### solving polynomial equations with to_poly_solve

I have a system of (quadratic) polynomial equations which I'd like to solve with Sage. Here and elsewhere I've seen references to the fact that `to_poly_solve`

is faster than just calling `solve`

in Sage.

Can I somehow tell `to_poly_solve`

that I'd like to work over `RR`

or `RDF`

instead of whatever symbolic ring it's using by default? The coefficients of the equations I'm using are floats, but my solutions are coming out as huge fractions. I imagine it would be faster if the solver were working over `RR`

.

EDIT: Upon further reflection, I suspect this is really a Maxima question, since the entire solving process is carried out there. Nevertheless, I would like to know if such functionality exists in Maxima and whether I can use it from Sage.

If you'd like to know, here are some test equations:

```
vars = var('a0, a1, a2, R')
eqs = []
eqs += [R^2 == a0^2 + 1.56835186587759*a0*a1 + 1.79444137578129*a0*a2 + a1^2 + 1.12952287573422*a1*a2 + a2^2 - 2*a0 - 1.56835186587759*a1 - 1.79444137578129*a2 + 1]
eqs += [R^2 == a0^2 + 1.56835186587759*a0*a1 + 1.79444137578129*a0*a2 + a1^2 + 1.12952287573422*a1*a2 + a2^2 - 1.56835186587759*a0 - 2*a1 - 1.12952287573422*a2 + 1]
eqs += [R^2 == a0^2 + 1.56835186587759*a0*a1 + 1.79444137578129*a0*a2 + a1^2 + 1.12952287573422*a1*a2 + a2^2 - 1.79444137578129*a0 - 1.12952287573422*a1 - 2*a2 + 1]
eqs += [a0 + a1 + a2 == 1]
maxima.load('topoly_solver')
meqs = maxima(eqs)
solns = meqs.solve(vars)
```

For example, here is the value of `a0`

and `R`

in this case:

```
sage: solns[0][0]
a0=-127216901572440833725/188266808300071978748
sage: solns[0][3]
R=9*sqrt(149350213112030161)/sqrt(47066702075017994687)
sage: solns[0][0].rhs().parent()
Maxima
```