How do you solve #2x^4+x^2-3=0#?

1 Answer
Feb 28, 2016

Answer:

Simplify by treating as a quadratic in #x^2# and factor by grouping.

Explanation:

This is a "quadratic in #x^2#" which we can factor by grouping:

#2x^4+x^2-3#

#=2(x^2)^2+(x^2)-3#

#=(2(x^2)^2-2(x^2))+(3(x^2)-3)#

#=2x^2(x^2-1)+3(x^2-1)#

#=(2x^2+3)(x^2-1)#

#=(2x^2+3)(x-1)(x+1)#

The remaining quadratic factor has no linear factors with Real coefficients, but it can be factored if we allow Complex coefficients:

#=(sqrt(2)x-sqrt(3)i)(sqrt(2)x+sqrt(3)i)(x-1)(x+1)#

Hence the roots of the equation are:

#x = +-1# and #x = +-sqrt(3)/sqrt(2)i = +-sqrt(6)/2i#