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

Aug 12, 2016

Use a trigonometric method to find roots:

${x}_{k} = \frac{4 \sqrt{3}}{3} \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{3 \sqrt{3}}{8}\right) + \frac{2 k \pi}{3}\right) \text{ }$ for $k = 0 , 1 , 2.$

#### Explanation:

$\textcolor{w h i t e}{}$

$f \left(x\right) = {x}^{3} - 4 x - 2$

$\textcolor{w h i t e}{}$
Descriminant

The discriminant $\Delta$ of a cubic polynomial in the form $a {x}^{3} + b {x}^{2} + c x + d$ is given by the formula:

$\Delta = {b}^{2} {c}^{2} - 4 a {c}^{3} - 4 {b}^{3} d - 27 {a}^{2} {d}^{2} + 18 a b c d$

In our example, $a = 1$, $b = 0$, $c = - 4$ and $d = - 2$, so we find:

$\Delta = 0 + 256 + 0 - 108 + 0 = 148$

Since $\Delta > 0$ this cubic has $3$ Real zeros.

Trigonometric solution

Use a substitution of the form $x = k \cos \theta$, choosing $k$ so that the resulting expression contains $4 {\cos}^{3} \theta - 3 \cos \theta = \cos 3 \theta$.

Let $k = \frac{4 \sqrt{3}}{3}$

Then:

$0 = {x}^{3} - 4 x - 2$

$= {k}^{3} {\cos}^{3} \theta - 4 k \cos \theta - 2$

$= \frac{64 \sqrt{3}}{9} {\cos}^{3} \theta - \frac{16 \sqrt{3}}{3} \cos \theta - 2$

$= \frac{16 \sqrt{3}}{9} \left(4 {\cos}^{3} \theta - 3 \cos \theta\right) - 2$

$= \frac{16 \sqrt{3}}{9} \cos 3 \theta - 2$

Hence:

$\cos 3 \theta = 2 \cdot \frac{9}{16 \sqrt{3}} = \frac{3 \sqrt{3}}{8}$

So:

$3 \theta = \pm {\cos}^{- 1} \left(\frac{3 \sqrt{3}}{8}\right) + 2 k \pi$

$\theta = \pm \frac{1}{3} {\cos}^{- 1} \left(\frac{3 \sqrt{3}}{8}\right) + \frac{2 k \pi}{3}$

$\cos \theta = \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{3 \sqrt{3}}{8}\right) + \frac{2 k \pi}{3}\right)$

Hence distinct roots:

${x}_{k} = \frac{4 \sqrt{3}}{3} \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{3 \sqrt{3}}{8}\right) + \frac{2 k \pi}{3}\right) \text{ }$ for $k = 0 , 1 , 2.$