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