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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
­
­
\(x+y=4+2\sqrt{xy}\)

\(x+y-2\sqrt{xy}=4\)

\((\sqrt{x} - \sqrt{y})^2=4\)­

Taking square root on both sides, we get

\(\sqrt{x} - \sqrt{y} = 2\)­

Note: We are ignoring the possibility of -2, as x > y­

\(\frac{\sqrt{2x}+\sqrt{2y}}{x-y}\)

\(\frac{\sqrt{2}(\sqrt{x}+\sqrt{y})}{(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}\)­

\(\frac{\sqrt{2}}{2}\)­

Am I missing something ?  :?
­i am also getting same.
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Fixed the options. Thank you both!
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gmatophobia
your solution is so elegant and faster to do.
I got the same answer
But it was so tedious
When you look at these algebraic equations, how finalise the method to go about it?
how do you decide which method to employ when faced with a question?
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Generally when I approach an algebra question, I'd spend 10 seconds to identify if the given expression can be modified to fit a formula like (a-b)^2, a^2-b^2, etc.
99% of the times it will fit into a formula.
But I'd suggest practice a lot to identify that.
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How come x−y is (x√+y√)(x√−y√) in denominator in the last 2 steps.
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Dp5Shivam
Bunuel
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
How come x−y is (x√+y√)(x√−y√) in denominator in the last 2 steps.

Check the highlighted part in the solution above.

\(x-y= (\sqrt{x})^2 - (\sqrt{y})^2 = (\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y})\)
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ive read the highlighted part before but howcome X-Y be replaced with Underoot X & underoot Y.
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Dp5Shivam
ive read the highlighted part before but howcome X-Y be replaced with Underoot X & underoot Y.

x can be written as \((\sqrt{x})^2\), similarly, y can be written as \((\sqrt{y})^2\). So:

\(x-y= (\sqrt{x})^2 - (\sqrt{y})^2 = (\sqrt{x} - \sqrt{y})(\sqrt{x} + \sqrt{y})\)
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