Ferrari's Method for QuarticsDate: 04/16/99 at 10:08:35 From: Stephane CATHELAIN Subject: Ferrari's method Could you briefly explain Ferrari's method for quartics or give me the location of a Web site where I can find more information on this method? Thanks. Date: 04/16/99 at 19:54:16 From: Doctor Anthony Subject: Re: Ferrari's method Ferrari (1540), a pupil of Cardano, was given a problem that could be stated in the following form: a+b+c = 10, a/b = b/c, ab = 6 which gives a = 6/b, c = b^2/a = b^3/6 and then 6/b + b + b^3/6 = 10 b^4 + 6b^2 - 60b + 36 = 0 a quartic without a third-degree term, which Ferrari rewrote (b^2 + 6)^2 = 6b^2 + 60b and again after introducing a new variable z (b^2 + 6 + z)^2 = (2z + 6)b^2 + 60b + (12z + z^2)......(1) For the righthand side of the equation to be a perfect square, the discriminant of the quadratic polynomial in b must equal zero. That is 4(2z+6)(z^2 + 12z) = 60^2 and this produces a cubic in z z^3 + 15z^2 + 36z - 450 = 0 By putting z = y-5 we get rid of the squared term and obtain y^3 - 39y - 380 = 0 which is the reducing cubic. From here you proceed to the solution of a cubic. Having solved for y you get the values of z and then the righthand side of (1) is a perfect square. You now take the square root of both sides and have a quadratic in b. There are other methods for dealing with quartics. The following is one such. Example: x^4 - 6x^3 + 15x^2 - 18x + 10 = 0 This is not a general method for a quartic but in the case of all real coefficients we know that it will have two pairs of conjugate roots. In looking for the quadratic factors it is reasonable with the given coefficients to assume that they must be of the form: (x^2 + ax + 5)(x^2 + bx + 2) and multiplying out we get x^4 + x^3(a+b) + x^2(2 + 5 + ab) + x(2a+5b) + 10 x^4 + x^3(a+b) + x^2(7+ab) + x(2a+5b) + 10 compare with x^4 + x^3(-6) + x^2(15) + x(-18) + 10 comparing the coefficients a+b = -6 ab = 8 2a+5b = -18 if a = -4, b = -2 these equations are all satisfied, so the quartic can be factorized into (x^2 -4x +5)(x^2 -2x + 2) = 0 and then solving the equation x^2 - 4x + 5 = 0 gives x = 2 +-i and solving " x^2 - 2x + 2 = 0 gives x = 1 +-i - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/