# How do you find all the real and complex roots of x^4-14x^2 +45=0?

Mar 3, 2016

Equation has all real roots given by $x = - \left\{\sqrt{5} , \sqrt{5} , - 3 , 3\right\}$

#### Explanation:

Let us first factorize x^4−14x^2+45, which can be done by splitting middle term into $- 9 {x}^{2}$ and $- 5 {x}^{2}$. Hence,

x^4−14x^2+45=0

$\Leftrightarrow$ x^4−9x^2-5x^2+45=0

$\Leftrightarrow$ x^2(x^2−9)-5(x^2-9)=0

$\Leftrightarrow$ (x^2-5)(x^2−9)=0 and using ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$, this becomes

$\Leftrightarrow$ $\left(x + \sqrt{5}\right) \left(x - \sqrt{5}\right) \left(x + 3\right) \left(x - 3\right) = 0$

or $x = - \left\{\sqrt{5} , \sqrt{5} , - 3 , 3\right\}$ and hence equation has all real roots.