JEE Main 2024 — 5 April, Shift 1 — Question 29

Case-1 :$\mathrm{x} \geq 0$

$x^{2}+2 x-5 x-5-1=0$

$x^{2}-3 x-6=0$

$\mathrm{x}=\frac{3 \pm \sqrt{9+24}}{2}=\frac{3 \pm \sqrt{33}}{2}$

One positive root

CASE -2 :$-1 \leq \mathrm{x}<0$

$-x^{2}-2 x-5 x-5-1=0$

$x^{2}+7 x+6=0$

$(x+6)(x+1)=0$

$\mathrm{x}=-1$

one root in range

CASE-3 :$-2 \leq \mathrm{x}<-1$

$x^{2}-2 x+5 x+5-1=0$

$x^{2}-3 x-4=0$

$(x-4)(x+1)=0$

No root in range

Case-4 :$\mathrm{x}<-2$

$\mathrm{x}^{2}+7 \mathrm{x}+4=0$

$x=\frac{-7 \pm \sqrt{49-16}}{2}=\frac{7 \pm \sqrt{33}}{2}$

one root in range

Total number of distinct roots are 3