Difference between revisions of "Quartic Equation"
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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== | ||
− | + | ||
===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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===Descartes' Solution=== | ===Descartes' Solution=== | ||
− | [ | + | [[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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==The Quartic Formula== | ==The Quartic Formula== | ||
− | Be prepared: This formula is <u>'''''really complicated.'''''</u> | + | 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> |
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. | 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 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>. | ||
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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 | + | 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> | ||
==External Links== | ==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
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.
Contents
Solving Quartic Equations
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 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.
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 and in the derivation; in the final formula we rewrite it in terms of and
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 rewrite these rather large expressions in terms of and We simplify the expression and get the quartic formula:
TLDR
Given the quartic equation the formula used to get the roots of is: