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16 Sep 2014, 00:58
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If \(x \gt y \gt 0\) then what is the value of \(\frac{\sqrt{2x}+\sqrt{2y}}{x-y}\)?
(1) \(x+y=4+2\sqrt{xy}\)
(2) \(x-y=9\)
(1) \(x+y=4+2\sqrt{xy}\)
(2) \(x-y=9\)
16 Sep 2014, 00:58
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Official Solution:
Given: \(\frac{\sqrt{2x}+\sqrt{2y}}{x-y}\). Factor out \(\sqrt{2}\) from the nominator and apply \(a^2-b^2=(a-b)(a+b)\) to the expression in the denominator: \(\frac{\sqrt{2}(\sqrt{x}+\sqrt{y})}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}=\frac{\sqrt{2}}{\sqrt{x}-\sqrt{y}}\). So we should find the value of \(\sqrt{x}-\sqrt{y}\).
(1) \(x+y=4+2\sqrt{xy}\). Rearrange: \(x-2\sqrt{xy}+y=4\), so \((\sqrt{x}-\sqrt{y})^2=4\), which gives \(\sqrt{x}-\sqrt{y}=2\) (note that since \(x-y \gt 0\) then the second solution \(\sqrt{x}-\sqrt{y}=-2\) is not valid). Sufficient.
(2) \(x-y=9\). Not sufficient.
Answer: A
Given: \(\frac{\sqrt{2x}+\sqrt{2y}}{x-y}\). Factor out \(\sqrt{2}\) from the nominator and apply \(a^2-b^2=(a-b)(a+b)\) to the expression in the denominator: \(\frac{\sqrt{2}(\sqrt{x}+\sqrt{y})}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}=\frac{\sqrt{2}}{\sqrt{x}-\sqrt{y}}\). So we should find the value of \(\sqrt{x}-\sqrt{y}\).
(1) \(x+y=4+2\sqrt{xy}\). Rearrange: \(x-2\sqrt{xy}+y=4\), so \((\sqrt{x}-\sqrt{y})^2=4\), which gives \(\sqrt{x}-\sqrt{y}=2\) (note that since \(x-y \gt 0\) then the second solution \(\sqrt{x}-\sqrt{y}=-2\) is not valid). Sufficient.
(2) \(x-y=9\). Not sufficient.
Answer: A
