Answer
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 {$\frac{1}{3}+\frac{2}{3}i,\frac{1}{3}-\frac{2}{3}i$}.
Work Step by Step
$9x^2 - 6x + 5 = 0$
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 {$\frac{1}{3}+\frac{2}{3}i,\frac{1}{3}-\frac{2}{3}i$}.
