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What are the general solutions to cubic and quartic polynomial equations?Cubic Equations
Let's say you have the equation
Let's introduce two new variables, t and u, defined by the equations u - t = q and tu = (p/3)^{3}.If you're really interested, you might want to know what t and u actually are in terms of p and q: -3qSqrt{3} +- Sqrt{4p^{3} + 27 q^{2}} t = --------------------------------- 6 Sqrt{3} 3qSqrt{3} +- Sqrt{4p^{3} + 27q^{2}} u = -------------------------------- 6 Sqrt{3}That +- is either + or - for BOTH of the equations at the same time. You can't have one as a + and the other as a - .
Anyway, y_{1} = CubeRoot{t} - CubeRoot{u} will be a solution of
Verify this result now (DON'T use the explicit formulas for t and u,
just plug CubeRoot{t} - CubeRoot{u} into the cubic polynomial), and
make sure you see why it works. To find the other two solutions of
our cubic polynomial (if there are any real ones) we could divide
As you can see, the complexity is much greater than in the Quadratic Formula. To try to go backward and come up with a closed form for the Cubic Formula in terms of the original a, b, c, d would be a real pain. For a nice reference about this problem, as well as some information about the Quartic, look at the 3-volume set called Mathematical Thought from Ancient to Modern Times by Morris Kline. It gives the math and the history together. For more about solving cubic polynomial equations, see Another Solution of the Cubic Equation, which describes an approach (submitted independently by Paul A. Torres and Robert A. Warren) based on "completing the cube." Quartic Equations - by Robert L. Ward, for the Math Forum
Simplifying the equationThere are several ways to solve quartic equations. They all start by dividing the equation by the leading coefficient, to make it of the form
x^{4} + a x^{3} + b x^{2} + c x + d = 0. There is an easy special case if d = 0, in which case you can factor the quartic into x times a cubic. The roots then are 0 and the roots of that cubic. (To solve the cubic equation, see above.) Substitute x = y - a/4, expand, and simplify, to get
y^{4} + e y^{2} + f y + g = 0 where
e = b - 3 a^{2}/8, At this point, there are two useful special cases.
Henceforth we can assume that d, f, and g are all nonzero.
First Solution MethodHere is one way to proceed from this point. The related auxiliary cubic equation
z^{3} + (e/2) z^{2} + ((e^{2}-4 g)/16) z - f^{2}/64 = 0 has three roots. Find those roots by solving this cubic (again, see above). None of these is zero because f isn't zero. Let p and q be the square roots of two of those roots (any choices of roots and signs will work), and set
r = -f/(8 p q). Then p^{2}, q^{2}, and r^{2} are the three roots of the above cubic. More important is the fact that the four roots of the original quartic are
x = p + q + r - a/4, This solution was discovered by Leonhard Euler (1707-1783).
Second Solution MethodHere is another way to proceed from the above point. The quartic in y must factor into two quadratics with real coefficients, since any complex roots must occur in conjugate pairs. Write
y^{4} + e y^{2} + f y + g = (y^{2} + h y + j) (y^{2} - h y + g/j) Since f and g aren't zero, neither j nor h is zero either. Without loss of generality, we can assume h > 0 (otherwise swap the two factors). Then, equating coefficients of y^{2} and y,
e = g/j + j - h^{2}, so
g/j + j = e + h^{2}, Adding and subtracting these two equations,
2 g/j = e + h^{2} + f/h, Multiplying these together,
4 g = e^{2} + 2 e h^{2} + h^{4} - f^{2}/h^{2}. Rearranging,
h^{6} + 2 e h^{4} + (e^{2}-4 g) h^{2} - f^{2} = 0. This is a cubic equation in h^{2}, with known coefficients, since we know e, f, and g. We can use the solution for cubic equations shown above to find a positive real value of h^{2}, whose existence is guaranteed by Descartes' Rule of Signs, and then take its positive square root. This value of h will give a value of j from 2 j = e + h^{2} - f/h. Once you know h and j, you know the quadratic factors of the quartic in y. Notice that you do not need all the roots h^{2} of the cubic, and that any positive real one will do. Further notice that h^{2} = 4 z from the previous section. Once you have factored the quartic into two quadratics, finishing the finding of the roots is simple, using the Quadratic Formula. Once the roots y are found, the corresponding x's are gotten from x = y - a/4.
ExamplesExample 1: Find the roots of x^{4} + 6 x^{3} - 5 x^{2} - 10 x - 3 = 0.
y^{4} + (-37/2) y^{2} + 32 y + (-231/16) = 0, Solving this cubic in h^{2}, we find
h^{2} = 16, h^{2} = (21 + sqrt[185])/2, h^{2} = (21 - sqrt[185])/2. We only need one of these, so we pick h^{2} = 16, so
h = 4, and one factorization is
(y^{2} + 4 y - 21/4) (y^{2} - 4 y + 11/4) = 0, or
(x^{2} + 7 x + 3) (x^{2} - x - 1) = 0. From here the rest is easy, and
x = (-7 +- sqrt[37])/2, x = (1 +- sqrt[5])/2, are the four roots of the original equation.
y^{4} + (-13/2) y^{2} + (-1) y + (-15/16) = 0, Solving this cubic in z, we find the roots are
z = (26-cbrt[124 (31-3 sqrt[93])]+cbrt[124 (31+3 sqrt[93])])/24 = 0.005468531191, approximately. Since the algebraic expressions for the roots z are rather complicated, we use numerical approximations from here on. Taking the square roots of the last two quantities, we find
p = 1.2869747589 + 0.1844947037 i, approximately. Then the four roots are
x = p + q + r - a/4, From here the rest is easy,
x = 1.1478990357, x = -1.5739495179 +/- 0.3689894075 i, and x = -4.0000000000, are the four approximate roots of the original equation. It is easy to check that x = -4 is an exact root, and x + 4 divides the original polynomial exactly. Incidentally, the exact expression for the irrational real root is
x = [-4+cbrt(4 [47-3 sqrt(93)])+cbrt(4 [47+3 sqrt(93)])]/6, and there are analogous expressions for the complex conjugate roots which involve w = (-1+sqrt[-3])/2 and w^{2} = (-1-sqrt[-3])/2, the complex cube roots of 1.
For some historical background, see Quadratic, cubic and quartic equations from the MacTutor Math History archives. For derivations and examples, see The Cubic Formula and The Quartic Formula from the S.O.S. Mathematics site.
For an explanation of Cardan's geometric solution method for cubic
equations, see In the Dr. Math archives, see: |
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