I think this work better suited for the purposes of a text-book than any other I have Been. AUGUSTUS W. SMITH LL.D., Professor of Mathematics and Astronomy in the Wesleyan University. Prof. Loomi*' Treatise on Algebra is an excellent elementary work. It is sufficiently extensive for ordinary purposes, and is characterized throughout by a happy combination of brevity and clear- ness. A. CASWELL, D.D., Professor of Mathematics and Natural Philosophy in Brown University. I have examined Prof. Loomis' new work on Algebra, and am highly pleased with it. For con ciseness and clearness of statement, and for its lucid explanation of elementary principles, it is de cidedly superior to any work with which I am acquainted. I hope it will be extensively used in all our public institutions. ALONZO GRAY, A.M., Professor in Brooklyn Female Academy. I have examined Prof. Loomis' Algebra carefully and with much interest, and am so perfectly satisfied with it, that I shall introduce it to my classes as soon as possible. It is just the work which I have been for a long time in search of. I am particularly delighted with the mode of treating the subject of logarithms, and, indeed, with the clearness of the investigations generally throughout the work. E. OTIS KENDALL, Professor of Mathematics and Astronomy in the Central High School of Philadelphia. I fully concur with Prof. Kendall in his opinion of Loony's' Algebra. SEARS C. WALKER, of the United States Coast Survey. A text-book like this of Prof. Loomis was much needed, and the desideratum is so well supplied, f hat I think it can not fail to commend itself at once to the favorable regard of those who are looking for the best work for college classes. I consider it decidedly the best book for college instruction that I am acquainted with on the subject, and it has been adopted as a text-book in our college by unanimous consent of the faculty. Prof. Loomis has been very happy in simplifying the more diffi- cult parts of the subject, especially on the theory of equations and on logarithms. JAMES NOONEY, A.M., Professor of Mathematics and Natural Philosophy in Western Reserve College. I have carefully examined Prof. Loomis' Algebra, and think it better adapted for a text-book lor college students than any other I have seen. C. GILL, Professor of Mathematics in St. Pants Col. Prof. Loomis seems very happily to have observed the proper medium between exuberance of ex- planation and demonstration on the one hand, which leaves little or nothing for the student himself K, do ; and a repulsive conciseness on the other, which discourages him, and gives him a disrelish fcr this portion of study. I have adopted it as a text-book in the Cornelius Institute, believing it to be better suited to youth who are preparing for college, than any other treatise on Algebra with waich 1 am acquainted. JOHN J. OWEN, D.D., Professor in the Free Academy. After a thorough examination of Prof. Loomis' work on Algebra, I have concluded to adopt it as a M t-book in this Institution. MARCUS CATLIN, A.M., Professor of Mathematics and Astronomy tw 23 -milton College 2 Critical Notices of Loomis' Algebra. - . - ' .1 JE >?? ^w** * * v - Prof. Loomis' work on Algebra is exceedingly wall adapted for the purposes of instruction. He has avoided the difficulties which result from too great conciseness, and aiming at the utmost rigor of demonstration ; and, at the same time, has furnished in his book a good and sufficient preparation for the subsequent parts of the mathematical course. I do not know of a treatise which, all things considered, keeps both these objects so steadily in view. J. WARD ANDREWS, A.M., Professor of Mathematics and Natural Philosophy in Marietta College. Prof. Loomis' work is well calculated to impart a clear and correct knowledge of the principles of Algebra. The rules are concise, yet sufficiently comprehensive, containing in few words all that is necessary, and nothing more; the absence of which quality mars many a scientific treatise. The collection of problems is peculiarly rich, adapted to impress the most important principles upon the youthful mind, and the student is led gradually and intelligently into the more interesting and higher departments of the science. JOHN BROCKLESBY, A.M., Professor of Mathematics and Natural Philosophy in Trinity College. I am much pleased with Prof. Loomis' Algebra. I think he has accomplished very happily the object he had in view, and has prepared a work remarkably well adapted for the use of college stu dents. E. S. SHELL, A.M., Professor of Mathematics and Natural Philosophy in Amherst College I am much pleased with Prof. Loomis' Algebra. The arrangement of the subjects Is, I think, an admirable one. The best proof I can give of the estimation in which I hold it is, that I am now hearing recitations in it the second time. JOHN TATLOCK, A.M., Professor of Mathematics in Williams College. I have examined Prof. Loomis' Algebra with great attention, and am so well pleased with its ar- larigement and execution throughout, that I have decided to adopt it as a text-book in this institu* tion. THOMAS E. SUDLKR, A.M., Professor of Mathematics in Dickinson College. Prof. Loomis has here aimed at exhibiting the first principles of Algebra in a form which, while evel with the capacities of ordinary students and the present state of the science, is fitted to elicit that degree of effort which educational purposes require. Throughout the work, whenever it can be done with advantage, the practice is followed of generalizing particular examples, or of extend- ing a question proposed relative to a particular quantity, to the class of quantities to which it be- longs ; a practice of obvious utility, as accustoming the student to pass from the particular to the general, and as fitted to impress a main distinction between the literal and numeral calculus. The general doctrine of Equations is expounded with clearness, and, we may add, with independence. The author has developed this subject in an order of his own. Theorems which find a place in other treatises are omitted, and what sometimes appears in a generic form, or in that of a corollary, be- comes specific, or assumes the place of a primary proposition. We venture to say that there will be but one opinion respecting the general character of the exposition. American Journal of Science and Arts. I regard Prof. Loomis' Algebra as altogether worthy of the high reputation its author deservedly enjoys. It possesses those qualities wl^ch are chiefly requisite in a college text-book. Its state- ments are clear and definite ; the more important principles are made so prominent as to arrest the pupil's attention ; and it conducts the pupil by a sure and easy path to those habits of generalization which the teacher of Algebra has so much difficulty in imparting to his pupils. JULIAN M. STUR- TEVANT, L.L.D., President of Illinois College. The arrangement of Prof. Loomis' Algebra is good i the doctrine of Equations is clearly presented, and the principle of generalization is ably developed in a manner well calculated to improve th youthful mind. W. P. ALRICH, Professor of Mathematics in Washington College, Pennsylvania. Prof. Loomis' Algebra is admirably got up. It is clear and simple in arrangement, and just the work for the class of learners for whom the author prepared it. The introduction of Horner's ad- mirable method for finding incoiraiensurable roots, and the section on Logarithms, render it superior to any text-book on the subjec; with which I am acquainted. Pres. COLLINS, Emory and Henry College, Virginia. We feel bound to express our conviction that this is a decidedly better text-book, especially for those not already far advanced in the study, than any other we have seen. It is carefully and lu- cidly arranged, and admirably enunciated and explained. Teacher's Journal. The present work is the fruit of long experience in teaching and diligent investigation of the sci race. The author has sought to avoid unnecessary prolixity on the one hand, and undue brevity o the other, and with the observance of this happy medium he has embodied all the latest itnpror* mants.- Methodist Quarterly Rtview. Critical Notices of Loomis' Geometry. 3 LOOMIS' GEOMETRY AND CONIC SECTIONS. Second Edition. 8vo, p. 226, Sheep, $1 00. Every page of this book bears marks of careful preparation. Only those propositions are selected which are most important in themselves, or which are indispensable in the demonstration of others. The propositions are all enunciated with studied precision and brevity. The demonstrations are complete without being encumbered with verbiage ; and, unlike many works we could mention, the diagrams are good representations of the objects intended. We believe this book will take its place among the best elementary works which our country has produced. American Review. Prof. Loomis' Geometry is characterized by the same neatness and elegance which were exhibited in his Algebra. While tlie logical form of argumentation peculiar to Playfair's Euclid is preserved, more completeness and symmetry is secured by additions in solid and spherical geometry, and by a different arrangement of the propositions. It will be a favorite with those who admire the chaste forms of argumentation of the old school ; and it is a question whether these are not the best for the purposes of mental discipline. Northern Christian Advocate. As a text-book for instruction, this work possesses advantages superior, in some respects, to any other work on the subject in our language. The arrangement is good, and the selection of proposi- tions so judiciously made as to comprise what is most valuable for the purposes of the student, both for intellectual culture and for a knowledge of geometry. Prof. Loomis has introduced some valu- able improvements, especially that of computing the area of a circle in a very simple and easy man ner, and that of shading the diagrams in solid geometry, which will greatly aid the student inform- ing his conceptions of solid angles and the positions of planes. Ohio Observer. The enunciations in Prof. Loomis' Geometry are concise and clear, and the processes neither too brief nor too diffuse. The part treating of solid Geometry is undoubtedly superior, in clearness and arrangement, to any other elementary treatise among us. New York Evangelist. Prof. Loomis has admirably harmonized the logical system of the Greek geometer with the more rapid processes of modern mathematicians. New York Observer. Having been requested, by a resolution of our Board of Trustees, to report such a course of mathe matical studies as I may deem best suited to our circumstances, I have selected Loomis' Geometry and Conic Sections as a part of the course. MATTHEW J. WILLIAMS, Professor of Mathematics in South Carolina College. Prof. Loomis has made Legendre's Geometry far more Euclidian, and therefore more valuable. Some of his demonstrations are decided improvements on those of Legendre. Professor C. DEWET, Rochester Collegiate Institute. This book is far in advance of Playfair's Euclid. It can not fail to come into general use. Albany Atlas. Particular attention has been paid by the author to the diagrams, reducing them to a uniformity of dimensions, and, of course, conveying clearer ideas of what they represent to the mind of the pupil. This book is worthy the attention of our high schools and colleges. Hartford Spectator. The propositions are all enunciated with clearness and brevity. It is a valuable work, and well deserving the attention of schools and mechanics. Farmer and Mechanic. An available text-book, which we have no doubt will become a standard authority in this depart- ment of instruction. Literary World. This work is admirably drawn up, and merits universal adoption in all schools where these branches of science are taught. Courier and Enquirer. These books are terse in style, clear in method, easy of comprehension, and perfectly free from that useless verbiage with which it is too much the fashion to load school-books under prctonse of explanation. Scott's Weekly Paper, Canada. Prof. Loomis is doing a valuable service to the cause of mathematical science by the course of text-books he is preparing. His writings in other departments of science are characterized by a remarkable clearness in the manner in which he exhibits truth, and his treatises on Algebra and Geometry bear evident marks of having emanated from the same mind. JAMES H. COFFIN, A.M., Professor of Mathematict in Lafayette College Prof. Loomis has made many improvements in Legendre's Geometry, retaining all the merits of that author without the defects. I shall adopt his work as a text-book in this college. THOMAS E SUDI.ER, A.M., Professor of Mathematict in Dickinson College. 4 Critical Notices of Loomis* Trigonometry and Tables. LOOMIS' TRIGONOMETRY AND TABLES. 8vo, p. 320, Sheep, $1 50. This work contains an exposition of the nature and properties of Logarithms , the principles of Plane Trigonometry ; the Mensuration of Surfaces and Solids ; the principles of Land Surveying, with a full description of the instruments em- ployed ; the Elements of Navigation, and of Spherical Trigonometry. The Tables furnish the Logarithms of Numbers to 10,000 ; logarithmic Sines and Tangents for every ten Seconds of the Quadrant ; natural Sines and Tangents for every Minute of the Quadrant ; a Traverse Table ; a Table of Meridional Parts, &c. The following are a few of the notices of this work which have been received by the publishers. Loomis' Trigonometry is well adapted to give the student that distinct knowledge of the princi- ples of the science so important in the further prosecution of the study of mathematics. The de- scription and representation of the instruments used in surveying, leveling, etc., are sufficient to prepare the student to make a practical application of the principles he has learned. The Tables are just the thing for college students. JOHN TATLOCK, A.M., Professor of Mathematics in Wil- liams College. Prof. Loomis has done up the work admirably. The brevity and clearness which characterize this excellent system of mathematical reasoning are the ne plus ultra for such a work. His Trigo- nometry will meet with the approval already accorded to his Algebra and Geometry. Professor C. DEWEY, Rochester Collegiate Institute. Loomis' Trigonometry is sufficiently extensive for collegiate purposes, and is every where clear and simple in its statements, without being redundant. The learner will here find what he really needs without being distracted by what is superfluous or irrelevant. A. CASWELL, D.D., Professor of Mathematics and Natural Philosophy in Brown University. Loomis' Tables are vastly better than those in common use. The extension of the sines and tan gents to ten seconds is a great improvement. The tables of natural sines are indispensable to a good understanding of Trigonometry ; and the natural tangents are exceedingly convenient in analytical geometry. J. WARD ANDREWS, A.M., Professor of Mathematics and Natural Philosophy in Mari- etta College. Loomis' Trigonometry and Tables are a great acquisition to mathematical schools. I know of no work in which the principles of Trigonometry are so well condensed and so admirably adapted to the course of instruction in the mathematical schools of our country. I shall adopt it as a text- book for instruction in this college. THOMAS E. SUDLER, A.M., Professor of Mathematics in Dick- inson College. Loomis' Trigonometry and Tables are both excellent works, and I shall recommend them at every opportunity which offers. JAMES CURLEY, Professor in Georgetown College. I am so much pleased with Prof. Loomis' Trigonometry that I intend to use it as a text-book in this college. JOHN BROCKLESBY, A.M., Professor of Mathematics in Trinity College. In this work the principles of Trigonometry and its applications are discussed with the same clear ness that characterizes the previous volumes. The portion appropriated to Mensuration, Surveying &c., will especially commend itself to teachers, by the judgment exhibited in the extent to whicl they are carried, and" the practically useful character of the matter introduced. What I have par icularly admired in this, as well as the previous volumes, is the constant recognition of the difficul ties, present and prospective, which are likely to embarrass the learner, and the skill and tact with which they are removed. The Logarithmic Tables will be found unsurpassed in practical conven- ience by any others of the same extent. AUGUSTUS W. SMITH, LL.D., Professor of Mathematics and Astronomy in the Wesleyan University. Prof. Loomis' text-books in Mathematics are models of neatness, precision, and practical adaptation to the wants of students. Methodist Quarterly Review. Valuable FOR COLLEGES, ACADEMIES, AND SCHOOLS, PUBLISHED BY HARPER & BROTHERS, NEW YORK, Liddell and Scott's Greek-English Lexicon. Based on the German Work of Passow. With Additions, &c., by HENRY DRISLER, under the Supervision of Professor ANTHON. Royal 8vo, Sheep extra, $5 00. Liddell and Scott's School Greek Lexicon ; Being an Abridgment of the above, by the Authors, with the Addition of a Second Part, viz., English-Greek. (In press ) Anthon's Classical Dictionary. 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