Precalculus (6th Edition)

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

Chapter 1 - Equations and Inequalities - 1.6 Other Types of Equations and Applications - 1.6 Exercises - Page 148: 92

Answer

$\{-13,1\}$

Work Step by Step

Step 1. Let $u=(2x-1)^{1/3}$, we have $u^2+2u-3=0$ Step 2. Factor and solve: $(u+3)(u-1)=0$ thus $u=-3, 1$ Step 3. For $u=-3$, we have $(2x-1)^{1/3}=-3$ or $2x-1=-27$ thus $x=-13$ Step 4. For $u=1$, we have $(2x-1)^{1/3}=1$ or $2x-1=1$ thus $x=1$ Step 5. Solution set $\{-13,1\}$
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