JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Basic Mathematics

Split at the critical points $x=-4$ and $x=-2$. For $x<-4$ the equation becomes $-x^2-7x+4=0$, i.e. $x^2+7x-4=0$, whose root $x=\frac{-7-\sqrt{65}}{2}\approx -7.5$ satisfies $x<-4$. For $-4\le x<-2$ it is $x^2+x+4=0$, with discriminant $1-16<0$. For $x\ge -2$ it is $x^2+7x+16=0$, with discriminant $49-64<0$. Only one real solution exists.