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For how many integer values of x, is |x – 6| > |3x + 6|?

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Veritas Prep GMAT Instructor
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Re: For how many integer values of x, is |x – 6| > |3x + 6|?  [#permalink]

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New post 05 Jun 2020, 04:43
1
Kritisood wrote:
VeritasKarishma wrote:
Bunuel wrote:
For how many integer values of x, is |x – 6| > |3x + 6|?

(A) 1

(B) 3

(C) 5

(D) 7

(E) Infinite


Think in terms of the definition of absolute values:

|x - 6| > 3*|x + 2|

Distance from 6 is greater than 3 times the distance from -2

-------- (-2) -------- (0) ------------------------------ (6) -------------

At point 0, the distance from 6 will be equal to 3 times the distance from -2. So on the left of 0, the distance from 6 will be greater.

The distance between -2 and 6 is 8. Distance from 6 will be equal to 3 times the distance from -2 at a point 4 units to the left of -2 i.e. -6. To its left, distance from 6 will be less. So between -6 to 0, distance from 6 is greater than 3 times the distance from -2. We have 5 integer values lying between these two.

Answer (C)


Hi VeritasKarishma i was trying to solve this problem by finding the critical points. the critical points I arrived at are -2 and 6. and I'm not able to find the answer through this. the critical points as mentioned by JeffTargetTestPrep are -6 and 0.

could you help me in where I am going wrong in deducing the critical points?

-- to elaborate how I got the critical points
x-6=0
x=6

and 3x+6=0
x=-2

https://gmatclub.com/forum/is-x-1-1-x-1 ... 82478.html
in this question the statement was |x + 1| = 2|x - 1| hence critical points mentioned are -1 and 1; using the same logic i arrived at the critical points here.

JeffTargetTestPrep VeritasKarishma i am really confused about this. could you please assist?


The transition points are -2 and 6 as you derived.
But note that the question has a negative between the terms:

|x – 6| - |3x + 6| > 0

If you are trying to use the same method that we do when the terms are added, you will not get the answer. Here, you need to use the method I have discussed above.
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Re: For how many integer values of x, is |x – 6| > |3x + 6|?  [#permalink]

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New post 02 Jul 2020, 06:19
I don't understand following problem:

When I raise both sides to 2 and get RHS to LHS I get the following inequality

(x-6)^2-(3x+6)^2>0
(x-6+3x+6)(x-6-3x-6)>0
(4x)(-2X-12)>0

So values for X will be 0 and -6

BUUT

Number line is

....+++.....-6..........---.............0...........+++..........

And as LHS should be BIGGER than 0 I could take infinitive values for X

Where am I wrong?
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Re: For how many integer values of x, is |x – 6| > |3x + 6|?  [#permalink]

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New post 06 Jul 2020, 00:50
chetan2u wrote:
Bunuel wrote:
For how many integer values of x, is |x – 6| > |3x + 6|?

(A) 1

(B) 3

(C) 5

(D) 7

(E) Infinite



Hi sobby,
The actual way would be...

Since both sides are positive, square both sides
\((x-6)^2>(3x+6)^2..........x^2-12x+36>9x^2+36x+36.........8x^2+48x<0\)
8x(x+6)<0.....

For 8x(x+6) to be NEGATIVE, one of 8x and x+6 will be negative and other positive..

If x is positive, 8x will be positive and x+6 will also be positive..
So x should be NEGATIVE..
Then 8x will be negative, and thus x+6 should be positive ..
For that \(x+6>0.....X>-6....\) But x<0.

So range becomes \(0>x>-6\)...
Values are \(-1,-2,-3,-4,-5\)
5 value
C

That's a nice approach,
I would like to add something too!
Here : x^2+6x<0 --> x^2+6x+9<9
Now take x^2+6x+9 = |x+3|^2 <9 .. after simplifying, we get -3 <x+3<3
= -6<x<0
5 integer values.
Hope this helps!
:D
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Re: For how many integer values of x, is |x – 6| > |3x + 6|?   [#permalink] 06 Jul 2020, 00:50

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