How do you find the complex roots of #a^4+a^2-2=0#?

1 Answer
Nov 24, 2016

Answer:

This quartic has Real roots #a=+-1# and Complex roots #a=+-sqrt(2)i#

Explanation:

We can write this quartic as a quadratic in #a^2#:

#(a^2)^2+(a^2)-2 = 0#

Note that the sum of the coefficients is #0#. That is:

#1+1-2 = 0#

Hence #a^2 = 1# is a solution and #(a^2-1)# a factor:

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

We can factor this into linear factors with Real or imaginary coefficients using the difference of squares identity:

#A^2-B^2=(A-B)(A+B)#

as follows:

#(a^2-1)(a^2+2) = (a^2-1^2)(a^2-(sqrt(2)i)^2)#

#color(white)((a^2-1)(a^2+2)) = (a-1)(a+1)(a-sqrt(2)i)(a+sqrt(2)i)#

So there are Real roots: #a=+-1# and Complex roots: #a=+-sqrt(2)i#