# How do you find all the real and complex roots of f(x)=x^3+x-3?

May 29, 2016

Use Cardano's method to find the Real zero:

${x}_{1} = \frac{1}{3} \left(\sqrt[3]{\frac{81 + 3 \sqrt{741}}{2}} + \sqrt[3]{\frac{81 - 3 \sqrt{741}}{2}}\right)$

and related Complex zeros.

#### Explanation:

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

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

So in our example, with $a = 1$, $b = 0$, $c = 1$ and $d = - 3$, we find:

$\Delta = 0 - 4 - 0 - 243 + 0 = - 247 < 0$

Since $\Delta < 0$ this cubic has one Real zero and a Complex conjugate pair of non-Real zeros.

It can be solved using Cardano's method.

Let $x = u + v$

Then:

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

Add the constraint $v = - \frac{1}{3 u}$ to eliminate the term in $\left(u + v\right)$ and get:

${u}^{3} - \frac{1}{27 {u}^{3}} - 3 = 0$

Multiply through by $27 {u}^{3}$ and rearrange a little to get:

$27 {\left({u}^{3}\right)}^{2} - 81 \left({u}^{3}\right) - 1 = 0$

Then use the quadratic formula to find:

${u}^{3} = \frac{81 \pm \sqrt{{81}^{2} - \left(4 \cdot 27 \cdot - 1\right)}}{2 \cdot 27}$

$= \frac{81 \pm \sqrt{6561 + 108}}{54}$

$= \frac{81 \pm \sqrt{6669}}{54}$

$= \frac{81 \pm 3 \sqrt{741}}{54}$

Since this is Real and the derivation was symmetric in $u$ and $v$m we can use one of these roots for ${u}^{3}$ and the other for ${v}^{3}$ to find the Real zero:

${x}_{1} = \frac{1}{3} \left(\sqrt[3]{\frac{81 + 3 \sqrt{741}}{2}} + \sqrt[3]{\frac{81 - 3 \sqrt{741}}{2}}\right)$

and Complex zeros:

${x}_{2} = \frac{1}{3} \left(\omega \sqrt[3]{\frac{81 + 3 \sqrt{741}}{2}} + {\omega}^{2} \sqrt[3]{\frac{81 - 3 \sqrt{741}}{2}}\right)$

${x}_{3} = \frac{1}{3} \left({\omega}^{2} \sqrt[3]{\frac{81 + 3 \sqrt{741}}{2}} + \omega \sqrt[3]{\frac{81 - 3 \sqrt{741}}{2}}\right)$

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