## College Algebra (10th Edition)

The solution set is {$-1,\frac{1}{3}$}.
$x^2+\frac{2}{3}x-\frac{1}{3}=0$ Move the constant to the right side $x^2+\frac{2}{3}x=\frac{1}{3}$ Add ($\frac{2}{3}\cdot\frac{1}{2})^2=\frac{1}{9}$ to both sides to complete the square. $x^2+\frac{2}{3}x+\frac{1}{9}=\frac{1}{3}+\frac{1}{9}$ $(x+\frac{1}{3})^2=\frac{4}{9}$ $x+\frac{1}{3}=\pm\sqrt {\frac{4}{9}}$ Use square root property. $x+\frac{1}{3}=\pm\frac{2}{3}$ $x=-\frac{1}{3}\pm\frac{2}{3}$ $x=-\frac{1}{3}+\frac{2}{3}$, $x=-\frac{1}{3}-\frac{2}{3 }$ $x=\frac{1}{3}$,$x=-\frac{3}{3}=-1$ The solution set is {$-1,\frac{1}{3}$}