How do you solve a system of quadratic equations? I was watching a video where a problem in Galois theory was posed such that it became necessary to tell if a certain element was a perfect square in a finite field extension of the rationals. By writing a general element in terms of its $\mathbb{Q}$-basis and squaring it, the result was a system of quadratic equations. In particular, it was system of three non-homogeneous equations in three unknowns.
The video deferred the solution to a computer algebra system, which felt unsatisfying. I was wondering what techniques would be useful for solving general equations of this kind. 
My only thought was that Groebner bases might be useful (and I'm sure that's how the computer algebra system solves it). But I don't know much about them, and I wonder how amenable the Buchberger algorithm is to solving it by hand.
 A: Consider a quadratic equation of form $ax^2 +by^2 +cxy+dx+ey+f=0$. The polynomial can be decomposed into two factors (polynomial) such as $ax+By+C$  and $dx+Ey+F$.
Therefore for a system of two quadratic equations we get four polynomials of first degree (i.e linear equations). The combination of these four equations give four systems of equations where the solutions satisfy the initial quadratic system of equations; look at this example:
Solve following system of quadratic equations:
$$\begin{cases}
2x^2+2y^2 +5xy-x+y-1=0 \\
-3x^2-2y^2+5xy+10x-8y-8=0 \\
\end{cases}
$$
The first equation can be written as:
$$(x+2y-1)(2x+y+1)=0$$
and the second one can be written as:
$$(-x+y+2)(3x-2y-4)=0$$
We get four systems of equations, which give following solutions:
$$\begin{cases}
x+2y-1=0 \\
-x+y+2=0 \\ 
\end{cases}
$$
give $x=5/3$ and $y=-1/3$
$$\begin{cases}
x+2y-1=0 \\
3x-2y-4=0 \\ 
\end{cases}
$$
give  $x=5/4$ and $y = -1/8$
$$\begin{cases}
2x+y+1=0 \\
-x+y+2=0 \\
\end{cases}  
$$
give  $x=1/3$ and $y=-5/3$ and
$$\begin{cases}
2x+y+1=0 \\
3x -2y-4=0 \\
\end{cases}
$$
give $x=2/7$ and $y=-11/7$
These are the solutions of the quadratic system of equations.
A: It seems systems like these can be solved with the use of resultants.
Say we have three polynomials $f(x, y, z) = g(x, y, z) = h(x, y, z) = 0$ in three variables.
We can treat each as a polynomial in the ring $R[x,y][z]$. We can use a resultant to form a new, simpler system $F(x, y) = Res(f, g) = 0$ and $G(x, y) = Res(g, h) = 0$. This effectively eliminates the $z$ variable from our system. 
Doing this again, we get a final polynomial $R(x) = Res(F, G)$ (taken with respect to $y$ this time over the ring $R[x]$). 
We solve $R(x) = 0$ to get all possible values of $x$.
Then, we go through each of these possible values and plug them into the system $F(x, y) = G(x, y) = 0$. This gives us the corresponding values for $y$.
And then, with values for $x$ and $y$ in hand, we solve the original system $f(x, y, z) = g(x, y, z) = h(x, y, z) = 0$ to get the corresponding $z$ values.
