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

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Answer 1 · 2016-11-24T22:08:48

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

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: $a=+-sqrt(2)i$

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