# What are the roots of x^3+52x^2+1060x-4624 = 0?

Jan 14, 2017

The real root is:

$x = \frac{2}{3} \left(- 26 + \sqrt[3]{21232 + 3 \sqrt{50275887}} + \sqrt[3]{21232 - 3 \sqrt{50275887}}\right)$

There are two related complex roots.

#### Explanation:

$f \left(x\right) = {x}^{3} + 52 {x}^{2} + 1060 x - 4624$

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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, $a = 1$, $b = 52$, $c = 1060$ and $d = - 4624$, so we find:

$\Delta = 3038214400 - 4764064000 + 2600685568 - 577297152 - 4587747840 = - 4290209024$

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.

$0 = 27 f \left(x\right) = 27 {x}^{3} + 1404 {x}^{2} + 28620 x - 124848$

$= {\left(3 x + 52\right)}^{3} + 1428 \left(3 x + 52\right) - 339712$

$= {t}^{3} + 1428 t - 339712$

where $t = \left(3 x + 52\right)$

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Cardano's method

We want to solve:

${t}^{3} + 1428 t - 339712 = 0$

Let $t = u + v$.

Then:

${u}^{3} + {v}^{3} + 3 \left(u v + 476\right) \left(u + v\right) - 339712 = 0$

Add the constraint $v = - \frac{476}{u}$ to eliminate the $\left(u + v\right)$ term and get:

${u}^{3} - \frac{107850176}{u} ^ 3 - 339712 = 0$

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

${\left({u}^{3}\right)}^{2} - 339712 \left({u}^{3}\right) - 107850176 = 0$

Use the quadratic formula to find:

${u}^{3} = \frac{339712 \pm \sqrt{{\left(- 339712\right)}^{2} - 4 \left(1\right) \left(- 107850176\right)}}{2 \cdot 1}$

$= \frac{339712 \pm \sqrt{115404242944 + 431400704}}{2}$

$= \frac{339712 \pm \sqrt{115835643648}}{2}$

$= \frac{339712 \pm 48 \sqrt{115835643648}}{2}$

$= 169856 \pm 24 \sqrt{115835643648}$

$= 8 \left(21232 \pm 3 \sqrt{50275887}\right)$

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:

${t}_{1} = 2 \sqrt[3]{21232 + 3 \sqrt{50275887}} + 2 \sqrt[3]{21232 - 3 \sqrt{50275887}}$

and related Complex roots:

${t}_{2} = 2 \omega \sqrt[3]{21232 + 3 \sqrt{50275887}} + 2 {\omega}^{2} \sqrt[3]{21232 - 3 \sqrt{50275887}}$

${t}_{3} = 2 {\omega}^{2} \sqrt[3]{21232 + 3 \sqrt{50275887}} + 2 \omega \sqrt[3]{21232 - 3 \sqrt{50275887}}$

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

Now $x = \frac{1}{3} \left(- 52 + t\right)$. So the roots of our original cubic are:

${x}_{1} = \frac{2}{3} \left(- 26 + \sqrt[3]{21232 + 3 \sqrt{50275887}} + \sqrt[3]{21232 - 3 \sqrt{50275887}}\right)$

${x}_{2} = \frac{2}{3} \left(- 26 + \omega \sqrt[3]{21232 + 3 \sqrt{50275887}} + {\omega}^{2} \sqrt[3]{21232 - 3 \sqrt{50275887}}\right)$

${x}_{3} = \frac{2}{3} \left(- 26 + {\omega}^{2} \sqrt[3]{21232 + 3 \sqrt{50275887}} + \omega \sqrt[3]{21232 - 3 \sqrt{50275887}}\right)$