# How do you solve 3n^4 - 4n^2=-1?

Apr 28, 2016

$n = 1$ is one solution
After further investigation I found that:
$\implies x = \pm 1 \text{ and } x = \pm \sqrt{\frac{1}{3}} = \pm \frac{\sqrt{3}}{3}$

#### Explanation:

Write as: ${n}^{2} \left(3 {n}^{2} - 4\right) = - 1$

Consider $3 - 4 = - 1$

Then ${n}^{2} \left(3 {n}^{2} - 4\right) \equiv 3 - 4 = - 1$

Thus ${n}^{2} \left(- 4\right) \equiv \left(- 4\right) \implies n = 1$
Also ${n}^{2} \left(3 {n}^{2}\right) \equiv \left(+ 3\right) \implies n = 1$

Conclusion is that $n = 1$

Now it is a matter of finding the others

Let ${x}^{2} = X$

Then we have:
$3 {X}^{2} - 4 X + 1 = 0$

$\implies \left(3 X - 1\right) \left(X - 1\right) = 0$

$X = 1 \text{ and } \frac{1}{3}$

But ${x}^{2} = X$

$\implies x = \pm 1 \text{ and } x = \pm \sqrt{\frac{1}{3}} = \pm \frac{\sqrt{3}}{3}$