Thanks to Robert Bradshaw and Jason Grout !
the problem is solved.
Eli
On Nov 26, 3:28 pm, Robert Bradshaw <[EMAIL PROTECTED]>
wrote:
> On Nov 26, 2008, at 12:21 PM, Jason Grout wrote:
>
>
>
> > Eli wrote:
> >> Hello,
> >> In the sage tutorial, I found how to solve equations:
>
> >> sage: x, b, c = var('x b c')
> >> sage: solve([x^2 + b*x + c == 0],x)
> >> [x == (-sqrt(b^2 - 4*c) - b)/2, x == (sqrt(b^2 - 4*c) - b)/2]
>
> >> However, I could not find how to assign the solution to some
> >> variable.
> >> That is, something that will do:
> >> assign the value (-sqrt(b^2 - 4*c) - b)/2 (first solution of the
> >> equation) to the variable X
> >> assign the value (sqrt(b^2 - 4*c) - b)/2 (second solution of the
> >> equation) to the variable Y
>
> >> How can this be done ?
>
> > Here is a session showing one way to accomplish that. The key is the
> > solution_dict argument.
>
> > sage: f=x^2+b*x+c == 0
> > sage: soln = f.solve(x,solution_dict=True)
> > sage: soln
> > [{x: (-sqrt(b^2 - 4*c) - b)/2}, {x: (sqrt(b^2 - 4*c) - b)/2}]
> > sage: soln[0][x]
> > (-sqrt(b^2 - 4*c) - b)/2
> > sage: soln[1][x]
> > (sqrt(b^2 - 4*c) - b)/2
> > sage: X=soln[0][x]
> > sage: Y=soln[1][x]
> > sage: X
> > (-sqrt(b^2 - 4*c) - b)/2
> > sage: Y
> > (sqrt(b^2 - 4*c) - b)/2
>
> You can also use the fact that it's a list of equations, which have
> rhs() and lhs() methods.
>
> sage: sage: solve([x^2 + b*x + c == 0],x)
> [x == (-sqrt(b^2 - 4*c) - b)/2, x == (sqrt(b^2 - 4*c) - b)/2]
> sage: all = solve([x^2 + b*x + c == 0],x)
> sage: all[0]
> x == (-sqrt(b^2 - 4*c) - b)/2
> sage: all[0].rhs()
> (-sqrt(b^2 - 4*c) - b)/2
>
> - Robert
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