Originally Posted by
e^(i*pi) If you like, it doesn't make a difference
x^2 = -\sqrt{22x}
Square both sides
$\displaystyle x^4 = 22x$
$\displaystyle x^4-22x=0$
$\displaystyle x(x^3-22)=0$
Either $\displaystyle x=0$ or $\displaystyle x^3-22=0$
You can use the difference of two cubes to solve
I get
$\displaystyle x=0$
$\displaystyle x = \sqrt [3]{22}$
$\displaystyle x=\frac{-\sqrt [3]{22}+i\,\sqrt{3\sqrt [3]{22^2}}}{2}$
$\displaystyle x=\frac{-\sqrt [3]{22}-i\,\sqrt{3\sqrt [3]{22^2}}}{2}$
Some of these answers may not be correct, sub into the original equation to check. Also the last two answers are complex