Answer
The solutions are $\frac{1}{4}±\frac{\sqrt{19}}{4}i$, and the solution set is {$\frac{1}{4}+\frac{\sqrt{19}}{4}i, \frac{1}{4}-\frac{\sqrt{19}}{4}i$}.
Work Step by Step
$4x^2 - 2x + 5 = 0$
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 {$\frac{1}{4}+\frac{\sqrt{19}}{4}i, \frac{1}{4}-\frac{\sqrt{19}}{4}i$}.
