Welcome to NRICH.

 
Quartic Equations
By Ian Sealey on Thursday, November 15, 2001 - 02:42 pm:

I came across a quartic equation while working on a solution to a physics problem, does anyone know how to solve them? Thanks.

By Kerwin Hui on Thursday, November 15, 2001 - 02:52 pm:

Have a look in AskedNRICH - Window cleaning problem.

Kerwin

By Henry Sealey on Thursday, November 15, 2001 - 03:47 pm:

Try this site:
http://members.tripod.com/l_ferrari/quartic_equation.htm

By Arun Iyer on Thursday, November 15, 2001 - 06:07 pm:

This site will surely cater to your needs.....
http://www.sosmath.com/algebra/factor/fac12/fac12.html

love arun

By Yatir Halevi on Friday, November 16, 2001 - 09:17 pm:

You can solve it in the following way:
our quartic equation:
(1)... x4+a1x3+a2x2+a 3x+a4=0
Now,
Let y1 be a real root of the cubic equation:
(2)... y3-a2y2+(a1a3-4a 4)y+(4a2a4-a32-a 12a4)=0
The solutions to the quartic equation are the 4 roots of:
(3)... z2+0.5{a1+/-sqrt(a12-4a 2+4y1)}z+0.5{y1+/-sqrt(y1 2-4a4)}=0

If all roots of (2) are real, computation is simplified by using that particular real root which produces all real coefficients in the quadratic equation (3).
x1+x2+x3+x4=-a1
x1x2+x2x3+x3x 4+x4x1+x1x3+x 2x4=a2
x1x2x3+x2x3x 4+x1x2x4+x1x3x 4=-a3
x1x2x3x4=a4
where x1,x2,x3,x4 are the four roots