JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Quadratic Equations and Inequalities

For $n$, solve $x^2-5|x+2|-4=0$. If $x\ge -2$: $x^2-5x-14=0\Rightarrow (x-7)(x+2)=0\Rightarrow x=7,-2$, both admissible. If $x<-2$: $x^2+5x+6=0\Rightarrow (x+2)(x+3)=0\Rightarrow x=-3$ (admissible; $x=-2$ is a boundary already counted). Thus $n=3$ with roots $-3,-2,7$.

For $m$, write $x^2-12x+31=-[x]$, i.e. $(x-6{)}^2-5=-[x]$. Since $(x-6{)}^2-5\ge -5$, a root needs $[x]\le 5$. Testing the intervals where the parabola is low enough: on $[4,5)$ it gives $(x-6{)}^2=1\Rightarrow x=5,7$ (both excluded); on $[5,6)$ it gives $(x-6{)}^2=0\Rightarrow x=6$ (excluded); on $[6,7)$ it needs $(x-6{)}^2=-1$ (impossible); elsewhere $(x-6{)}^2-5>-[x]$. No interval yields a valid root, so $m=0$.

Therefore $m^2+mn+n^2=0+0+9=9.$