# How do you solve 64x^4-81>0?

$- \frac{3 \sqrt{2}}{2} < x < \frac{3 \sqrt{2}}{2}$

and

$- \frac{3 i \sqrt{2}}{2} < x < \frac{3 i \sqrt{2}}{2}$

#### Explanation:

Let's first see that all of the terms are perfect squares:

64x^2=(8x^2)^2; 81=9^2

If we assign $A = 8 x , B = 9$, then we'll have a left hand side that is in the form of:

$\left({A}^{2} - {B}^{2}\right)$

which can be factored as

$\left(A - B\right) \left(A + B\right)$

Substituting back in:

$\left(8 {x}^{2} - 9\right) \left(8 {x}^{2} + 9\right) > 0$

And now let's solve the inequality. First we change the $>$ for an $=$, we get

$\left(8 {x}^{2} - 9\right) \left(8 {x}^{2} + 9\right) = 0$

we get:

$8 {x}^{2} - 9 = 0 \implies {x}^{2} = \frac{9}{8} \implies x = \pm \sqrt{\frac{9}{8}} = \pm \frac{3}{\sqrt{2}} = \pm \frac{3 \sqrt{2}}{2}$

$8 {x}^{2} = 0 \implies {x}^{2} = - \frac{9}{8} \implies x = \pm \sqrt{- \frac{9}{8}} = \pm \frac{3 i}{\sqrt{2}} = \pm \frac{3 i \sqrt{2}}{2}$

as points of significance.

Working with the roots, if we set $x = 0$, we can see that it isn't valid for the original equation:

$64 {\left(0\right)}^{4} - 81 > 0 \implies - 81 > 0 \textcolor{w h i t e}{00} \textcolor{red}{\text{X}}$

and so our solution lies outside the points (as opposed to inside), and so the final answer is:

$- \frac{3 \sqrt{2}}{2} < x < \frac{3 \sqrt{2}}{2}$

and

$- \frac{3 i \sqrt{2}}{2} < x < \frac{3 i \sqrt{2}}{2}$