How do you solve #2x^4+x^2-3=0#?
1 Answer
Feb 28, 2016
Simplify by treating as a quadratic in
Explanation:
This is a "quadratic in
#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#
