What are the solutions of the equation x4 + 3x2 + 2 = 0? Use u substitution to solve. mc018-1.jpg and x = ±1 mc018-2.jpg and x = ±i mc018-3.jpg and x = ±i mc018-4.jpg and x = ±1

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]