Question

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

Answer 1

Let's replace `x^2=u` remembering that in the end `x=+-sqrtu`

Explanation:

Then we first solve for `u^2-12u+11=0->`
`(u-1)(u-12)=0->u=1oru=12->`

There are now four solutions:
`x=+-sqrt1=+-1or x=+-sqrt12=+-2sqrt3`

We'll have to examine what happens between and outside those points.
If we take any value between `-2sqrt3 and -1` the outcome will be negative, between `-1and+1` it's positive, and between `+1 and +2sqrt3` it will be negative again.
`<-2sqrt3and >+2sqrt3` the outcome will be positive.

Conclusion:
`-2sqrt3 < x<-1or +1 < x<+2sqrt3`

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