Answer
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 {$\frac{1}{3}, -1$}.
