# How do you solve 2x^4-128=0?

Jan 10, 2017

The roots of this quartic equation are:

$\pm 2 \sqrt{2} \text{ }$ and $\text{ } \pm 2 \sqrt{2} i$

#### Explanation:

The difference of squares identity can be written:

${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$

So we find:

$0 = 2 {x}^{4} - 128$

$\textcolor{w h i t e}{0} = 2 \left({x}^{4} - 64\right)$

$\textcolor{w h i t e}{0} = 2 \left({\left({x}^{2}\right)}^{2} - {8}^{2}\right)$

$\textcolor{w h i t e}{0} = 2 \left({x}^{2} - 8\right) \left({x}^{2} + 8\right)$

$\textcolor{w h i t e}{0} = 2 \left({x}^{2} - {\left(2 \sqrt{2}\right)}^{2}\right) \left({x}^{2} - {\left(2 \sqrt{2} i\right)}^{2}\right)$

$\textcolor{w h i t e}{0} = 2 \left(x - 2 \sqrt{2}\right) \left(x + 2 \sqrt{2}\right) \left(x - 2 \sqrt{2} i\right) \left(x + 2 \sqrt{2} i\right)$

So the roots of the quartic equation are:

$\pm 2 \sqrt{2} \text{ }$ and $\text{ } \pm 2 \sqrt{2} i$