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

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 {$\frac{1}{4}+\frac{1}{4}i,\frac{1}{4}-\frac{1}{4}i$}.
$8x^2 - 4x + 1 = 0$ 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 {$\frac{1}{4}+\frac{1}{4}i,\frac{1}{4}-\frac{1}{4}i$}.