## Intermediate Algebra for College Students (7th Edition)

Add $\frac{1}{9}$ to both sides of the equation to complete the square. The solutions are $\frac{1}{3}±\frac{2}{3}i$, and the solution set is {$\frac{1}{3}+\frac{2}{3}i,\frac{1}{3}-\frac{2}{3}i$}.
$9x^2 - 6x + 5 = 0$ Divide the whole equation by $9$. Thus, the equation becomes: $x^2 - \frac{2}{3}x + \frac{5}{9} = 0$ Subtract $\frac{5}{9}$ from both sides to isolate the binomial $x^2 - \frac{2}{3}x$. $x^2 - \frac{2}{3}x + \frac{5}{9} - \frac{5}{9} = 0- \frac{5}{9}$ $x^2 - \frac{2}{3}x = - \frac{5}{9}$ The coefficient of the x-term is $- \frac{2}{3}$. Half of $- \frac{2}{3}$ is $- \frac{1}{3}$, and $(-\frac{1}{3})^2 = \frac{1}{9}$. Thus, add $\frac{1}{9}$ to both sides of the equation to complete the square. $x^2 - \frac{2}{3}x +\frac{1}{9} = - \frac{5}{9} +\frac{1}{9}$ $x^2 - \frac{2}{3}x +\frac{1}{9} = - \frac{4}{9}$ $(x-\frac{1}{3})^2 = - \frac{4}{9}$ $x-\frac{1}{3} = \sqrt {- \frac{4}{9}}$ or $x-\frac{1}{3} = -\sqrt {- \frac{4}{9}}$ $x = \frac{1}{3}+\sqrt {- \frac{4}{9}}$ or $x = \frac{1}{3} -\sqrt {- \frac{4}{9}}$ $x = \frac{1}{3}+\sqrt \frac{4}{9}\sqrt {-1}$ or $x = \frac{1}{3} -\sqrt \frac{4}{9}\sqrt {-1}$ $x = \frac{1}{3}+\frac{2}{3}i$ or $x = \frac{1}{3}-\frac{2}{3}i$ The solutions are $\frac{1}{3}±\frac{2}{3}i$, and the solution set is {$\frac{1}{3}+\frac{2}{3}i,\frac{1}{3}-\frac{2}{3}i$}.