Answer
Add $\frac{1}{16}$ to both sides of the equation to complete the square.
The solutions are $\frac{1}{4}±\frac{1}{4}i$, and the solution set is {$\frac{1}{4}+\frac{1}{4}i,\frac{1}{4}-\frac{1}{4}i$}.
Work Step by Step
$8x^2 - 4x + 1 = 0$
Divide the whole equation by $8$. Thus, the equation becomes:
$x^2 - \frac{1}{2}x + \frac{1}{8} = 0$
Subtract $\frac{1}{8}$ from both sides to isolate the binomial $x^2 - \frac{1}{2}x$.
$x^2 - \frac{1}{2}x + \frac{1}{8} - \frac{1}{8} = 0 - \frac{1}{8}$
$x^2 - \frac{1}{2}x = - \frac{1}{8}$
The coefficient of the x-term is $- \frac{1}{2}$. Half of $- \frac{1}{2}$ is $- \frac{1}{4}$, and $(-\frac{1}{4})^2 = \frac{1}{16}$.
Thus, add $\frac{1}{16}$ to both sides of the equation to complete the square.
$x^2 - \frac{1}{2}x + \frac{1}{16} = - \frac{1}{8} +\frac{1}{16}$
$x^2 - \frac{1}{2}x + \frac{1}{16} = - \frac{1}{16}$
$(x-\frac{1}{4})^2 = - \frac{1}{16}$
$x-\frac{1}{4} = \sqrt {-\frac{1}{16}}$ or $x-\frac{1}{4} = -\sqrt {-\frac{1}{16}}$
$x = \frac{1}{4}+\sqrt {-\frac{1}{16}}$ or $x = \frac{1}{4}-\sqrt {-\frac{1}{16}}$
$x = \frac{1}{4}+\sqrt \frac{1}{16}\sqrt {-1}$ or $x = \frac{1}{4}-\sqrt \frac{1}{16}\sqrt {-1}$
$x = \frac{1}{4}+\frac{1}{4}i$ or $x = \frac{1}{4}-\frac{1}{4}i$
The solutions are $\frac{1}{4}±\frac{1}{4}i$, and the solution set is {$\frac{1}{4}+\frac{1}{4}i,\frac{1}{4}-\frac{1}{4}i$}.
