Roots of quadratic vector equation Given $$A_{i j k}X_j X_k + B_{ij} X_j + C_i = 0$$ where $A_{ijk}$, $B_{ij}$, and $C_i$ are arbitrary real numbers for all $i$, $j$, $k$ which are $N$-dimensional indices, such that $A_{ijk}=A_{ikj}$ holds. There is a summation for repeated indices, so this is $N$ equations with $N$ unknowns. Is it possible to determine $\vec{X}$ analytically?
If a full solution does not exist, is there a solution for special cases? For example


*

*if $B_{ij}=0$ for all $i$ and $j$, or

*if $C_{i}=0$ for all $i$, or

*if the $A_{ijk}, B_{ij}, C_i$ constant coefficients are independent of $i$ for each fixed $i$.
If such an expression exists, it would allow one to give a second order Newton-Raphson algorithm. 
 A: This belongs to the class of "nonsymmetric algebraic Ricatti equations", which can be solved following the method explained in Nonsymmetric algebraic Riccati equations and Wiener–Hopf factorization for M-matrices (2001) and in Federico Poloni's paper on Quadratic vector equations (2010).
A: There are many solutions --- up to $2^N$ in general. Do you need a specific one?
Otherwise, some of the special cases you suggest are simple: for instance, if $C_i=0$ for all $i$ then all zeros is a trivial solution. Or, if all entries are independent from $i$, then you get $N$ copies of the same equation.
In general, as far as I know, a closed-form solution does not exist. Note that by introducing a few auxiliary variables you can turn any polynomial equation in this form, so this form is quite general.
My paper cited in the other answer studies numerical methods for this equation under some particular sign conditions ($A_{ijk} \geq 0$, $C_i \geq 0$, and $-B$ is an $M$-matrix.) In most cases the go-to algorithm is Newton's method, so if you were planning to use this to speed up another Newton iteration then probably you are out of luck. :)
