Is x>y?

claudayst wrote:Is x > y?

1) x + y + 2√(xy) = 0
2) x² + y² = 0

Statement 1: x + y = -2√(xy)
Squaring both sides, (x + y)² = [-2√(xy)]²
--> x² + 2xy + y² = 4xy
--> x² - 2xy + y² = 0
--> (x - y)² = 0
--> (x - y) = 0
--> x = y

Sufficient

Statement 2: As x² and y² cannot be negative, only way (x² + y²) can be equal to zero if x = y = 0

Sufficient

The correct answer is D.

rohankamath619 wrote:My approach was a little different. Anju, can you please let me know if this is correct?

x + y + 2√(xy) = 0 can also be expressed as (√x+√y)²=0 That would leave us with:
√x+√y = 0
√x = - √y
x = y

Although you've reached the correct conclusion, this is not mathematically correct.
By definition, √x is the positive square root of x (if x ≠ 0).
Hence, it is not possible that √x = -√y = -(some positive number, again if y ≠ 0)

Now, where did you go wrong?
Your initial interpretation "x + y + 2√(xy) = 0 can also be expressed as (√x+√y)²=0" is correct. But this could lead to only one possible solution : as √x and √y are non-negative, both x and y must be equal to zero.

But by doing so, we are discarding a large number of possible solutions which is x = y as I have deduced. When did we discarded those solutions? As soon as we introduced √x and √y.

Let me explain with an example...
If we assume x = y = -1, then √x and √y are not defined (in the scope of GMAT).
But, x + y + 2√(xy) = (-1) + (-1) + 2*√[(-1)*(-1)] = -2 + 2√1 = 0

In fact, for statement 1 to be true, x and y must be equal to some non-positive integer. If, x = y = some positive integer, then x + y + 2√(xy) is always greater than zero.

Hope that helps.