> -----Message d'origine-----
> De : maxima-bounces at math.utexas.edu
> [mailto:maxima-bounces at math.utexas.edu] De la part de Richard Fateman
> Envoy? : samedi 9 septembre 2006 21:03
> ? : 'laurent couraud'
> Cc : maxima at math.utexas.edu
> Objet : Re: [Maxima] solve(a + x - sqrt(x) ,x)
>
>
> %i3) a+x-sqrt(x);
> (%o3) x - sqrt(x) + a
> (%i4) solve(%,sqrt(x));
> (%o4) [sqrt(x) = x + a]
> (%i5) %^2;
> 2
> (%o5) [x = (x + a) ]
> (%i6) solve(%,x);
> 2 a + sqrt(1 - 4 a) - 1 - 2 a + sqrt(1 - 4 a) + 1
> (%o6) [x = - -----------------------, x = -------------------------]
> 2 2
> (%i7)
>
This is what i do actualy. But for a more complex case this is not so easy.
For exemple:
(P1(x) + sqrt( P2(x) ) )/(P3(x) + sqrt(P4(x)) ) = P5(x)
Where P1, P2, P3, P4, P5 are polynomial.
And may be the more complicated case that i have found is when we have
A system of equation where more than one unknown are of degree 2
One of the simplest case of this is :
(%i1) [a1 * x^2 + b1 * y + c1 = 0, a2 * y^2 + b2 * x + c2 = 0]$
(%i2) solve(%, [x,y]);
(%o2) []
> You have to check for extraneous roots, in general.
>
> Could solve have done this automatically?
>
> It would not be hard to change the program to separate out
> sqrt and square both sides (etc.) but I don't know of a clean
> way to remove extraneous roots.
Yes, I understands the problem. my experiment has to show to me that
often this method of resolution gives solutions strictly identical.
solve can at least eliminate these redundancies.
Laurent.
>
> > -----Original Message-----
> > From: maxima-bounces at math.utexas.edu
> > [mailto:maxima-bounces at math.utexas.edu] On Behalf Of laurent couraud
> > Sent: Saturday, September 09, 2006 11:32 AM
> > To: maxima at math.utexas.edu
> > Subject: solve(a + x - sqrt(x) ,x)
> >
> > Hello,
> >
> > What is the way to solve this with Maxima?
> >
> > (%i1) solve(a + x - sqrt(x) ,x);
> > (%o1) [x = sqrt(x) - a]
> >
> > is there a way (may be some switch) to force Maxima to
> > eliminate sqrt(x).
> >
> > Thank.
> >
> >
> > Laurent.
> >
>
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