How do you solve the quadratic with complex numbers given x^4+13x^2+36=0?

Sep 28, 2016

$x = \pm 2 i$ or $\pm 3 i$

Explanation:

${x}^{4} + 13 {x}^{2} + 36 = 0$

Let $z = {x}^{2} \to {z}^{2} + 13 z + 36 = 0$

$\left(z + 4\right) \left(z + 9\right) = 0$

Hence: $z = - 4$ or $- 9$

$\therefore {x}^{2} = - 4$ or $- 9$

$x = \sqrt{- 4}$ or $\sqrt{- 9}$

$x = \pm 2 i$ or $\pm 3 i$
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