Machinery's Handbook, 31st Edition
Equation Solving 33 is x 2 – r 2 = 0, and then factor as the difference of squares (see Polynomials on page 28): ( x + r )( x – r ) = 0, hence, x = – r and r . Verifying this to be the factorization of the difference of squares is simply a matter of applying the distributive property of multiplication over addition to each term: ( x + r )( x – r ) = xx – xr + xr – rr = x 2 – r 2 As previously explained, this process is called expanding by FOIL, for the F irst, O uter, I nner, and L ast terms of each binomial, which are multiplied in that order to get the prod- uct. From the zero property of multiplication, ( x + r )( x – r ) = 0 implies x + r = 0 or x – r = 0, thus x = – r or x = r . If a quadratic cannot be factored as a difference of squares, then factorization is ap- proached as shown in Factoring Polynomials on page 29. Given a quadratic equation in the form ax 2 + bx + c = 0, first obtain the product ac from the coefficients a and c ; then determine two numbers whose product is ac and whose sum is b. Example: Find the solution to x 2 – 5 x + 6 = 0 by factoring. In this example, a = 1, b = –5, c = 6 and ac = (1)(6) = 6. The factors of 6 whose sum is –5 are –2 and –3. The equation is factored as x 2 – 5 x + 6 = ( x – 2)( x – 3) = 0. Then by the zero prop- erty of multiplication, the roots of the equation are x = 2 and x = 3. The parabola intersects the x -axis at these two values of x . A more difficult example has a leading coefficient of x other than 1. Example: Factor 8 x 2 + 22 x + 5 = 0 and find the values of x that satisfy the equation. Solution: Here, a = 8, b = 22, and c = 5. Therefore, ac = 8 × 5 = 40, and ac is positive, so we are looking for two factors of ac , namely f 1 and f 2 , such that f 1 + f 2 = 22. The possible combinations of numbers with product of 40 are 20 and 2, 8 and 5, 4 and 10, and 40 and 1. The requirements that satisfy a sum of 22 are 20 and 2, since 20 × 2 = 40 and 20 + 2 = 22. Hence: 8 x 2 22 x 5 + + = 0 8 x 2 20 x 2 x 5 + + + = 0 4 x 2 x +5 ( ) 1 2 x +5 ( ) + = 0 2 x +5 ( ) 4 x +1 ( ) = 0 On the second line, the common factor of 4 x in the first two terms is factored out, so the common binomial factor of 2 x + 5 is then apparent, to be factored out of the larger terms. Checking the answer is a matter of simply remultiplying the factors to produce the original expression. Because the product of the two factors equals zero, each of the factors also equals zero. Thus, 2 x + 5 = 0 and 4 x +1 = 0. Rearranging and solving, x = - 5 ⁄ 2 or x = - 1 ⁄ 4 . Example: Factor 8 x 2 + 3 x - 5 = 0 and find the solutions of the equation. Solution: Here a = 8, b = 3, c = - 5, and ac = 8 × ( - 5) = - 40. The required numbers must have a product of –40 and a sum of 3. As in the previous example, the possible combinations are 20 and –2, –20 and 2, –8 and 5, 8 and –5, 40 and –1, and –40 and 1. Only 8 and –5 satisfy the requirements because 8 × (–5) = –40, and 8 + ( - 5) = 3. Notice that 3 x in the first line is thus written as 8 x – 5 x in the second line, making it possible to rearrange and simplify the expression. 8 x 2 3 x + – 5 = 0 8 x 2 8 x 5 x + – – 5 = 0 8 x x +1 ( ) 5 x +1 ( ) – = 0 x +1 ( ) 8 x – 5 ( ) = 0 Solving, for x + 1 = 0, x = - 1; and, for 8 x - 5 = 0, x = 5 ⁄ 8 .
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