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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 thee 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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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?" –  user126 Aug 5 '10 at 23:40
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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! –  ja72 Aug 16 '11 at 2:00

9 Answers 9

up vote 44 down vote accepted

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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You beat me to the answer by a few seconds :). –  Akhil Mathew Jul 27 '10 at 18:45
    
+1, I like the poster a lot. thx for the link. –  Chao Xu Jul 27 '10 at 18:59
    
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. –  J. M. Aug 6 '10 at 0:43
    
@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.) –  Isaac Aug 6 '10 at 0:48
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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

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

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

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

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 $A=\frac{3}{2},B=-4,C=\frac{9}{16},D=\frac{9}{16}$. 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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Right under equation (2) you have $$C=\frac{\varepsilon }{\alpha }-\frac{\beta \delta }{4\alpha ^{2}}+\frac{ \beta ^{2}c}{16\alpha ^{3}}-\frac{3\beta ^{4}}{256\alpha ^{4}}.$$ What does the little $c$ stand for? –  Fmonkey2001 Sep 30 at 1:14
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@Fmonkey2001 Thanks! It should be $\gamma$ instead of $c$. Fixed. –  Américo Tavares Sep 30 at 11:58

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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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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"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? –  Zach Conn Dec 8 '10 at 20:35
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@Zach: Umemura shows how to use (multidimensional) theta functions for representing roots of polynomials. –  J. M. Aug 16 '11 at 4:47

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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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 (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:

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

Descartes-Euler-Cardano:

$$y^3 + \left(2b - \frac34 a^2\right)y^2 + \left(\frac3{16}a^4 - a^2b + ac + b^2 - 4d\right)y + abc - \frac{a^6}{64} + \frac{a^4b}{8} - \frac{a^3c}{4} - \frac{a^2b^2}{4} - c^2 = 0$$

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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http://planetmath.org/encyclopedia/QuarticFormula.html This has the whole thing written out... Might take a few hours to actually input it.

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Basically, there is a formula for the quartic equation. There is also great process to use as well explained by MrYouMath on Youtube. I started making a calculator using Microsoft Excel to program this method. Yet sometimes you won't need to learn that long and tedious formula to solve certain quartic equations. If the quartic is in the form of ax^4+bx^2+c=0 we can substitute g=x^2 to get a quadratic. Then we take the square root of the solutions. Don't forget that x^2=b has 2 solutions, the positive and the negative (Although the square root of B considers only the positive one). However for some reason the general form of an polynomial equation with degree p, can only be solved when p is the member of the set {1,2,3,4}. Yet we can solve the quintic by graphing it. We choose one of the x-intercepts ( let's call it "r" ) and divide the quantic by x-r. This gives us a quartic equation which we can solve no matter what the new coefficients are.

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Here is a MathJax tutorial: meta.math.stackexchange.com/questions/5020/… . Please rewrite your answer. –  Ludolila Oct 19 at 12:39

I am a fan of Descartes' Quadratic Factorization technique. Relatively simple process to follow. Given the quartic: $$ax^4 + bx^3 + cx^2 + dx + e =0$$ We may convert this quartic into a depressed monic quartic. First we depress it by substituting $x = z-\frac{b}{4a}$. We arrive at the quartic: $$az^4 + Bz^2 + Cz + D =0$$ This quartic, now in $z$, is merely a horizontal shift from the original quartic in $x$. Next we divide through by the lead coefficient $a$ to make this quartic monic. $$z^4 + pz^2 + qz + r = 0$$

There is absolutely no loss in generality. All of these constants are computed from the original coefficients.

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 the depressed monic quartic can be factored into: $$(z^2 + mz+n)(z^2+sz+t)=0$$ Where $m,n,s,t\in\mathbb{R}$.

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. 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: $$(z^2 + mz+n)(z^2-mz+\frac{r}{n})=0$$ 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 this second equation can be solved for $n$ readily in terms of $m$. This can be utilized later when we know the $m$. We can find the $m$ by taking these latest two equations and multiplying them and thus eliminating the unknown $n$: $$ 4r = (p+m^2)^2 - (\frac{q}{m})^2 = m^4 + 2pm^2 + p^2 - \frac{q^2}{m^2}$$

Multiply this through by $m^2$ and rearrange: $$ m^6 + 2pm^4 + (p^2-4r)m^2 - q^2 = 0$$ This is still rather ugly, 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. The problem has been reduced from that of finding the roots of a quartic to that of finding the roots of a cubic. In fact, all quartic root finding methods require finding the roots of a cubic first, whether its plainly obvious or not, just as finding the roots of a cubic involve solving a quadratic. Probably not the answer you were looking for.

Anyway, solve the cubic in $w$. Then with that recall the quadratic substitution $w=m^2$ and solve for $m$. Then recall that $ n = \frac{1}{2}(p+m^2) -\frac{q}{2m}$. 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 $z$. But dont forget about the original quartic... we had a horizontal shift $x = z-\frac{b}{4a}$.

Be sure to check your answers. You may have redundant or superfluous solutions.

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