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Question 7

The product of the distinct roots of $$\mid x^2 - x - 6 \mid = x + 2$$ is

Solution

We have, $$\mid x^2 - x - 6 \mid = x + 2$$

=> |(x-3)(x+2)|=x+2

For x<-2, (3-x)(-x-2)=x+2

=> x-3=1 =>x=4 (Rejected as x<-2)

For -2$$\le\ $$x<3, (3-x)(x+2)=x+2 =>x=2,-2

For x$$\ge\ $$3, (x-3)(x+2)=x+2 =>x=4

Hence the product =4*-2*2=-16

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