# How do you find the complex roots of 4m^4+17m^2+4=0?

Nov 26, 2016

The roots are:

$m = \pm \frac{1}{2} i \text{ }$ and $\text{ } m = \pm 2 i$

#### Explanation:

$4 {m}^{4} + 17 {m}^{2} + 4 = 0$

Treating this as a quadratic in ${m}^{2}$, we can use an AC method to factor it into quadratics, then use the difference of squares identity with Complex coefficients.

First find a pair of factors of $A C = 4 \cdot 4 = 16$ with sum $B = 17$.

The pair $16 , 1$ works.

Use this pair to split the middle term, then factor by grouping:

$4 {m}^{4} + 17 {m}^{2} + 4 = 4 {m}^{4} + 16 {m}^{2} + {m}^{2} + 4$

$\textcolor{w h i t e}{4 {m}^{4} + 17 {m}^{2} + 4} = \left(4 {m}^{4} + 16 {m}^{2}\right) + \left({m}^{2} + 4\right)$

$\textcolor{w h i t e}{4 {m}^{4} + 17 {m}^{2} + 4} = 4 {m}^{2} \left({m}^{2} + 4\right) + 1 \left({m}^{2} + 4\right)$

$\textcolor{w h i t e}{4 {m}^{4} + 17 {m}^{2} + 4} = \left(4 {m}^{2} + 1\right) \left({m}^{2} + 4\right)$

$\textcolor{w h i t e}{4 {m}^{4} + 17 {m}^{2} + 4} = \left({\left(2 m\right)}^{2} - {i}^{2}\right) \left({m}^{2} - {\left(2 i\right)}^{2}\right)$

$\textcolor{w h i t e}{4 {m}^{4} + 17 {m}^{2} + 4} = \left(2 m - i\right) \left(2 m + i\right) \left(m - 2 i\right) \left(m + 2 i\right)$

Hence zeros:

$m = \pm \frac{1}{2} i \text{ }$ and $\text{ } m = \pm 2 i$