Re: cubic polynomial
- To: mathgroup at smc.vnet.net
- Subject: [mg22684] Re: cubic polynomial
- From: "ERIK DOFS" <erik.dofs at swipnet.se>
- Date: Sat, 18 Mar 2000 01:27:59 -0500 (EST)
- Organization: A Customer of Tele2
- References: <8aqs6m$ac8@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Just a check question: Should the last term be -3(c^2)? If so, you have a very simple analytic solution, X=-3c Regards Erik D. E.Carletti1 skrev i meddelandet <8aqs6m$ac8 at smc.vnet.net>... > >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? > >Thank your for your help > >Elena Carletti >Financial Markets Group >LSE >Houghton Street >London WC2A 2AE > >Tel. 0044 (0)20 7955 7896 >Fax 0044 (0) 20 7242 1006 > >