# How do you find all the real and complex roots of 2x^4 + 3x^3 - x^2 + 5 = 0?

Nov 20, 2016

See explanation for a sketch of how to solve this algebraically...

#### Explanation:

$f \left(x\right) = 2 {x}^{4} + 3 {x}^{3} - {x}^{2} + 5$

This is an example of the worst possible case in quartic equations.

The full solution is too long, so I will just sketch it here...

• Use a Tschirnhaus transformation to simplify the quartic to one with no cube term:

$2048 f \left(x\right) = 4096 {x}^{4} + 6144 {x}^{3} - 2048 {x}^{2} + 10240$

$\textcolor{w h i t e}{2048 f \left(x\right)} = {\left(8 x + 3\right)}^{4} - 86 {\left(8 x + 3\right)}^{2} + 408 \left(8 x + 3\right) + 9709$

$\textcolor{w h i t e}{2048 f \left(x\right)} = {t}^{4} - 86 {t}^{2} + 408 t + 9709$

where $t = 8 x + 3$

• Consider a factorisation of the form $\left({t}^{2} - a t + b\right) \left({t}^{2} + a t + c\right)$ and equate coefficients to get:

$\left\{\begin{matrix}b + c = {a}^{2} - 86 \\ b - c = \frac{408}{a} \\ b c = 9709\end{matrix}\right.$

• Use ${\left(b + c\right)}^{2} = {\left(b - c\right)}^{2} + 4 b c$ to derive a cubic in ${a}^{2}$:

${\left({a}^{2}\right)}^{3} - 172 {\left({a}^{2}\right)}^{2} - 31440 \left({a}^{2}\right) - 166464 = 0$

• Use a Tschirnhaus transformation to simplify the cubic to one with no square term:

$27 \left({\left({a}^{2}\right)}^{3} - 172 {\left({a}^{2}\right)}^{2} - 31440 \left({a}^{2}\right) - 166464\right)$

$= 27 {\left({a}^{2}\right)}^{3} - 46644 {\left({a}^{2}\right)}^{2} - 848880 \left({a}^{2}\right) - 4494528$

$= {\left(3 {a}^{2} - 172\right)}^{3} - 371712 \left(3 {a}^{2} - 172\right) - 63340544$

$= {s}^{3} - 371712 s - 63340544$

where $s = 3 {a}^{2} - 172$

• This cubic has $3$ Real zeros, so use a trigonometric substitution of the form $s = k \cos \theta$ with $k = 704$

$0 = {s}^{3} - 371712 s - 63340544$

$\textcolor{w h i t e}{0} = {704}^{3} {\cos}^{3} \theta - 371712 \cdot 704 \cos \theta - 63340544$

$\textcolor{w h i t e}{0} = 87228416 \left(4 {\cos}^{3} \theta - 3 \cos \theta\right) - 63340544$

$\textcolor{w h i t e}{0} = 87228416 \cos 3 \theta - 63340544$

$\textcolor{w h i t e}{0} = 32768 \left(2662 \cos 3 \theta - 1933\right)$

Hence:
$s = 704 \cos \left(\frac{1}{3} {\cos}^{- 1} \left(\frac{1933}{2662}\right) + \frac{2 k \pi}{3}\right)$ for $k = 0 , 1 , 2$

The positive Real root occurs for $k = 0$, so choose that to get:

a = sqrt(1/3( 704 cos(1/3cos^(-1)(1933/2662))+172)

• Going back to our simultaneous equations in $a , b$ and $c$ we find:

$b = \frac{1}{2} \left({a}^{2} - 86 + \frac{408}{a}\right)$

$c = \frac{1}{2} \left({a}^{2} - 86 - \frac{408}{a}\right)$

• Hence we have two quadratic equations to solve:

${t}^{2} - a t + \frac{1}{2} \left({a}^{2} - 86 + \frac{408}{a}\right) = 0$

${t}^{2} + a t + \frac{1}{2} \left({a}^{2} - 86 - \frac{408}{a}\right) = 0$

• Solve these quadratic equations using the quadratic formula to find solutions to ${t}^{4} - 86 {t}^{2} + 408 t + 9709 = 0$

• Reverse the initial Tschirnhaus transformation using $x = \frac{1}{8} \left(t - 3\right)$ to find the four (Complex) roots of the original quartic.