Bunuel wrote:

Is x > y?

(1) \(x + 2\sqrt{xy} + y = 0\) --> \((\sqrt{x}+\sqrt{y})^2=0\) --> \(\sqrt{x}+\sqrt{y}=0\). The sum of two non-negative values is 0, only when both are 0, so x=y=0: x is not greater than y. Sufficient.

(2) x^2 - y^2 = 0 --> x^2 = y^2 --> |x| = |y|. x can be more than y (x=1, y=-1), less than y (x=-1, y=1) or equal to y (x=y=1). Not sufficient.

Answer: A.

Hi

BunuelFrom my point of view, in statement 1 we know that xy >= 0, but we can't conclude that x>=0 and y>=0 . Thus, writing \(\sqrt{x}\) and \(\sqrt{y}\) seems not quite right.

Can you please check it?

Good catch. You are absolutely right. Edited the post. Thank you.