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For real non-zero $2\times 2$ matrices, can we say:
For any $A,B$, there are at most two matrices $X$ such that $XX + AX + B =0$
Is there a way to see this without going in the direction of writing the open form result of $XX + AX + B$ and solve each element is equal to zero, etc.?
No. E.g. over $\mathbb R$, we have $\pmatrix{1&t\\ 0&-1}^2=I$ for any $t$. It also follows that $$ \pmatrix{1&t\\ 0&-1}^2+\pmatrix{0&0\\ 0&1}\pmatrix{1&t\\ 0&-1}+\pmatrix{-1&0\\ 0&0}=0 $$ for any $t$.
Edit. Another counterexample (inspired by Dietrich Burde's answer). Let $u,v$ be any two vectors such that $v^Tu=0$. Then $X=uv^T$ is a solution to the equation $X^2=0$. In particular, if the underlying field is $GF(2)$ (the field consisting of only two elements $0$ and $1$, with $1+1=0$), the equation $Y^2=0$ for a $2\times2$ matrix $Y$ has exactly four solutions $$ Y\in\left\{\pmatrix{0&0\\ 0&0},\ \pmatrix{0&1\\ 0&0},\ \pmatrix{0&0\\ 1&0},\ \pmatrix{1&1\\ 1&1}\right\}. $$ And so does the equation $X^2=I$ over $GF(2)$: just put $X=Y+I$.
No, take for example $A=0$ and $$ B=\begin{pmatrix} 0 & 1 \cr 0 & 0 \end{pmatrix}. $$ Then $X^2=-B$ has no solution, since $X$ is nilpotent, so that $X^2=0$ but $B\neq 0$.
