How do you solve #x^4 + x^2 = 1 #?

Question

10677 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-14T20:24:07

Solve as a quadratic in #x^2# using the quadratic formula, then take square roots...

#x^4+x^2=1#

Subtract #1# from both sides to get:

#x^4+x^2-1 = 0#

Writing #x^4 = (x^2)^2# we have:

#(x^2)^2+(x^2)-1 = 0#

This is in the form #aX^2+bX+c = 0# with #X=x^2#, #a=1#, #b=1# and #c=-1#.

We can use the quadratic formula to find:

#x^2 = (-b+-sqrt(b^2-4ac))/(2a)#

#=(-1+-sqrt(1^2-(4*1*-1)))/(2*1)#

#=(-1+-sqrt(5))/2#

So:

#x = +-sqrt((-1+sqrt(5))/2)#

Or:

#x = +-sqrt((-1-sqrt(5))/2) = +-sqrt((1+sqrt(5))/2)i#