Ignoring the x = 0 root, you are asking for the real roots of (%beta*x^3-2*%beta*x^2+%beta*x-3*x+2) = 0.
When %beta = 1 there are three real roots; when %beta = -2, there is one real root:
(%i39) (%beta*x^3-2*%beta*x^2+%beta*x-3*x+2);
(%o39) %beta*x^3-2*%beta*x^2+%beta*x-3*x+2
(%i40) subst(%beta=1,%);
(%o40) x^3-2*x^2-2*x+2
(%i41) nroots(%,minf,inf);
(%o41) 3
(%i44) subst(%beta = -2,%o39);
(%o44) -2*x^3+4*x^2-5*x+2
(%i45) nroots(%,minf,inf);
(%o45) 1
Maxima is unable to easily determine the number of real roots. Just looking at solutions
for which freeof(%i, rectform(xxx)) is true isn't going to work. I think you'll to give
axima plenty of help.
--Barton
-----maxima-bounces at math.utexas.edu wrote: -----
>I'm?trying?the?following:
>urb(x):=x*ui/(x*(1-x)*%beta+1);
>deltaurb(x):=ui*x-urb(x);
>diff(deltaurb(x),x);
>solve(%,x);
>
>How?can?I?make?Maxima?show?only?the?"real"?(not?complex)?solutions?
>
>
>hugo
>