# How do you solve 3x^4+12x^2-15=0?

Oct 1, 2015

Assuming we are restricted to Real (non-Complex) solutions:
$\textcolor{w h i t e}{\text{XXX}} x = \pm 1$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} 3 {x}^{4} + 12 {x}^{2} - 15 = 0$

Dividing both sides by $3$
$\textcolor{w h i t e}{\text{XXX}} {x}^{4} + 4 {x}^{2} - 5 = 0$

Temporarily substituting $q$ for ${x}^{2}$
$\textcolor{w h i t e}{\text{XXX}} {q}^{2} + 4 q - 5 = 0$

Factoring the left side
$\textcolor{w h i t e}{\text{XXX}} \left(q + 5\right) \left(q - 1\right) = 0$

With potential solutions
$\textcolor{w h i t e}{\text{XXX}} q = - 5$ and $q = 1$

If $q = - 5$
then
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} = - 5$
which is impossible for Real values of $x$

If $q = 1$
then
$\textcolor{w h i t e}{\text{XXX}} {x}^{2} = 1$

$\rightarrow \textcolor{w h i t e}{\text{X}} x = \pm 1$