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. Also, if you only want the final results, the "TLDR" subsections give these results.

Solving Quartic Equations

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

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

$U^3 + 2pU^2 + (p^2 - 4r)U - q^2 = 0$

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

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 $U^3 + 2pU^2 + (p^2 - 4r)U - q^2, u = \sqrt{U}, s = -u, t = \frac{u^3 + pu + q}{2u}, v = t - \frac{q}{u}$ (or subtract the two equations to obtain $v = \frac{u^3 + pu - q}{2u}$ ). 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.

The Quartic Formula

Be prepared: This formula is TOTO SLOT really complicated.

I also don't suggest memorizing this formula, since it is too complex to do so. Even if you can, it is very hard to use. You should be better off if you follow the process and break everything into easy steps.

We are going to keep using $p, q,$ and $r$ in the derivation; in the final formula we rewrite it in terms of $a, b,$ and $c.$

So, we start with $y^4 + py^2 + qy + r = 0$ .

We factor it into two quadratics: $(y^2 + sy + t)(y^2 + uy + v)$ .

We have obtained $u^2(p + u^2)^2 - q^2 = 4u^2r$ . With $U$ being a solution to $U^3 + 2pU^2 + (p^2 - 4r)U - q^2$ , $U =$ , according to the cubic formula,

${\tiny{\sqrt[3 \text{ }]{\frac{\left(\frac{2p^3 - 72pr + 27q^2}{27}\right) \pm \sqrt{\frac{3\left(\frac{-2p^3 + 72pr - 27q^2}{9}\right)^2 + 4\left(\frac{p^2 + 12r}{3}\right)^3\left(\frac{-2p^3 + 72pr - 27q^2}{27}\right)}{27}}}{2}} - \sqrt[3 \text{ }]{\frac{\left(\frac{-2p^3 + 72pr - 27q^2}{27}\right) \pm \sqrt{\frac{3\left(\frac{-2p^3 + 72pr - 27q^2}{9}\right)^2 + 4\left(\frac{p^2 + 12r}{3}\right)^3\left(\frac{-2p^3 + 72pr + 27c}{27}\right)}{27}}}{2}} - \frac{2p}{3}}}$

Already messy. Therefore, ${\tiny{u = \sqrt{\sqrt[3 \text{ }]{\frac{\left(\frac{2p^3 - 72pr + 27q^2}{27}\right) \pm \sqrt{\frac{3\left(\frac{-2p^3 + 72pr - 27q^2}{9}\right)^2 + 4\left(\frac{p^2 + 12r}{3}\right)^3\left(\frac{-2p^3 + 72pr - 27q^2}{27}\right)}{27}}}{2}} - \sqrt[3 \text{ }]{\frac{\left(\frac{-2p^3 + 72pr - 27q^2}{27}\right) \pm \sqrt{\frac{3\left(\frac{-2p^3 + 72pr - 27q^2}{9}\right)^2 + 4\left(\frac{p^2 + 12r}{3}\right)^3\left(\frac{-2p^3 + 72pr + 27c}{27}\right)}{27}}}{2}} - \frac{2p}{3}}}}$

${\tiny{s = \sqrt{\sqrt[3 \text{ }]{\frac{\left(\frac{-2p^3 + 72pr - 27q^2}{27}\right) \pm \sqrt{\frac{3\left(\frac{-2p^3 + 72pr - 27q^2}{9}\right)^2 + 4\left(\frac{p^2 + 12r}{3}\right)^3\left(\frac{-2p^3 + 72pr - 27q^2}{27}\right)}{27}}}{2}} - \sqrt[3 \text{ }]{\frac{\left(\frac{2p^3 - 72pr + 27q^2}{27}\right) \pm \sqrt{\frac{3\left(\frac{2p^3 - 72pr + 27q^2}{9}\right)^2 + 4\left(\frac{p^2 + 12r}{3}\right)^3\left(\frac{-2p^3 + 72pr + 27c}{27}\right)}{27}}}{2}} - \frac{2p}{3}}}}$

$t = \tiny{\sqrt[3 \text{ }]{\frac{\left(\frac{2p^3 - 72pr + 27q^2}{27}\right) \pm \sqrt{\frac{3\left(\frac{-2p^3 + 72pr - 27q^2}{9}\right)^2 + 4\left(\frac{p^2 + 12r}{3}\right)^3\left(\frac{-2p^3 + 72pr - 27q^2}{27}\right)}{27}}}{2}} - \sqrt[3 \text{ }]{\frac{\left(\frac{-2p^3 + 72pr - 27q^2}{27}\right) \pm \sqrt{\frac{3\left(\frac{-2p^3 + 72pr - 27q^2}{9}\right)^2 + 4\left(\frac{p^2 + 12r}{3}\right)^3\left(\frac{-2p^3 + 72pr + 27c}{27}\right)}{27}}}{2}} + \frac{p}{3}} + \tiny{\frac{q}{{\tiny{\sqrt{\sqrt[3 \text{ }]{\frac{\left(\frac{2p^3 - 72pr + 27q^2}{27}\right) \pm \sqrt{\frac{3\left(\frac{-2p^3 + 72pr - 27q^2}{9}\right)^2 + 4\left(\frac{p^2 + 12r}{3}\right)^3\left(\frac{-2p^3 + 72pr - 27q^2}{27}\right)}{27}}}{2}} - \sqrt[3 \text{ }]{\frac{\left(\frac{-2p^3 + 72pr - 27q^2}{27}\right) \pm \sqrt{\frac{3\left(\frac{-2p^3 + 72pr - 27q^2}{9}\right)^2 + 4\left(\frac{p^2 + 12r}{3}\right)^3\left(\frac{-2p^3 + 72pr + 27c}{27}\right)}{27}}}{2}} - \frac{2p}{3}}}}}}$

$v = \tiny{\sqrt[3 \text{ }]{\frac{\left(\frac{2p^3 - 72pr + 27q^2}{27}\right) \pm \sqrt{\frac{3\left(\frac{-2p^3 + 72pr - 27q^2}{9}\right)^2 + 4\left(\frac{p^2 + 12r}{3}\right)^3\left(\frac{-2p^3 + 72pr - 27q^2}{27}\right)}{27}}}{2}} - \sqrt[3 \text{ }]{\frac{\left(\frac{-2p^3 + 72pr - 27q^2}{27}\right) \pm \sqrt{\frac{3\left(\frac{-2p^3 + 72pr - 27q^2}{9}\right)^2 + 4\left(\frac{p^2 + 12r}{3}\right)^3\left(\frac{-2p^3 + 72pr + 27c}{27}\right)}{27}}}{2}} + \frac{p}{3}} - \tiny{\frac{q}{{\tiny{\sqrt{\sqrt[3 \text{ }]{\frac{\left(\frac{2p^3 - 72pr + 27q^2}{27}\right) \pm \sqrt{\frac{3\left(\frac{-2p^3 + 72pr - 27q^2}{9}\right)^2 + 4\left(\frac{p^2 + 12r}{3}\right)^3\left(\frac{-2p^3 + 72pr - 27q^2}{27}\right)}{27}}}{2}} - \sqrt[3 \text{ }]{\frac{\left(\frac{-2p^3 + 72pr - 27q^2}{27}\right) \pm \sqrt{\frac{3\left(\frac{-2p^3 + 72pr - 27q^2}{9}\right)^2 + 4\left(\frac{p^2 + 12r}{3}\right)^3\left(\frac{-2p^3 + 72pr + 27c}{27}\right)}{27}}}{2}} - \frac{2p}{3}}}}}}$

$x = \frac{\pm_1 {\tiny{\sqrt{\sqrt[3 \text{ }]{\frac{\left(\frac{2p^3 - 72pr + 27q^2}{27}\right) \pm \sqrt{\frac{3\left(\frac{-2p^3 + 72pr - 27q^2}{9}\right)^2 + 4\left(\frac{p^2 + 12r}{3}\right)^3\left(\frac{-2p^3 + 72pr - 27q^2}{27}\right)}{27}}}{2}} - \sqrt[3 \text{ }]{\frac{\left(\frac{-2p^3 + 72pr - 27q^2}{27}\right) \pm \sqrt{\frac{3\left(\frac{-2p^3 + 72pr - 27q^2}{9}\right)^2 + 4\left(\frac{p^2 + 12r}{3}\right)^3\left(\frac{-2p^3 + 72pr + 27c}{27}\right)}{27}}}{2}} - \frac{2p}{3}}}}\pm_2\sqrt{{\tiny{\sqrt[3 \text{ }]{\frac{\left(\frac{2p^3 - 72pr + 27q^2}{27}\right) \pm \sqrt{\frac{3\left(\frac{-2p^3 + 72pr - 27q^2}{9}\right)^2 + 4\left(\frac{p^2 + 12r}{3}\right)^3\left(\frac{-2p^3 + 72pr - 27q^2}{27}\right)}{27}}}{2}} - \sqrt[3 \text{ }]{\frac{\left(\frac{-2p^3 + 72pr - 27q^2}{27}\right) \pm \sqrt{\frac{3\left(\frac{-2p^3 + 72pr - 27q^2}{9}\right)^2 + 4\left(\frac{p^2 + 12r}{3}\right)^3\left(\frac{-2p^3 + 72pr + 27c}{27}\right)}{27}}}{2}} - \frac{2p}{3}}} + 2p - \frac{2q}{{\tiny{\sqrt{\sqrt[3 \text{ }]{\frac{\left(\frac{2p^3 - 72pr + 27q^2}{27}\right) \pm \sqrt{\frac{3\left(\frac{-2p^3 + 72pr - 27q^2}{9}\right)^2 + 4\left(\frac{p^2 + 12r}{3}\right)^3\left(\frac{-2p^3 + 72pr - 27q^2}{27}\right)}{27}}}{2}} - \sqrt[3 \text{ }]{\frac{\left(\frac{-2p^3 + 72pr - 27q^2}{27}\right) \pm \sqrt{\frac{3\left(\frac{-2p^3 + 72pr - 27q^2}{9}\right)^2 + 4\left(\frac{p^2 + 12r}{3}\right)^3\left(\frac{-2p^3 + 72pr + 27c}{27}\right)}{27}}}{2}} - \frac{2p}{3}}}}}}}{2} - \frac{b}{4a}$

Then we rewrite these rather large expressions in terms of $a, b,$ and $c.$ We simplify the expression and get the quartic formula:

$x=-\frac{b}{4a}\pm\left(\frac{1}{2}\sqrt{\frac{3b^2-8ac}{12a^2}+\frac{1}{3a}\left(\sqrt[3]{\frac{2c^3-9bcd+27b^2e+27ad^2-72ace+\sqrt{\left(2c^3-9bcd+27b^2e+27ad^2-72ace\right)^2-4\left(c^2-3bd+12ae\right)^3}}{2}}+\frac{c^2-3bd+12ae}{\sqrt[3]{\frac{2c^3-9bcd+27b^2e+27ad^2-72ace+\sqrt{\left(2c^3-9bcd+27b^2e+27ad^2-72ace\right)^2-4\left(c^2-3bd+12ae\right)^3}}{2}}}\right)}\right)\pm\frac{1}{2}\sqrt{-4\left(\frac{1}{2}\sqrt{\frac{3b^2-8ac}{12a^2}+\frac{1}{3a}\left(\sqrt[3]{\frac{2c^3-9bcd+27b^2e+27ad^2-72ace+\sqrt{\left(2c^3-9bcd+27b^2e+27ad^2-72ace\right)^2-4\left(c^2-3bd+12ae\right)^3}}{2}}+\frac{c^2-3bd+12ae}{\sqrt[3]{\frac{2c^3-9bcd+27b^2e+27ad^2-72ace+\sqrt{\left(2c^3-9bcd+27b^2e+27ad^2-72ace\right)^2-4\left(c^2-3bd+12ae\right)^3}}{2}}}\right)}\right)^2-\frac{8ac-3b^2}{4a^2}+\frac{\frac{b^3-4abc+8a^2d}{8a^3}}{\frac{1}{2}\sqrt{\frac{3b^2-8ac}{12a^2}+\frac{1}{3a}\left(\sqrt[3]{\frac{2c^3-9bcd+27b^2e+27ad^2-72ace+\sqrt{\left(2c^3-9bcd+27b^2e+27ad^2-72ace\right)^2-4\left(c^2-3bd+12ae\right)^3}}{2}}+\frac{c^2-3bd+12ae}{\sqrt[3]{\frac{2c^3-9bcd+27b^2e+27ad^2-72ace+\sqrt{2c^3-9bcd+27b^2e+27ad^2-72ace^2-4\left(c^2-3bd+12ae\right)^3}}{2}}}\right)}}}$

TLDR

Given the quartic equation $f(x)=ax^4 + bx^3 + cx^2 + dx + e,$ the formula used to get the $4$ roots of $f(x)$ is:

$x=-\frac{b}{4a}\pm\left(\frac{1}{2}\sqrt{\frac{3b^2-8ac}{12a^2}+\frac{1}{3a}\left(\sqrt[3]{\frac{2c^3-9bcd+27b^2e+27ad^2-72ace+\sqrt{\left(2c^3-9bcd+27b^2e+27ad^2-72ace\right)^2-4\left(c^2-3bd+12ae\right)^3}}{2}}+\frac{c^2-3bd+12ae}{\sqrt[3]{\frac{2c^3-9bcd+27b^2e+27ad^2-72ace+\sqrt{\left(2c^3-9bcd+27b^2e+27ad^2-72ace\right)^2-4\left(c^2-3bd+12ae\right)^3}}{2}}}\right)}\right)\pm\frac{1}{2}\sqrt{-4\left(\frac{1}{2}\sqrt{\frac{3b^2-8ac}{12a^2}+\frac{1}{3a}\left(\sqrt[3]{\frac{2c^3-9bcd+27b^2e+27ad^2-72ace+\sqrt{\left(2c^3-9bcd+27b^2e+27ad^2-72ace\right)^2-4\left(c^2-3bd+12ae\right)^3}}{2}}+\frac{c^2-3bd+12ae}{\sqrt[3]{\frac{2c^3-9bcd+27b^2e+27ad^2-72ace+\sqrt{\left(2c^3-9bcd+27b^2e+27ad^2-72ace\right)^2-4\left(c^2-3bd+12ae\right)^3}}{2}}}\right)}\right)^2-\frac{8ac-3b^2}{4a^2}+\frac{\frac{b^3-4abc+8a^2d}{8a^3}}{\frac{1}{2}\sqrt{\frac{3b^2-8ac}{12a^2}+\frac{1}{3a}\left(\sqrt[3]{\frac{2c^3-9bcd+27b^2e+27ad^2-72ace+\sqrt{\left(2c^3-9bcd+27b^2e+27ad^2-72ace\right)^2-4\left(c^2-3bd+12ae\right)^3}}{2}}+\frac{c^2-3bd+12ae}{\sqrt[3]{\frac{2c^3-9bcd+27b^2e+27ad^2-72ace+\sqrt{2c^3-9bcd+27b^2e+27ad^2-72ace^2-4\left(c^2-3bd+12ae\right)^3}}{2}}}\right)}}}$

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Quartic Equation

Contents

Solving Quartic Equations

Bringing it down to a depressed quartic

TLDR

Descartes' Solution

Solve for t and v

TLDR

The Quartic Formula

TLDR

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