YOU TRY 3 EXAMPLE 4 When using substitution, make special notations so that you do not forget to substitute back to the original variables. YOU TRY 4 Solve. a) r4 13r2 36 0 b) c2/3 4c1/3 5 0 3 Solve an Equation in Quadratic Form Using Substitution The equations in Example 3 can also be solved using a method called substitution. We will illustrate the method in Example 4. Solve x4 10x2 9 0 using substitution. Solution x4 10x2 9 0 T x4 (x2)2 To rewrite x4 10x2 9 0 in quadratic form, let u x2. If u x2, then u2 x4. x4 10x2 9 0 u2 10u 9 0 Substitute u2 for x4 and u for x2. (u 9)(u 1) 0 Solve by factoring. b R u 9 0 or u 1 0 Set each factor equal to 0. u 9 or u 1 Solve for u. Be careful! u 9 and u 1 are not the solutions to x4 10x2 9 0. We still need to solve for x. Above we let u x2. To solve for x, substitute 9 for u and solve for x, and then substitute 1 for u and solve for x. u x2 u x2 Substitute 9 for u. 9 x2 1 x2 Substitute 1 for u. Square root property 3 x 1 x Square root property The solution set is 53, 1, 1, 36. This is the same as the result we obtained in Example 3a). Solve by substitution. a) r4 13r2 36 0 b) c2/3 4c1/3 5 0 If, after substitution, an equation cannot be solved by factoring, we can use the quadratic formula. 640 CHAPTER 10 Quadratic Equations and Functions www.mhhe.com/messersmith
messersmith_power_intermediate_algebra_1e_ch4_7_10
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