ALGEBRA. Addition.-Add a and b. Ans. a+b. Add a, b, and -c. Ans. a+b-c. Add 2 and 3a. Ans. a. Add 2ab,- 3ab, c, 3c. Ans. ab 4c. Subtraction.-Subtract a from b. Ans. b -a. Subtract a from -- b. Ans.ba. Subtract b+c from a. Ans. a b-c. Subtract 3a2b-9c from 4a2b+c. Ans. a2b+10c. RULE: Change the signs of the subtrahend and proceed as in addition. Multiplication.-Multiply a by b. Ans. a2bab2. Ans. ab. Multiply ab by a+b. a by b. Multiply a + b by a+b. Ans. (a+b)(a+b) =a2+2ab+b2. Multiply -a by - b. Ans. ab. Multiply Ans. - ab. Like signs give plus, unlike signs minus. Powers of numbers.-The product of two or more powers of any number is the number with an exponent equal to the sum of the powers: a2 × a3 = α5; a2b2 × ab = a3b3; 7ab x 2ac = -14 a2bc. To multiply a polynomial by a monomial, multiply each term of the polynomial by the monomial and add the partial products: (6a - 3b) x 3c = 18ac 9bc. To multiply two polynomials, multiply each term of one factor by each term of the other and add the partial products: (5a - 6b) × (3a — 4b) 15a38ab +24b2. The square of the sum of two numbers = sum of their squares + twice their product. The square of the difference of two numbers the sum of their squares twice their product. The product of the sum and difference of two numbers = the difference of their squares: (a + b)2 =a2+2ab+b2; (a b)2=a2 (a + b) x (a· b) = a2 - b2. 2ab+b2; The square of half the sums of two quantities is equal to their product plus the square of half their difference: (a + b) 2 2 = ab + (a - b)2. 2 The square of the sum of two quantities is equal to four times their prod ucts, plus the square of their difference: (a + b)2 = 4ab + (a - b)2 The sum of the squares of two quantities equals twice their product, plus the square of their difference: a2 + b2 = 2ab + (a - b)2. The square of a trinomial = the square of each term + twice the product of each term by each of the terms that follow it: (a+b+c)2 = a2+b+ c2+2ab+2ac2bc; (a - b − c)2 = a2 + b2 + c2 - 2ab2ac+2bc. The square of (any number+) square of the number + the number +4 =the number X (the number + 1) +14; (a+1⁄2)2= a2+a+1⁄44, =a(a+1)+1⁄44. (41⁄2)2=42+4+4=4×5+4=2014. The product of any number +2 by any other number += product of the numbers + half their sum +14. (a+) × b +1⁄2) = ab +1⁄2(a+b)+1⁄4. 4% × 6% = 4 × 6 + 1⁄41⁄2(4 + 6) + 4 = 24 + 5 + 4 = 294. Square, cube, 4th power, etc., of a binomial a+b. (a + b)2 = a2+2ab+b2; (a + b)3 = a3 +3a2b+3ab2 +b3; In each case the number of terms is one greater than the exponent of the power to which the binomial is raised. 2. In the first term the exponent of a is the same as the exponent of the power to which the binomial is raised, and it decreases by 1 in each succeeding term. 3. b appears in the second term with the exponent 1, and its exponent increases by 1 in each succeeding term. 4. The coefficient of the first term is 1. 5. The coefficient of the second term is the exponent of the power to which the binomial is raised. 6. The coefficient of each succeeding term is found from the next preceding term by multiplying its coefficient by the exponent of a, and dividing the product by a number greater by 1 than the exponent of b. (S Binomial Theorem, below.) Parentheses.-When a parenthesis is preceded by a plus sign it may be removed without changing the value of the expression: a + b + (a + b) = 2a +26. When a parenthesis is preceded by a minus sign it may be removed if we change the signs of all the terms within the parenthesis: 1 − (a - b - c) = 1 − a+b+c. When a parenthesis is within a parenthesis remove the inner one first: a · [b - {c - (α − e) } ] = = a - [b-c+d- e] = α b+c-d+e. [b-{c=d+e}] A multiplication sign, X, has the effect of a parenthesis, in that the operation indicated by it must be performed before the operations of addition or subtraction. a + b xa + b = a + ab +b; while (a + b) × (a + b) = a2 + 2ab+b2, and (a + b) × a + b = a2 + ab + b. Division. The quotient is positive when the dividend and divisor have like signs, and negative when they have unlike signs: abc÷b = ac; abc-bac. To divide a monomial by a monomial, write the dividend over the divisor with a line between them. If the expressions have common factors, remove the common factors: To divide a polynomial by a monomial, divide each term of the polynomial by the monomial: (8ab - 12ac)+4a= 2b - 3c. To divide a polynomial by a polynomial, arrange both dividend and divisor in the order of the ascending or descending powers of some common letter, and keep this arrangement throughout the operation. Divide the first term of the dividend by the first term of the divisor, and write the result as the first term of the quotient. Multiply all the terms of the divisor by the first term of the quotient and subtract the product from the dividend. If there be a remainder, consider it as a new dividend and proceed as before: (a2 — b2) ÷ (a + b). The difference of two equal odd powers of any two numbers is divisible by their difference and also by their sum: (as b3) + (a - b) = a2 + ab + b2; (a3 — b3) ÷ (a + b) = a2 - ab + b2. The difference of two equal even powers of two numbers is divisible by their difference and also by their sum: (a2 — b2) ÷ (a - b) = a + b. The sum of two equal even powers of two numbers is not divisible by either the difference or the sum of the numbers; but when the exponent of each of the two equal powers is composed of an odd and an even factor, the sum of the given power is divisible by the sum of the powers expressed by the even factor. Thus x + y is not divisible by x + y or by x-y, but is divisible by x2 + y2. Simple equations.-An equation is a statement of equality between two expressions; as, a + b = c +d. A simple equation, or equation of the first degree, is one which contains only the first power of the unknown quantity. If equal changes be made (by addition, subtraction, multiplication, or division) in both sides of an equation, the results will be equal. Any term may be changed from one side of an equation to another, provided its sign be changed: a+b=c+d; a=c+db. To solve an equation having one unknown quantity, transpose all the terms involving the unknown quantity to one side of the equation, and all the other terms to the other side; combine like terms, and divide both sides by the coefficient of the unknown quantity. Solve 8x 29 = 26 - 3x. 8x+3x=29+ 26; 11x 55; x = 5, ans. Simple algebraic problems containing one unknown quantity are solved by making x = the unknown quantity, and stating the conditions of the problem in the form of an algebraic equation, and then solving the equation. What two numbers are those whose sum is 48 and difference 14? Let x = the smaller number, x + 14 the greater. x+x+14= 48. 2x = 34, x = 17; x+14= 31, ans. Find a number whose treble exceeds 50 as much as its double falls short of 40. Let x = the number. 3x 50402x; 5x = 90; x= 18, ans. Prov ing, 54 50 40 36. Equations containing two unknown quantities.-If one equation contains two unknown quantities, x and y, an indefinite number of pairs of values of x and y may be found that will satisfy the equation, but if a second equation be given only one pair of values can be found that will satisfy both equations. Simultaneous equations, or those that may be satisfied by the same values of the unknown quantities, are solved by combining the equations so as to obtain a single equation containing only one unknown quantity. This process is called elimination. Elimination by addition or subtraction-Multiply the equation by such numbers as will make the coefficients of one of the unknown quanti ties equal in the resulting equation. Add or subtract the resulting equations according as they have unlike or like signs. 2x + 3y - 14 4x-5y= 3 11y = Solve 43. Subtract: 11; y = 1. Substituting value of y in first equation, 2x+3= 7; x = 2. Elimination by substitution.-From one of the equations obtain the value of one of the unknown quantities in terms of the other. Substitutute for this unknown quantity its value in the other equation and reduce the resulting equations. 8 3y 2 Substitute this value in (2): 8(3 − 3Y) + 7y = 7; 3 2 whence y = -2. Substitute this value in (1): 2o - 6 = 8; x = 7. Elimination by comparison.-From each equation obtain the value of one of the unknown quantities in terms of the other. Forin an equation from these equal values, and reduce this equation. Substitute this value of y in (1): 2x +9 = 11; x = 1. If three simultaneous equations are given containing three unknown quantities, one of the unknown quantities must be eliminated between two pairs of the equations; then a second between the two resulting equations. Quadratic equations.—A quadratic equation contains the square of the unknown quantity, but no higher power. A pure quadratic contains the square only; an affected quadratic both the square and the first power. To solve a pure quadratic, collect the unknown quantities on one side, and the known quantities on the other; divide by the coefficient of the unknown quantity and extract the square root of each side of the resulting equation. Solve 3x2 150. 3x2 = 15; x2 = 5; x = A root like V5, which is indicated, but which can be found only approximately, is called a surd. Solve 3x2+ 150. 3x2 = - 15; 25; x = N-5. The square root of - 5 cannot be found even approximately, for the square of any number positive or negative is positive; therefore à root which is indicated, but cannot be found even approximately, is called imaginary. To solve an affected quadratic.-1. Convert the equation into the forni a2x2+2abx= c, multiplying or dividing the equation if necessary, so as to make the coefficient of x2 a square number. 2. Complete the square of the first member of the equation, so as to convert it to the form of a2x2 + 2abx + b2, which is the square of the binomial arb, as follows: add to each side of the equation the square of the quotient obtained by dividing the second term by twice the square root of the first term. 3. Extract the square root of each side of the resulting equation. Solve 3x2 - 4x 32. To make the coefficient of x2 a square number, multiply by 3: 9x2 - 12x= 96; 12x + (2 × 3x) = 2; 22 = 4. Complete the square: 92* 10, whence x = 4 or 2 2/3. The square root of 100 is either + 10 or 10, since the square of - 10 as well as + 102 100. Problems involving quadratic equations have apparently two solutions, as a quadratic has two roots. Sometimes both will be true solutions, but generally one only will be a solution and the other be inconsistent with the conditions of the problem. The sum of the squares of two consecutive positive numbers is 481. Find the numbers. Let x = one number, x + 1 the other. x2 + (x + 1)2 = 481. 2x2 + 2x + 1 = 481. x2+x=240. Completing the square, x2 + x +0.25 the root we obtain x + 0.5 ± 15.5; x = 15 or 16. = 240.25. Extracting The positive root gives for the numbers 15 and 16. The negative root 16 is inconsistent with the conditions of the problem. Quadratic equations containing two unknown quantities require different methods for their solution, according to the form of the equations. For these methods reference must be made to works on algebra. n Theory of exponents.-Wa when n is a positive integer is one of n equal factors of a. Wam means a is to be raised to the mth power and the nth root extracted. m (a) means that the nth root of a is to be taken and the result raised to the mth power. Wa m = m =an. When the exponent is a fraction, the numerator indicates a power, and the denominator a root. a = √a® = a3; αξ To extract the root of a quantity raised to an indicated power, divide the exponent by the index of the required root; as, Subtracting 1 from the exponent of a is equivalent to dividing by a : 1 a2 - 1 = a1 = a; a11 = ao = = 1; a° -1 = a -1 = ; a -1 -1 = a -2 = α a' 1 a2 A number with a negative exponent denotes the reciprocal of the number with the corresponding positive exponent. A factor under the radical sign whose root can be taken may, by having the root taken, be removed from under the radical sign: A factor outside the radical sign may be raised to the corresponding power and placed under it: Binomial Theorem.-To obtain any power, as the nth, of an expression of the form + a The following laws hold for any term in the expansion of (a + x)n. The exponent of a is n minus the exponent of x. The last factor of the numerator is greater by one than the exponent of a. The exponent of a will be n − (r− 1), or nr +1. GEOMETRICAL PROBLEMS. FIG. 1. A FIG. 3. 3 FIG. 4. A YE FIG. 5. FIG. 6. D B IC B 1. To bisect a straight line, or an arc of a circle (Fig. 1).— From the ends A, B, as centres, describe arcs intersecting at C and D, and draw a line through C and D which will bisect the line at E or the arc at F. 2. To draw a perpendicular to a straight line, or a radial line to a circular arc.-Same as in Problem 1. CD is perpendicular to the line AB, and also radial to the arc. 3. To draw a perpendicular to a straight line from a given point in that line (Fig. 2).—With any radius, from the given point A in the line B C, cut the line at B and C. With a longer radius describe arcs from B and C, cutting each other at D, and draw the perpendicular D A. 4. From the end 4 of a given line A D to erect a perpendicular A E (Fig. 3).-From any centre F, above A D, describe a circle passing through the given point A, and cutting the given line at D. Draw D F and produce it to cut the circle at E, and draw the perpendicular A E. Second Method (Fig. 4).-From the given point A set off a distance A E equal to three parts, by any scale; and on the centres A and E, with radii of four and five parts respectively, describe arcs intersecting at C. Draw the perpendicular A C. NOTE. This method is most useful on very large scales, where straight edges are inapplicable. Any multiples of the numbers 3, 4, 5 may be taken with the same effect as 6, 8, 10, or 9, 12, 15. 5. To draw a perpendicular to a straight line from any point without it (Fig. 5.)-Froi the point A, with a sufficient radius cut the given line at F and G, and from these points describe arcs cutting at E. Draw the perpendicular A E. 6. To draw a straight line parallel to a given line, at a given distance apart (Fig. 6).— From the centres A, B, in the given line, with the given distance as radius, describe arcs C, D, and draw the parallel lines CD touching the arcs. |