By the way: For %beta > 0, the polynomial discriminant tells us that
(%beta*x^3-2*%beta*x^2+%beta*x-3*x+2) = 0 has three real roots; for %beta
< 0,
there is exactly one real root. For %beta > 0, the roots might not appear
to be real,
but they are (keeping it real is hard to do when you use radicals to solve
a cubic).
(%i65) %beta*x^3-2*%beta*x^2+%beta*x-3*x+2$
(%i66) sol : solve(%,x)$
(%i67) subst(%beta=1,%)$
(%i68) rectform(float(%));
(%o68) [x=0.68889218253402,x=-1.170086486626034,x=2.481194304092016]
(%i69) subst(%beta=-1,sol)$
(%i70) rectform(float(%));
(%o70)
[x=1.633170240915238*%i+0.68055154026432,x=0.68055154026432-1.633170240915238*%i,x=0.63889691947135]
--Barton
maxima-bounces at math.utexas.edu wrote on 11/29/2010 10:09:47 AM:
> [image removed]
>
> Re: [Maxima] [newbie] removing complex solutions
>
> Barton Willis
>
> to:
>
> Hugo Coolens
>
> 11/29/2010 10:09 AM
>
> Sent by:
>
> maxima-bounces at math.utexas.edu
>
> Cc:
>
> maxima-bounces, "hugo.coolens", maxima mailing list
>
> By default, Maxima assumes that all variables are real; the command
> assume(%beta > 0)
> tells Maxima that %beta is positive.
>
> The solve function makes little use of the assumed facts. Thus doing
> assume(%beta > 0)
> will not expunge the non-real solutions.
>
> --Barton
>
>
> > "hugo.coolens", maxima mailing list
> >
> > Can I tell Maxima somewhere/somehow that %beta is real and positive
and
> > thus make my original question solvable by Maxima?
>
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