Answer
$\left\{\dfrac{1}{4}-\dfrac{1}{4}i,\dfrac{1}{4}+\dfrac{1}{4}i \right\}$
Work Step by Step
The given equation has $a=8,b=-4$ and $c=1$.
The discriminant is
$=b^2-4ac$
$=(-4)^2-4(8)(1)$
$=16-32$
$=-16$
Use Quadratic formula.
$x=\dfrac{-b\pm \sqrt {b^2-4ac}}{2a}$
Substitute the values of $a, b, c, $ and discriminant to obtain:
$x=\dfrac{-(-4)\pm \sqrt {(-16)}}{2(8)}$
Simplify.
$x=\dfrac{4\pm \sqrt{16(-1)}}{16}$
$x=\dfrac{4\pm 4i}{16}$
Factor out $4$.
$x=\dfrac{4(1\pm i)}{16}$
Cancel common factor $4$.
$x=\dfrac{1\pm i}{4}$
The equation has two solutions: $\frac{1}{4}-\frac{i}{4}$ and $\frac{1}{4}+\frac{i}{4}$.
Check:
For $x=\frac{1}{4}-\frac{i}{4}$.
$=8(\frac{1}{4}-\frac{i}{4})^2-4(\frac{1}{4}-\frac{i}{4})+1$
$=8(\frac{1}{16}-2\cdot \frac{i}{16}+\frac{i^2}{16})-1+i+1$
$=\frac{1}{2}-i+\frac{i^2}{2}-1+i+1$
$=\frac{1}{2}+\frac{i^2}{2}$
$=\frac{1}{2}-\frac{1}{2}$
$=0$
For $x=\frac{1}{4}+\frac{i}{4}$.
$=8(\frac{1}{4}+\frac{i}{4})^2-4(\frac{1}{4}+\frac{i}{4})+1$
$=8(\frac{1}{16}+2\cdot \frac{i}{16}+\frac{i^2}{16})-1-i+1$
$=\frac{1}{2}+i+\frac{i^2}{2}-1-i+1$
$=\frac{1}{2}+\frac{i^2}{2}$
$=\frac{1}{2}-\frac{1}{2}$
$=0$
Hence, the solution set is $\left\{\dfrac{1}{4}-\dfrac{1}{4}i,\dfrac{1}{4}+\dfrac{1}{4}i \right\}$.
