# How do you solve for x in the equation x^4 - 27x^2 + 50 = 0?

Jul 21, 2016

$\left({x}^{2} - 25\right) \left({x}^{2} - 2\right) = 0$

${x}^{2} - 25 = 0 \mathmr{and}$x^2 - 2 =0 #

$x = \pm 5 \mathmr{and} \pm \sqrt{2}$

Hopefully this helps!

Jul 21, 2016

$x = \pm \sqrt{2} , \mathmr{and} x = \pm 5$

#### Explanation:

This equation is called a bi-quadratic equation.
Call ${x}^{2} = X$ and solve the quadratic equation:
$f \left(X\right) = {X}^{2} - 27 X + 50 = 0.$
Find 2 numbers (real roots) knowing their sum (-b = 27) and their product (c = 50). They are: 2 and 25.
$X = {x}^{2} = 2$--> $x = \pm \sqrt{2}$
$X = {x}^{2} = 25$ --> $x = \pm 5$