JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Basic Mathematics

The critical points of $x|x+3|+|x-1|-2=0$ are $x=-3$ and $x=1$. For $x<-3$ it becomes $x^2+4x+1=0$, whose root $x=-2-\sqrt{3}$ lies in the interval, giving one solution. For $-3\le x<1$ it becomes $x^2+2x-1=0$, whose roots $-1\pm \sqrt{2}$ both lie in the interval, giving two solutions. For $x\ge 1$ it becomes $x^2+4x-3=0$, whose roots $-2\pm \sqrt{7}$ are both outside the interval, giving none. The number of real solutions is $3$.