# How do you solve the quadratic using the quadratic formula given x^4+13x^2+36=0 over the set of complex numbers?

Feb 15, 2017

$\pm 2 i \mathmr{and} \pm 3 i$

#### Explanation:

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

assuming $y = {x}^{2}$,

${\left({x}^{2}\right)}^{2} + 13 {x}^{2} + 6 = {y}^{2} + 13 y + 36 = 0$

using $y = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$, where $a = 1 , b = 13 \mathmr{and} c = 36$

$y = \frac{- 13 \pm \sqrt{{13}^{2} - 4 \left(1\right) \left(36\right)}}{2 \left(1\right)}$

$y = \frac{- 13 \pm \sqrt{169 - 144}}{2}$

$y = \frac{- 13 \pm \sqrt{25}}{2}$

$y = \frac{- 13 \pm 5}{2}$

$y = \frac{- 13 + 5}{2} = - 4$, $y = \frac{- 13 - 5}{2} = - 9$

when $y = - 4 , {x}^{2} = - 4$

$x = \sqrt{- 4} = \sqrt{4 \cdot \left(- 1\right)} = \sqrt{4} \cdot \sqrt{- 1} = \pm 2 i$

when $y = - 9 , {x}^{2} = - 9$

$x = \sqrt{- 9} = \sqrt{9 \cdot \left(- 1\right)} = \sqrt{9} \cdot \sqrt{- 1} = \pm 3 i$