How do you solve $x^{4} - 12 x^{2} + 11 < 0$ $x^4-12x^2+11<0$ ?

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How do you solve $x^{4} - 12 x^{2} + 11 < 0$ $x^4-12x^2+11<0$ ?

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Answer 1 · 2016-10-03T11:44:32

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

Explanation:

Then we first solve for $u^{2} - 12 u + 11 = 0 \to$ $u^2-12u+11=0->$
$(u - 1) (u - 12) = 0 \to u = 1 or u = 12 \to$ $(u-1)(u-12)=0->u=1oru=12->$

There are now four solutions:
$x = \pm \sqrt{1} = \pm 1 or x = \pm \sqrt{12} = \pm 2 \sqrt{3}$ $x=+-sqrt1=+-1or x=+-sqrt12=+-2sqrt3$

We'll have to examine what happens between and outside those points.
If we take any value between $- 2 \sqrt{3} and - 1$ $-2sqrt3 and -1$ the outcome will be negative, between $- 1 and + 1$ $-1and+1$ it's positive, and between $+ 1 and + 2 \sqrt{3}$ $+1 and +2sqrt3$ it will be negative again.
$< - 2 \sqrt{3} and > + 2 \sqrt{3}$ $<-2sqrt3and >+2sqrt3$ the outcome will be positive.

Conclusion:
$- 2 \sqrt{3} < x < - 1 or + 1 < x < + 2 \sqrt{3}$ $-2sqrt3 < x<-1or +1 < x<+2sqrt3$

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

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How do you solve $x^{4} - 12 x^{2} + 11 < 0$ $x^4-12x^2+11<0$ ?

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