How do you find all the real and complex roots of $x^{4} + 3 x^{2} + 2 = 0$ $x^4 + 3x^2 + 2 = 0$ ?

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Answer 1 · 2016-02-06T02:40:43

$x = \pm i, \pm i \sqrt{2}$ $x=+-i,+-isqrt2$

Note that $x^{2} + 3 x + 2$ $x^2+3x+2$ can be factored into $(x + 1) (x + 2)$ $(x+1)(x+2)$ , so

$x^{4} + 3 x^{2} + 2 = (x^{2} + 1) (x^{2} + 2)$ $x^4+3x^2+2=(x^2+1)(x^2+2)$

So, we have the equation

$(x^{2} + 1) (x^{2} + 2) = 0$ $(x^2+1)(x^2+2)=0$

Set both of these equal to $0$ $0$ :

$x^{2} + 1 = 0$ $x^2+1=0$

$x^{2} = - 1$ $x^2=-1$

$x = \pm i$ $x=+-i$

and

$x^{2} + 2 = 0$ $x^2+2=0$

$x^{2} = - 2 i$ $x^2=-2i$

$x = \pm i \sqrt{2}$ $x=+-isqrt2$

The equation has $4$ $4$ complex roots and no real roots (meaning it never crosses the $x$ $x$ -axis:

graph{x^4+3x^2+2 [-13.83, 14.65, -2.31, 11.93]}

How do you find all the real and complex roots of x^4 + 3x^2 + 2 = 0x4+3x2+2=0x^4 + 3x^2 + 2 = 0?