Intermediate Algebra for College Students (7th Edition)

The solution set is {$\frac{1}{3}, -1$}.
$g(x) = (x +\frac{1}{3})^2$, $g(x) = \frac{4}{9}$ 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 {$\frac{1}{3}, -1$}.