How do you solve x^3+2x^2 = 2 ?

Feb 4, 2017

Use Cardano's method to find real root:

${x}_{1} = \frac{1}{3} \left(- 2 + \sqrt[3]{19 + 3 \sqrt{33}} + \sqrt[3]{19 - 3 \sqrt{33}}\right)$

and related Complex roots.

Explanation:

First subtract $2$ from both sides to get this equation into standard form:

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

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

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

$\Delta = 0 + 0 + 64 - 108 + 0 = - 44$

Since $\Delta < 0$ this cubic has $1$ Real zero and $2$ non-Real Complex zeros, which are Complex conjugates of one another.

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Tschirnhaus transformation

To make the task of solving the cubic simpler, we make the cubic simpler using a linear substitution known as a Tschirnhaus transformation.

$0 = 27 \left({x}^{3} + 2 {x}^{2} - 2\right)$

$\textcolor{w h i t e}{0} = 27 {x}^{3} + 54 {x}^{2} - 54$

$\textcolor{w h i t e}{0} = {\left(3 x + 2\right)}^{3} - 12 \left(3 x + 2\right) - 38$

$\textcolor{w h i t e}{0} = {t}^{3} - 12 t - 38$

where $t = \left(3 x + 2\right)$

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Cardano's method

We want to solve:

${t}^{3} - 12 t - 38 = 0$

Let $t = u + v$.

Then:

${u}^{3} + {v}^{3} + 3 \left(u v - 4\right) \left(u + v\right) - 38 = 0$

Add the constraint $v = \frac{4}{u}$ to eliminate the $\left(u + v\right)$ term and get:

${u}^{3} + \frac{64}{u} ^ 3 - 38 = 0$

Multiply through by ${u}^{3}$ and rearrange slightly to get:

${\left({u}^{3}\right)}^{2} - 38 \left({u}^{3}\right) + 64 = 0$

Use the quadratic formula to find:

${u}^{3} = \frac{38 \pm \sqrt{{\left(- 38\right)}^{2} - 4 \left(1\right) \left(64\right)}}{2 \cdot 1}$

$= \frac{38 \pm \sqrt{1444 - 256}}{2}$

$= \frac{38 \pm \sqrt{1188}}{2}$

$= \frac{38 \pm 6 \sqrt{33}}{2}$

$= 19 \pm 3 \sqrt{33}$

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

${t}_{1} = \sqrt[3]{19 + 3 \sqrt{33}} + \sqrt[3]{19 - 3 \sqrt{33}}$

and related Complex roots:

${t}_{2} = \omega \sqrt[3]{19 + 3 \sqrt{33}} + {\omega}^{2} \sqrt[3]{19 - 3 \sqrt{33}}$

${t}_{3} = {\omega}^{2} \sqrt[3]{19 + 3 \sqrt{33}} + \omega \sqrt[3]{19 - 3 \sqrt{33}}$

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

Now $x = \frac{1}{3} \left(- 2 + t\right)$. So the roots of our original cubic are:

${x}_{1} = \frac{1}{3} \left(- 2 + \sqrt[3]{19 + 3 \sqrt{33}} + \sqrt[3]{19 - 3 \sqrt{33}}\right)$

${x}_{2} = \frac{1}{3} \left(- 2 + \omega \sqrt[3]{19 + 3 \sqrt{33}} + {\omega}^{2} \sqrt[3]{19 - 3 \sqrt{33}}\right)$

${x}_{3} = \frac{1}{3} \left(- 2 + {\omega}^{2} \sqrt[3]{19 + 3 \sqrt{33}} + \omega \sqrt[3]{19 - 3 \sqrt{33}}\right)$