College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Section 1.6 - Other Types of Equations and Applications - Summary Exercises on Solving Equations - Page 136: 10


$x=-2$ or $x=1$

Work Step by Step

We solve: $(2x+1)^{2}=9$ We take the square root of both sides: $\sqrt{(2x+1)^{2}}=\pm\sqrt{9}$ $2x+1=\pm\sqrt{9}$ $2x+1=\pm 3$ $2x=-1\pm 3$ $x=\frac{-1\pm 3}{2}$ $x=-\frac{4}{2}=-2$ or $x=\frac{2}{2}=1$ $x=-2$ or $x=1$
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