Re: cubic polynomial
- To: mathgroup at smc.vnet.net
- Subject: [mg22681] Re: [mg22646] cubic polynomial
- From: Hartmut Wolf <hwolf at debis.com>
- Date: Sat, 18 Mar 2000 01:27:55 -0500 (EST)
- Organization: debis Systemhaus
- References: <200003161411.JAA10248@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
E.Carletti1 schrieb:
>
> I have a problem: I want to solve this equation:
>
> x^3+(3c-1)(x^2)-4cx-4(c^2)=0 with respect to x.
>
> It has to have a real solution because it is continuous
> in x. c is a positive parameter.
> If i solve it numerically, plugging numbers for c, then it is fine. But
> I would like an analytical solution: in this case i get only one
> solution which should give me a real value for x (the other two are
> imaginary) but
> it has a square root with all negative members (all terms with c, which
> is positive, with a sign - in front of it, so immaginary). How is that
> possible? What procedure does mathematica use to solve cubic expression?
>
> How can I express the expression in a nicer way to get rid of these
> negative terms? My feeling is that the programm is not able to simplify
> the expression for the solution.
> Could you please help me? if I write the expression I find in the paper
> I am writing, noone will believe it is real!
> And even more funny, if I plug number into the solution I find for x, it
> comes the same number as I plug numbers directly into the function I
> want to solve, except for the last part which is an imaginary number
> which shoult tend to zero.
> What is going on?
>
Elena,
if you cut and pasted directly from Mathematica your problem perhaps is
due to a typo 'cx'. If you write 'c x' you'll get a perfect symbolical
solution
sols = Solve[x^3 + (3c - 1)(x^2) - 4c x - 4(c^2) == 0, x]
and you may plot
Plot[x /. sols[[1]] , {c, -1, 1}]
or any different range.
Hartmut
- References:
- cubic polynomial
- From: "E.Carletti1" <E.Carletti1@lse.ac.uk>
- cubic polynomial