We have the following system of quadratic equations in $\mathrm x \in \mathbb R^n$

$$\mathrm A (\mathrm x \circ \mathrm x) + \mathrm B \mathrm x + \mathrm c = 0_m$$

where $\mathrm A \in \mathbb R^{m \times n}$, $\mathrm B \in \mathbb R^{m \times n}$ and $\mathrm c \in \mathbb R^m$ are given, and $\circ$ denotes the entry-wise product.

Let $\mathrm a_i, \mathrm b_i \in \mathbb R^n$ denote the $i$-th rows of matrices $\rm A$ and $\rm B$, respectively. Let $c_i$ denote the $i$-th entry of vector $\rm c$. Hence, the system of quadratic equations can be rewritten as follows

$$\mathrm x^\top \mbox{diag} (\mathrm a_i) \, \mathrm x + \mathrm b_i^\top \mathrm x + c_i = 0, \qquad i \in [m]$$

Note that

$$\begin{aligned} \mathrm x^\top \mbox{diag} (\mathrm a_i) \, \mathrm x + \mathrm b_i^\top \mathrm x + c_i &= \begin{bmatrix} \mathrm x\\ 1\end{bmatrix}^\top \begin{bmatrix} \mbox{diag} (\mathrm a_i) & \frac 12 \mathrm b_i\\ \frac 12\mathrm b_i^\top & c_i\end{bmatrix} \begin{bmatrix} \mathrm x\\ 1\end{bmatrix}\\ &= \mbox{tr} \left( \begin{bmatrix} \mbox{diag} (\mathrm a_i) & \frac 12 \mathrm b_i\\ \frac 12\mathrm b_i^\top & c_i\end{bmatrix} \begin{bmatrix} \mathrm x\\ 1\end{bmatrix} \begin{bmatrix} \mathrm x\\ 1\end{bmatrix}^\top \right)\\ &= \mbox{tr} \left( \begin{bmatrix} \mbox{diag} (\mathrm a_i) & \frac 12 \mathrm b_i\\ \frac 12\mathrm b_i^\top & c_i\end{bmatrix} \begin{bmatrix} \mathrm x \mathrm x^\top & \mathrm x\\ \mathrm x^\top & 1\end{bmatrix} \right)\end{aligned}$$

Let

$$\mathrm M_i := \begin{bmatrix} \mbox{diag} (\mathrm a_i) & \frac 12 \mathrm b_i\\ \frac 12\mathrm b_i^\top & c_i\end{bmatrix}$$

The original system of $m$ quadratic equations in $\mathrm x \in \mathbb R^n$ can then be replaced by the following system of $m$ *linear* equations in symmetric positive semidefinite matrix $\mathrm Y \in \mathbb R^{(n+1) \times (n+1)}$

$$\langle \mathrm M_i , \mathrm Y \rangle = 0, \qquad i \in [m]$$

plus the constraint $\mbox{rank}(\mathrm Y) = 1$ and the equality constraint $Y_{n+1,n+1} = 1$.

The rank constraint is a major difficulty. Let us discard that constraint and minimize the rank instead

$$\begin{array}{ll} \text{minimize} & \mbox{rank}(\mathrm Y)\\ \text{subject to} & \langle \mathrm M_i , \mathrm Y \rangle = 0, \qquad i \in [m]\\ & Y_{n+1,n+1} = 1\\ & \mathrm Y \succeq \mathrm O_{n+1}\end{array}$$

Since the rank is *non*-convex, let us use the nuclear norm, which is a convex proxy for rank. Hence, we obtain the following **convex** optimization problem

$$\begin{array}{ll} \text{minimize} & \| \mathrm Y \|_*\\ \text{subject to} & \langle \mathrm M_i , \mathrm Y \rangle = 0, \qquad i \in [m]\\ & Y_{n+1,n+1} = 1\\ & \mathrm Y \succeq \mathrm O_{n+1}\end{array}$$

Let $\bar{\rm Y}$ denote the minimal solution of the convex program above. There are two cases to consider:

If $\mbox{rank} (\bar{\rm Y}) = 1$, find $\bar{\mathrm x} \in \mathbb R^n$ such that $$\begin{bmatrix} \bar{\mathrm x}\\ 1\end{bmatrix} \begin{bmatrix} \bar{\mathrm x}\\ 1\end{bmatrix}^\top = \bar{\mathrm Y}$$ Note that $\bar{\mathrm x}$ is a solution of the original system of quadratic equations.

If $\mbox{rank} (\bar{\rm Y}) \neq 1$, try *another* approach.