# Solve ax^4+bx^3+cx^2+dx+e=0?

Mar 18, 2018

A quick sketch...

#### Explanation:

Given:

$a {x}^{4} + b {x}^{3} + c {x}^{2} + \mathrm{dx} + e = 0 \text{ }$ with $a \ne 0$

This gets messy quite quickly, so I will just give a sketch of one method...

Multiply by $256 {a}^{3}$ and substitute $t = \left(4 a x + b\right)$ to get a depressed monic quartic of the form:

${t}^{4} + p {t}^{2} + q t + r = 0$

Note that since this has no term in ${t}^{3}$, it must factor in the form:

${t}^{4} + p {t}^{2} + q t + r = \left({t}^{2} - A t + B\right) \left({t}^{2} + A t + C\right)$

$\textcolor{w h i t e}{{t}^{4} + p {t}^{2} + q t + r} = {t}^{4} + \left(B + C - {A}^{2}\right) {t}^{2} + A \left(B - C\right) t + B C$

Equating coefficients and rearranging a little, we have:

$\left\{\begin{matrix}B + C = {A}^{2} + p \\ B - C = \frac{q}{A} \\ B C = d\end{matrix}\right.$

So we find:

${\left({A}^{2} + p\right)}^{2} = {\left(B + C\right)}^{2}$

$\textcolor{w h i t e}{{\left({A}^{2} + p\right)}^{2}} = {\left(B - C\right)}^{2} + 4 B C$

$\textcolor{w h i t e}{{\left({A}^{2} + p\right)}^{2}} = {q}^{2} / {A}^{2} + 4 d$

Multiplying out, multiplying by ${A}^{2}$ and rearranging slightly, this becomes:

${\left({A}^{2}\right)}^{3} + 2 p {\left({A}^{2}\right)}^{2} + \left({p}^{2} - 4 d\right) \left({A}^{2}\right) - {q}^{2} = 0$

This "cubic in ${A}^{2}$" has at least one real root. Ideally it has a positive real root that yields two possible real values for $A$. Regardless, any root of the cubic will do.

Given the value of $A$, we have:

$B = \frac{1}{2} \left(\left(B + C\right) + \left(B - C\right)\right) = \frac{1}{2} \left({A}^{2} + p + \frac{q}{A}\right)$

$C = \frac{1}{2} \left(\left(B + C\right) - \left(B - C\right)\right) = \frac{1}{2} \left({A}^{2} + p - \frac{q}{A}\right)$

Hence we get two quadratics to solve.