Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 1 - Review Exercises - Page 154: 95


x = 1, 3

Work Step by Step

Solve the individual equaions obtained after breaking the expression into two separate equations. Case 1: $x^2-3=-2x$ $x^2+2x-3=0$ (x+3)(x-1)=0 x = -3, 1 We plug in the Case 1 equation to see -3 is extraneous solution. Hence x=1. Case 2: $x^2-3=2x$ $x^2-2x-3=0$ (x-3)(x+1)=0 x = 3, -1 Plug in the Case 2 equation to see -1 is extraneous solution. Hence x=3
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