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

Question

Avery Alexander · Accepted Answer

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#. The quadratic formula can be used to determine: #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#

Eva Abramson · Answer

undefined To solve the equation $ x^4 + x^2 = 1 $, you can rearrange it into a quadratic equation by substituting $ y = x^2 $. This gives $ y^2 + y - 1 = 0 $. Then, you can solve for $ y $ using the quadratic formula:

$$ y = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $$

where $ a = 1 $, $ b = 1 $, and $ c = -1 $. Substituting these values into the formula yields:

$$ y = \frac{{-1 \pm \sqrt{{1^2 - 4(1)(-1)}}}}{{2(1)}} $$
$$ y = \frac{{-1 \pm \sqrt{{1 + 4}}}}{2} $$
$$ y = \frac{{-1 \pm \sqrt{5}}}{2} $$

So, $ x^2 = \frac{{-1 \pm \sqrt{5}}}{2} $. Taking the square root of both sides gives:

$$ x = \pm \sqrt{\frac{{-1 \pm \sqrt{5}}}{2}} $$

Hence, there are four solutions to the equation.