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

1 Answer
Oct 3, 2016

Answer:

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]}