# Solve the equation x^4+4x^3+8=0?

## Obtain the resolvent cubics, by Descartes' method and by Ferrari's method, of the equation ${x}^{4} + 4 {x}^{3} + 8 = 0$. Are the cubics the same? Further, Use either method to obtain the roots of this equation.

Mar 24, 2017

See below.

#### Explanation:

Consider the polynomial

${x}^{4} + 2 a {x}^{3} + b = 0$

and also the expansion

${\left({x}^{2} + a x + \xi\right)}^{2} = {x}^{4} + 2 a {x}^{3} + \left({a}^{2} + 2 \xi\right) {x}^{2} + 2 a \xi x + {\xi}^{2}$

(here $\xi$ is a dummy parameter)

now substituting ${x}^{4} + 2 a {x}^{3} = - b$ we obtain

${\left({x}^{2} + a x + \xi\right)}^{2} = \left({a}^{2} + 2 \xi\right) {x}^{2} + 2 a \xi x + {\xi}^{2} - b$

and now choosing properly $\xi$ to make $\left({a}^{2} + 2 \xi\right) {x}^{2} + 2 a \xi x + {\xi}^{2} - b$ a square with

$2 {\xi}^{3} - 2 b \xi - {a}^{2} b = 0$

or

${\xi}^{3} - 8 \xi - 16 = 0$

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Note that solving $\left({a}^{2} + 2 \xi\right) {x}^{2} + 2 a \xi x + {\xi}^{2} - b = 0$ or

$x = \frac{- a \xi \pm \sqrt{{a}^{2} b + 2 b \xi - 2 {\xi}^{3}}}{{a}^{2} + 2 \xi}$ if we choose
${\xi}_{0}$ such that ${a}^{2} b + 2 b {\xi}_{0} - 2 {\xi}_{0}^{3} = 0$ then

$\left({a}^{2} + 2 {\xi}_{0}\right) {x}^{2} + 2 a {\xi}_{0} x + {\xi}_{0}^{2} - b = \left({a}^{2} + 2 {\xi}_{0}\right) {\left(x + a {\xi}_{0}\right)}^{2}$
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Now solving for $\xi$

$\xi = {\xi}_{0} = \frac{1}{3} {\left(216 - 24 \sqrt{57}\right)}^{\frac{1}{3}} + \frac{2 {\left(9 + \sqrt{57}\right)}^{\frac{1}{3}}}{3} ^ \left(\frac{2}{3}\right)$ which is the only real root.

Now

$\left({a}^{2} + 2 {\xi}_{0}\right) {x}^{2} + 2 a {\xi}_{0} x + {\xi}_{0}^{2} - b = \left({a}^{2} + 2 {\xi}_{0}\right) {\left(x + a {\xi}_{0}\right)}^{2}$

and the former quartic remains

${\left({x}^{2} + a x + {\xi}_{0}\right)}^{2} = \left({a}^{2} + 2 {\xi}_{0}\right) {\left(x + a {\xi}_{0}\right)}^{2}$ or

${\left({x}^{2} + a x + {\xi}_{0}\right)}^{2} - \left({a}^{2} + 2 {\xi}_{0}\right) {\left(x + a {\xi}_{0}\right)}^{2} = 0$ or

$\left({x}^{2} + a x + {\xi}_{0} + \sqrt{{a}^{2} + 2 {\xi}_{0}} \left(x + a {\xi}_{0}\right)\right) \left({x}^{2} + a x + {\xi}_{0} - \sqrt{{a}^{2} + 2 {\xi}_{0}} \left(x + a {\xi}_{0}\right)\right) = 0$

resulting in the resolution of

$\left\{\begin{matrix}{x}^{2} + \left(a + \sqrt{{a}^{2} + 2 {\xi}_{0}}\right) x + {\xi}_{0} + a {\xi}_{0} \sqrt{{a}^{2} + 2 {\xi}_{0}} = 0 \\ {x}^{2} + \left(a - \sqrt{{a}^{2} + 2 {\xi}_{0}}\right) x + {\xi}_{0} - a {\xi}_{0} \sqrt{{a}^{2} + 2 {\xi}_{0}} = 0\end{matrix}\right.$

This is left as an exercise for the proficient reader.

NOTE: This is the so called Ferrari's method.