Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 9 - Section 9.1 - Solving Quadratic Equations by the Square Root Property - Exercise Set - Page 631: 23


$x=\left\{ -\dfrac{5}{3},\dfrac{1}{3} \right\}$

Work Step by Step

Taking the square root of both sides (the Square Root Property), the solutions to the given equation, $ (3x+2)^2=9 ,$ are \begin{array}{l}\require{cancel} (3x+2)=\pm\sqrt{9} \\\\ 3x+2=\pm\sqrt{(3)^2} \\\\ 3x+2=\pm3 \\\\ 3x=-2\pm3 \\\\ x=\dfrac{-2\pm3}{3} .\end{array} Hence, $ x=\left\{ -\dfrac{5}{3},\dfrac{1}{3} \right\} .$
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