# How do you solve x^4-12x^2+11<0?

Oct 3, 2016

Let's replace ${x}^{2} = u$ remembering that in the end $x = \pm \sqrt{u}$

#### Explanation:

Then we first solve for ${u}^{2} - 12 u + 11 = 0 \to$
$\left(u - 1\right) \left(u - 12\right) = 0 \to u = 1 \mathmr{and} u = 12 \to$

There are now four solutions:
$x = \pm \sqrt{1} = \pm 1 \mathmr{and} x = \pm \sqrt{12} = \pm 2 \sqrt{3}$

We'll have to examine what happens between and outside those points.
If we take any value between $- 2 \sqrt{3} \mathmr{and} - 1$ the outcome will be negative, between $- 1 \mathmr{and} + 1$ it's positive, and between $+ 1 \mathmr{and} + 2 \sqrt{3}$ it will be negative again.
$< - 2 \sqrt{3} \mathmr{and} > + 2 \sqrt{3}$ the outcome will be positive.

Conclusion:
$- 2 \sqrt{3} < x < - 1 \mathmr{and} + 1 < x < + 2 \sqrt{3}$

graph{x^4-12x^2+11 [-52.03, 52.04, -26, 26]}