ANSWER
[tex]x = \pm \: i[/tex]
or
[tex] x = \pm \: \sqrt{2} i[/tex]
EXPLANATION
We want to solve the equation,
[tex] {x}^{4} + 3 {x}^{2} + 2 = 0[/tex]
We rewrite the equation to get,
[tex]( {x}^{2} ) ^{2} + 3 {x}^{2} + 2 = 0[/tex]
We can now let
[tex]y = {x}^{2} [/tex]
We substitute this value to get,
[tex] {y}^{2} + 3y + 2 = 0[/tex]
We factor this to get,
[tex](y + 1)(y + 2) = 0[/tex]
This implies that,
[tex]y = - 1[/tex]
or
[tex]y = - 2[/tex]
But
[tex]x = \pm \: \sqrt{y} [/tex]
Thus implies that,
[tex]x = \pm \: \sqrt{ - 1} [/tex]
[tex]\Rightarrow x = \pm \: i[/tex]
or
[tex]x = \pm \: \sqrt{ - 2} [/tex]
[tex]\Rightarrow x = \pm \: \sqrt{2} i[/tex]
