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Question Stats:
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25%
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If \(x > y > 0\) and \(x+y=4+2\sqrt{xy}\), what is the value of \(\frac{\sqrt{2x}+\sqrt{2y}}{x-y}\)?
A. \(-2\)
B. \(-\frac{\sqrt{2}}{2}\)
C. \(\frac{\sqrt{2}}{2}\)
D. \(\sqrt{2}\)
E. \(2\)
A. \(-2\)
B. \(-\frac{\sqrt{2}}{2}\)
C. \(\frac{\sqrt{2}}{2}\)
D. \(\sqrt{2}\)
E. \(2\)
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02 Jun 2024, 07:30
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Official Solution:
If \(x > y > 0\) and \(x+y=4+2\sqrt{xy}\), what is the value of \(\frac{\sqrt{2x}+\sqrt{2y}}{x-y}\)?
A. \(-2\)
B. \(-\frac{\sqrt{2}}{2}\)
C. \(\frac{\sqrt{2}}{2}\)
D. \(\sqrt{2}\)
E. \(2\)
Start with the question: \(\frac{\sqrt{2x}+\sqrt{2y}}{x-y}=?\)
Factor out \(\sqrt{2}\) from the numerator and apply the difference of squares formula, \(a^2-b^2=(a-b)(a+b)\), to the expression in the denominator:
\(\frac{\sqrt{2}(\sqrt{x}+\sqrt{y})}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}=\frac{\sqrt{2}}{\sqrt{x}-\sqrt{y}}\). Essentially, all we need to find is the value of \(\sqrt{x}-\sqrt{y}\).
Next, consider the equation \(x+y = 4 + 2\sqrt{xy}\):
\(x-2\sqrt{xy}+y=4\);
\((\sqrt{x}-\sqrt{y})^2=4\);
\(\sqrt{x}-\sqrt{y} = 2\) (since \(x-y > 0\), the second solution \(\sqrt{x}-\sqrt{y} = -2\) is not valid).
Therefore,\(\frac{\sqrt{2}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{2}}{2}\).
Answer: C
If \(x > y > 0\) and \(x+y=4+2\sqrt{xy}\), what is the value of \(\frac{\sqrt{2x}+\sqrt{2y}}{x-y}\)?
A. \(-2\)
B. \(-\frac{\sqrt{2}}{2}\)
C. \(\frac{\sqrt{2}}{2}\)
D. \(\sqrt{2}\)
E. \(2\)
Start with the question: \(\frac{\sqrt{2x}+\sqrt{2y}}{x-y}=?\)
Factor out \(\sqrt{2}\) from the numerator and apply the difference of squares formula, \(a^2-b^2=(a-b)(a+b)\), to the expression in the denominator:
\(\frac{\sqrt{2}(\sqrt{x}+\sqrt{y})}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}=\frac{\sqrt{2}}{\sqrt{x}-\sqrt{y}}\). Essentially, all we need to find is the value of \(\sqrt{x}-\sqrt{y}\).
Next, consider the equation \(x+y = 4 + 2\sqrt{xy}\):
\(x-2\sqrt{xy}+y=4\);
\((\sqrt{x}-\sqrt{y})^2=4\);
\(\sqrt{x}-\sqrt{y} = 2\) (since \(x-y > 0\), the second solution \(\sqrt{x}-\sqrt{y} = -2\) is not valid).
Therefore,\(\frac{\sqrt{2}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{2}}{2}\).
Answer: C
General Discussion
15 May 2024, 02:37
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Bunuel
If \(x > y > 0\) and \(x+y=4+2\sqrt{xy}\), what is the value of \(\frac{\sqrt{2x}+\sqrt{2y}}{x-y}\)?
A. -4
B. -2
C. 1
D. 2
E. 4
A. -4
B. -2
C. 1
D. 2
E. 4
\(x+y=4+2\sqrt{xy}\)
\(x+y-2\sqrt{xy}=4\)
\((\sqrt{x} - \sqrt{y})^2=4\)
Taking square root on both sides, we get
\(\sqrt{x} - \sqrt{y} = 2\)
Note: We are ignoring the possibility of -2, as x > y
\(\frac{\sqrt{2x}+\sqrt{2y}}{x-y}\)
\(\frac{\sqrt{2}(\sqrt{x}+\sqrt{y})}{(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}\)
\(\frac{\sqrt{2}}{2}\)
Option C
