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Quartic Equation

A general quartic equation (also called a Biquadratic Equation) is a fourth-order Polynomial of the form

z^4+a_3z^3+a_2z^2+a_1z+a_0 = 0.
\end{displaymath} (1)

The Roots of this equation satisfy Newton's Relations:
$ x_1+x_2+x_3+x_4 = -a_3\quad$ (2)
$ x_1x_2+x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4 = a_2\quad$ (3)
$ x_1x_2x_3+x_2x_3x_4+x_1x_2x_4+x_1x_3x_4 = -a_1\quad$ (4)
$ x_1x_2x_3x_4 = a_0,\quad$ (5)
where the denominators on the right side are all $a_4\equiv 1$.

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

\end{displaymath} (6)

(Smith 1994, p. 207).

The $x^3$ term can be eliminated from the general quartic (1) by making a substitution of the form

z\equiv x-\lambda,
\end{displaymath} (7)

$ +(a_0-a_1\lambda+a_2\lambda^2-a_3\lambda^3+\lambda^4).\quad$ (8)
Letting $\lambda=a_3/4$ so
z\equiv x-{\textstyle{1\over 4}}a_3
\end{displaymath} (9)

then gives
\end{displaymath} (10)

$\displaystyle p$ $\textstyle \equiv$ $\displaystyle a_2-{\textstyle{3\over 8}}{a_3}^2$ (11)
$\displaystyle q$ $\textstyle \equiv$ $\displaystyle a_1-{\textstyle{1\over 2}}a_2a_3+{\textstyle{1\over 8}}{a_3}^3$ (12)
$\displaystyle r$ $\textstyle \equiv$ $\displaystyle a_0-{\textstyle{1\over 4}}a_1a_3+{\textstyle{1\over 16}}a_2{a_3}^2-{\textstyle{3\over 256}}{a_3}^4.$ (13)

Adding and subtracting $x^2u+u^2/4$ to (10) gives

(x^4+x^2u+{\textstyle{1\over 4}}u^2)-x^2u-{\textstyle{1\over 4}}u^2+px^2+qx+r=0,
\end{displaymath} (14)

which can be rewritten
(x^2+{\textstyle{1\over 2}}u)^2-[(u-p)x^2-qx+({\textstyle{1\over 4}}u^2-r)]=0
\end{displaymath} (15)

(Birkhoff and Mac Lane 1965). The first term is a perfect square $P^2$, and the second term is a perfect square $Q^2$ for those $u$ such that
q^2=4(u-p)({\textstyle{1\over 4}}u^2-r).
\end{displaymath} (16)

This is the resolvent Cubic, and plugging a solution $u_1$ back in gives
\end{displaymath} (17)

so (15) becomes
(x^2+{\textstyle{1\over 2}}u_1+Q)(x^2+{\textstyle{1\over 2}}u_1-Q),
\end{displaymath} (18)

$\displaystyle Q$ $\textstyle \equiv$ $\displaystyle Ax-B$ (19)
$\displaystyle A$ $\textstyle \equiv$ $\displaystyle \sqrt{u_1-p}$ (20)
$\displaystyle B$ $\textstyle \equiv$ $\displaystyle -{q\over 2A}.$ (21)

Let $y_1$ be a Real Root of 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.
\end{displaymath} (22)

The four Roots are then given by the Roots of the equation

x^2+{\textstyle{1\over 2}}(a_3\pm \sqrt{{a_3}^2-4a_2}+4y_1)+{\textstyle{1\over 2}}(y_1\mp\sqrt{{y_1}^2-4a_0}\,) = 0,
\end{displaymath} (23)

which are
$\displaystyle z_1$ $\textstyle =$ $\displaystyle -{\textstyle{1\over 4}}a_3 +{\textstyle{1\over 2}}R+{\textstyle{1\over 2}}D$ (24)
$\displaystyle z_2$ $\textstyle =$ $\displaystyle -{\textstyle{1\over 4}}a_3 +{\textstyle{1\over 2}}R-{\textstyle{1\over 2}}D$ (25)
$\displaystyle z_3$ $\textstyle =$ $\displaystyle -{\textstyle{1\over 4}}a_3 -{\textstyle{1\over 2}}R+{\textstyle{1\over 2}}E$ (26)
$\displaystyle z_4$ $\textstyle =$ $\displaystyle -{\textstyle{1\over 4}}a_3 -{\textstyle{1\over 2}}R-{\textstyle{1\over 2}}E,$ (27)

$\displaystyle R$ $\textstyle \equiv$ $\displaystyle \sqrt{{\textstyle{1\over 4}}{a_3}^2-a_2+y_1}$ (28)
$\displaystyle D$ $\textstyle \equiv$ $\displaystyle \left\{\begin{array}{ll} \sqrt{{\textstyle{3\over 4}}{a_3}^2-R^2-...
...{a_3}^2-2a_2+2\sqrt{{y_1}^2-4a_0}}& \mbox{}\\  \ R=0& \mbox{}\end{array}\right.$  
$\displaystyle E$ $\textstyle \equiv$ $\displaystyle \left\{\begin{array}{ll} \sqrt{{\textstyle{3\over 4}}{a_3}^2-R^2-...
...a_3}^2-2a_2-2\sqrt{{y_1}^2-4a_0}}& \mbox{}\\  \ R=0.& \mbox{}\end{array}\right.$  

Another approach to solving the quartic (10) defines

$\displaystyle \alpha$ $\textstyle =$ $\displaystyle (x_1+x_2)(x_3+x_4)=-(x_1+x_2)^2$ (31)
$\displaystyle \beta$ $\textstyle =$ $\displaystyle (x_1+x_3)(x_2+x_4)=-(x_1+x_3)^2$ (32)
$\displaystyle \gamma$ $\textstyle =$ $\displaystyle (x_1+x_4)(x_2+x_3)=-(x_2+x_3)^2,$ (33)

where use has been made of
\end{displaymath} (34)

(which follows since $a_3=0$), and
$\displaystyle h(x)$ $\textstyle =$ $\displaystyle (x-\alpha)(x-\beta)(x-\gamma)$ (35)
  $\textstyle =$ $\displaystyle x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\alpha\gamma+\beta\gamma)x-\alpha\beta\gamma.$  

Comparing with

$\displaystyle P(x)$ $\textstyle =$ $\displaystyle x^3+px^2+qx+r$ (37)
  $\textstyle =$ $\displaystyle (x-x_1)(x-x_2)(x-x_3)(x-x_4)$ (38)
  $\textstyle =$ $\displaystyle x^3+\left({\,\prod_{i\not=j}^4 x_ix_j}\right)x^2+(x_1+x_2)(x_1+x_3)(x_2+x_3)x$  
  $\textstyle \phantom{=}$ $\displaystyle -x_1x_2x_3(x_1+x_2+x_3),$ (39)

\end{displaymath} (40)

Solving this Cubic Equation gives $\alpha$, $\beta$, and $\gamma$, which can then be solved for the roots of the quartic $x_i$ (Faucette 1996).

See also Cubic Equation, Discriminant (Polynomial), Quintic Equation


Abramowitz, M. and Stegun, C. A. (Eds.). Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 9th printing. New York: Dover, pp. 17-18, 1972.

Berger, M. §16.4.1- in Geometry I. New York: Springer-Verlag, 1987.

Beyer, W. H. CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, p. 12, 1987.

Birkhoff, G. and Mac Lane, S. A Survey of Modern Algebra, 3rd ed. New York: Macmillan, pp. 107-108, 1965.

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.

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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© 1996-9 Eric W. Weisstein