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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