# Is there a general formula for solving 4th degree equations?

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
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

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

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

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

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

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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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
@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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