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Statement 1 Does't give you a final value of |x-2|. I tried to rearange the inequality but you can not multiply by \(-|x|x^3\) because the term can be positive or negative (you don't know). Therefore the whole Statement 1 just tells you that some Division with X is > 0 and is therefore insufficient because it doesn't provide a VALUE for |x-2|.

Statement 2 x can be 2 or -2. Which is insufficient.

If you combine Statement 1 and 2 you know that in order to be true, Statement 1 needs a positive denominator which is just possible if X equals -2 (x^3 would then be -8 and the denominator would be -8*-2 = 16.

Answer Choice C _________________

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hi, (1) \(\frac{(|x+1|)}{(-|x|x^3)}>0\)... we know that for fraction to be greater than 0 both deno and numerator should be of same sign.... numeratos has to be +ive as it is mod.. so -|x|x^3 should be positive so x is a negative integer.. no other info ... insuff

(2) \(x^4 = 16\) x can be 2 or -2... insuff

combined we get x as -2 .. suff ans C
_________________

In order to find the value of |x-2|, we first need to know the value of x. From question statement, we know that x is not equal to 0 or -1. Let's see if St. 1 and/or 2 can get us a unique value of x.

Analyzing St. 1 Independently

\(\frac{(|x+1|)}{(-|x|x^3 )}>0\)

Multiplying both sides by a positive number doesn’t impact the sign of inequality. So, multiplying both sides by \(\frac{(|x|)}{(|x+1|)}\), we get:

\(\frac{(-1)}{x^3} >0\)

(Note that we could multiply both sides by this number because we were told that x ≠ -1. Therefore, x + 1 ≠ 0 (division by 0 is not defined))

Now, multiplying both sides by -1 will reverse the sign of inequality. We get:

\(\frac{1}{x^3}<0\)

\(x^3\) will have the same +/- sign as x

This means, \(1/x\) is negative.

This means, x is negative.

This information is not sufficient to find the value of x.

Analyzing St. 2 independently \(x^4 = 16\)

That is, \(x^4 - 2^4 = 0\) \((x^2 + 2^2)(x+2)(x-2) = 0\)

This gives us, x = 2 or -2

Since we don’t get a unique value of x, not sufficient.

Re: If x is not equal to 0 or -1, what is the value of |x-2|? [#permalink]

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28 May 2016, 03:26

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Re: If x is not equal to 0 or -1, what is the value of |x-2|? [#permalink]

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11 Jul 2016, 08:05

from the 1 statement we come to know that x is negative as we have denominator as negative.So the expression to be grater than 0 x^3 should be negative.then only it will br=e greater than zero.We do not get any definite value except x is negative.So NS

From2 statement we get that x could be 2 or -2 and both give different answers when put inside the expression.So NS

Combined we get x is negative which is only -2.And we get definite value.So answer is C.

Re: If x is not equal to 0 or -1, what is the value of |x-2|? [#permalink]

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11 Aug 2017, 21:37

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If x is not equal to 0 or -1, what is the value of |x-2|? [#permalink]

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17 Aug 2017, 18:28

I know this is unnecessary, but can anyone confirm that the below is correct if it was necessary to do so. Trying to make sure I understand the 3 step approach to absolute modulus.

For statement 1:

Since l x+1 l is always positive, then l x+ 1 l >0

critical point is -1,

a) so if x>-1, then x+1 is positive, so x+1>0, and x>-1, b) if x<-1, then X+1 is negative, so -(x+1)>0, and x<-1,

so x>-1 or x<-1 right?

Denominator needs to have the same sign as the numerator, so x must be negative for denominator to be positive. so x>-1, but doesn't solve what lx-2l is.

2) x^4=16, x= 2 or -2, insuff.

c: x>-1, so x=2. Suff.

Did I handle the splitting up the modulus correctly in statement 1? Thank you! Kudos to whoever answers.