Solving 3d degree polynomial

*To*: mathgroup at smc.vnet.net*Subject*: [mg90778] Solving 3d degree polynomial*From*: Floris Zoutman <fzoutman at hotmail.com>*Date*: Thu, 24 Jul 2008 04:51:15 -0400 (EDT)

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

**Follow-Ups**:**Re: Solving 3d degree polynomial***From:*Murray Eisenberg <murray@math.umass.edu>