From: Michael Jørgensen Subject: Re: 4th degree equation Date: Fri, 08 Oct 1999 11:26:12 +0200 Newsgroups: sci.math To: carel I quite like the following alternate solution: The equation x^4 + p*x^2 + q*x + r = 0 has the solution 2*x = sqrt(z1) + sqrt(z2) + sqrt(z3), where z1, z2, and z3 are the three solutions of the cubic equation z^3 + 2*p*z^2 + (p^2-4*r)*z - q^2 = 0. You have to be careful when choosing the signs of the three sqrt's in the expression for 2*x: They must satisfy the following additional constraint: sqrt(z1) * sqrt(z2) * sqrt(z3) = -q. There are a total of four different combinations of signs that satisfy this constraint. This in turn leads to the four solutions of the original quartic! I believe this solution is attributed to Faucette, due to an article in Amer. Math. Monthly (vol 103) in 1996. -Michael. P.S. To prove the above formula, you start by noticing that z1 = (x2+x3)^2, z2 = (x1+x3)^2, and z3 = (x1+x2)^2, where x1, x2, and x3 are three of the four roots of the original quartic. The fourth root, x4, satisfies x4=-(x1+x2+x3), because there is no cubic term (x^3) in the original quartic equation. carel wrote: > I dont know how others did it, but I used the general solution of the third > and complex roots to solve the fourth. > > See attachment. > > Carel > > karouri@my-deja.com wrote in article <7thlf1$a7e$1@nnrp1.deja.com>... > > Simply put, I would like to know what is the algebraic exact solution > > of the general fourth degree equation? If possible, I would like to > > know the history of the discovery too. > > > > > > Sent via Deja.com http://www.deja.com/ > > Before you buy. > > > > Name: Quadric.doc > Quadric.doc Type: Microsoft Word Document (application/msword) > Encoding: x-uuencode