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Quartic Equation


A quartic equation is a fourth-order polynomial equation of the form

 z^4+a_3z^3+a_2z^2+a_1z+a_0=0.
(1)

While some authors (Beyer 1987b, p. 34) use the term "biquadratic equation" as a synonym for quartic equation, others (Hazewinkel 1988, Gellert et al. 1989) reserve the term for a quartic equation having no cubic term, i.e., a quadratic equation in x^2.

Ferrari was the first to develop an algebraic technique for solving the general quartic, which was stolen and published in Cardano's Ars Magna in 1545 (Boyer and Merzbach 1991, p. 283). The Wolfram Language can solve quartic equations exactly using the built-in command Solve[a4 x^4 + a3 x^3 + a2 x^2 + a1 x + a0 == 0, x]. The solution can also be expressed in terms of Wolfram Language algebraic root objects by first issuing SetOptions[Roots, Quartics -> False].

The roots of this equation satisfy Vieta's formulas:

x_1+x_2+x_3+x_4=-a_3
(2)
x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4=a_2
(3)
x_1x_2x_3+x_2x_3x_4+x_1x_2x_4+x_1x_3x_4=-a_1
(4)
x_1x_2x_3x_4=a_0,
(5)

where the denominators on the right side are all a_4=1. Writing the quartic in the standard form

 x^4+px^2+qx+r=0,
(6)

the properties of the symmetric polynomials appearing in Vieta's formulas then give

z_1^2+z_2^2+z_3^2+z_4^2=-2p
(7)
z_1^3+z_2^3+z_3^3+z_4^3=-3q
(8)
z_1^4+z_2^4+z_3^4+z_4^4=2p^2-4r
(9)
z_1^5+z_2^5+z_3^5+z_4^5=5pq.
(10)

Eliminating p, q, and r, respectively, gives the relations

 z_1z_2(p+z_1^2+z_1z_2+z_2^2)-r=0
(11)
 z_1^2z_2(z_1+z_2)-qz_1-r=0
(12)
 q+pz_2+z_2^3=0,
(13)

as well as their cyclic permutations.

Ferrari was the first to develop an algebraic technique for solving the general quartic. He applied his technique (which was stolen and published by Cardano) to the equation

 x^4+6x^2-60x+36=0
(14)

(Smith 1994, p. 207).

The x^3 term can be eliminated from the general quartic (◇) by making a substitution of the form

 z=x-lambda,
(15)

so

 x^4+(a_3-4lambda)x^3+(a_2-3a_3lambda+6lambda^2)x^2 
 +(a_1-2a_2lambda+3a_3lambda^2-4lambda^3)x+(a_0-a_1lambda+a_2lambda^2-a_3lambda^3+lambda^4).
(16)

Letting lambda=a_3/4 so

 z=x-1/4a_3
(17)

then gives the standard form

 x^4+px^2+qx+r=0,
(18)

where

p=a_2-3/8a_3^2
(19)
q=a_1-1/2a_2a_3+1/8a_3^3
(20)
r=a_0-1/4a_1a_3+1/(16)a_2a_3^2-3/(256)a_3^4.
(21)

The quartic can be solved by writing it in a general form that would allow it to be algebraically factorable and then finding the condition to put it in this form. The equation that must be solved to make it factorable is called the resolvent cubic. To do this, note that the quartic will be factorable if it can be written as the difference of two squared terms,

 P^2-Q^2=(P+Q)(P-Q).
(22)

It turns out that a factorization of this form can be obtained by adding and subtracting x^2u+u^2/4 (where u is for now an arbitrary quantity, but which will be specified shortly) to equation (◇) to obtain

 (x^4+x^2u+1/4u^2)-x^2u-1/4u^2+px^2+qx+r=0.
(23)

This equation can be rewritten

 (x^2+1/2u)^2-[(u-p)x^2-qx+(1/4u^2-r)]=0
(24)

(Birkhoff and Mac Lane 1966). Note that the first term is immediately a perfect square P^2 with

 P=x^2+1/2u,
(25)

and the second term will be a perfect square Q^2 if u is chosen to that the square can be completed in

 Q^2=(u-p)(x^2-q/(u-p)x+(1/4u^2-r)/(u-p)).
(26)

This means we want

 Q^2=(u-p)(x-sqrt((1/4u^2-r)/(u-p)))^2
(27)

which requires that

 2sqrt((1/4u^2-r)/(u-p))=q/(u-p),
(28)

or

 q^2=4(u-p)(1/4u^2-r).
(29)

This is the resolvent cubic.

Since an analytic solution to the cubic is known, we can immediately solve algebraically for one of the three solution of equation (29), say u_1, and plugging equation (29) into equation (26) then gives

 Q=Ax-q/(2A)
(30)

with

 A=sqrt(u_1-p).
(31)

Q therefore is linear in x and P is quadratic in x, so each term P+Q and P-Q is quadratic and can be solved using the quadratic formula, thus giving all four solutions to the original quartic.

Explicitly, plugging p, q, and r back into (◇) gives

 u^3+(3/8a_3^2-a_2)u^2+(3/(64)a_3^4-1/4a_2a_3^2+a_1a_3-4a_0)u+(1/(512)a_3^6-1/(64)a_2a_3^4+1/8a_1a_3^3-3/2a_0a_3^2+4a_0a_2-a_1^2).
(32)

This can be simplified by making the substitution

 u=y-1/8a_3^2,
(33)

which gives the resolvent cubic equation

 y^3-a_2y^2+(a_1a_3-4a_0)y+(4a_2a_0-a_1^2-a_3^2a_0)=0.
(34)

Let y_1 be a real root of (34), then the four roots of the original quartic are given by the roots of the equation

 x^2+1/2(a_3+/-sqrt(a_3^2-4a_2+4y_1))x+1/2(y_1+/-sqrt(y_1^2-4a_0))=0,
(35)

which are

z_1=-1/4a_3+1/2R+1/2D
(36)
z_2=-1/4a_3+1/2R-1/2D
(37)
z_3=-1/4a_3-1/2R+1/2E
(38)
z_4=-1/4a_3-1/2R-1/2E,
(39)

where

R=sqrt(1/4a_3^2-a_2+y_1)
(40)
D={sqrt(3/4a_3^2-R^2-2a_2+1/4(4a_3a_2-8a_1-a_3^3)R^(-1)) for R!=0; sqrt(3/4a_3^2-2a_2+2sqrt(y_1^2-4a_0)) for R=0
(41)
E={sqrt(3/4a_3^2-R^2-2a_2-1/4(4a_3a_2-8a_1-a_3^3)R^(-1)) for R!=0; sqrt(3/4a_3^2-2a_2-2sqrt(y_1^2-4a_0)) for R=0
(42)

(Abramowitz and Stegun 1972, p. 17; Beyer 1987a, p. 12). This result is sometimes known as the quartic formula.

Another approach to solving the quartic (◇) defines

alpha=(x_1+x_2)(x_3+x_4)=-(x_1+x_2)^2
(43)
beta=(x_1+x_3)(x_2+x_4)=-(x_1+x_3)^2
(44)
gamma=(x_1+x_4)(x_2+x_3)=-(x_2+x_3)^2,
(45)

where the second forms follow from

 x_1+x_2+x_3+x_4=-a_3=0,
(46)

and defining

h(x)=(x-alpha)(x-beta)(x-gamma)
(47)
=x^3-(alpha+beta+gamma)x^2+(alphabeta+alphagamma+betagamma)x-alphabetagamma.
(48)

This equation can be written in terms of the original coefficients p, q, and r as

 h(x)=x^3-2px^2+(p^2-4r)x+q^2.
(49)

The roots of this cubic equation then give alpha, beta, and gamma, and the equations (◇) to (◇) can be solved for the four roots x_i of the original quartic (Faucette 1996).


See also

Biquadratic Equation, Cubic Equation, Polynomial Discriminant, Quadratic Equation, Quartic Formula, Quintic Equation

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References

Abramowitz, M. and Stegun, I. A. (Eds.). "Solutions of Quartic Equations." §3.8.3 in Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, pp. 17-18, 1972.Berger, M. §16.4.1-16.4.11.1 in Geometry I. New York:Springer-Verlag, 1987.Beyer, W. H. CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, p. 12, 1987a.Beyer, W. H. Handbook of Mathematical Sciences, 6th ed. Boca Raton, FL: CRC Press, 1987b.Birkhoff, G. and Mac Lane, S. A Survey of Modern Algebra, 5th ed. New York: Macmillan, pp. 107-108, 1996.Borwein, P. and Erdélyi, T. "Quartic Equations." §1.1.E.1e in Polynomials and Polynomial Inequalities. New York:Springer-Verlag, p. 4, 1995.Boyer, C. B. and Merzbach, U. C. A History of Mathematics, 2nd ed. New York: Wiley, pp. 286-287, 1991.Ehrlich, G. §4.16 in Fundamental Concepts of Abstract Algebra. Boston, MA: PWS-Kent, 1991.Faucette, W. M. "A Geometric Interpretation of the Solution of the General Quartic Polynomial." Amer. Math. Monthly 103, 51-57, 1996.Gellert, W.; Gottwald, S.; Hellwich, M.; Kästner, H.; and Künstner, H. (Eds.). VNR Concise Encyclopedia of Mathematics, 2nd ed. New York: Van Nostrand Reinhold, 1989.Hazewinkel, M. (Managing Ed.). Encyclopaedia of Mathematics: An Updated and Annotated Translation of the Soviet "Mathematical Encyclopaedia." Dordrecht, Netherlands: Reidel, 1988.MathPages. "Reducing Quartics to Cubics." http://www.mathpages.com/home/kmath296.htm.Smith, D. E. A Source Book in Mathematics. New York: Dover, 1994.van der Waerden, B. L. §64 in Algebra, Vol. 1. New York:Springer-Verlag, 1993.

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Quartic Equation

Cite this as:

Weisstein, Eric W. "Quartic Equation." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/QuarticEquation.html

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