When you are computing with sqrt(polynomial) there is a problem of
interpretation.
For example,
sqrt( x^2+2*x+1) could be x+1 or -(x+1), and therefore your original
equation may have
two different meanings. When you check the solution, which do you use?
> -----Original Message-----
> From: maxima-bounces at math.utexas.edu
> [mailto:maxima-bounces at math.utexas.edu] On Behalf Of laurent couraud
>
> >
>
> 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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> >
>
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