Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 8 - Section 8.1 - The Square Root Property and Completing the Square - Exercise Set - Page 593: 60


The solution set is {$\frac{1}{3}, -1$}.

Work Step by Step

$g(x) = (x +\frac{1}{3})^2$, $g(x) = \frac{4}{9}$ For $g(x) = (x +\frac{1}{3})^2$, substitute $g(x)$ with $ \frac{4}{9}$ $ \frac{4}{9} = (x +\frac{1}{3})^2$ Using the Square Root Property $u^2 = d$, then $u = \sqrt d$ or $u = - \sqrt d$. Thus, $x +\frac{1}{3} = \sqrt \frac{4}{9}$ or $x+\frac{1}{3} = -\sqrt \frac{4}{9}$ $x = -\frac{1}{3}+ \sqrt \frac{4}{9}$ or $x =-\frac{1}{3} -\sqrt \frac{4}{9}$ $x = -\frac{1}{3} + \frac{2}{3}$ or $x = -\frac{1}{3} - \frac{2}{3}$ $x = \frac{1}{3}$ or $x = -1$ The solution set is {$\frac{1}{3}, -1$}.
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