Is there a unique solution for this quadratic matrix equation? Here is the quadratic matrix equation I've been looking at lately:
$$ Q_{r,r}=A_{r,r}X_{r,r}^2+B_{r,r}X_{r,r}+C_{r,r}=0_{r,r} $$
Note that $A, B, C,$ and $X$ are $r \times r$ matrices. $A$ contains known elements, $B$ contains known elements, $C$ contains known elements, and $X$ contains the unknown elements that you are solving for. $ 0_{r,r} $ is just the $r \times r$ null matrix.
Is there any solution for $X$ in terms of $A, B,$ and $C$ (making no easy assumptions)? (e.g. $X$ is a diagonal matrix, $A=B=C$, or anything of that sort.)
I have tried to solve this and nothing has worked out. I attempted solving it generally by manipulating the matrices in variable form (i.e. actually writing out the matrices $A, B, C,$ and $X$ in variables) and finding a unique solution for all of the elements of $X$ in terms of the elements of $A, B,$ and $C$. That didn't work out beyond the case of $r=1$.
Trying to solve it by looking at $r$ at different values did not work out either; I ended up with very abysmal equations at just $r=2$. I don't know exactly how to make this appealing to the denizens of math.stackexchange, but it (as far as I know) isn't a heavily studied problem.
There is a very high possibility that I've just been doing elementary techniques and nothing of note, so I hope someone or a group of people could shed light on this.
 A: You could note that the matrix similarity 
\begin{align}
  \pmatrix{I & \mathbf{0} \\ X & -I}\pmatrix{\mathbf{0} & I \\ -C & -B} & \pmatrix{I & \mathbf{0} \\ X & -I} & \\
  =\pmatrix{\mathbf{0} & I \\ C & X + B} & \pmatrix{I & \mathbf{0} \\ X & -I} & \\ 
 =\pmatrix{ X & -I \\  X^2 + BX + C& -X - B} & \\
\end{align}
gives your equation in $X$, and if the equation is solved, then the matrix $\pmatrix{0 & I \\ -C & -B}$ is block diagonalized. Also note that there is a closed form solution to bring (almost) any matrix into the form $\pmatrix{0 & I \\ -C & -B}$ through a similarity transform.  I do not know what this form is called in the literature, but I like to call it the block companion form. Here is how to do it
\begin{align}
  \pmatrix{G^{-1} & \mathbf{0} \\ G^{-1}M & I}\pmatrix{M & G \\ F & D}  \pmatrix{G & \mathbf{0} \\ -G^{-1}MG & I} & \\
  =\pmatrix{G^{-1}M & I \\ G^{-1}M^2 + F & G^{-1}MG + D}  \pmatrix{G & \mathbf{0} \\ -G^{-1}MG & I} & \\ 
 =\pmatrix{ \mathbf{0} & I \\  G^{-1}M^2G + FG - G^{-1}M^2G - DG^{-1}MG & D+G^{-1}MG} & \\
 =\pmatrix{ \mathbf{0} & I \\  (F-DG^{-1}M)G & D+G^{-1}MG} & \\
 =\pmatrix{0 & I \\ -C & -B} & \\
\end{align}
If a solution to $X^2 + BX + C = \mathbf{0}$ were possible in a closed form here, then it could split the eigenvalue problem in half, be applied recursively and thus have a closed form solution to the eigenvalue problem. The existence of such a solution has already been dis-proven by the Abel-Ruffini Theorem.
Since diagonalization is a difficult problem without a closed form solution, it is no wonder you had such difficulty finding such a solution. Maybe this is why the equation is related to what is named the quadratic eigenvalue problem as J.M. noted.
A: An answer by year!
Consider the Riccati equation (1) $XAX+XB+CX+D=0_n$ where $X\in M_n(\mathbb{C})$ is unknown and $A,B,C,D\in M_n(\mathbb{C})$ are generic given matrices. (that is, the entries of these matrices are independent commutative indeterminates OR, more simply, there are no algebraic relations with coefficients in $\mathbb{Q}$ linking the entries of $A,B,C,D$).
Then (1) can be reduced to an equation in the form (2) $X^2+AX+B=0_n$ where $A,B$ are generic matrices. This equation has exactly $\binom{2n}{n}$ solutions and solving (2) can be reduced to solve a polynomial equation $P(u)=0$ with $degree(P)=2n$ and Galois group $S_{2n}$.
Note that , in your equation (3) $AX^2+BX+C=0_n$, if $A$ is generic, then $A$ is invertible and (3) has in fact the form (2).
