# How do you solve the quadratic using the quadratic formula given x^4+16x^2-225=0 over the set of complex numbers?

Oct 8, 2016

$x \in \left\{- 5 i , 5 i , - 3 , 3\right\}$

#### Explanation:

Using the quadratic formula, along with that $\sqrt{- x} = i \sqrt{x}$ for any $x \in \mathbb{R}$:

${x}^{4} + 16 {x}^{2} - 225 = 0$

$\implies {\left({x}^{2}\right)}^{2} + 16 \left({x}^{2}\right) - 225 = 0$

$\implies {x}^{2} = \frac{- 16 \pm \sqrt{{16}^{2} - 4 \left(1\right) \left(- 225\right)}}{2 \left(1\right)}$

$= \frac{- 16 \pm 34}{2}$

$= - 8 \pm 17$

$\implies {x}^{2} = - 25 \mathmr{and} {x}^{2} = 9$

$\implies x = \pm \sqrt{- 25} \mathmr{and} x = \pm \sqrt{9}$

$\implies x = \pm 5 i \mathmr{and} x = \pm 3$

$\therefore x \in \left\{- 5 i , 5 i , - 3 , 3\right\}$