If x is not equal to 0 or -1, what is the value of |x-2|?

Question

If x is not equal to 0 or -1, what is the value of |x - 2|?

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

(2)  x^4 = 16

This is Question 1 for the e-GMAT Question Series on Absolute Value.

Provide your solution below. Kudos for participation. Happy Solving! :-D

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The e-GMAT Team

chetan2u · Accepted Answer

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

EgmatQuantExpert · Answer

Official Explanation

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.

Combining St. 1 and St. 2

From St. 1, x is negative
From St. 2, x = 2 or -2

Combining both, x = -2

Sufficient.

reto · Answer

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

cutegirlsimran · Answer

Statement 1 says x is negative. 
Statement 2 gives 2 and -2 as values. 
Combining the two we get x as -2

So Answer must be C

UJs · Answer

Stmt 1
fraction is > 0 then Numerator & Denominator need to be same sign.

Numerator |x-2| > 0

Denominator \(-|x|x^3 > 0\) for this ( \(x^3 < 0\)) , implies x < 0 insufficient.

Stmt 2
x can be 2 or -2. insufficient.

combine 1 and 2 we get the answer, from 1 we know x < 0, and from 2 we know (x = 2 or -2)

Ans : C

kplusprinja1988 · Answer

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.

Let me know if i am wrong.

brandon7 · Answer

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.

Mo2men · Answer

X is NOT equal to -1 or 0.

|x-2| = ???

Strategically I will start with statement 2 to have sense of numbers.

Statement 2

x^4 = 16

x =2.....|2-2|= 0
or 
x =-2...|-2-2|= -4

Hence it gives Two different values

Insufficient

Statement 1

From what we have let's test the answers we got:

Let x =2......... \frac{(|x+1|)}{(-|x|x^3)}>0 ..When x is positive, then this will make Statement 1 invalid........ Discard any positive number .

Let x =-2......... \frac{(|x+1|)}{(-|x|x^3)}>0  ..Any negative number will make make Statement 1 valid... But No certain value .

Insufficient

Combine 1 & 2

From 1: We know that we need a negative number to make make it valid

From 2: We got -2

Hence we reach certain value

Answer: C

energetics · Answer

If x is not equal to 0 or -1, what is the value of |x-2|?

(2)  x^4 = 16 
Go to statement 2 first since it's easier and can provide hints.
x = 2, -2 
So |x-2| = 0 or 4, Not sufficient.

(1)  \frac{(|x+1|)}{(-|x|x^3)}>0 
1. multiply by -1
|x+1|/|x|*x^3 < 0 
2. Draw number line or test values algebraically, taking into account constraints in prompt (x≠0, -1)
Test x=1 ... 2/1*1 < 0, NO
Test x=-1/2 ... (1/2) / -(1/8) = -8 < 0, YES
Test x=-2 ... 1 / -(1/4) = -4 < 0, YES
So we have several possible solutions for x, not sufficient.

(3) Combining the two we can see that x=-2 is a valid option and x=2 is not, so |x-2| = 4, sufficient.

GDT · Answer

chetan2u ,  VeritasKarishma

In statement 2, can't we take 4th root of both sides and get x= +2 as answer

KarishmaB · Answer

When you take the fourth root, you will get 
|x| = 2
which will give x = 2 or -2

Note that  \sqrt{x^2} = |x|

CEdward · Answer

If x is not equal to 0 or -1, what is the value of |x-2|?

(1) (|x+1|)(−|x|x^3)>0
(2) x^4 = 16

St1: (|x+1|)(−|x|x^3)>0
Open the modulus
x + 1 / (-x) x (x^3) <--- We see right away that x MUST BE < -1
Pick values
x = -2 ---> 1/16
x = -3 ---> 2/81

St2: x^4 = 16
\sqrt{x^4} = \sqrt{4√16}
|x| = 2
x = -2,2
Insufficient

Combo pack:
We know from st1 that x <-1 so x = -2

Sufficient.

happypuppy · Answer

Since (1) is an in-equality, 
 \frac{(|x+1|)}{(-|x|x^3)}>0 
We have range of x. For ex, (1) is true for -10, -20, -100 etc. Hence, can't get one value of x.

(2)  x^4 = 16 
This is an equation. We have +2 or -2 as real solutions.

Combining, (1) and (2), -2 works on (1) to make it > 0. 
This means, (C)

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If x is not equal to 0 or -1, what is the value of |x-2|?

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