In mathematics, a quartic equation is one which can be expressed as a quartic function equaling zero. The general form of a quartic equation is
$ax^{4}+bx^{3}+cx^{2}+dx+e=0\,$
where a ≠ 0.
The quartic is the highest order polynomial equation that can be solved by radicals in the general case (i.e., one in which the coefficients can take any value).
Historyedit
Lodovico Ferrari is attributed with the discovery of the solution to the quartic in 1540, but since this solution, like all algebraic solutions of the quartic, requires the solution of a cubic to be found, it could not be published immediately.^{[1]} The solution of the quartic was published together with that of the cubic by Ferrari's mentor Gerolamo Cardano in the book Ars Magna (1545).
The proof that this was the highest order general polynomial for which such solutions could be found was first given in the Abel–Ruffini theorem in 1824, proving that all attempts at solving the higher order polynomials would be futile. The notes left by Évariste Galois before his death in a duel in 1832 later led to an elegant complete theory of the roots of polynomials, of which this theorem was one result.^{[2]}
Solving a quartic equation, special casesedit
Consider a quartic equation expressed in the form $a_{0}x^{4}+a_{1}x^{3}+a_{2}x^{2}+a_{3}x+a_{4}=0$:
There exists a general formula for finding the roots to quartic equations, provided the coefficient of the leading term is non-zero. However, since the general method is quite complex and susceptible to errors in execution, it is better to apply one of the special cases listed below if possible.
If the constant term a_{4} = 0, then one of the roots is x = 0, and the other roots can be found by dividing by x, and solving the resulting cubic equation,
$a_{0}x^{3}+a_{1}x^{2}+a_{2}x+a_{3}=0.\,$
Evident roots: 1 and −1 and −kedit
Call our quartic polynomial Q(x). Since 1 raised to any power is 1,
$Q(1)=a_{0}+a_{1}+a_{2}+a_{3}+a_{4}\ .$
Thus if $\ a_{0}+a_{1}+a_{2}+a_{3}+a_{4}=0\ ,$Q(1) = 0 and so x = 1 is a root of Q(x). It can similarly be shown that if $\ a_{0}+a_{2}+a_{4}=a_{1}+a_{3}\ ,$x = −1 is a root.
In either case the full quartic can then be divided by the factor (x − 1) or (x + 1) respectively yielding a new cubic polynomial, which can be solved to find the quartic's other roots.
If $\ a_{1}=a_{0}k\ ,$$\ a_{2}=0\$ and $\ a_{4}=a_{3}k\ ,$ then $\ x=-k\$ is a root of the equation. The full quartic can then be factorized this way:
Alternatively, if $\ a_{1}=a_{0}k\ ,$$\ a_{3}=a_{2}k\ ,$ and $\ a_{4}=0\ ,$ then x = 0 and x = −k become two known roots. Q(x) divided by x(x + k) is a quadratic polynomial.
Biquadratic equationsedit
A quartic equation where a_{3} and a_{1} are equal to 0 takes the form
$a_{0}x^{4}+a_{2}x^{2}+a_{4}=0\,\!$
and thus is a biquadratic equation, which is easy to solve: let $z=x^{2}$, so our equation turns to
$a_{0}z^{2}+a_{2}z+a_{4}=0\,\!$
which is a simple quadratic equation, whose solutions are easily found using the quadratic formula:
$z^{2}+(a_{1}/a_{0})z+(a_{2}/a_{0}-2m)=0$ (a quadratic in z = x + m/x)
Multiple rootsedit
If the quartic has a double root, it can be found by taking the polynomial greatest common divisor with its derivative. Then they can be divided out and the resulting quadratic equation solved.
In general, there exist only four possible cases of quartic equations with multiple roots, which are listed below:^{[3]}
Multiplicity-4 (M4): when the general quartic equation can be expressed as $a(x-l)^{4}=0$, for some real number$l$. This case can always be reduced to a biquadratic equation.
Multiplicity-3 (M3): when the general quartic equation can be expressed as $a(x-l)^{3}(x-m)=0$, where $l$ and $m$ are a couple of two different real numbers. This is the only case that can never be reduced to a biquadratic equation.
Double Multiplicity-2 (DM2): when the general quartic equation can be expressed as $a(x-l)^{2}(x-m)^{2}=0$, where $l$ and $m$ are a couple of two different real numbers or a couple of non-real complex conjugate numbers. This case can also always be reduced to a biquadratic equation.
Single Multiplicity-2 (SM2): when the general quartic equation can be expressed as $a(x-l)^{2}(x-m)(x-n)=0$, where $l$, $m$, and $n$ are three different real numbers or $l$ is a real number and $m$ and $n$ are a couple of non-real complex conjugate numbers. This case is divided into two subcases, those that can be reduced to a biquadratic equation and those in which this is impossible.
So, if the three non-monic coefficients of the depressed quartic equation, $x^{4}+px^{2}+qx+r=0$, in terms of the five coefficients of the general quartic equation are given as follows: $p={\frac {8ac-3b^{2}}{8a^{2}}}$, $q={\frac {b^{3}-4abc+8a^{2}d}{8a^{3}}}$ and $r={\frac {16ab^{2}c-64a^{2}bd-3b^{4}+256a^{3}e}{256a^{4}}}$, then the criteria to identify a priori each case of quartic equations with multiple roots and their respective solutions are exposed below.
M4. The general quartic equation corresponds to this case whenever $p=q=r=0$, so the four roots of this equation are given as follows: $x_{1}=x_{2}=x_{3}=x_{4}=-{\frac {b}{4a}}$.
M3. The general quartic equation corresponds to this case whenever $p^{2}=-12r>0$ and $27q^{2}=-8p^{3}>0$, so the four roots of this equation are given as follows: $x_{1}=x_{2}=x_{3}={\sqrt {-{\frac {p}{6}}}}-{\frac {b}{4a}}$ and $x_{4}=-{\sqrt {-{\frac {3p}{2}}}}-{\frac {b}{4a}}$ , whether $q>0$; otherwise, $x_{1}=x_{2}=x_{3}=-{\sqrt {-{\frac {p}{6}}}}-{\frac {b}{4a}}$ and $x_{4}={\sqrt {-{\frac {3p}{2}}}}-{\frac {b}{4a}}$.
DM2. The general quartic equation corresponds to this case whenever $p^{2}=4r>0=q$, so the four roots of this equation are given as follows: $x_{1}=x_{3}={\sqrt {-{\frac {p}{2}}}}-{\frac {b}{4a}}$ and $x_{2}=x_{4}=-{\sqrt {-{\frac {p}{2}}}}-{\frac {b}{4a}}$.
Biquadratic SM2. The general quartic equation corresponds to this subcase of the SM2 equations whenever $p\neq q=r=0$, so the four roots of this equation are given as follows: $x_{1}=x_{2}=-{\frac {b}{4a}}$, $x_{3}={\sqrt {-p}}-{\frac {b}{4a}}$ and $x_{4}=-{\sqrt {-p}}-{\frac {b}{4a}}$.
Non-Biquadratic SM2. The general quartic equation corresponds to this subcase of the SM2 equations whenever $(p^{2}+12r)^{3}=[p(p^{2}-36r)+{\frac {27}{2}}q^{2}]^{2}>0\neq {q}$, so the four roots of this equation are given by the following formula:^{[4]}$x={\frac {1}{2}}\left[\xi {\sqrt {s_{1}}}\pm {\sqrt {2{\biggl (}s_{2}-{\frac {\xi q}{\sqrt {s_{1}}}}{\biggr )}}}\right]-{\frac {b}{4a}}$, where $s_{1}={\frac {9q^{2}-32pr}{p^{2}+12r}}>0$, $s_{2}=-{\frac {2p(p^{2}-4r)+9q^{2}}{2(p^{2}+12r)}}\neq 0$ and $\xi =\pm 1$.
The general caseedit
To begin, the quartic must first be converted to a depressed quartic.
Converting to a depressed quarticedit
Let
$\ Ax^{4}+Bx^{3}+Cx^{2}+Dx+E=0\$
(1')
be the general quartic equation which it is desired to solve. Divide both sides by A,
If $\ b=0\$ then we have the special case of a biquadratic equation, which is easily solved, as explained above. Note that the general solution, given below, will not work for the special case $\ b=0\ .$ The equation must be solved as a biquadratic.
In either case, once the depressed quartic is solved for u, substituting those values into
$\ x=u-{B \over 4A}\$
produces the values for x that solve the original quartic.
Solving a depressed quartic when b ≠ 0edit
After converting to a depressed quartic equation
$u^{4}+au^{2}+bu+c=0$
and excluding the special case b = 0, which is solved as a biquadratic, we assume from here on that b ≠ 0 .
We will separate the terms left and right as
$u^{4}=-au^{2}-bu-c$
and add in terms to both sides which make them both into perfect squares.
Using another y from among the three roots of the cubic simply causes these same four values of u to appear in a different order. The solutions of the cubic are:
using any one of the three possible cube roots. A wise strategy is to choose the sign of the square-root that makes the absolute value of w as large as possible.
Otherwise, the depressed quartic can be solved by means of a method discovered by Lodovico Ferrari. Once the depressed quartic has been obtained, the next step is to add the valid identity
The effect has been to fold up the u^{4} term into a perfect square: (u^{2} + a)^{2}. The second term, au^{2} did not disappear, but its sign has changed and it has been moved to the right side.
The next step is to insert a variable y into the perfect square on the left side of equation (2), and a corresponding 2y into the coefficient of u^{2} in the right side. To accomplish these insertions, the following valid formulas will be added to equation (2),
The objective now is to choose a value for y such that the right side of equation (3) becomes a perfect square. This can be done by letting the discriminant of the quadratic function become zero. To explain this, first expand a perfect square so that it equals a quadratic function:
The quadratic function on the right side has three coefficients. It can be verified that squaring the second coefficient and then subtracting four times the product of the first and third coefficients yields zero:
This is a cubic equation in y. Solve for y using any method for solving such equations (e.g. conversion to a reduced cubic and application of Cardano's formula). Any of the three possible roots will do.
Folding the second perfect squareedit
With the value for y so selected, it is now known that the right side of equation (3) is a perfect square of the form
(This is correct for both signs of square root, as long as the same sign is taken for both square roots. A ± is redundant, as it would be absorbed by another ± a few equations further down this page.)
Remember: The two $\pm _{s}$ come from the same place in equation (5'), and should both have the same sign, while the sign of $\pm _{t}$ is independent.
Summary of Ferrari's methodedit
Given the quartic equation
$Ax^{4}+Bx^{3}+Cx^{2}+Dx+E=0,\,$
its solution can be found by means of the following calculations:
The two ±_{s} must have the same sign, the ±_{t} is independent. To get all roots, compute x for ±_{s},±_{t} = +,+ and for +,−; and for −,+ and for −,−. This formula handles repeated roots without problem.
Ferrari was the first to discover one of these labyrinthine solutions^{[citation needed]}. The equation which he solved was
$x^{4}+6x^{2}-60x+36=0$
which was already in depressed form. It has a pair of solutions which can be found with the set of formulas shown above.
Ferrari's solution in the special case of real coefficientsedit
If the coefficients of the quartic equation are real then the nested depressed cubic equation (5) also has real coefficients, thus it has at least one real root.
where P and Q are given by (5) has the properties that
$C\left({a \over 3}\right)={-b^{2} \over 8}<0$ and
$\lim _{v\to \infty }C(v)=\infty ,$
where a and b are given by (1).
This means that (5) has a real root greater than $a \over 3$,
and therefore that (4) has a real root greater than $-a \over 2$.
Using this root the term ${\sqrt {a+2y}}$ in (8) is always real,
which ensures that the two quadratic equations (8) have real coefficients.^{[5]}
Obtaining alternative solutions the hard wayedit
It could happen that one only obtained one solution through the formulae above, because not all four sign patterns are tried for four solutions, and the solution obtained is complex. It may also be the case that one is only looking for a real solution. Let x_{1} denote the complex solution. If all the original coefficients A, B, C, D and E are real—which should be the case when one desires only real solutions – then there is another complex solution x_{2} which is the complex conjugate of x_{1}. If the other two roots are denoted as x_{3} and x_{4} then the quartic equation can be expressed as
$(x-x_{1})(x-x_{2})(x-x_{3})(x-x_{4})=0,\,$
but this quartic equation is equivalent to the product of two quadratic equations:
One of these two solutions should be the desired real solution.
Alternative methodsedit
Quick and memorable solution from first principlesedit
Most textbook solutions of the quartic equation require a substitution that is hard to memorize. Here is an approach that makes it easy to understand. The job is done if we can factor the quartic equation into a product of two quadratics. Let
This is harder to solve than it looks, but if we start again with a depressed quartic where $b=0$, which can be obtained by substituting $(x-b/4)$ for $x$, then $r=-p$, and:
The symmetries in this solution are easy to see. There are three roots of the cubic, corresponding to the three ways that a quartic can be factored into two quadratics, and choosing positive or negative values of $p$ for the square root of $P$ merely exchanges the two quadratics with one another.
Galois theory and factorizationedit
The symmetric groupS_{4} on four elements has the Klein four-group as a normal subgroup. This suggests using a resolvent whose roots may be variously described as a discrete Fourier transform or a Hadamard matrix transform of the roots.
Suppose r_{i} for i from 0 to 3 are roots of
then since the transformation is an involution, we may express the roots in terms of the four s_{i} in exactly the same way. Since we know the value s_{0} = −b/2, we really only need the values for s_{1}, s_{2} and s_{3}. These we may find by expanding the polynomial
This polynomial is of degree six, but only of degree three in z^{2}, and so the corresponding equation is solvable. By trial we can determine which three roots are the correct ones, and hence find the solutions of the quartic.
We can remove any requirement for trial by using a root of the same resolvent polynomial for factoring; if w is any root of (3), and if
We therefore can solve the quartic by solving for w and then solving for the roots of the two factors using the quadratic formula.
Approximate methodsedit
The methods described above are, in principle, exact root-finding methods. It is also possible to use successive approximation methods which iteratively converge towards the roots, such as the Durand–Kerner method. Iterative methods are the only ones available for quintic and higher-order equations, beyond trivial or special cases.
^Chávez-Pichardo, Mauricio; Martínez-Cruz, Miguel A.; Trejo-Martínez, Alfredo; Martínez-Carbajal, Daniel; Arenas-Resendiz, Tanya (July 2022). "A Complete Review of the General Quartic Equation with Real Coefficients and Multiple Roots". Mathematics. 10 (14): 2377. doi:10.3390/math10142377. ISSN 2227-7390.
^Chávez-Pichardo, Mauricio; Martínez-Cruz, Miguel A.; Trejo-Martínez, Alfredo; Vega-Cruz, Ana Beatriz; Arenas-Resendiz, Tanya (March 2023). "On the Practicality of the Analytical Solutions for all Third- and Fourth-Degree Algebraic Equations with Real Coefficients". Mathematics. 11 (6): 1447. doi:10.3390/math11061447. ISSN 2227-7390.
^Carstensen, Jens, Komplekse tal, First Edition, (Systime 1981), ISBN 87-87454-71-8. (in Danish)