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Re: Solving 3d degree polynomial

  • To: mathgroup at
  • Subject: [mg90812] Re: [mg90778] Solving 3d degree polynomial
  • From: Murray Eisenberg <murray at>
  • Date: Fri, 25 Jul 2008 06:13:41 -0400 (EDT)
  • Organization: Mathematics & Statistics, Univ. of Mass./Amherst
  • References: <>
  • Reply-to: murray at

When you fill in those specific values of the parameters, are you 
supplying exact rationals or other exact numbers, with square-roots, 
e.g., or are you supplying decimals?

Floris Zoutman wrote:
> I have a function
> f(q)=2n(n+1)q^3-(1+t)(2n+1)q^2+2(t-(n+1)A)q+(1+t)A
> in which n>1, -1<A<1, and 0<t<1 are parameters. If I fill in appropriate values for the parameters before:
> Solve[Y==0,q] it gives me 3 real roots.
> If instead I ask for a symbolic solution before filling in the same values of the parameters, Mathematica gives almost the same roots, but now there is an imaginary part to the solution. It is of the order 10^(-16). Why are the roots complex if I find a solution before filling in parameters, but real if I fill in parameters before finding the solution?
> Thank you for your answer
> Floris

Murray Eisenberg                     murray at
Mathematics & Statistics Dept.
Lederle Graduate Research Tower      phone 413 549-1020 (H)
University of Massachusetts                413 545-2859 (W)
710 North Pleasant Street            fax   413 545-1801
Amherst, MA 01003-9305

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