# Difference between revisions of "Quartic Equation"

Vincentwant (talk | contribs) (Created page with "A quartic equation is an algebraic equation of the form <math>ax^4 + bx^3 + cx^2 + dx + e = 0.</math> These types of equations are extremely hard to solve; however, there ar...") |
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Now you have a depressed quartic: <math>y^4 + py^2 + qy + r = 0</math> where <math>p = \left(\frac{8ac - 3b^2}{8a^2}\right)</math>, <math>q = \left(\frac{b^3 - 4abc + 8a^2d}{8a^3}\right)</math> and <math>r = \left(\frac{-3b^4 + 16ab^2c - 64a^2bd + 256a^3e}{256a^4}\right)</math>. | Now you have a depressed quartic: <math>y^4 + py^2 + qy + r = 0</math> where <math>p = \left(\frac{8ac - 3b^2}{8a^2}\right)</math>, <math>q = \left(\frac{b^3 - 4abc + 8a^2d}{8a^3}\right)</math> and <math>r = \left(\frac{-3b^4 + 16ab^2c - 64a^2bd + 256a^3e}{256a^4}\right)</math>. | ||

− | === | + | ====TLDR?==== |

− | + | The new depressed quartic is | |

+ | <math>y^4 + py^2 + qy + r = 0</math> where <math>p = \left(\frac{8ac - 3b^2}{8a^2}\right)</math>, <math>q = \left(\frac{b^3 - 4abc + 8a^2d}{8a^3}\right)</math> and <math>r = \left(\frac{-3b^4 + 16ab^2c - 64a^2bd + 256a^3e}{256a^4}\right)</math>. | ||

+ | ===Descartes' Solution=== | ||

+ | |||

+ | [https://en.wikipedia.org/wiki/René_Descartes René Descartes] thought of factoring the depressed quartic into two [[quadratic Equations|quadratics]]: <math>y^4 + py^2 + qy + r = (y^2 + sy + t)(y^2 + uy + v)</math>. Expanding the [[RHS|right-hand side]] gives <math>y^4 + sy^3 + ty^2 + uy^3 + suy^2 + tuy + vy^2 + svy + tv</math>, simplifying to <math>y^4 + (s + u)y^3 + (t + v + su)y^2 + (sv + tu)y + tv</math>. [[Equating coefficients]] gives the following [[system of equations]]: | ||

+ | |||

+ | <math>\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}</math> | ||

+ | |||

+ | from which we derive <math>s = -u</math> and [[substitution|substitute]] this: | ||

+ | |||

+ | <math>\begin{cases} p + u^2 = t + v \\ q = u(t - v) \\ r = tv \end{cases}</math> | ||

+ | |||

+ | Now eliminate <math>t</math> and <math>v</math> by doing the following: | ||

+ | |||

+ | <cmath>\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*}</cmath> | ||

+ | |||

+ | Substitute <math>U = u^2</math> to get | ||

+ | |||

+ | <math>U(p + U)^2 - q^2 = 4Ur</math> | ||

+ | |||

+ | <math>U^3 + 2pU^2 + p^2U - q^2</math> | ||

+ | |||

+ | This can be solved via the [[cubic Equation|cubic formula.]] After <math>U</math> is obtained, we have <math>u = \sqrt{u}</math> and can now solve for <math>s</math>, <math>t</math> and <math>v</math>: | ||

+ | |||

+ | ====Solve for s==== | ||

+ | <math>s = -u</math> | ||

+ | |||

+ | ====Solve for t and v==== | ||

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

+ | |||

+ | Now that both factors have been obtained, we can solve for <math>y</math> 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; add <math>\frac{b}{4a}</math> to each of the solutions to obtain the solutions to the original quartic. |

## Revision as of 17:09, 10 December 2020

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 . You now have:

Now you 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; add to each of the solutions to obtain the solutions to the original quartic.