SECTION A.6 Solving Equations A51 NOTE There is no loss in generality to assume that >a 0, since if <a 0 we can multiply both sides by −1 to obtain an equivalent equation with a positive leading coefficient. j + = ± = − ± = − + = − = − − = − x x x x 5 2 3 2 5 2 3 2 5 2 3 2 1 or 5 2 3 2 4 The solution set is { } − − 4, 1. 6 Solve Quadratic Equations Using The Quadratic Formula The method of completing the square can be used to obtain a general formula for solving the quadratic equation + + = ≠ ax bx c a 0 0 2 As in Example 7, begin by rearranging the terms as + = − > ax bx c a 0 2 Since > a 0, divide both sides by a to get + = − x b a x c a 2 Now the coefficient of x2 is 1. To complete the square on the left side, add the square of 1 2 of the coefficient of x; that is, add ( ) ⋅ = b a b a 1 2 4 2 2 2 to both sides. Then ( ) + + = − + = − x b a x b a b a c a x b a b ac a 4 4 2 4 4 2 2 2 2 2 2 2 2 b a c a b a ac a b ac a 4 4 4 4 4 4 2 2 2 2 2 2 2 − = − = − (4) Provided that − ≥ b ac 4 0, 2 the Square Root Method can be used to get + = ± − + = ± − = − ± − = − ± − x b a b ac a x b a b ac a x b a b ac a b b ac a 2 4 4 2 4 2 2 4 2 4 2 2 2 2 2 2 What if − b ac 4 2 is negative? Then equation (4) states that the left expression (a real number squared) equals the right expression (a negative number). Since this occurence is impossible for real numbers, the quadratic equation has no real solution if − < b ac 4 0. 2 (We discuss quadratic equations for which − < b ac 4 0 2 in detail in the next section.) The square root of a quotient equals the quotient of the square roots. Also, a a ca 4 2 sin 0. 2 = > Add b a2 − to both sides. Combine the quotients on the right. The solution of the equation in Example 7 can also be obtained by factoring. Rework Example 7 using factoring. Now Work problem 107
RkJQdWJsaXNoZXIy NjM5ODQ=