MathGroup Archive 2010

[Date Index] [Thread Index] [Author Index]

Search the Archive

Re: quartic equation

  • To: mathgroup at smc.vnet.net
  • Subject: [mg108743] Re: quartic equation
  • From: Bob Hanlon <hanlonr at cox.net>
  • Date: Tue, 30 Mar 2010 04:59:54 -0500 (EST)

soln = Solve[x^4 + a*x^3 + b*x^2 + c*x + d == 0, x];

Length[soln]

4

Looking at the first of the four roots

x /. soln[[1]]

-((1/2)*Sqrt[(1/(3*2^(1/3)))*
              (Sqrt[(27*a^2*d - 9*a*b*c + 2*b^3 - 72*b*d + 
                            27*c^2)^2 - 4*(-(3*a*c) + b^2 + 12*d)^3] + 
                   27*a^2*d - 9*a*b*c + 2*b^3 - 72*b*d + 27*c^2)^
                (1/3) + (2^(1/3)*(-(3*a*c) + b^2 + 12*d))/
              (3*(Sqrt[(27*a^2*d - 9*a*b*c + 2*b^3 - 72*b*d + 
                               27*c^2)^2 - 4*(-(3*a*c) + b^2 + 12*d)^3] + 
                      27*a^2*d - 9*a*b*c + 2*b^3 - 72*b*d + 27*c^2)^
                   (1/3)) + a^2/4 - (2*b)/3]) - 
   (1/2)*Sqrt[-((1/(3*2^(1/3)))*
              (Sqrt[(27*a^2*d - 9*a*b*c + 2*b^3 - 72*b*d + 
                            27*c^2)^2 - 4*(-(3*a*c) + b^2 + 12*d)^3] + 
                   27*a^2*d - 9*a*b*c + 2*b^3 - 72*b*d + 27*c^2)^
                (1/3)) - (2^(1/3)*(-(3*a*c) + b^2 + 12*d))/
           (3*(Sqrt[(27*a^2*d - 9*a*b*c + 2*b^3 - 72*b*d + 
                            27*c^2)^2 - 4*(-(3*a*c) + b^2 + 12*d)^3] + 
                   27*a^2*d - 9*a*b*c + 2*b^3 - 72*b*d + 27*c^2)^
                (1/3)) + a^2/2 - (-a^3 + 4*a*b - 8*c)/
           (4*Sqrt[(1/(3*2^(1/3)))*
                    (Sqrt[(27*a^2*d - 9*a*b*c + 2*b^3 - 72*b*d + 
                                  27*c^2)^2 - 4*(-(3*a*c) + b^2 + 12*d)^3] + 
                         27*a^2*d - 9*a*b*c + 2*b^3 - 72*b*d + 27*c^2)^
                      (1/3) + (2^(1/3)*(-(3*a*c) + b^2 + 12*d))/
                    (3*(Sqrt[(27*a^2*d - 9*a*b*c + 2*b^3 - 72*b*d + 
                                     27*c^2)^2 - 4*(-(3*a*c) + b^2 + 12*d)^
                                    3] + 27*a^2*d - 9*a*b*c + 2*b^3 - 
                            72*b*d + 27*c^2)^(1/3)) + a^2/4 - (2*b)/3]) - 
         (4*b)/3] - a/4

Bob Hanlon

---- Leslaw Bieniasz <nbbienia at cyf-kr.edu.pl> wrote: 

=============

Hi,

I am totally unexperienced in MATHEMATICA, and I am looking for some 
advice. I need to solve a quartic equation symbolically, that is to factor
a fourth order polynomial given in the power base. The coefficients
of the polynomial are complicated expressions of a number of parameters,
and I need to obtain expressions for the roots as functions of these
parameters. Is there any way to do this? I would appreciate if
anybody can send me some simple example code showing how this sort
of problems can be solved.

Leslaw




  • Prev by Date: Re: Mathematica starts badly 1 time over 5 in OSX
  • Next by Date: Re: priorities between @, @@ and //
  • Previous by thread: Re: a little nasty problem?
  • Next by thread: Re: quartic equation