Question:
The product of the distinct roots of $\left|x^{2}-x-6\right|=x+2$ is
- $-8$
- $-24$
- $-4$
- $-16$
Correct Answer: Option: 4
${{x}^{2}}-x-6=(x+2)(x-3)$
Case 1: ${{x}^{2}}-x-6<0$
i.e. (x+2)(x-3)<0
$\Rightarrow -2<x<3$ and $\left| {{x}^{2}}-x-6 \right|=-\left( {{x}^{2}}-x-6 \right)$
Therefore, $\left|x^{2}-x-6\right|=x+2$
$\begin{align} & =-(x+2)(x-3)=x+2 \\ & \Rightarrow (x-3)=-1\Rightarrow x=2 \\ \end{align}$Case 2:${{x}^{2}}-x-6\ge 0$
i.e. $\left( x+2 \right)\left( x-3 \right)\ge 0$
$\Rightarrow x\le -2\ or\ x\ge 3$
Checking for boundary conditions:
For x=-2, $\left|x^{2}-x-6\right|=x+2$, therefore, x=-2 is also the root. But for x=3, $\left| {{x}^{2}}-x-6 \right|\ne x+2$.
Hence x=3 is NOT the root.
And for the interval$x<-2\ or\ x>3$ the expression$\left| {{x}^{2}}-x-6 \right|={{x}^{2}}-x-6$
Therefore, $\left|x^{2}-x-6\right|=x+2$
$\begin{align} & =(x+2)(x-3)=x+2 \\ & \Rightarrow (x-3)=1\Rightarrow x=4 \\ \end{align}$Therefore, the root are -2, 2, and 4. So the required product = (2)(-2)(4)=-16
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