Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 2 - Linear and Quadratic Functions - Section 2.8 Equations and Inequalities Involving the Absolute Value Function - 2.8 Assess Your Understanding - Page 182: 38

Answer

$-4,-3,0,1$.

Work Step by Step

Step 1. The original equation can be separated into two cases: $x^2+3x-2=2$ or $x^2+3x-2=-2$. Step 2. For $x^2+3x-2=-2$, we have$x^2+3x=0 \longrightarrow (x)(x+3)=0 \longrightarrow x=-3,0$. Step 3. For $x^2+3x-2=2$, we have $x^2+3x-4=0 \longrightarrow (x+4)(x-1)=0 \longrightarrow x=-4,1$.
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