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

The solutions are $\frac{1}{4}±\frac{\sqrt{19}}{4}i$, and the solution set is {$\frac{1}{4}+\frac{\sqrt{19}}{4}i, \frac{1}{4}-\frac{\sqrt{19}}{4}i$}.
$4x^2 - 2x + 5 = 0$ Divide the whole equation by $4$. Thus, the equation becomes: $x^2 - \frac{1}{2}x +\frac{5}{4} = 0$ Subtract $\frac{5}{4}$ from both sides to isolate the binomial $x^2 - \frac{1}{2}x$. $x^2 - \frac{1}{2}x +\frac{5}{4} - \frac{5}{4} = 0 - \frac{5}{4}$ $x^2 - \frac{1}{2}x = - \frac{5}{4}$ 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. $x^2 - \frac{1}{2}x + \frac{1}{16}= - \frac{5}{4} + \frac{1}{16}$ $x^2 - \frac{5}{2}x +\frac{1}{16}= - \frac{19}{16}$ $(x-\frac{1}{4})^2 = - \frac{19}{16}$ $x-\frac{1}{4} = \sqrt {- \frac{19}{16}}$ or $x-\frac{1}{4} = -\sqrt {- \frac{19}{16}}$ $x = \frac{1}{4} +\sqrt {- \frac{19}{16}}$ or $x =\frac{1}{4} -\sqrt {- \frac{19}{16}}$ $x = \frac{1}{4}+\sqrt \frac{19}{16}\sqrt {-1}$ or $x = \frac{1}{4} -\sqrt \frac{19}{16}\sqrt {-1}$ $x =\frac{1}{4}+\frac{\sqrt{19}}{4}i$ or $x = \frac{1}{4}-\frac{\sqrt{19}}{4}i$ The solutions are $\frac{1}{4}±\frac{\sqrt{19}}{4}i$, and the solution set is {$\frac{1}{4}+\frac{\sqrt{19}}{4}i, \frac{1}{4}-\frac{\sqrt{19}}{4}i$}.