lakshya wrote:
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.
(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.
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