quartic equation
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From: Leslaw Bieniasz on 29 Mar 2010 07:55 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
From: Bob Hanlon on 30 Mar 2010 05:59 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(a)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
From: dh on 30 Mar 2010 06:01
Hi Leslaw, assume ci is the coefficient of x^i. Then we can write: poly = c0 + c1 x + c2 x^2 + c3 x^3 + c4 x^4; Solve[poly == 0, x] This gives the roots in terms of ci. Daniel On 29.03.2010 13:55, Leslaw Bieniasz 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 > -- Daniel Huber Metrohm Ltd. Oberdorfstr. 68 CH-9100 Herisau Tel. +41 71 353 8585, Fax +41 71 353 8907 E-Mail:<mailto:dh(a)metrohm.com> Internet:<http://www.metrohm.com> |