On 10/13/2013 09:00 PM, Stavros Macrakis wrote:
>
> I'll guess that these are
http://en.m.wikipedia.org/wiki/Casus_irreducibilis, where an exact real
solution can only be expressed with complex expressions. For approximate
(numeric) results, you can use realroots.
>
> -s
Yes, that's it. If the solution set is three irrational (real) numbers,
then a solution using radicals only requires complex radicals. Maybe
Maxima and Mathematica prefer this form because it seems more consistent
to use radicals if possible (or maybe it's accidental). I suppose Maxima
could use a function to convert to trig functions. I don't know if
something similar applies to quartic equations and some solutions of
higher order equations.
> On Oct 13, 2013 1:12 PM, "Joerg Rauh" <jrgrauh at yahoo.com> wrote:
>
> ("5.28.0-2","2012-08-27 23:16:48","i686-pc-mingw32","GNU Common Lisp
(GCL)","GCL 2.6.8")
>
> Dear Maxima Supporter,
> out of wxMaxima 12.04.0 I had Maxima solve for x:
> ((40-3*x)*(20-2*x)*x)/2-500
> Here are the results:
>
[x=(-(sqrt(3)*%i)/2-1/2)*((250*sqrt(101)*%i)/3^(7/2)+25750/729)^(1/3)+(1300*((sqrt(3)*%i)/2-1/2))/(81*((250*sqrt(101)*%i)/3^(7/2)+25750/729)^(1/3))+70/9,
>
x=((sqrt(3)*%i)/2-1/2)*((250*sqrt(101)*%i)/3^(7/2)+25750/729)^(1/3)+(1300*(-(sqrt(3)*%i)/2-1/2))/(81*((250*sqrt(101)*%i)/3^(7/2)+25750/729)^(1/3))+70/9,
>
x=((250*sqrt(101)*%i)/3^(7/2)+25750/729)^(1/3)+1300/(81*((250*sqrt(101)*%i)/3^(7/2)+25750/729)^(1/3))
>
> It missed the real results: x=1.7407 and x=6.23415, which I found here:
>
http://www.wolframalpha.com/input/?i=solve%28%28%2840-3*x%29%2F2%29*%2820-2*x%29*x-500%2Cx%29
>
> Is there anything I can do differently to make it find the real solutions?
> Thank you and kind regards
>
> Joerg
>
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