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

|x^2 − x − 6| = x + 2
--> |(x - 3)(x + 2)l = x + 2
--> ±(x - 3)(x + 2) = x + 2

Case 1: (x - 3)(x + 2) = x + 2
--> (x - 3)(x + 2) - (x + 2) = 0
--> (x + 2)(x - 4) = 0
--> Roots are -2 & 4

Case 2: (x - 3)(x + 2) = - (x + 2)
--> (x - 3)(x + 2) + (x + 2) = 0
--> (x + 2)(x - 2) = 0
--> Roots are -2 & 2

Roots = {-2, 2, 4}
--> Product of the roots = -2*2*4 = -16

Option B

positive case
x^2-x-6=(x-3)(x+2)
range x>=3 or x<=-2
x^2-x-6=x+2, x^2-2x-8=0
(x-4)(x+2)=0, x=4,-2

negative case
range -2<x<3
x^2-x-6=-(x+2), x^2-4=0
(x-2)(x+2)=0, x=-2,2

product roots: 4*-2*2=-16

Ans (B)

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

If (x+2)(x-3)>=0 --> x<=-2 or x>=3,
(x+2)(x-3-1)=0
x=-2 or x=4 (both are OK)

If (x+2)(x-3)<0 --> -2<x<3,
(x+2)(-x+3-1)=0
x=-2 (excluded) or x=2 (OK)

Product of distinct roots = (-2)(2)(4) = -16

FINAL ANSWER IS (B)

Posted from my mobile device

Usually in those simple looking equations you can transform them to an even more simpler equation
Thus
x^2-x-6 can become
X^2-3x+2x-6x=x(x-3)+2(x-3)=(x+2)(x-3)

we get
|(x+2)(x-3)|=x+2
or
|(x+2)(x-3)|-(x+2)=0
we open the || with + and one time with -
so we get
(1)(x+2)(x-2)=0 and (2) (x+2)(x-4)=0

Roots are 2,-2 and 4
product is -16

Answer (B)

|x2−x−6|=x+2
Case 1 : If (x2-x-6) is positive,
x2−x−6=x+2
x2-2x-8=0
(x-4)(x+2)=0
Roots x=4, x=-2

Case 2 : If (x2-x-6) is Negative,
-x2+x+6=x+2
-x2+4=0
(x-2)(x+2)=0
Roots x=2, x=-2

So the final roots of x are -2, 2 and 4
Product of the roots = -16 (Option B)

So we have to find the distinct roots of the equation and then multiply them to get the answer.

|x2−x−6|=x+2|x2−x−6|=x+2 is

We combine the functions "x^2−x−6" = "x+2" into x^2-2x-8=0 we can simplify this into (x+2)*(x-4).

So -2 & 4 are the solutions for this equation, then their product will be solution D. -8

Edit: If you did the same than me you are wrong, I did not considered absoluteness of the first equation for what 2 is a valid solution as |-4|=4.

Regards,
Pablo

The product of the distinct roots of
\(| x^{2} —x —6 | = x+2\) is

Case1: \(x^{2} —x —6 = x+2\)
\(x^{2} —2x —8 = 0\)
\((x—4)(x+2) = 0\)
\(x= 4\) and \(x=—2\)
Need to check answers:
x= 4 —> | 16–4–6| = 4+2 —Ok
x= —2 —> | 4+2 —6| = —2+2 —Ok

Case2: \(x^{2} —x—6 = —x—2\)
\((x —2)(x+2 ) = 0\)
\(x= 2\) and \(x= —2\)
Need to check answers:
x= 2 —> | 4–2 —6| = 2+2 —Ok

Well, the product of the distinct roots —> \(4*2(—2) = —16 \)

Answer ( B)

Posted from my mobile device

Ans is -16

roots: -2,4,2

I tried solving this with the wavy line method. So as to save time.. But i dont know how to post picture here as this is my first post.
Can someone solve via that method too.

Also to clarify, whenever we have X^2 in the absolute or non-absolute equation. the critical value of the equation is omitted right?

Hi,
Why cannot we square both sides and cancel terms?
|(x-3)(x+2)|=(x+2)
Squaring both sides
(X-3)^2 (x+2)^2 = (x+2)^2
(X-3)^2 = 1
X=4

What should I have not done to miss out two distinct roots? Please help Bunuel.

Posted from my mobile device

Because x + 2 could be 0 and we cannot divide by 0. Here you loos one root x = -2.

Next, (x - 3)^2 = 1 has two roots x = 4 and x = 2.

Hi Bunuel
Thank you. I missed x+2 could be zero part. *facepalm*
Second, x-3 could also be <0.

Your explanations are always simple, crisp and clear. I am sure it’s not just me who scrolls through comment section to look for your solution. Great job.

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

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

|x-3|= (x-3) if x>=3, -(x+3) if x<3
|x+2|= (x+2) if x>=-2, -(x+2) if x<-2

In a number line we have <------------(-2)-----0--------3------>
We have to check the case when

1) x>=3

|x-3||x+2|=x+2 becomes
(x-3)(x+2)=(x+2)
x^2-x-6=x+2
x^2-2x-8=0
(x-4)(x+2)=0
x=4, x=-2
Since x>=3 we keep x=4 as a solution, discard x=-2

2) x>=-2 and x<3
|x-3||x+2|=x+2 becomes
-(x-3)(x+2)=x+2
-(x^2-x-6)=x+2
x^2-x-6=-x-2
x^2-4=0
x^2=4
x=2 and x=-2
We keep both solutions since they are in the relevant interval

3) x<-2
|x-3||x+2|=x+2 becomes
-(x-3)*-(x+2)=x+2
(x-3)(x+2)=x+2
Doing the same work as case 1 we get
x=4, x=-2
Since both solutions are out of our relevant range of x-values we ignore them.


Therefore our solutions are
x=4, 2, -2


Their product is
4*2*-2= -16

Final Answer: B

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