# How do you find the roots of 2x^4+13x^2+17x-12=0?

Jan 14, 2017

$x = \frac{1}{4} \left(a \pm \sqrt{- {a}^{2} - 52 - \frac{136}{a}}\right)$

$x = \frac{1}{4} \left(- a \pm \sqrt{- {a}^{2} - 52 + \frac{136}{a}}\right)$

where

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

#### Explanation:

$f \left(x\right) = 2 {x}^{4} + 13 {x}^{2} + 17 x - 12$

To cut a long story a little shorter:

Use the rational roots theorem to find possible rational roots:

$\pm \frac{1}{2} , \pm 1 , \pm \frac{3}{2} , \pm 2 , \pm 3 , \pm 4 , \pm 6 , \pm 12$

None of these are roots, so $f \left(x\right)$ has no rational zeros.

Use Descartes' Rule of Signs to find that $f \left(x\right)$ has exactly one positive real zero, one negative real zero and a complex conjugate pair of non-real zeros.

This is a typical nasty quartic, except that it has two things going for it:

• There is no term in ${x}^{3}$, so we do not need to do a linear Tschirnhaus transformation to simplify it.

• The cubic resolvent is reducible. That is, it has one real zero and two complex zeros, so is expressible in terms of real square and cube roots.

Let's take the plunge and solve it algebraically:

$0 = 2 f \left(x\right)$

$\textcolor{w h i t e}{0} = 4 {x}^{4} + 26 {x}^{2} + 34 x - 24$

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

$\textcolor{w h i t e}{0} = 4 {x}^{4} + \left(2 \left(b + c\right) - {a}^{2}\right) {x}^{2} + a \left(b - c\right) x + b c$

Equating coefficients and rearranging a little:

$\left\{\begin{matrix}b + c = {a}^{2} / 2 + 13 \\ b - c = \frac{34}{a} \\ b c = - 24\end{matrix}\right.$

Hence:

${\left({a}^{2} / 2 + 13\right)}^{2} = {\left(b + c\right)}^{2} = {\left(b - c\right)}^{2} + 4 b c = {\left(\frac{34}{a}\right)}^{2} + 4 \left(- 24\right)$

That is:

${a}^{4} / 4 + 13 {a}^{2} + 169 = \frac{1156}{a} ^ 2 - 96$

Multiply through by $4 {a}^{2}$ and rearrange slightly to get:

${\left({a}^{2}\right)}^{3} + 52 {\left({a}^{2}\right)}^{2} + 1060 \left({a}^{2}\right) - 4624 = 0$

Solve this cubic (see https://socratic.org/s/aBjKeuBA) to find the real root:

${a}^{2} = \frac{2}{3} \left(- 26 + \sqrt[3]{21232 + 3 \sqrt{50275887}} + \sqrt[3]{21232 - 3 \sqrt{50275887}}\right) \approx 3.659$

Since this is also positive, its square roots are real. Without loss of generality we can use the positive one:

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

From our previous simultaneous equations we can find:

$b = {a}^{2} / 4 + \frac{13}{2} + \frac{17}{a}$

$c = {a}^{2} / 4 + \frac{13}{2} - \frac{17}{a}$

Which leaves us two quadratic equations to solve:

$2 {x}^{2} - a x + \left({a}^{2} / 4 + \frac{13}{2} + \frac{17}{a}\right) = 0$

$2 {x}^{2} + a x + \left({a}^{2} / 4 + \frac{13}{2} - \frac{17}{a}\right) = 0$

Using the quadratic formula we find solutions:

$x = \frac{1}{4} \left(a \pm \sqrt{- {a}^{2} - 52 - \frac{136}{a}}\right)$

$x = \frac{1}{4} \left(- a \pm \sqrt{- {a}^{2} - 52 + \frac{136}{a}}\right)$