# Difference between revisions of "Quartic Equation"

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<math>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}</math> | <math>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}</math> | ||

− | + | Then we substitute each of the <math>p, q, r</math>s for those rather large expressions. I'm not going to write it all out, but you get how complicated it would be. |

## Latest revision as of 15:38, 18 January 2021

A **quartic equation** is an algebraic equation of the form

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.

## Contents

## 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 Divide both sides by a: Now, convert to a depressed quartic by substituting . We now have:

Now we have a depressed quartic: where , and .

#### TLDR?

The new depressed quartic is where , and .

### Descartes' Solution

René Descartes thought of factoring the depressed quartic into two quadratics: . Expanding the right-hand side gives , simplifying to . Equating coefficients gives the following system of equations:

from which we derive and substitute this:

Now eliminate and by doing the following:

Substitute to get

This can be solved via the cubic formula. After is obtained, we have and can now solve for , and :

#### Solve for s

#### Solve for t and v

We have the system of equations . We can obtain and . Similarly, .

Now that both factors have been obtained, we can solve for 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 to each of the solutions to obtain the solutions to the original quartic.

#### TLDR?

is a nonzero solution to the cubic (or subtract the two equations to obtain ). The solutions to the depressed quartic are subtract from each of the roots to obtain the roots of the original quartic.

#### Roots Rewritten

The roots are (s must be the same sign).

## The Quartic Formula

Be prepared: This formula is **really complicated.**

We are going to keep using in the derivation; we are going to substitute them into the final formula.

So, we start with .

We factor it into two quadratics: .

We have obtained . With being a solution to , , according to the cubic formula,

Already messy. Therefore,

Then we substitute each of the s for those rather large expressions. I'm not going to write it all out, but you get how complicated it would be.