Answer
The solutions are $x=0$, $x=2$ and $x=-1\pm\sqrt{3}i$
Work Step by Step
$2x^{4}=16x$
Take $16x$ to the left side of the equation:
$2x^{4}-16x=0$
Take out common factor $2x$:
$2x(x^{3}-8)=0$
Factor the binomial inside the parentheses:
$2x(x-2)(x^{2}+2x+4)=0$
Set all three factors equal to $0$ and solve each individual equation separately:
$2x=0$
$x=\dfrac{0}{2}$
$x=0$
$x-2=0$
$x=2$
$x^{2}+2x+4=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
In this case, $a=1$, $b=2$ and $c=4$
Substitute the known values into the formula and evaluate:
$x=\dfrac{-2\pm\sqrt{2^{2}-4(1)(4)}}{2(1)}=\dfrac{-2\pm\sqrt{4-16}}{2}=\dfrac{-2\pm\sqrt{-12}}{2}=...$
$...=\dfrac{-2\pm2\sqrt{3}i}{2}=-1\pm\sqrt{3}i$
The solutions are $x=0$, $x=2$ and $x=-1\pm\sqrt{3}i$
