# Difference between revisions of "Quartic Equation"

A quartic equation is an algebraic equation of the form

$ax^4 + bx^3 + cx^2 + dx + e = 0.$

These types of equations are extremely hard to solve; however, there are very clever methods for solving them by bringing it down to a cubic. I am going to list the simplest of the five.

## Solving Quartic Equations

Look in the "TLDR" section for the final result of each step.

### Bringing it down to a depressed quartic

Start with the equation $ax^4 + bx^3 + cx^2 + dx + e = 0.$ Divide both sides by a: $x^4 + \frac{b}{a}x^3 + \frac{c}{a}x^2 + \frac{d}{a}x + \frac{e}{a} = 0$ Now, convert to a depressed quartic by substituting $x = y - \frac{b}{4a}$. We now have:

$\left(y - \frac{b}{4a}\right)^4 + \frac{b}{a}\left(y - \frac{b}{4a}\right)^3 + \frac{c}{a}\left(y - \frac{b}{4a}\right)^2 + \frac{d}{a}\left(y - \frac{b}{4a}\right) + \frac{e}{a} = 0$

$y^4 - \left(\frac{b}{a}\right)y^3 + \left(\frac{3b^2}{8a^2}\right)y^2 - \left(\frac{b^3}{16a^3}\right)y + \left(\frac{b^4}{256a^4}\right) + \left(\frac{b}{a}\right)y^3 - \left(\frac{6b^2}{8a^2}\right)y^2 + \left(\frac{3b^3}{16a^3}\right)y - \left(\frac{4b^4}{256a^4}\right) + \left(\frac{8ac}{8a^2}\right)y^2 - \left(\frac{8abc}{16a^3}\right)y$ $+ \left(\frac{16ab^2c}{256a^4}\right) + \left(\frac{16a^2d}{16a^3}\right)y - \left(\frac{64a^2bd}{256a^4}\right) + \left(\frac{256a^3e}{256a^4}\right) = 0$

$y^4 + \left(\frac{8ac - 3b^2}{8a^2}\right)y^2 + \left(\frac{b^3 - 4abc + 8a^2d}{8a^3}\right)y + \left(\frac{-3b^4 + 16ab^2c - 64a^2bd + 256a^3e}{256a^4}\right) = 0$

Now we have a depressed quartic: $y^4 + py^2 + qy + r = 0$ where $p = \left(\frac{8ac - 3b^2}{8a^2}\right)$, $q = \left(\frac{b^3 - 4abc + 8a^2d}{8a^3}\right)$ and $r = \left(\frac{-3b^4 + 16ab^2c - 64a^2bd + 256a^3e}{256a^4}\right)$.

#### TLDR?

The new depressed quartic is $y^4 + py^2 + qy + r = 0$ where $p = \left(\frac{8ac - 3b^2}{8a^2}\right)$, $q = \left(\frac{b^3 - 4abc + 8a^2d}{8a^3}\right)$ and $r = \left(\frac{-3b^4 + 16ab^2c - 64a^2bd + 256a^3e}{256a^4}\right)$.

### Descartes' Solution

René Descartes thought of factoring the depressed quartic into two quadratics: $y^4 + py^2 + qy + r = (y^2 + sy + t)(y^2 + uy + v)$. Expanding the right-hand side gives $y^4 + sy^3 + ty^2 + uy^3 + suy^2 + tuy + vy^2 + svy + tv$, simplifying to $y^4 + (s + u)y^3 + (t + v + su)y^2 + (sv + tu)y + tv$. Equating coefficients gives the following system of equations:

$\begin{cases} s + u = 0 \text{ since the } y^3 \text{ term is 0} \\ p = t + v + su \\ q = sv + tu \\ r = tv \end{cases}$

from which we derive $s = -u$ and substitute this:

$\begin{cases} p + u^2 = t + v \\ q = u(t - v) \\ r = tv \end{cases}$

Now eliminate $t$ and $v$ by doing the following:

\begin{align*} u^2(p + u^2)^2 - q^2 &= u^2(t + v)^2 - u^2(t - v)^2 \text{ by substitution}\\ &= u^2((t + v)^2 - (t - v)^2) \text{ by factoring}\\ &= u^2(t + v + t - v)(t + v - t + v) \text{ by difference of squares}\\ &= u^2(2t)(2v) \\ &= 4u^2tv \\ &= 4u^2r \end{align*}

Substitute $U = u^2$ to get

$U(p + U)^2 - q^2 = 4Ur$

$U^3 + 2pU^2 + p^2U - q^2$

This can be solved via the cubic formula. After $U$ is obtained, we have $u = \sqrt{U}$ and can now solve for $s$, $t$ and $v$:

#### Solve for s

$s = -u$

#### Solve for t and v

We have the system of equations $\begin{cases} p + u^2 = t + v \\ \frac{q}{u} = t - v \end{cases}$. We can obtain $p + u^2 + \frac{q}{u} = 2t$ and $t = \frac{u^3 + pu + q}{2u}$. Similarly, $v = t - \frac{q}{u}$.

Now that both factors have been obtained, we can solve for $y$ by using the quadratic formula on each of the factors. The two solutions for the quadratics combined form the four solutions of the depressed quartic; subtract $\frac{b}{4a}$ to each of the solutions to obtain the solutions to the original quartic.

#### TLDR?

$U$ is a nonzero solution to the cubic $U^3 + 2pU^2 + p^2U - q^2, u = \sqrt{U}, s = -u, t = \frac{u^3 + pu + q}{2u}, v = t - \frac{q}{u}.$ The solutions to the depressed quartic are $\frac{-u \pm \sqrt{u^2 - 4v}}{2} \text{ and } \frac{-s \pm \sqrt{s^2 - 4t}}{2},$ subtract $\frac{b}{4a}$ from each of the roots to obtain the roots of the original quartic.

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