SECTION A.6 Solving Equations A49 so = = x x 1 3 or 1 3 Solve for x. This equation has only the repeated solution 1 3 . The solution set is { } 1 3 . Now Work problem 69 4 Solve Quadratic Equations Using the Square Root Method Suppose that we wish to solve the quadratic equation = x p 2 (2) = > = = − x p p x p x p If and 0, then or . 2 (3) where > p 0 is a real number. Proceeding as in the earlier examples: ( )( ) − = − + = = = − x p x p x p x p x p 0 0 or 2 Put in standard form. Factor (over the real numbers). Solve. The result follows: When statement (3) is used, it is called the Square Root Method . In statement (3), note that the equation = x p 2 has two solutions: = x p and = − x p. Usually these solutions are abbreviated as = ± x p, which is read as “ x equals plus or minus the square root of p.” For example, the two solutions of the equation = x 4 2 are = ± x 4 Use the Square Root Method. and, since = 4 2, = ± x 2 The solution set is { } −2, 2 . Solving Quadratic Equations by Using the Square Root Method Solve each equation. (a) = x 5 2 (b) ( ) − = x 2 16 2 Solution EXAMPLE 6 (a) = = ± = = − x x x x 5 5 5 or 5 2 Use the Square Root Method. The solution set is { } − 5, 5. (b) ( ) − = − = ± − = − = − − = − = − = = − x x x x x x x x 2 16 2 16 2 16 or 2 16 2 4 2 4 6 2 2 Use the Square Root Method. The solution set is { } −2, 6 . Now Work problem 97
RkJQdWJsaXNoZXIy NjM5ODQ=