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: 59


The solution set is {$1, -\frac{1}{5}$}.

Work Step by Step

$g(x)=(x-\frac{2}{5})^2, g(x)=\frac{9}{25}$ For $g(x)=(x-\frac{2}{5})^2$, substitute $g(x)$ with $\frac{9}{25}$ $\frac{9}{25} = (x-\frac{2}{5})^2$ Using the Square Root Property $u^2 = d$, then $u = \sqrt d$ or $u = - \sqrt d$. Thus, $x-\frac{2}{5} = \sqrt \frac{9}{25}$ or $x-\frac{2}{5} = -\sqrt \frac{9}{25}$ $x = \frac{2}{5}+ \sqrt \frac{9}{25}$ or $x =\frac{2}{5} -\sqrt \frac{9}{25}$ $x = \frac{2}{5} + \frac{3}{5}$ or $x = \frac{2}{5} - \frac{3}{5}$ $x = 1$ or $x = -\frac{1}{5}$ The solution set is {$1, -\frac{1}{5}$}.
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