JEE Main 2024 — 5 April, Shift 2 — Question 30

$$x|x+5|+2|x+7|-2=0$$

The critical points are $x=-7$ and $x=-5$.

Case 1: $x \ge -5$

$$|x+5|=x+5,\quad |x+7|=x+7$$ $$x(x+5)+2(x+7)-2=0$$ $$x^2+7x+12=0$$ $$(x+3)(x+4)=0 \Rightarrow x=-3,-4$$

Both values satisfy $x \ge -5$.

Case 2:$-7 \le x < -5$

$$|x+5|=-(x+5),\quad |x+7|=x+7$$ $$x(-x-5)+2(x+7)-2=0$$ $$-x^2-3x+12=0 \Rightarrow x^2+3x-12=0$$ $$x=\frac{-3\pm\sqrt{57}}{2}$$

Only

$$x=\frac{-3-\sqrt{57}}{2}$$

lies in the interval $[-7,-5)$.

Case 3:$x<-7$

$$|x+5|=-(x+5),\quad |x+7|=-(x+7)$$ $$x(-x-5)-2(x+7)-2=0$$ $$-x^2-7x-16=0$$

The discriminant is negative, so there are no real solutions.

Number of real solutions $=3$