Solving a system of generic quadratic forms Suppose we have a system of $n$ quadratic forms, 
$$x^T A_1 x = 0 \\ x^T A_2 x = 0 \\ ... \\ x^T A_n x = 0$$
where $x\in \mathbb{R}^d$ and $A_i \in \mathbb{R}^{d \times d}$ symmetric, $d > n$.
What is the best way to solve this system without any assumptions on the $A_i$? Can this even be done in general?
What kind of assumptions on the $A_i$ would make this problem feasible/easier?
 A: The theoretical idea is as follows. One has $n$ homogeneous equations of degree $2$ in $d$ unknowns. We break the homogeneity adding the equation $f(x)=x^Tx=1$.
Assume that the $(A_i)_i$ are generic (there exist no algebraic relations linking the entries of these matrices; or randomly choose the $(A_i)_i$). In particular, the $(A_i)$ are invertible.
The gradients of the $f_i(x)=x^TA_ix$ are $2A_ix$ and $\nabla f(x)=2x$. Let $x$ be a solution and suppose that there are $(u_i)_i,u$, not all zero, s.t. $\sum_iu_iA_ix+ux=0$; that implies $u=0$; let $rank((A_ix)_i)=r<n$. Assume, for example, that $(A_ix)_{i\leq r}$ is a free system and let $\Pi=span((A_ix)_i)$. Then, the condition (concerning the $(A_j)_{j>r}$): for every $j>r,A_jx\in\Pi$ can be written in $(d-r)(n-r)$ independent relations in the $(x_i)$. On the other hand, the $r$ independent relations for every $i\leq r, x^TA_ix=0$ concern only the $(A_i)_{i\leq r}$. Finally, such a solution $x$ must satisfy a system of $(d-r)(n-r)+r$ independent   HOMOGENEOUS equations; yet, $(d-r)(n-r)+r\leq d$ implies that $r=n-1$ and $(d-r)(n-r)+r=d$. Thus, the sole solution is $x=0$, a contradiction.
Conclusion. When the $(A_i)_i$ are generic, the variety $V$, defined by the $n+1$ relations $x^TA_ix=0,x^Tx=1$, is the empty set or has dimension $d-n-1$. 
Remark. If one of the $(A_i)$ is $>0$ or $<0$, then $V=\emptyset$.
Practically, $V$ is defined by $n+1$ equations and you cannot do better, except if you calculate a Grobner basis; yet, if $d$ is large, then it is hopeless. Finally, the most interesting case occurs when $n=d-1$; then one has a finite number of solutions that can be numerically calculated. Generically, there are $2^d$ complex solutions; of course,they are not all real. I have an example, for $n=3,d=4$, with $4$ real solutions (opposite by pairs).  
EDIT. Answer to madison54 . Over $\mathbb{R}$, the variety $V$ is not connected, even if you identify -by a quotient- $x$ and $-x$ (perhaps it is true over $\mathbb{C}$ ?). I used Grobner's method for $n=2,d=4$; then (generically) the system is not solvable by radicals. That implies that there does not exist any algebraic parametrization of $V$.
When $n=d-1$, you can solve the system by Grobner's method until $d=10$; otherwise, use numerical methods.
If $n<d-1$ and you know a numerical solution $x_0\in V$, then you can deduce a local solution $x_0+U$ where $U$ is a small disc with center $0$ in the tangent space $T_{x_0}V$, which is the orthogonal of $span(A_1x_0,\cdots,A_nx_0,x_0)$.
