RE: Exact real solutions of cubic equations
- To: mathgroup at smc.vnet.net
- Subject: [mg46956] RE: Exact real solutions of cubic equations
- From: David.Cousens at csiro.au
- Date: Wed, 17 Mar 2004 02:29:09 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
Jonas, If you get the solutions in symbolic form and then use ComplexExpand to get the real and Imaginary components in symbolic form, the imaginary part will or should evaluate to zero when you substitute the values for a and b. The expressions are a bit longwinded but work. Mathematica can't determine if the roots are real or not until the values are substituted. Cheers Dr David Cousens Phone 61-(0)7-33274564 CSIRO Exploration and Mining Fax 61-(0)7-33274455 QCAT, Technology Court, Email:David.Cousens at csiro.au Pullenvale 4069, Brisbane, Qld., Australia -----Original Message----- From: JonasB at iui.se [mailto:JonasB at iui.se] To: mathgroup at smc.vnet.net Subject: [mg46956] [mg46932] Exact real solutions of cubic equations Hello, I would like to find the _exact_ real roots of some cubic polynomials. Mathematica seems to have problems determining that a root is real Solve[1 + a s + b s^2 + s^3 == 0, s] results in three complex solutions for a = -4 and b = 3. FullSimplify does not help, either it does nothing or it gets stuck, depending on the values of a and b. I can of course evaluate the solution numerically, but that is not what I want. Does anyone know of a package that can simplify expressions with complex numbers? Jonas