162 CHAPTER 1 Equations and Inequalities EXAMPLE 8 Solving Equations Quadratic in Form Solve each equation. (a) 1x + 122/3 - 1x + 121/3 - 2 = 0 (b) 6x-2 + x-1 = 2 SOLUTION (a) 1x + 122/3 - 1x + 121/3 - 2 = 0 1x + 122/3 = 31x + 121/342, so let u = 1x + 121/3. u2 - u - 2 = 0 1u - 221u + 12 = 0 Factor. u - 2 = 0 or u + 1 = 0 Zero-factor property u = 2 or u = -1 Solve each equation. 1x + 121/3 = 2 or 1x + 121/3 = -1 Replace u with (x + 1)1/3. 31x + 121/343 = 23 or 31x + 121/343 = 1-123 Cube each side. x + 1 = 8 or x + 1 = -1 Apply the exponents. x = 7 or x = -2 Proposed solutions Don’t forget this step. Both proposed solutions check, so the solution set is 5-2, 76. (b) 6 x-2 + x-1 = 2 6x-2 + x-1 - 2 = 0 Subtract 2 from each side. 6u2 + u - 2 = 0 Let u = x-1. Then u2 = x-2. 13u + 2212u - 12 = 0 Factor. 3u + 2 = 0 or 2u - 1 = 0 Zero-factor property u = - 2 3 or u = 1 2 Solve each equation. x-1 = - 2 3 or x-1 = 1 2 Replace u with x-1. x = - 3 2 or x = 2 x-1 is the reciprocal of x. Both proposed solutions check, so the solution set is E - 3 2 , 2F. S Now Try Exercises 93 and 99. Don’t stop here. Substitute for u. CAUTION When using a substitution variable in solving an equation that is quadratic in form, do not forget the step that gives the solution in terms of the original variable. CHECK 1x + 122/3 - 1x + 121/3 - 2 = 0 Original equation 17 + 122/3 - 17 + 121/3 - 2≟0 Let x = 7. 82/3 - 81/3 - 2≟0 4 - 2 - 2≟0 0 = 0 ✓ True 1-2 + 122/3 - 1-2 + 121/3 - 2≟0 Let x = -2. 1-122/3 - 1-121/3 - 2≟0 1 + 1 - 2≟0 0 = 0 ✓ True
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