Given a set of multivariate, quadratic, non-homogeneous equations, is there a way to decide how many non-negative roots it have?
All the coefficients are real numbers.
The number of variables is the same as the number of equations, and all the equations are independent.
Some of the equations might actually be linear equations.
By "non-negative" solution, I mean a solution in which all the variables take non-negative values.
By some "physical" reasoning, we know it must have at least one non-negative solution.
I know generally a set of n quadratic equations with n variables has at most $2^n$ distinct roots.
The background of this problem is: the set of quadratic equations is the right-hand side of the chemical rate equation. By equating is to zero, the steady-state case is being considered. As we only take one-body or two-body reactions into account, the degree of the equations are at most two. As the abundance of the molecules cannot be negative, we only care about the non-negative solutions.
The number of variables can be up to 1000, so simple numerical test is not practical.
I am not a math student, and I am not sure whether this kind of question is allowed here.
Not efficiently, at least not unless the problem has some additional structure which can be exploited. The set of mixed Nash equilibria of a two-player game can be written as the nonnegative solutions of such a polynomial system. In general it is #P-hard to count the equilibria of such a game (Conitzer and Sandholm). It is even PPAD-hard (still thought to be polynomial time unsolvable) to compute a single equilibrium, although one is guaranteed to exist by Nash's Theorem.
You may want to investigate connections with Nash equilibria further -- if your problem is somehow equivalent to this one then you will have a lot of established results and techniques to build on. Also people would probably find a connection between game theory and chemistry surprising. Or if your problem is easier, looking at Nash equilibria might help you figure out what extra features your problem has that makes it solvable efficiently. If it is harder or incomparable it might still be interesting to know that.