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

Question

Competition Mode Question

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

A. -24 
B. -16
C. -12
D. -8 
E. -4

Ravixxx · Accepted Answer

This is an equation with Absolute Values, so we must consider: 
1) x^2 - x - 6 = x+2
2) x^2 - x - 6 = -(x+2)

Thus: 
1) x^2 - x - 6 = x+2
 x^2 -2x-8 = 0

Sum product, find two numbers: 
x1+x2 = -2 
x1*x2 = -8

(x-4)(x+2) = 0
Two solution: 
x=4 V x=-2

2) x^2-x-6= -(x+2)
 x^2= 4
 x= +-2

x: {4;-2;2}

So: (4)*(-2)*(2) = -16

nick1816 · Answer

|x^2−x−6| = x+2

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

Case 1. when x ≤-2 or x≥3

(x-3)(x+2)≥0

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

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

(x+2)(x-3-1) = 0

x= -2 or 4

Case 2. when -2 < x <3

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

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

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

(x+2)(-x+3-1) = 0

x= -2(Not in range) or 2

Distinct roots = -2, 2 and 4

Product = -2*2*4= -16

CareerGeek · Answer

|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

exc4libur · Answer

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)

freedom128 · Answer

|(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

madzaka · Answer

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)

SaravananA · Answer

|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)

pabpinor · Answer

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

lacktutor · Answer

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)

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debjit1990 · Answer

Ans is -16

roots: -2,4,2

ashishakonnur · Answer

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?

Doer01 · Answer

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 .

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Bunuel · Answer

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.

Doer01 · Answer

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.

chacinluis · Answer

|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

bumpbot · Answer

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

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