Let m and n be the numbers of real roots of the quadratic equations x 2 - 12 x + [ x ] + 31 = 0 and x 2 - 5…

Given,

$x^2-12x+\left[x\right]+31=0$

$\Rightarrow \underset{\ge -5}{\underbrace{x^2-12x+31}}=-\left[x\right]$

Now from above equation we say that, it could have its solution in $[5,6)$ but it does not exist as at $x=5$ as LHS$=1$,

So no solution, hence $m=0$

Now solving,

$x^2-5\left|x+2\right|-4=0$

Taking Case $1$ when $x\ge -2$ we get,

$x^2-5\left(x+2\right)-4=0$

$\Rightarrow x^2-5x-14=0\Rightarrow x=7,-2$

Now taking Case $2$ when $x<-2$ we get,

$x^2+5x+10-4=0$

$\Rightarrow x=-2,-3$

So, total $3$ solution i.e., $x=-3, -2, 7$

Hence, $n=3$

So, the value of $m^2+mn+n^2=0+0+3^2=9$