Answer
$\left\{\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i,\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i\right\}$
Work Step by Step
The given quadratic equation has $a=1,b=-1$ and $c=1$.
The discriminant is
$=b^2-4ac$
$=(-1)^2-4(1)(1)$
$=1-4$
$=-3$
Use Quadratic formula.
$x=\dfrac{-b\pm \sqrt {b^2-4ac}}{2a}$
Substitute the values of $a, b, c, $ and discriminant.
$x=\dfrac{-(-1)\pm \sqrt {-3}}{2(1)}$
$x=\dfrac{-(-1)\pm \sqrt {3(-1)}}{2(1)}$
$x=\dfrac{1\pm \sqrt{3}i}{2}$
The equation has two solutions: $\frac{1}{2}-\frac{\sqrt{3}i}{2}$ and $\frac{1}{2}+\frac{\sqrt{3}i}{2}$.
Check:
For $x=\frac{1}{2}-\frac{\sqrt{3}i}{2}$.
$=(\frac{1}{2}-\frac{\sqrt{3}i}{2})^2-1(\frac{1}{2}-\frac{\sqrt{3}i}{2})+1$
$=\frac{1}{4}-2\cdot \frac{\sqrt3i}{4}+\frac{3i^2}{4}-\frac{1}{2}+\frac{\sqrt{3}i}{2}+1$
$=\frac{1}{4}- \frac{\sqrt3i}{2}+\frac{3i^2}{4}-\frac{1}{2}+\frac{\sqrt{3}i}{2}+1$
$=\frac{3}{4}+\frac{3i^2}{4}$
$=\frac{3}{4}-\frac{3}{4}$
$=0$
For $x=\frac{1}{2}+\frac{\sqrt{3}i}{2}$.
$=(\frac{1}{2}+\frac{\sqrt{3}i}{2})^2-1(\frac{1}{2}+\frac{\sqrt{3}i}{2})+1$
$=\frac{1}{4}+2\cdot \frac{\sqrt3i}{4}+\frac{3i^2}{4}-\frac{1}{2}-\frac{\sqrt{3}i}{2}+1$
$=\frac{1}{4}+ \frac{\sqrt3i}{2}+\frac{3i^2}{4}-\frac{1}{2}-\frac{\sqrt{3}i}{2}+1$
$=\frac{3}{4}+\frac{3i^2}{4}$
$=\frac{3}{4}-\frac{3}{4}$
$=0$
Hence, the solution set is $\left\{\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i,\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i\right\}$.
