Re: Solving 3d degree polynomial

*To*: mathgroup at smc.vnet.net*Subject*: [mg90812] Re: [mg90778] Solving 3d degree polynomial*From*: Murray Eisenberg <murray at math.umass.edu>*Date*: Fri, 25 Jul 2008 06:13:41 -0400 (EDT)*Organization*: Mathematics & Statistics, Univ. of Mass./Amherst*References*: <200807240851.EAA18857@smc.vnet.net>*Reply-to*: murray at math.umass.edu

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 math.umass.edu 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

**References**:**Solving 3d degree polynomial***From:*Floris Zoutman <fzoutman@hotmail.com>