# How do you find the roots of x^3-x^2-17x+15=0?

Jun 6, 2017

$x \approx 6.0664 \mathmr{and} 0.9336270$

#### Explanation:

${x}^{3} - {x}^{2} - 17 x + 15 = 0$

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

${x}^{2} - 17 x + 15 = 0$

Now we can use the quadratic equation to solve.

$a {x}^{2} + b x + c = 0$

${x}^{2} - 17 x + 15 = 0$

$a = 1$
$b = - 17$
$c = 15$

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- - 17 \pm \sqrt{- {17}^{2} - 4 \times 1 \times 15}}{2 \times 1}$

We can simplify some things that we don't need to make the equation more simple.

$x = \frac{\cancel{- -} 17 \pm \sqrt{- {17}^{2} - 4 \cancel{\times 1} \times 15}}{2 \cancel{\times 1}}$

$x = \frac{17 \pm \sqrt{- {17}^{2} - 4 \times 15}}{2}$

$x = \frac{17 \pm \sqrt{289 - 4 \times 15}}{2}$

$x = \frac{17 \pm \sqrt{289 - 60}}{2}$

$x = \frac{17 \pm \sqrt{229}}{2}$

$x = \frac{17 \pm 15.132746}{2}$

${x}_{1} = \frac{17 + 15.132746}{2}$

${x}_{1} = \frac{32.132746}{2}$

 color(blue)(x_1 = 16.0664

${x}_{2} = \frac{17 - 15.132746}{2}$

${x}_{2} = \frac{1.86725405}{2}$

 color(blue)(x_2 = 0.9336270

Jun 6, 2017

Three real roots:

${x}_{n} = \frac{1}{3} \left(1 + 4 \sqrt{13} \cos \left(\frac{1}{3} {\cos}^{- 1} \left(- \frac{125 \sqrt{13}}{1352}\right) + \frac{2 n \pi}{3}\right)\right)$

for $n = 0 , 1 , 2$

#### Explanation:

Given:

${x}^{3} - {x}^{2} - 17 x + 15 = 0$

First use a Tschirnhaus transformation:

$0 = 27 \left({x}^{3} - {x}^{2} - 17 x + 15\right)$

$\textcolor{w h i t e}{0} = 27 {x}^{3} - 27 {x}^{2} - 459 x + 405$

$\textcolor{w h i t e}{0} = {\left(3 x - 1\right)}^{3} - 156 \left(3 x - 1\right) + 250$

$\textcolor{w h i t e}{0} = {t}^{3} - 156 t + 250 \text{ }$ where $\text{ } t = 3 x - 1$

Let:

$k = 2 \sqrt{- \frac{\textcolor{b l u e}{- 156}}{3}} = 4 \sqrt{13}$

Note that:

${k}^{3} / 4 = 52 k = 208 \sqrt{13}$

Substitute:

$t = k \cos \theta$

Then our equation becomes:

$0 = {\left(k \cos \theta\right)}^{3} - 156 \left(k \cos \theta\right) + 250$

$\textcolor{w h i t e}{0} = 208 \sqrt{13} \left(4 {\cos}^{3} \theta - 3 \cos \theta\right) + 250$

$\textcolor{w h i t e}{0} = 208 \sqrt{13} \cos 3 \theta + 250$

So:

$\cos 3 \theta = - \frac{250}{208 \sqrt{13}} = - \frac{125 \sqrt{13}}{1352}$

So:

$3 \theta = \pm {\cos}^{- 1} \left(- \frac{125 \sqrt{13}}{1352}\right) + 2 n \pi$

So:

$\cos \theta = \cos \left(\frac{1}{3} {\cos}^{- 1} \left(- \frac{125 \sqrt{13}}{1352}\right) + \frac{2 n \pi}{3}\right)$

So:

${t}_{n} = 4 \sqrt{13} \cos \left(\frac{1}{3} {\cos}^{- 1} \left(- \frac{125 \sqrt{13}}{1352}\right) + \frac{2 n \pi}{3}\right)$

with distinct roots for $n = 0 , 1 , 2$

Hence solutions of the original cubic:

${x}_{n} = \frac{1}{3} \left(1 + 4 \sqrt{13} \cos \left(\frac{1}{3} {\cos}^{- 1} \left(- \frac{125 \sqrt{13}}{1352}\right) + \frac{2 n \pi}{3}\right)\right)$

for $n = 0 , 1 , 2$

${x}_{0} \approx 4.19825$

${x}_{1} \approx - 4.07504$

${x}_{2} \approx 0.876781$

graph{x^3-x^2-17x+15 [-10, 10, -52, 52]}

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

I suspect there is a typo in the question. If the sign on the ${x}^{2}$ term was $+$ instead of $-$, then we find:

$0 = {x}^{3} + {x}^{2} - 17 x + 15$

$\textcolor{w h i t e}{0} = \left(x - 1\right) \left({x}^{2} + 2 x - 15\right)$

$\textcolor{w h i t e}{0} = \left(x - 1\right) \left(x + 5\right) \left(x - 3\right)$

and hence roots $1$, $- 5$ and $3$.