Is x² + y² -2xy? (1) xy 0 (2) x - y = 4

Question

Is x2 + y2 > -2xy?

(1) xy > 0
(2) x - y = 4

gmatophobia · Answer

x^2 + y^2 > -2xy

x^2 + y^2 +2xy > 0

(x+y)^2 > 0

The square of any real number is always non-negative. Hence if  x + y 
eq 0 ,  (x+y)^2 > 0

Statement 1

(1) xy > 0

x and y have the same sign and the numbers are not equal to zero. Hence, we can conclude that  (x+y)^2 > 0 .

The statement is sufficient.

We can eliminate B, C, and E.

Statement 2

(2) x - y = 4

The statement tells that the distance between x and y on a number line is 4. That is the numbers are 4 units apart from each other.

However, we don't know if the numbers are on the same side with respect to zero or on the opposite sides.

For example: x = 2 ; y = -2

x + y = 0; in that case  (x+y)^2 = 0

x = 4 ; y = 0

x + y > 0; in that case  (x+y)^2 > 0

As we don't have a definite answer, the statement is not sufficient to answer the question.

Option A

BrentGMATPrepNow · Answer

Target question :    Is x² + y² > -2xy?  
This is a good candidate for rephrasing the target question. 
Add 2xy to both sides to get:  Is x² + 2xy + y² > 0? 
Factor to get:  Is (x + y)² > 0? 
REPHRASED target question:   Is (x + y)² > 0?

IMPORTANT: Notice that (x + y)² is almost always greater than 0. The only time (x + y)² isn't greater than 0, is when x + y = 0.

Statement 1 : xy > 0  
This means x and y are both the SAME SIGN, which means there are 2 cases to consider:
 Case a : x and y are both positive, which means x + y > 0, which means the answer to the REPHRASED target question is  YES, (x + y)² is greater than 0 
 Case b : x and y are both negative, which means x + y < 0, which means the answer to the REPHRASED target question is  YES, (x + y)² is greater than 0 
Since we get the same answer ( YES, (x + y)² is greater than 0 ) in both cases, we can answer the  REPHRASED target question  with certainty. 
Statement 1 is SUFFICIENT

Statement 2 : x - y = 4 
There are several values of x and y that satisfy statement 2. Here are two: 
 Case a : x = 6 and y = 2, which means (x + y)² = (6 + 2)² = 8² = 64. In this case, the answer to the REPHRASED target question is  YES, (x + y)² is greater than 0 
 Case b : x = 2 and y = -2, which means (x + y)² = (2 + -2)² = 0² = 0. In this case, the answer to the REPHRASED target question is  NO, (x + y)² is NOT greater than 0 
Since we can’t answer the  REPHRASED target question  with certainty, statement 2 is NOT SUFFICIENT

Answer: A

VIDEO ON REPHRASING THE TARGET QUESTION: 
 https://www.youtube.com/watch?v=MyG1K3ee69w

TheBipedalHorse · Answer

Is x² + y² > -2xy?

(1) xy > 0
(2) x - y = 4

Statement 1 --->  xy>0 
LHS of the equation will always be non negative. If xy>0, that means -2xy will always be negative. Adding two squared terms to give a negative term will always be greater or equal to 0 (unless we are talking of imaginary numbers). Hence statement 1 is sufficient

Statement 2 --->  x-y=4 
Again, LHS is always non negative. -2xy can be negative, zero or positive depending on what values of x and y you choose. Since there are infinite number of solutions for x-y=4, we cannot say for sure if -2xy would be negative, zero or positive

To explain above logic further
if x-y=4, we can have solutions where x and y are of the same sign (either both positive or both negative) or of different sign (one of x and y are positive and the other is negative)
In both the cases -2xy have different signs. Furthermore, either x or y can be zero as well.

Hence statement 1 alone is sufficient

bumpbot · Answer

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Is x² + y² > -2xy? (1) xy > 0 (2) x - y = 4

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