Is |x^2 + y^2| |x^2 - y^2|? (1) x y (2) x 0

Question

Is  |x^2 + y^2| > |x^2 - y^2| ?

(1)  x > y 
(2)  x > 0

Bunuel · Accepted Answer

If you square, the questions becomes: is x^2y^2>0 ? Now, this holds true if xy eq{0} . But even when we consider the statements together we cannot say whether y=0 or not. Therefore the answer is E. Similar questions to practice: https://gmatclub.com/forum/is-x-y-x-y-123108.html https://gmatclub.com/forum/if-a-y-z-b-is ... 82673.html https://gmatclub.com/forum/is-x-y-146991.html https://gmatclub.com/forum/is-x-y-x-y-1- ... 32654.html https://gmatclub.com/forum/is-x-y-x-y-137050.html https://gmatclub.com/forum/is-x-y-x-z-1- ... 86132.html Hope it helps.

sujoykrdatta · Answer

Since square of any number is ALWAYS non-negative, we can say that:

|x^2 + y^2| > |x^2 - y^2|
will always be true as long as x and y are non-negative
 
As soon as x = 0 Or y = 0 Or x = y = 0, we have: 
|x^2 + y^2| = |x^2 - y^2|

Example: 
If y=0: |x^2 + 0^2| = |x^2 - 0^2| = x²
If x=0: |x^2 + y^2| = |x^2 - y^2| = y²

Thus, we need to ensure neither x nor y is zero.

Statement 1: x > y
Not sufficient to say anything since either term can be zero - insufficient

Statement 2: x > 0 
Not sufficient to say anything about y since it can be zero - insufficient

Combining: 
Not sufficient to say anything since y can still be zero - insufficient

Example:
x = 4, y = 0: 
|x^2 + y^2| = |x^2 - y^2| = 16

x = 4, y = 1: 
|x^2 + y^2| = 17, that is greater than |x^2 - y^2| = 15

Answer E

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Showmeyaa · Answer

Given:
 |x^2 + y^2| > |x^2 - y^2| 
 (x^2 + y^2)^{2} > (x^2 - y^2)^{2} 
 (x^2 + y^2)^{2} - (x^2 - y^2)^{2} > 0 
 (x^2 + y^2 + x^2 - y^2)(x^2 + y^2 - x^2 + y^2) > 0 
 4x^2y^2 > 0 
 x^2y^2 >0 
Is  x^2y^2 > 0  (If we can prove x is not equal to 0 and y is not equal to 0, we should be good)

1) x > y:
 Case1:  
x = 1, y = 0 
 x^2y^2 = 0 
 Case2: 
x = 2, y = 1
 x^2y^2 > 0 
Hence, Insufficient.
2) x > 0:
 Case1: 
x = 1, y = 0
 x^2y^2 = 0 
 Case2: 
x = 2, y = 1
 x^2y^2 > 0 
Hence, Insufficient.
Combining 1 & 2:
x > 0
x > y
 Case1: 
x = 1, y = 0
 x^2y^2 = 0 
 Case2: 
x = 2, y = 1
 x^2y^2 > 0 
Hence, Insufficient.
 Option E. 
PS: If option B was y > 0, then the answer would have been C. As x > y > 0.

chetan2u · Answer

The inequality will always be true, unless one of the two or both are 0.

Combined 
We can still not say whether y=0.

When y=0
 |x^2+0|=|x^2-0| , so answer is No.

When x=2, and y=1, answer is yes. 
Insufficient

E

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Is |x^2 + y^2| > |x^2 - y^2|? (1) x > y (2) x > 0

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