JEE Main · 2023 · 15 Apr 2023 · Shift 1 · Basic Mathematics

Split by the signs inside the modulus. For $x\ge 0$, $x^2-5(x+2)+6=0$ gives $x^2-5x-4=0$, whose only non-negative root is $x=\frac{5+\sqrt{41}}{2}$. For $-2\le x<0$, $-x^2-5(x+2)+6=0$ gives $x^2+5x+4=0$, i.e. $x=-1$ or $x=-4$, of which only $x=-1$ lies in the interval. For $x<-2$, $-x^2+5(x+2)+6=0$ gives $x^2-5x-16=0$, whose only root below $-2$ is $x=\frac{5-\sqrt{89}}{2}$. There are three real roots in all.