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Difference between revisions of "Quartic Equation"

m (S has already been solved; we don't need to solve for it again.)
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<math>ax^4 + bx^3 + cx^2 + dx + e = 0.</math>
 
<math>ax^4 + bx^3 + cx^2 + dx + e = 0.</math>
  
−
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 Equation|cubic]]. I am going to list the simplest of the five.
+
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 Equation|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==
 
==Solving Quartic Equations==
−
Look in the "TLDR" subsections for the final results of each step.
+
 
 
===Bringing it down to a depressed quartic===
 
===Bringing it down to a depressed quartic===
 
Start with the equation <math>ax^4 + bx^3 + cx^2 + dx + e = 0.</math>
 
Start with the equation <math>ax^4 + bx^3 + cx^2 + dx + e = 0.</math>
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Now we 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 we 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>.
  
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====TLDR?====
+
====TLDR====
  
 
The new depressed quartic is  
 
The new depressed quartic is  
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===Descartes' Solution===
 
===Descartes' Solution===
  
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[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 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]]:
+
[[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 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>
 
<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>
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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; subtract <math>\frac{b}{4a}</math> to each of the solutions to obtain the solutions to the original quartic.
 
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; subtract <math>\frac{b}{4a}</math> to each of the solutions to obtain the solutions to the original quartic.
  
−
====TLDR?====
+
====TLDR====
 
<math>U</math> is a nonzero solution to the cubic <math>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}</math> (or subtract the two equations to obtain <math>v = \frac{u^3 + pu - q}{2u}</math>). The solutions to the depressed quartic are <math>\frac{-u \pm \sqrt{u^2 - 4v}}{2} \text{ and } \frac{-s \pm \sqrt{s^2 - 4t}}{2},</math>  
 
<math>U</math> is a nonzero solution to the cubic <math>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}</math> (or subtract the two equations to obtain <math>v = \frac{u^3 + pu - q}{2u}</math>). The solutions to the depressed quartic are <math>\frac{-u \pm \sqrt{u^2 - 4v}}{2} \text{ and } \frac{-s \pm \sqrt{s^2 - 4t}}{2},</math>  
 
subtract <math>\frac{b}{4a}</math> from each of the roots to obtain the roots of the original quartic.
 
subtract <math>\frac{b}{4a}</math> from each of the roots to obtain the roots of the original quartic.
  
−
====Roots Rewritten====
+
==The Quartic Formula==
−
The roots are <math>\frac{\pm_1 u \pm_2 \sqrt{u^2 + 2p \pm_1 \frac{2q}{u}}}{2}</math> (<math>\pm_n</math>s must be the same sign).
 
  
−
==The Quartic Formula==
+
Be prepared: This formula is [https://artofproblemsolving.com/wiki/index.php/TOTO_SLOT_:_SITUS_TOTO_SLOT_MAXWIN_TERBAIK_DAN_TERPERCAYA TOTO SLOT] <u>'''''really complicated.'''''</u>
  
−
Be prepared: This formula is <u>'''''really complicated.'''''</u>
+
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 <math>p, q, r</math> in the derivation; we are going to substitute them into the final formula.
+
We are going to keep using <math>p, q,</math> and <math>r</math> in the derivation; in the final formula we rewrite it in terms of <math>a, b,</math> and <math>c.</math>
  
 
So, we start with <math>y^4 + py^2 + qy + r = 0</math>.
 
So, we start with <math>y^4 + py^2 + qy + r = 0</math>.
Line 93: Line 92:
 
<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 and simplify.
+
Then we rewrite these rather large expressions in terms of <math>a, b,</math> and <math>c.</math> We simplify the expression and get the quartic formula:
 +
 
 +
<math>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)}}}</math>
 +
 
 +
===TLDR===
 +
 
 +
Given the quartic equation <math>f(x)=ax^4 + bx^3 + cx^2 + dx + e,</math> the formula used to get the <math>4</math> roots of <math>f(x)</math> is:
  
 
<math>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)}}}</math>
 
<math>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)}}}</math>
−
https://www.quora.com/What-is-the-general-formula-for-quartic-equation
+
 
 +
==External Links==
 +
[https://www.quora.com/What-is-the-general-formula-for-quartic-equation Quora]
 +
 
 +
[https://en.wikipedia.org/wiki/Quartic_function Wikipedia]
 +
 
 +
==See Also==
 +
 
 +
[[Cubic Equation]]
 +
 
 +
[[Quadratic Formula]]

Latest revision as of 14:56, 23 October 2024

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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See Also

Cubic Equation

Quadratic Formula

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