Let m and $\mathrm{n}$ be the numbers of real roots of the quadratic equations $x^{2}-12 x+[x]+31=0$ and $x^{2}-5|x+2|-4=0$ respectively, where $[x]$ denotes the greatest integer $\leq x$. Then $\mathrm{m}^{2}+\mathrm{mn}+\mathrm{n}^{2}$ is equal to __________.

The givne eqn is : $x^2-12 x+[x]+31=0$ <br/><br/>$$ \begin{aligned} & \Rightarrow\{x\}-x=x^2-12 x+31 \\\\ & \Rightarrow\{x\}=x^2-11 x+31 \end{aligned} $$ <br/><br/>So, $0 \leq x^2-11 x+31<1$ <br/><br/>$$ \begin{aligned} & \Rightarrow x^2-11 x+30 \leq 0 \\\\ & \Rightarrow(x-5)(x-6)<0 \\\\ & \Rightarrow x \in(5,6) \\\\ & \therefore[x]=5 \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \therefore x^2-12 x+5+31=0 \\\\ & \Rightarrow x^2-12 x+36=0 \\\\ & \Rightarrow(x-6)^2=0 \Rightarrow x=6 \end{aligned} $$ <br/><br/>Hence, $x \in \phi$ <br/><br/>$(\because x \in(5,6))$ <br/><br/>$\therefore m=0$ <br/><br/>Another equation is $x^2-5[x+2]-4=0$ <br/><br/><b>Case I :</b> $x \geq-2$ <br/><br/>$x^2-5 x-14=0 \Rightarrow x=7,-2$ <br/><br/><b>Case II :</b> $x<-2$ <br/><br/>$$ \begin{aligned} & x^2+5 x+6=0 \Rightarrow x=-3-2 \\\\ & \therefore x \in\{-3,-2,7\} \end{aligned} $$ <br/><br/>$\therefore n=3$ <br/><br/>Hence, $m^2+m x+n^2=0+0+9=9$