If x > y > 0 and x + y = 4 + 2(xy)^(1/2), what is the value of

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