# How do you solve 0 = 4x^3 + 2x-1?

Mar 10, 2016

Check there are no rational roots, then use Cardano's method to find the Real and two Complex roots.

#### Explanation:

Let $f \left(x\right) = 4 {x}^{3} + 2 x - 1$

By the rational root theorem, any roots of $4 {x}^{3} + 2 x - 1 = 0$ will be expressible in the form $\frac{p}{q}$ for integers $p$ and $q$ where $p$ is a divisor of the constant term $- 1$ and $q$ a divisor of the coefficient $4$ of the leading term.

So the only possible rational roots are:

$\pm \frac{1}{4}$, $\pm \frac{1}{2}$, $\pm 1$

We find $f \left(x\right) \ne 0$ for all of these values, so $f \left(x\right) = 0$ has no rational roots.

Use Cardano's method, substituting $x = u + v$ to get:

$4 {u}^{3} + 4 {v}^{3} + 2 \left(6 u v + 1\right) \left(u + v\right) - 1 = 0$

Add the constraint $v = - \frac{1}{6 u}$ to eliminate the term in $\left(u + v\right)$ to get:

$0 = 4 {u}^{3} - \frac{4}{6 u} ^ 3 - 1 = 4 {u}^{3} - \frac{1}{54 {u}^{3}} - 1$

Multiply through by $54 {u}^{3}$ to get:

$216 {\left({u}^{3}\right)}^{2} - 54 \left({u}^{3}\right) - 1 = 0$

Use the quadratic formula to find:

${u}^{3} = \frac{54 \pm \sqrt{{54}^{2} + 4 \cdot 216}}{2 \cdot 216}$

$= \frac{54 \pm \sqrt{3780}}{432}$

$= \frac{54 \pm 6 \sqrt{105}}{432}$

$= \frac{27 \pm 3 \sqrt{105}}{216}$

Due to the symmetry of the derivation in $u$ and $v$, we can take one of these roots as ${u}^{3}$ and the other as ${v}^{3}$ to find the Real root:

${x}_{1} = \frac{1}{6} \left(\sqrt[3]{27 + 3 \sqrt{105}} + \sqrt[3]{27 - 3 \sqrt{105}}\right)$

and Complex roots:

${x}_{2} = \frac{1}{6} \left(\omega \sqrt[3]{27 + 3 \sqrt{105}} + {\omega}^{2} \sqrt[3]{27 - 3 \sqrt{105}}\right)$

${x}_{3} = \frac{1}{6} \left({\omega}^{2} \sqrt[3]{27 + 3 \sqrt{105}} + \omega \sqrt[3]{27 - 3 \sqrt{105}}\right)$

where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive Complex cube root of $1$.