QJEEM24Ma8614: JEE Main 2024 Mathematics - Class 11 JEEM24Ma8614 Solution: The number of distinct real roots of the equation , is _______

Let's analyze the equation $|x+1||x+3|-4|x+2|+5=0$ based on different intervals of $x$ .

(I) If x < -3 :

$\begin{aligned} & |x+1| = -(x+1), \\ & |x+3| = -(x+3), \\ & |x+2| = -(x+2), \\ & -(x+1)*(-(x+3)) - 4(-(x+2)) + 5 = 0 \\ & x^2 + 4x + 3 + 4x + 8 + 5 = 0 \\ & x^2 + 8x + 16 = 0 \\ & \Rightarrow x = -4 \quad (\text{one solution}) \end{aligned}$

(II) If -3 \leq x < -2 :

$\begin{aligned} & |x+1| = -(x+1), \\ & |x+3| = -(x+3), \\ & |x+2| = -(x+2), \\ & -(x+1)*(-(x+3)) - 4(-(x+2)) + 5 = 0 \\ & x^2 - 10 = 0 \\ & \Rightarrow x = \pm \sqrt{10} \\ & \text{(Do not satisfy } -3 \leq x < -2) \end{aligned}$

(III) If -2 \leq x < -1 :

$\begin{aligned} & |x+1|=-(x+1), \\ & |x+3|=x+3, \\ & |x+2|=-(x+2), \\ & -(x+1)(x+3)-4(-(x+2))+5=0 \\ & -x^2-4x-3-4x-8+5=0 \\ & -x^2-8x-6=0 \\ & \Rightarrow x^2+8x+6=0 \\ & x=\frac{-8 \pm 2\sqrt{10}}{2} = -4 \pm \sqrt{10} \end{aligned}$

(IV) If $x \geq -1$ :

$\begin{aligned} & |x+1|=x+1, \\ & |x+3|=x+3, \\ & |x+2|=x+2, \\ & (x+1)(x+3)-4(x+2)+5=0 \\ & x^2+4x+3-4x-8+5=0 \\ & x^2=0 \\ & \Rightarrow x=0 \quad (\text{one solution}) \end{aligned}$

$\Rightarrow$ The number of distinct real roots are two : $x = -4$ and $x = 0$ .