Determine the common zero set of $n$ quadratic equations in $n$ variables? Note, for the case $n=2$ the questions is already answered here.

Notation:
Let $\mathbf{x}:= (x_1, \ldots, x_n)$.
For $k = 1, \ldots, n$ I have the n equations:
$$f_k(\mathbf{x})=
c_{k, 1, 1} x_1^2 + c_{k, 1, 2} x_1 x_2 + \ldots + c_{k, n, n} x_n^2 \stackrel{!}{=}0
$$
with all coefficients $c_{1, 1, 1}, \ldots, c_{n, n, n}$ given.
I am interested in the set of common zeros for all $f_k,$ with $k=1, \ldots, n$, i.e. the set
$$Z:=\{\mathbf{x}\in \mathbb{R}^n | f_k(\mathbf{x})= 0  \enspace \forall \enspace k = 1,\ldots, n\}.$$
Obviously we have $0\in Z$ is always a solution.
Further, if $\mathbf{x} \in Z$, then $\lambda \mathbf{x} \in Z$ for all $\lambda \in \mathbb R$ .
The approach for $n=2$ was (see here) to factor each quadratic equation:
$$c_{k,1,1} x_1^2 + c_{k,1,2} x_1 x_2 + c_{k,2,2} x_2^2 = (p_k x_1 + q_k x_2)(r_k x_1 + s_k x_2) $$ and then compare the subspaces (straight lines) given by the factors for each $k$.
The free parameters $p_k, q_k, r_k, s_k$ can be determined by equating coefficients.
In $n$ dimensions this would look like:
$$
c_{k, 1, 1} x_1^2 + c_{k, 1, 2} x_1 x_2 + \ldots + c_{k, n, n} x_n^2 \stackrel{!}{=} \left(\sum_{i=1}^n a_{k,i} x_i\right)\left(\sum_{j=1}^n b_{k,j} x_j\right).
$$
However, from my perspective the problem is, that there are in general less free parameters than conditions: for each fixed $k$ we have $\frac{n(n+1)}{2}$ coefficients $c_{k, 1, 1}, \ldots, c_{k, n, n}$ but only $2n$ free parameters $a_{k,1}, \ldots, a_{k, n}$ and $b_{k,1}, \ldots, b_{k, n}$.
Question: Is there another approach to determine the set of zeros of $f_k$ and thus finally the common set of zeros for all $k$?
 A: This is not an easy problem, and we can no longer couch it in the elementary language of the $n=2$ case.

Why is it hard?
Since we're doing nontrivial algebraic geometry, we should describe it in the complex projective setting.  If you want to return to the real affine setting you originally asked the question in, you could then just throw away any points at infinity or points with complex coordinates once you were done.
What you're trying to do is intersect a bunch of quadric hypersurfaces in $\mathbb{P}^{n-1}$. In the case $n=2$ we were working on $\mathbb{P}^1$, the projective line, and so a "quadric hypersurface" was just a pair of points and working out what could happen was not difficult.  For $n>2$, though, the situation is qualitatively different: a quadric hypersurface could be a pair of hyperplanes, but more likely it will irreducible.  When $n=3$ the only irreducible quadrics are the conic sections, which are smooth, but once $n \geq 4$ irreducible conics can be either smooth or singular.  So the possibilities are fairly complicated.

If you had an arbitrary system of polynomials and wanted an exact solution you'd need to use Grobner bases.  If you wanted numeric solutions, you could use homotopy continuation.
However, there are apparently specialized algorithms that work specifically in the case you're discussing, namely quadratic systems over the reals.  (I imagine these make use of various real tricks, like the fact that $f(x)^2 + g(x)^2 = 0$ iff $f(x)=g(x)=0$, which obviously don't hold over more general fields.)  See:
https://mathoverflow.net/questions/153436/can-you-efficiently-solve-a-system-of-quadratic-multivariate-polynomials
