The product of the distinct roots of ∣x ^2 - x - 6∣ = x + 2 is

CAT 2019 Shift 1 Paper

Question : 98 of 100

Marks: +1, -0

The product of the distinct roots of

∣x ^2 - x - 6∣ = x + 2

Solution:

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

Case 1:

x^{2}-x-6<0

i.e.

(x+2)(x-3)<0

$⇒-2 Therefore,

|x^{2}-x-6|=x+2

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

⇒(x-3)=-1 ⇒ x=2

Case

2: x^{2}-x-6 ≥ 0

i.e.

(x+2)(x-3) ≥ 0

⇒ x ≤-2

x ≥ 3

Checking for boundary conditions:
For

x=-2,|x^{2}-x-6|=x+2,

therefore,

x=-2

is also the root. But for

x=3,|x^{2}-x-6| ≠ x+2

.
Hence

{x}=3

is NOT the root.
And for the interval

x<-2

x>3

the expression

|x^{2}-x-6|=x^{2}-x-6

Therefore,

|x^{2}-x-6|=x+2

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

⇒(x-3)=1 ⇒ x=4

Therefore, the root are

-2,2,

and 4 . So the required product

=(2)(-2)(4)=-16

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