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There is a general formula for solving quadratic equations, namely the Quadratic Formula.

For third degree equations of the form $ax^3+bx^2+cx+d=0$, there is a set of three equations: one for each root.

Is there a general formula for solving equations of the form $ax^4+bx^3+cx^2+dx+e=0$ ?

How about for higher degrees? If not, why not?

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    $\begingroup$ Could you change the title to "Is there a general formula for solving 4th degree polynomial equations" or "Is there a general formula for solving quartic equations?" $\endgroup$ – user126 Aug 5 '10 at 23:40
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    $\begingroup$ Put a*x^4+b*x^3+c*x^2+d*x +e = 0 in wolfram alpha (wolframalpha.com) and then sit back and watch the fireworks! $\endgroup$ – ja72 Aug 16 '11 at 2:00
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    $\begingroup$ The general quartic can be reduced to just solving a quadratic. See this simple answer. $\endgroup$ – Tito Piezas III Apr 4 '15 at 12:50
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    $\begingroup$ See upload.wikimedia.org/wikipedia/commons/thumb/9/99/… $\endgroup$ – Watson Dec 24 '16 at 17:26

13 Answers 13

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There is, in fact, a general formula for solving quartic (4th degree polynomial) equations. As the cubic formula is significantly more complex than the quadratic formula, the quartic formula is significantly more complex than the cubic formula. Wikipedia's article on quartic functions has a lengthy process by which to get the solutions, but does not give an explicit formula.

Beware that in the cubic and quartic formulas, depending on how the formula is expressed, the correctness of the answers likely depends on a particular choice of definition of principal roots for nonreal complex numbers and there are two different ways to define such a principal root.

There cannot be explicit algebraic formulas for the general solutions to higher-degree polynomials, but proving this requires mathematics beyond precalculus (it is typically proved with Galois Theory now, though it was originally proved with other methods). This fact is known as the Abel-Ruffini theorem.

Also of note, Wolfram sells a poster that discusses the solvability of polynomial equations, focusing particularly on techniques to solve a quintic (5th degree polynomial) equation. This poster gives explicit formulas for the solutions to quadratic, cubic, and quartic equations.

edit: I believe that the formula given below gives the correct solutions for x to $ax^4+bx^3+cx^2+d+e=0$ for all complex a, b, c, d, and e, under the assumption that $w=\sqrt{z}$ is the complex number such that $w^2=z$ and $\arg(w)\in(-\frac{\pi}{2},\frac{\pi}{2}]$ and $w=\sqrt[3]{z}$ is the complex number such that $w^3=z$ and $\arg(w)\in(-\frac{\pi}{3},\frac{\pi}{3}]$ (these are typically how computer algebra systems and calculators define the principal roots). Some intermediate parameters $p_k$ are defined to keep the formula simple and to help in keeping the choices of roots consistent.

Let: \begin{align*} p_1&=2c^3-9bcd+27ad^2+27b^2e-72ace \\\\ p_2&=p_1+\sqrt{-4(c^2-3bd+12ae)^3+p_1^2} \\\\ p_3&=\frac{c^2-3bd+12ae}{3a\sqrt[3]{\frac{p_2}{2}}}+\frac{\sqrt[3]{\frac{p_2}{2}}}{3a} \end{align*} $\quad\quad\quad\quad$

\begin{align*} p_4&=\sqrt{\frac{b^2}{4a^2}-\frac{2c}{3a}+p_3} \\\\ p_5&=\frac{b^2}{2a^2}-\frac{4c}{3a}-p_3 \\\\ p_6&=\frac{-\frac{b^3}{a^3}+\frac{4bc}{a^2}-\frac{8d}{a}}{4p_4} \end{align*}

Then: $$\begin{align} x&=-\frac{b}{4a}-\frac{p_4}{2}-\frac{\sqrt{p_5-p_6}}{2} \\\\ \mathrm{or\ }x&=-\frac{b}{4a}-\frac{p_4}{2}+\frac{\sqrt{p_5-p_6}}{2} \\\\ \mathrm{or\ }x&=-\frac{b}{4a}+\frac{p_4}{2}-\frac{\sqrt{p_5+p_6}}{2} \\\\ \mathrm{or\ }x&=-\frac{b}{4a}+\frac{p_4}{2}+\frac{\sqrt{p_5+p_6}}{2} \end{align}$$

(These came from having Mathematica explicitly solve the quartic, then seeing what common bits could be pulled from the horrifically-messy formula into parameters to make it readable/useable.)

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    $\begingroup$ You beat me to the answer by a few seconds :). $\endgroup$ – Akhil Mathew Jul 27 '10 at 18:45
  • $\begingroup$ The computational effort is simplified greatly if you "depress" the quartic (i.e. remove the cubic term through a change of variables) by appealing to Vieta's formulae (the mean of the roots is -b/(4a)) first. If you will notice, all the roots of the original equation have a -b/(4a) term added in to compensate for this preliminary translation. $\endgroup$ – J. M. is a poor mathematician Aug 6 '10 at 0:43
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    $\begingroup$ @J. Mangaldan: Absolutely. My goal (in the edit) was to create a fully-general formula that could be applied straight away; it is not at all illustrative of how to get such a formula. (Your observation is true for the nth-degree polynomial equation formula for n=2, 3, and 4: each formula has a -b/(na) term common to every solution corresponding to the depression.) $\endgroup$ – Isaac Aug 6 '10 at 0:48
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    $\begingroup$ The formula given above is not correct. One can try some examples: x^4-1,x^4+x^2 and x^4+5*x^2+4. $\endgroup$ – user14620 Aug 15 '11 at 15:02
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    $\begingroup$ Wikipedia's article has an explicit formula now. $\endgroup$ – Lee Sleek May 4 '13 at 18:34
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$$x_1=-\frac{b}{4 a}-\frac{1}{2} \sqrt{\frac{\sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}{3 \sqrt[3]{2} a}+\frac{b^2}{4 a^2}-\frac{2 c}{3 a}+\frac{\sqrt[3]{2} \left(c^2-3 b d+12 a e\right)}{3 a \sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}}-\frac{1}{2} \sqrt{-\frac{\sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}{3 \sqrt[3]{2} a}+\frac{b^2}{2 a^2}-\frac{4 c}{3 a}-\frac{\sqrt[3]{2} \left(c^2-3 b d+12 a e\right)}{3 a \sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}-\frac{-\frac{b^3}{a^3}+\frac{4 c b}{a^2}-\frac{8 d}{a}}{4 \sqrt{\frac{\sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}{3 \sqrt[3]{2} a}+\frac{b^2}{4 a^2}-\frac{2 c}{3 a}+\frac{\sqrt[3]{2} \left(c^2-3 b d+12 a e\right)}{3 a \sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}}}}$$

$$x_2=-\frac{b}{4 a}-\frac{1}{2} \sqrt{\frac{\sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}{3 \sqrt[3]{2} a}+\frac{b^2}{4 a^2}-\frac{2 c}{3 a}+\frac{\sqrt[3]{2} \left(c^2-3 b d+12 a e\right)}{3 a \sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}}+\frac{1}{2} \sqrt{-\frac{\sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}{3 \sqrt[3]{2} a}+\frac{b^2}{2 a^2}-\frac{4 c}{3 a}-\frac{\sqrt[3]{2} \left(c^2-3 b d+12 a e\right)}{3 a \sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}-\frac{-\frac{b^3}{a^3}+\frac{4 c b}{a^2}-\frac{8 d}{a}}{4 \sqrt{\frac{\sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}{3 \sqrt[3]{2} a}+\frac{b^2}{4 a^2}-\frac{2 c}{3 a}+\frac{\sqrt[3]{2} \left(c^2-3 b d+12 a e\right)}{3 a \sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}}}}$$

$$x_3=-\frac{b}{4 a}+\frac{1}{2} \sqrt{\frac{\sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}{3 \sqrt[3]{2} a}+\frac{b^2}{4 a^2}-\frac{2 c}{3 a}+\frac{\sqrt[3]{2} \left(c^2-3 b d+12 a e\right)}{3 a \sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}}-\frac{1}{2} \sqrt{-\frac{\sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}{3 \sqrt[3]{2} a}+\frac{b^2}{2 a^2}-\frac{4 c}{3 a}-\frac{\sqrt[3]{2} \left(c^2-3 b d+12 a e\right)}{3 a \sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}+\frac{-\frac{b^3}{a^3}+\frac{4 c b}{a^2}-\frac{8 d}{a}}{4 \sqrt{\frac{\sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}{3 \sqrt[3]{2} a}+\frac{b^2}{4 a^2}-\frac{2 c}{3 a}+\frac{\sqrt[3]{2} \left(c^2-3 b d+12 a e\right)}{3 a \sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}}}}$$

$$x_4=-\frac{b}{4 a}+\frac{1}{2} \sqrt{\frac{\sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}{3 \sqrt[3]{2} a}+\frac{b^2}{4 a^2}-\frac{2 c}{3 a}+\frac{\sqrt[3]{2} \left(c^2-3 b d+12 a e\right)}{3 a \sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}}+\frac{1}{2} \sqrt{-\frac{\sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}{3 \sqrt[3]{2} a}+\frac{b^2}{2 a^2}-\frac{4 c}{3 a}-\frac{\sqrt[3]{2} \left(c^2-3 b d+12 a e\right)}{3 a \sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}+\frac{-\frac{b^3}{a^3}+\frac{4 c b}{a^2}-\frac{8 d}{a}}{4 \sqrt{\frac{\sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}{3 \sqrt[3]{2} a}+\frac{b^2}{4 a^2}-\frac{2 c}{3 a}+\frac{\sqrt[3]{2} \left(c^2-3 b d+12 a e\right)}{3 a \sqrt[3]{2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e+\sqrt{\left(2 c^3-9 b d c-72 a e c+27 a d^2+27 b^2 e\right)^2-4 \left(c^2-3 b d+12 a e\right)^3}}}}}}$$

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    $\begingroup$ I think purpose of maths is to make less effort to arrive the same result. this is the opposite case $\endgroup$ – Fennekin Feb 9 '17 at 17:31
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    $\begingroup$ I've condensed the post by using $\gamma$. (The page was taking a long time to load.) $\endgroup$ – Tito Piezas III Jul 2 '17 at 5:02
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    $\begingroup$ @Fennekin, but it’s still cool just to see how crazy massive it is. $\endgroup$ – Benjamin Thoburn Mar 11 at 18:29
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Edited in response to Quonux's comments.

Yes. As an answer I will use a shorter version of this Portuguese post of mine, where I deduce all the formulae. Suppose you have the general quartic equation (I changed the notation of the coefficients to Greek letters, for my convenience):

$$\alpha x^{4}+\beta x^{3}+\gamma x^{2}+\delta x+\varepsilon =0.\tag{1}$$

If you make the substitution $x=y-\frac{\beta }{4\alpha }$, you get a reduced equation of the form

$$y^{4}+Ay^{2}+By+C=0\tag{2},$$

with

$$A=\frac{\gamma }{\alpha }-\frac{3\beta ^{2}}{8\alpha ^{2}},$$

$$B=\frac{\delta }{\alpha }-\frac{\beta \gamma }{2\alpha ^{2}}+\frac{\beta^3 }{ 8\alpha^3 },$$

$$C=\frac{\varepsilon }{\alpha }-\frac{\beta \delta }{4\alpha ^{2}}+\frac{ \beta ^{2}\gamma}{16\alpha ^{3}}-\frac{3\beta ^{4}}{256\alpha ^{4}}.$$

After adding and subtracting $2sy^{2}+s^{2}$ to the LHS of $(2)$ and rearranging terms, we obtain the equation

$$\underset{\left( y^{2}+s\right) ^{2}}{\underbrace{y^{4}+2sy^{2}+s^{2}}}-\left[ \left( 2s-A\right) y^{2}-By+s^{2}-C\right] =0. \tag{2a}$$

Then we factor the quadratic polynomial $$\left(2s-A\right) y^{2}-By+s^{2}-C=\left(2s-A\right)(y-y_+)(y-y_-)$$ and make $y_+=y_-$, which will impose a constraint on $s$ (equation $(4)$). We will get:

$$\left( y^{2}+s+\sqrt{2s-A}y-\frac{B}{2\sqrt{2s-A}}\right) \left( y^{2}+s- \sqrt{2s-A}y+\frac{B}{2\sqrt{2s-A}}\right) =0,$$ $$\tag{3}$$

where $s$ satisfies the resolvent cubic equation

$$8s^{3}-4As^{2}-8Cs+\left( 4AC-B^{2}\right) =0.\tag{4}$$

The four solutions of $(2)$ are the solutions of $(3)$:

$$y_{1}=-\frac{1}{2}\sqrt{2s-A}+\frac{1}{2}\sqrt{-2s-A+\frac{2B}{\sqrt{2s-A}}}, \tag{5}$$

$$y_{2}=-\frac{1}{2}\sqrt{2s-A}-\frac{1}{2}\sqrt{-2s-A+\frac{2B}{\sqrt{2s-A}}} ,\tag{6}$$

$$y_{3}=\frac{1}{2}\sqrt{2s-A}+\frac{1}{2}\sqrt{-2s-A-\frac{2B}{\sqrt{2s-A}}} ,\tag{7}$$

$$y_{4}=\frac{1}{2}\sqrt{2s-A}-\frac{1}{2}\sqrt{-2s-A-\frac{2B}{\sqrt{2s-A}}} .\tag{8}$$

Thus, the original equation $(1)$ has the solutions $$x_{k}=y_{k}-\frac{\beta }{4\alpha }.\qquad k=1,2,3,4\tag{9}$$

Example: $x^{4}+2x^{3}+3x^{2}-2x-1=0$

$$y^{4}+\frac{3}{2}y^{2}-4y+\frac{9}{16}=0.$$

The resolvent cubic is

$$8s^{3}-6s^{2}-\frac{9}{2}s-\frac{101}{8}=0.$$

Making the substitution $s=t+\frac{1}{4}$, we get

$$t^{3}-\frac{3}{4}t-\frac{7}{4}=0.$$

One solution of the cubic is

$$s_{1}=\left( -\frac{q}{2}+\frac{1}{2}\sqrt{q^{2}+\frac{4p^{3}}{27}}\right) ^{1/3}+\left( -\frac{q}{2}-\frac{1}{2}\sqrt{q^{2}+\frac{4p^{3}}{27}}\right) ^{1/3}-\frac{b}{3a},$$

where $a=8,b=-6,c=-\frac{9}{2},d=-\frac{101}{8}$ are the coefficients of the resolvent cubic and $p=-\frac{3}{4},q=-\frac{7}{4}$ are the coefficients of the reduced cubic. Numerically, we have $s_{1}\approx 1.6608$.

The four solutions are :

$$x_{1}=-\frac{1}{2}\sqrt{2s_{1}-A}+\frac{1}{2}\sqrt{-2s_{1}-A+\frac{2B}{ \sqrt{2s-A}}}-\frac{\beta }{4\alpha },$$

$$x_{2}=-\frac{1}{2}\sqrt{2s_{1}-A}-\frac{1}{2}\sqrt{-2s_{1}-A+\frac{2B}{ \sqrt{2s-A}}}-\frac{\beta }{4\alpha },$$

$$x_{3}=\frac{1}{2}\sqrt{2s_{1}-A}+\frac{1}{2}\sqrt{-2s_{1}-A-\frac{2B}{ \sqrt{2s-A}}}-\frac{\beta }{4\alpha },$$

$$x_{4}=\frac{1}{2}\sqrt{2s_{1}-A}-\frac{1}{2}\sqrt{-2s_{1}-A-\frac{2B}{ \sqrt{2s-A}}}-\frac{\beta }{4\alpha },$$

with $A=\frac{3}{2},B=-4,C=\frac{9}{16}$. Numerically we have $x_{1}\approx -1.1748+1.6393i$, $x_{2}\approx -1.1748-1.6393i$, $x_{3}\approx 0.70062$, $x_{4}\approx -0.35095$.

Another method is to expand the LHS of the quartic into two quadratic polynomials, and find the zeroes of each polynomial. However, this method sometimes fails. Example: $x^{4}-x-1=0$. If we factor $x^{4}-x-1$ as $x^{4}-x-1=\left( x^{2}+bx+c\right) \left( x^{2}+Bx+C\right) $ expand and equate coefficients we will get two equations, one of which is $-1/c-c^{2}\left( 1+c^{2}\right) ^{2}+c=0$. This is studied in Galois theory.

The general quintic is not solvable in terms of radicals, as well as equations of higher degrees.

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  • $\begingroup$ Great answer! Howevern, could you give an example of a quartic for which the method described in the last paragraph doesn't work? $\endgroup$ – Heinz Doofenschmirtz Jul 28 '15 at 8:19
  • $\begingroup$ where did the D = 9/16 come from? I think its an error from an old version? $\endgroup$ – Quonux May 8 '16 at 21:03
  • $\begingroup$ furthermore in your example i don't get -4 for B, i get -4.75 ... -2.0/1.0 - (2.0*3.0)/(2.0*1.0*1.0) + 2.0/(8.0*1.0) $\endgroup$ – Quonux May 8 '16 at 21:18
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    $\begingroup$ because the formula for B is wrong, the last part must be written with a exponent of 3! $\endgroup$ – Quonux May 10 '16 at 6:08
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    $\begingroup$ furthermore, in equation (4) the c must be a big C, in my code i called my input variables a,b,c,d,e, so its the perfect storm $\endgroup$ – Quonux May 10 '16 at 7:43
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There most certainly is, but its ugly, complicated, and not worth memorizing. People know about it and have quoted or cited it for you, but really they would never use it. If you want something actually useful for pen-and-paper solutions, you might want to understand the actual theory behind the solution. I will provide one method below.

The Quartic Formula is just the end result of this methodology, written in terms of the original coefficients. Because of that, the method is far easier to remember than the formula, which is why I find it annoying when people cite just the formula and tell you, "dont bother, use a computer instead." A pen-paper solution is not complicated, just time consuming.

There are three methods for solving Quartics that I both know and know of:

  • Descartes' Quadratic Factorization
  • Euler's Method
  • Ferrari's Method

Ferrari's Method is historically the first method discovered. Euler's Method looks a lot like Cardano's Method for the Cubic. But I am partial to Descartes' Quadratic Factorization technique. Its a relatively simple process to follow and is what I will be using below. If you want to see how the others work, let me know.

All of the above methods start out the same: depression (removing the $n-1$ term, in this case the cubic term) and normalization (making the lead coefficient 1). They all end up roughly the same place too: solving a cubic equation. So you need to be prepared for that.


Depress and Normalize

Given the arbitrary quartic: $$ax^4 + bx^3 + cx^2 + dx + e =0$$ Where $a,b,c,d,e\in\mathbb{R}$.

We must convert this quartic into a depressed monic quartic. First we depress it by substituting $x = z-\frac{b}{4a}$. We arrive at the quartic: $$az^4 + Bz^2 + Cz + D =0$$ For some $B,C,D\in\mathbb{R}$. This quartic, now in $z$, is merely a horizontal shift from the original quartic in $x$. All points have shifted, therefore the roots have shifted, but by the same constant. Also, $B,C,D$ are computed from the previous $a,b,c,d,e$ and have no dependence on $x,z$. Whats important is that there is no $z^3$ term. Next we divide through by the lead coefficient $a$ to make this quartic monic (normalize). $$z^4 + pz^2 + qz + r = 0\;\;\;\;\;\;\;\;\;\;\;\;(1)$$

For some $p,q,r\in\mathbb{R}$. This has no effect on the root positions; all the lead coefficient did was scale up the non-zero values. There is absolutely no loss in generality with respect to the zeros. All of these constants, $p,q,r$, can be computed from the original coefficients, $a,b,c,d,e$. There is still no cubic term and now there is no lead coefficient either.

What might be interesting to note is what has happened to the polynomial. We started out with 5 arbitrary constants and have reduced it to 3, by normalizing the lead and removing the cubic term. We originally had arbitrary values of $a,b,c,d,e\in\mathbb{R}$, and now we have arbitrary values $p,q,r\in\mathbb{R}$. Although the latter three are computed from the original five, they have arbitrary values, and there is no loss in generality. This is a significant simplification of the problem. The non-existence of the cubic term will prove vital.

Everything so far has simply been the setup: writing the polynomial in reduced monic form. Next we implement Descartes Factorization method.


Descartes Factorization Method

We must assume that all of the coefficients are real, $p,q,r\in\mathbb{R}$. This is a required condition to make the methodology work. The reason is because now all solutions with non-zero imaginary components come in complex conjugate pairs. Big deal? It allows us to group two solutions together, even if they are purely real, into quadratic factors with real coefficients. We know that all real-valued monic quartics can be factored into: $$(z^2 + mz+n)(z^2+sz+t)=0\;\;\;\;\;\;\;\;\;\;\;\;(2)$$ Where $m,n,s,t\in\mathbb{R}$. Clearly the task at hand is to find the values of these constants. If we can convert (1) into (2) by determining quadratic factors that satisfies the equation, we can find the roots with the quadratic formula applied to those quadratic factors.

What we do is distribute these quadratic factors out into normal polynomial form and you get: $$z^4 + (m+s)z^3 + (t+n+ms)z^2 + (mt+ns)z + nt = 0$$ Which we compare term-by-term to the depressed monic in (1). You get this system of equations: $$ m+s=0 \\ n+t+ms=p \\mt+ns=q \\ nt=r$$ Its plain to see then that $s = -m$ and that $t=\frac{r}{n}$ from the first and fourth equations. Our quadratic factors can be rewritten: $$(z^2 + mz+n)(z^2-mz+\frac{r}{n})=0\;\;\;\;\;\;\;\;\;\;\;\;(3)$$ We only have two unknowns in this factorization: $m$ and $n$. The remaining two of the four simultaneous equations can be rewritten: $$ \frac{r}{n}+n-m^2 = p \\ m(\frac{r}{n} -n)=q$$ By moving the $m$ to the right-hand side: $$ \frac{r}{n}+n = p+m^2 \\ \frac{r}{n} -n=\frac{q}{m}$$

From here we form two new equations by adding and subtracting the previous two. By adding we get: $$ 2\frac{r}{n} = p+m^2 + \frac{q}{m}$$ By subtracting we get: $$ 2n = p+m^2 -\frac{q}{m}$$ Notice that both of these equations can be solved for $n$ and $\frac{r}{n}$ readily in terms of $m$, both of which appear in the quadratic factors of (3). These can be utilized later once we know the $m$ to complete the quadratic factor.

We can find the $m$ by taking these latest two equations and multiplying them, therefore eliminating the unknown $n$. Notice that $n$ is in the numerator in one and in the denominator in the other. Thus: $$ 4r = (p+m^2)^2 - (\frac{q}{m})^2 = m^4 + 2pm^2 + p^2 - \frac{q^2}{m^2}$$

Thus we are essentially down to $m$ as our last unknown. Everything else is known in terms of $m$, and $m$ is the only unknown in the above equation. One unknown, one remaining equation. Multiply the equation through by $m^2$ and rearrange: $$ m^6 + 2pm^4 + (p^2-4r)m^2 - q^2 = 0$$ This is still rather ugly. It's a sextic, not a quartic, which is worse. But notice that the powers of $m$ are all even. We can substitute $w=m^2$ and arrive at: $$ w^3 + 2pw^2 + (p^2 - 4r)w - q^2 =0$$

And thus we are essentially done. We are left with a cubic polynomial in $w$, which is solvable with its own techniques. Techniques which I only assume you already know about if you are trying to solve quartics. Just like with quartics, as you know already, there do exist cubic formulae, but I do recommend learning the methods behind them.

If you need help with cubics, I recommend Cardano's method (the original solution) or Vieta's Trigonometric Solution (my preferred). There is also Completing the Cube, a nice proof of concept but Id never use it. Feel free to ask a separate question for a cubic and I will be happy to answer.

The point is that the problem has been reduced from that of finding the roots of a quartic to that of finding the roots of a cubic. A simpler problem! That's usually how it goes. All quartic root finding methods require finding the roots of a cubic first, obvious or not. Just as finding the roots of a cubic involve solving a quadratic. Hope this works for you.

Anyway, solve the cubic in $w$. Then with that recall the quadratic substitution $w=m^2$ and solve for $m$. Then recall that we previously had equations for $ 2n$ and $2\frac{r}{n}$. And now you know all of the unknown terms in the quadratic factorization $(z^2 + mz + n)(z^2 - mz + \frac{r}{n})=0$.

Still not done. Each of these quadratic factors must now be solved using the quadratic formula, and you have solutions in $z$. This solves the depressed monic quartic we started Descartes Quadratic Factorization method with.


Finally

Dont forget about the original quartic we had at the very beginning, prior to the depression and normalization. We had introduced a horizontal shift of $x = z-\frac{b}{4a}$. Doing this last bit will solve the original quartic in terms of $x$, which is the solution you want.

You are going to arrive at a set of solutions when done. Be sure to check your answers. You may have redundant or superfluous solutions.

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Yes, there is a quartic formula.

There is no such solution by radicals for higher degrees. This is a result of Galois theory, and follows from the fact that the symmetric group $S_5$ is not solvable. It is called Abel's theorem. In fact, there are specific fifth-degree polynomials whose roots cannot be obtained by using radicals from $\mathbb{Q}$.

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We can reduce the problem of solving the general quartic to merely solving a quadratic. Given,

$$x^4+ax^3+bx^2+cx+d=0$$

Then the four roots are,

$$x_{1,2} = -\frac{a}{4}+\frac{\color{red}\pm\sqrt{u}}{2}\color{blue}+\frac{1}{4}\sqrt{3a^2-8b-4u+\frac{-a^3+4ab-8c}{\color{red}\pm\sqrt{u}}}\tag1$$

$$x_{3,4} = -\frac{a}{4}+\frac{\color{red}\pm\sqrt{u}}{2}\color{blue}-\frac{1}{4}\sqrt{3a^2-8b-4u+\frac{-a^3+4ab-8c}{\color{red}\pm\sqrt{u}}}\tag2$$

where,

$$u = \frac{a^2}{4}-\frac{2b}{3} +\frac{1}{3}\left(v_1^{1/3}\zeta_3+\frac{b^2 - 3 a c + 12 d}{v_1^{1/3}\zeta_3}\right)\tag{3a}$$

or alternatively,

$$u = \frac{a^2}{4}-\frac{2b}{3} +\frac{1}{3}\left(\color{blue}{v_1^{1/3}+v_2^{1/3}}\right)\tag{3b}$$

with $v_1$ any non-zero root of the quadratic,

$$v^2 + (-2 b^3 + 9 a b c - 27 c^2 - 27 a^2 d + 72 b d)v + (b^2 - 3 a c + 12 d)^3 = 0$$

and a chosen cube root of unity $\zeta_3^3 = 1$ such that $u$ is also non-zero. (Normally, use $\zeta_3=1$, but not when $a^3-4ab+8c = 0$.)

P.S. This is essentially the method used by Mathematica, though much simplified for aesthetics.

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  • $\begingroup$ I noticed that by defining $x=(v_1+v_2)/2$ and $y=(v_1-v_2)/2$ one can rewrite $v_1^{1/3}+v_2^{1/3}=2(b^2-3ac+12d)^{1/2}\cos\left(\frac{2}{3}\arctan\left(((x^2+y^2)^{1/2}-x)/y\right)\right)$ - which is true at least if the square root in $v_i$ turns out imaginary. Now I wonder if there is any trigonometric identity that can resolve the $\cos(2/3\arctan(...))$ into something rational involving roots? $\endgroup$ – Kagaratsch Sep 11 '15 at 1:06
  • $\begingroup$ @Status: No, it does not imply $x^4=0$. I've clarified the post. If $v_1, v_2$ are zero, then the expression in blue in $(3b)$ is also zero, but one can still use this $u$ to find the roots you mentioned. (The case when $u, v_1, v_2$ are all zero is degenerate and the quartic has 4 identical roots.) To explain the two forms, the advantage of $(3a)$ is there is only one complex cube root $v_1^{1/3}$, while the advantage of $(3b)$ is there is no variable denominator (hence no division by zero). $\endgroup$ – Tito Piezas III Sep 21 '17 at 12:00
  • $\begingroup$ Solving the quartic requires the resolution of a cubic (itself requiring the resolution of a quadratic and possibly trigonometric functions), then of two quadratics. This is provably unavoidable. $\endgroup$ – Yves Daoust Oct 26 '17 at 18:42
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What has not been mentioned thus far is that one can in fact use any number of "auxiliary cubics" in the solution of the quartic equation. Don Herbison-Evans, in this page (Wayback link, as the original page has gone kaput), which was adapted from his technical report, mention five such possible auxiliary cubics.

Given the quartic equation

$$x^4 + ax^3 + bx^2 + cx + d = 0$$

the five possible auxiliary cubics are referred to in the document as

Christianson-Brown:

$$y^3 +\frac{4a^2b - 4b^2 - 4ac + 16d - \frac34a^4}{a^3 - 4ab + 8c}y^2 + \left(\frac3{16}a^2 - \frac{b}{2}\right)y - \frac{1}{64}(a^3 + 4a b - 8c) = 0$$

Descartes-Euler-Cardano:

$$y^3 + \left(2b - \frac34 a^2\right)y^2 + \left(\frac3{16}a^4 - a^2b + ac + b^2 - 4d\right)y - \frac{1}{64}(a^3 + 4a b - 8c)^2 = 0$$

Ferrari-Lagrange

$$y^3 + by^2 + (ac - 4d)y + a^2d + c^2 - 4bd = 0$$

Neumark

$$y^3 - 2by^2 + (ac + b^2 - 4d)y + a^2d - abc + c^2 = 0$$

Yacoub-Fraidenraich-Brown

$$(a^3 - 4ab + 8c)y^3 + (a^2b - 4b^2 + 2ac + 16d)y^2 + (a^2c - 4bc + 8ad)y + a^2d - c^2 = 0 $$

See the page for how to obtain the quadratics that will yield the solutions to the original quartic equation from a root of the auxiliary cubic.

Let me also mention this old ACM algorithm in Algol. Netlib has a C implementation of that algorithm.

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  • $\begingroup$ That page is no longer accessible. However, Don Herbison-Evans' paper, "Finding Real Roots of Quartics" can be found here. $\endgroup$ – Tito Piezas III Mar 24 '16 at 1:15
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    $\begingroup$ @Tito, luckily the Wayback Machine still has it! $\endgroup$ – J. M. is a poor mathematician Jan 8 '17 at 15:09
  • $\begingroup$ Interesting content in that article. Thanks. Im wondering if you are aware that some of these are not really unique. If you depress the quartic and make $a=0$ then the YFB and CB equations are identical, and the DEC and N are horizontal reflections of one another. After a simple depression there are only three unique equations to concern yourself with. $\endgroup$ – CogitoErgoCogitoSum Jul 1 '17 at 2:39
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Regarding the inability to solve the quintic, this is sort-of true and sort-of false. No, there is no general solution in terms of $+$, $-$, $\times$ and $\div$, along with $\sqrt[n]{}$. However, if you allow special theta values (a new operation, not among the standard ones!) then yes, you can actually write down the solutions of arbitrary polynomials this way. Also, you can do construct lengths equal to the solutions by intersecting lower degree curves (for a quintic, you can do so with a trident and a circle.)

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    $\begingroup$ "However, if you allow special theta values (a new operation, not among the standard ones!) then yes, you can actually write down the solutions of arbitrary polynomials this way." Do you know of a reference which expands on this point? $\endgroup$ – Zach Conn Dec 8 '10 at 20:35
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    $\begingroup$ @Zach: Umemura shows how to use (multidimensional) theta functions for representing roots of polynomials. $\endgroup$ – J. M. is a poor mathematician Aug 16 '11 at 4:47
  • $\begingroup$ On the topic of solving the quintic I recommend the paper by Peter Doyle and Curt McMullen, "Solving the quintic by iteration", Acta Math. 163 (1989), no. 3-4, 151–180. $\endgroup$ – Lee Mosher Jul 1 '17 at 23:31
  • $\begingroup$ @Lee, as the paper's title itself notes, the method presented is an iterative method, and would not furnish a symbolic solution. $\endgroup$ – J. M. is a poor mathematician Nov 14 '17 at 4:16
  • $\begingroup$ In the spirit of this answer, I'm only pointing out (and hence it was just a comment) another method in the topic of "solving" polynomial equations. $\endgroup$ – Lee Mosher Nov 14 '17 at 13:17
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To your question, "if not, why not", I could say this (which I think goes a little further that the previous answers). A polynomial $p(x) \in \mathbb{Q}[x]$ of degree $n$ has a Galois group $G = G(p)$ attached to it. This is a subgroup of the symmetric group $S_{n},$ and this identification comes about because $G$ is a group of permutations of the roots of $p(x)$. The equation $p(x)$ is solvable by radicals if and only if $G$ is a solvable group, which is a key theorem of Galois theory. When $n \leq 4$, the symmetric group $S_n$ is itself a solvable group, so all its subgroups are solvable, and the group $G$ must be solvable. As remarked in earlier answers, when $n \geq 5,$ the group $S_n$ is never solvable. This does not mean that the polynomial $p(x)$ is never solvable by radicals, but (depending what its Galois group is), we can not be sure that it is (and there are explicit examples when it is not for each $n \geq 5$). I do make the point though that the solvability or otherwise of $p(x)$ by radicals really depends on the factorization of $p(x)$ as a product of irreducible polynomials, rather than just the degree of $p(x)$. If $p(x)$ is a product of irreducible polynomials which each have degree at most 4, then its Galois group is solvable, so $p(x)$ is solvable by radicals.

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Contrary to common opinion, the resolution of the quartic with real coeficients is not so difficult.

I take for granted that depression and reduction of the equation by a linear change of variable

$$\alpha x^4+\beta x^3+\gamma x^2+\delta x+\epsilon=0\to x^4+px^2+qx+r=0$$ are known.

Then we factor this depressed quadrinomial as

$$(x^2+ax+b)(x^2-ax+c)=x^4+(-a^2+b+c)x^2+a(c-b)x+bc$$ and identify the coefficients.

We can express $b$ and $c$ in terms of $a^2$ and form their product:

$$c-b=\frac qa,\\b+c=p+a^2,\\4r=4bc=4\left(p+a^2-\frac qa\right)\left(p+a^2+\frac qa\right)=(p^2+a^2)^2-\frac{q^2}{a^2}.$$

This is a cubic equation in $a^2$, and fortunately it always has a positive solution.

Then after resolution, from $a$ we obtain $b$ and $c$, and the roots of the two quadratic trinomials.

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  • $\begingroup$ @ Yves: May I kindly ask for a more detailed explanation, since $a$, $b$, $c$ in the quartic and the deflated quadrinomial are different? $\endgroup$ – justik Aug 13 '18 at 15:59
  • $\begingroup$ @justik: I have updated the notation. $\endgroup$ – Yves Daoust Aug 13 '18 at 16:29
  • $\begingroup$ @ Yves: Thanks for the updated notation. A linear change represents here the substitution $x=y-b/4$? Does it lead to the next substitution $z=a^2$, where $f(z)=4r$ is the cubic for $z$? I am having some problems with the quartic factorization in my question. Maybe, you will find this question interesting :-). Thanks. $\endgroup$ – justik Aug 13 '18 at 17:08
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Yes, there is such a formula as the "Quartic Formula." However, it is too complicated for practical use, so if someone needs to find a root to one of these equations, one may use the Rational Root Theorem, Factor Theorem. If the roots are approximated, then one would use Newton's Method (studied in calculus.) Newton's method is how commercial and scientific computer programs approximate roots of all equations (not just polynomials).

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If anyone is interested, there is a quartic equation calculator with a method and complete steps to solving the roots in the link here.

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  • $\begingroup$ Which method would be easiest to implement given that one is interested in real roots only? Does that simplify things at all? $\endgroup$ – Jim White Aug 29 '17 at 11:42
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There is up to fourth grade. From there, no. The demonstration that there can not be is Group Theory and it is something strong.

The general formula for solving 4th grade equations, too.

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