Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 1 - Equations and Inequalities - 1.8 Absolute Value Equations and Inequalities - 1.8 Exercises - Page 169: 84

Answer

$\{-2, 3 \}$

Work Step by Step

For $x^2-x=6$, we have $x^2-x-6=0$ or $(x+2)(x-3)=0$ which gives $x=-2, 3$ or in set form $\{-2, 3 \}$
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