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CAT 2019 Quant Question with Solution 18

Question:
The product of the distinct roots of $\left|x^{2}-x-6\right|=x+2$ is

1. $-8$
2. $-24$
3. $-4$
4. $-16$

${{x}^{2}}-x-6=(x+2)(x-3)$

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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