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Bunuel
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M41-28 13 May 2024, 07:31
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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. \(2\)
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C
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M41-28 13 May 2024, 07:31
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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
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M41-28 03 May 2025, 13:18
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Bunuel,

I have a question. According to the rules of absolute values, sqrt{x^2}would be lxl, which could be either negative or positive. In this problem, sqrt{4} = 2,not -2. Hence why the answer is C. Is this because x>y>0?

Thanks!
Quote:
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
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M41-28 03 May 2025, 13:28
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braddouglas10
Bunuel,

I have a question. According to the rules of absolute values, sqrt{x^2}would be lxl, which could be either negative or positive. In this problem, sqrt{4} = 2,not -2. Hence why the answer is C. Is this because x>y>0?

Thanks!
Quote:
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
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1. Even roots cannot give negative result.

\(\sqrt{...}\) is the square root sign, a function (called the principal square root function), which cannot give negative result. So, this sign (\(\sqrt{...}\)) always means non-negative square root.


The graph of the function f(x) = √x

Notice that it's defined for non-negative numbers and is producing non-negative results.

TO SUMMARIZE:
When the GMAT (and generally in math) provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).

2. Original expression issue.

We have\((\sqrt{x}-\sqrt{y})^2=4\). When taking the square root we get \(|\sqrt{x}-\sqrt{y}|=2\). This implies that \(\sqrt{x}-\sqrt{y}=2\) or \(\sqrt{x}-\sqrt{y}=-2\). However, since \(x-y > 0\), then \(\sqrt{x}-\sqrt{y}>0\), thus the second solution \(\sqrt{x}-\sqrt{y} = -2\) is invalid and we are left with \(\sqrt{x}-\sqrt{y}=2\) only.

Hope it helps.
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M41-28 03 May 2025, 14:00
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That did help. Thank you, Bunuel!
Bunuel
braddouglas10
Bunuel,

I have a question. According to the rules of absolute values, sqrt{x^2}would be lxl, which could be either negative or positive. In this problem, sqrt{4} = 2,not -2. Hence why the answer is C. Is this because x>y>0?

Thanks!
Quote:
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

1. Even roots cannot give negative result.

\(\sqrt{...}\) is the square root sign, a function (called the principal square root function), which cannot give negative result. So, this sign (\(\sqrt{...}\)) always means non-negative square root.


The graph of the function f(x) = √x

Notice that it's defined for non-negative numbers and is producing non-negative results.

TO SUMMARIZE:
When the GMAT (and generally in math) provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).

2. Original expression issue.

We have\((\sqrt{x}-\sqrt{y})^2=4\). When taking the square root we get \(|\sqrt{x}-\sqrt{y}|=2\). This implies that \(\sqrt{x}-\sqrt{y}=2\) or \(\sqrt{x}-\sqrt{y}=-2\). However, since \(x-y > 0\), then \(\sqrt{x}-\sqrt{y}>0\), thus the second solution \(\sqrt{x}-\sqrt{y} = -2\) is invalid and we are left with \(\sqrt{x}-\sqrt{y}=2\) only.

Hope it helps.
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Danielmm
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M41-28 14 Sep 2025, 05:16
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Can you elaborate the logic of factoring the denominator?

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
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Bunuel
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M41-28 14 Sep 2025, 10:54
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Danielmm
Can you elaborate the logic of factoring the denominator?


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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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