Answer
The solution set for the quadratic equation, ${{x}^{2}}=4x-8$ is $\left\{ 2\pm 2i \right\}$.
Work Step by Step
Consider quadratic equation,
${{x}^{2}}=4x-8$
Convert the equation into standard form by adding $4x+8$ on both sides.
$\begin{align}
& {{x}^{2}}=4x-8 \\
& {{x}^{2}}-4x+8=4x-8-4x+8 \\
& {{x}^{2}}-4x+8=0
\end{align}$
Compare the equation with the standard quadratic equation $a{{x}^{2}}+bx+c=0\text{ , }\left( a\ne 0 \right)$.
Here, $a=1,\text{ }b=-4\text{ and }c=8$
Apply the quadratic formula
$x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Substitute 1 for a, $-4$ for b and 8 for c.
$x=\frac{-\left( -4 \right)\pm \sqrt{{{\left( -4 \right)}^{2}}-4\left( 1 \right)\left( 8 \right)}}{2\left( 1 \right)}$
Simplify the radical.
$\begin{align}
& x=\frac{-\left( -4 \right)\pm \sqrt{{{\left( -4 \right)}^{2}}-4\left( 1 \right)\left( 8 \right)}}{2\left( 1 \right)} \\
& =\frac{4\pm \sqrt{16-32}}{2} \\
& =\frac{4\pm \sqrt{-16}}{2}
\end{align}$
As $i=\sqrt{-1}$
Therefore,
$\begin{align}
& x=\frac{4\pm \sqrt{16}\sqrt{-1}}{2} \\
& =\frac{4\pm 4i}{2} \\
& =\frac{2\left( 2\pm 2i \right)}{2} \\
& =2\pm 2i
\end{align}$
The solutions of the quadratic are complex conjugates of each other.
Hence, the solution set for the quadratic equation ${{x}^{2}}=4x-8$ is $\left\{ 2\pm 2i \right\}$.
