Is (x – 2)^2 x^2? (1) x^2 x (2) 1/x 0

Question

Is  (x – 2)^2 > x^2 ?

(1)  x^2 > x

(2)  \frac{1}{x} > 0

Bunuel · Accepted Answer

Is (x - 2)^2 > x^2?

Let's work on the stem first:

Is \((x-2)^2>x^2\)? --> Is \(x^2-4x+4>x^2\)? --> Is \(x<1\)? So this is what the question is basically asking.

(1) \(x^2>x\) --> \(x(x-1)>0\) --> \(x<0\) OR \(x>1\). Two ranges, not sufficient.

(2) \(\frac{1}{x}>0\) --> \(x>0\). Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is \(x>1\). Hence the answer to the question is no. Sufficient.

Answer: C.

srini123 · Answer

Is (x – 2)^2 > x^2?
(A) x^2 > x
(B) 1/x > 0

Answer: C

A) is not sufficient by itself as x can be positive or negative, but x cannot be a fraction i.e., x is not between -1 and 1 (inclusive)
the question stem is true when x is negative and false when x is positive so insufficient

B) from this , x must be positive or x>0

when x is 1/2 , question stem is TRUE, when x is >= 1 its false; so B by itself insufficient

taking together we know that x is > 1 and the question stem is always FAlse... hence answer C

honey86 · Answer

Need some help here.

Is (x - 2)^2 > x^2?

Let's work on the stem first:

Is (x-2)^2>x^2? --> Is x^2-4x+4>x^2? --> Is x<1? So this is what the question is basically asking.

(1) x^2>x --> x(x-1)>0 --> x<0 OR x>1. Two ranges, not sufficient. As per me it should be x>0 and x>1. I know this is wrong as I tried with 1/2 as a value of x which falsifies the equation.

Per my understanding,
if the equation were x(x-1) = 0, x can have two values i.e. 0 & 1. Please help me understand if I am missing out something.

(2) \frac{1}{x}>0 --> x>0. Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is x>1. Hence the answer to the question is no. Sufficient.

Answer: C.

Bunuel · Answer

x>0 and x>1 does not make any sense. x(x-1)>0 --> the "roots" are 0 and 1, this gives us 3 ranges: x<0; 01. Next, test some extreme value for x: if x is some large enough number, say 10, then both multiples will be positive which gives the positive result for the whole expression, so when x>1 the expression is positive. Now the trick: as in the 3rd range expression is positive then in the 2nd it'll be negative and finally in the 1st it'll be positive again: + - + . So, the ranges when the expression is positive are: x<0 and x>1. Check this for more: x2-4x-94661.html#p731476 Hope it helps.

honey86 · Answer

Got it!
Thanks a lot Bunuel. You have explained it very well on the other link that you posted.

Will work on more such problems.

Sailesh1986 · Answer

How about this case? Since its not mentioned x an is integer or a fraction, let x = 3/2 So stmt becomes (3/2 - 2)^2 > (3/2) which is false. Ans C is when 0 < x < 1 But when x > 1, then ans should be E

Bunuel · Answer

x cannot be less than or equal to 1, because when we combine the statements we get that x>1.

US09 · Answer

Why not A?

In 1, if x^2>x, then wouldn't x>1?

chetan2u · Answer

(x – 2)^2 > x^2.....x^2-4x+4>x^2....4x<4....x<1 So Q basically asks us - Is x<1? (1) x^2 > x .. x^2 > x....x^2-x>0...x(x-1)>0 .. so if x>0, x-1>0 or x>1...NO if x<0, x-1<0 or x<1...YES insuff (2) \frac{1}{x} > 0 .. \frac{1}{x} > 0 .. this tells us that x >0 insuff combined x>0, so x>1 ans is NO suff C

amanvermagmat · Answer

Hi Urvashi

If x^2 > x , then it could mean two things:
Either x > 1 Or 
x < 0 (x can take any negative value which would make its square positive, positive is always greater than negative). 
Thats why first statement alone is NOT sufficient.

BrentGMATPrepNow · Answer

Target question : Is (x - 2)² > x²? This is a great candidate for rephrasing the target question . Take (x - 2)² > x² and expand the left side to get: x² - 4x + 4 > x² Subtract x² from both sides to get: -4x + 4 > 0 Add 4x to both sides to get: 4 > 4x Divide both sides by 4 to get: 1 > x REPHRASED target question : Is x < 1? Aside: the video below has tips on rephrasing the target question Statement 1 : x² > x Subtract x from both sides of the inequality to get: x² - x > 0 Factor to get: x(x - 1) > 0 This means that EITHER x < 0 OR x > 1 So there are two possible cases to consider. case a : If x < 0, then it IS the case that x < 1 case b : If x > 1, then it is NOT the case that x < 1 Since we cannot answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT Statement 2 : (1/x) > 0 If 1 divided by x equals some POSITIVE value, we can conclude that x is POSITIVE If x is POSITIVE, then x could be greater than 1 , or x could be less than 1 Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT Statements 1 and 2 combined Statement 1 tells us that that EITHER x < 0 OR x > 1 Statement 2 tells us that x is POSITIVE So, we can eliminate the possibility that x < 0 This means it MUST be the case that x > 1 So, we can conclude that x is NOT less than 1 Since we can answer the REPHRASED target question with certainty, the combined statements are SUFFICIENT Answer: C Cheers, Brent

chetan2u · Answer

Is (x – 2)^2 > x^2 ? (x^2-4x+4)>(x)^2………..4x<4………x<1 So we are looking for whether x is less than 1. (1) x^2 > x…..x(x-1)>0 If x>0, then x-1>0 or x>1….x>1 If x<0, then x-1<0 or x<1….x<0 We cannot say whether x<1. Insufficient (2) \frac{1}{x} > 0 Since 1>0, the denominator x is also >0. Again, we cannot say whether x<1. Insufficient Combined x>0, and from statement I we get x>1. Sufficient as answer is NO. C

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