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605-655 (Medium)|   Algebra|      
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Bunuel
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If x - y = x^(1/2) + y^(1/2), what is x^(1/2) in terms of y? 10 Apr 2023, 10:45
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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45% (medium)

Question Stats:

68% (02:13) correct 32% (02:43) wrong based on 322 sessions

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If \(x-y=\sqrt{x}+\sqrt{y}\), what is \(\sqrt{x}\) in terms of \(y\)?

A. \(1+\sqrt{y}\)
B. \(1-\sqrt{y}\)
C. \(\sqrt{y}-1\)
D. \(1-y\)
E. \(1+y\)
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A
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If x - y = x^(1/2) + y^(1/2), what is x^(1/2) in terms of y? 10 Apr 2023, 13:07
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Bunuel
If \(x-y=\sqrt{x}+\sqrt{y}\), what is \(\sqrt{x}\) in terms of \(y\)?

A. \(1+\sqrt{y}\)
B. \(1-\sqrt{y}\)
C. \(\sqrt{y}-1\)
D. \(1-y\)
E. \(1+y\)
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\(x-y=\sqrt{x}+\sqrt{y}\)

\((\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})- (\sqrt{x}+\sqrt{y}) = 0\)

\((\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y} - 1) = 0\)

\(\sqrt{x} = 1 + \sqrt{y}\)

Option A
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If x - y = x^(1/2) + y^(1/2), what is x^(1/2) in terms of y? 06 Aug 2024, 06:21
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\((\sqrt{x} + \sqrt{y})(\sqrt{­x} - \sqrt{y}) = (\sqrt{x} + \sqrt{y})\)

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

\((\sqrt{­x} - \sqrt{y}) = 1\)
\(\sqrt{­x} = 1 + \sqrt{y}\)­
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If x - y = x^(1/2) + y^(1/2), what is x^(1/2) in terms of y? 01 Sep 2024, 06:34
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Please note that √x and √y will be positive. (even root)

x−y=√x+√y

Rearranging the equation, we get

x- √x = y + √y
√x (√x - 1) = √y(√y+1)
On comparing, we can say that √x= √y + 1. Hence, the answer.
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If x - y = x^(1/2) + y^(1/2), what is x^(1/2) in terms of y? 01 Sep 2024, 06:45
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Bunuel
If \(x-y=\sqrt{x}+\sqrt{y}\), what is \(\sqrt{x}\) in terms of \(y\)?

A. \(1+\sqrt{y}\)
B. \(1-\sqrt{y}\)
C. \(\sqrt{y}-1\)
D. \(1-y\)
E. \(1+y\)
\(x-y=\sqrt{x}+\sqrt{y}\)

\((\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})- (\sqrt{x}+\sqrt{y}) = 0\)

\((\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y} - 1) = 0\)

\(\sqrt{x} = 1 + \sqrt{y}\)

Option A
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­Hi gmatophobia
Can you please let me know why did you pull all the various to LHS? We can directly divide and get the answer. Is it done to check all possible roots?
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If x - y = x^(1/2) + y^(1/2), what is x^(1/2) in terms of y? 02 Sep 2024, 00:04
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btsaami

gmatophobia

Bunuel
If \(x-y=\sqrt{x}+\sqrt{y}\), what is \(\sqrt{x}\) in terms of \(y\)?

A. \(1+\sqrt{y}\)
B. \(1-\sqrt{y}\)
C. \(\sqrt{y}-1\)
D. \(1-y\)
E. \(1+y\)
\(x-y=\sqrt{x}+\sqrt{y}\)

\((\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})- (\sqrt{x}+\sqrt{y}) = 0\)

\((\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y} - 1) = 0\)

\(\sqrt{x} = 1 + \sqrt{y}\)

Option A
­Hi gmatophobia
Can you please let me know why did you pull all the various to LHS? We can directly divide and get the answer. Is it done to check all possible roots?
Show more
The question should have specified that x and y are positive numbers. In this case, we could safely divide \((\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})\) by \((\sqrt{x} + \sqrt{y})\) and get \(\sqrt{x} - \sqrt{y} = 1\).

However, without that condition, we cannot divide \((\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})\) by \((\sqrt{x} + \sqrt{y})\) because \((\sqrt{x} + \sqrt{y})\) could be 0 when x = y = 0, and division by 0 is not allowed. So, we should follow the approach used in the solution, which gives \((\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y} - 1) = 0\). This, in turn, gives two possibilities: \(\sqrt{x} + \sqrt{y} = 0\) or \(\sqrt{x} - \sqrt{y} - 1 = 0\). Thus, the correct answer to the question would be \(\sqrt{x} = -\sqrt{y}\) or \(\sqrt{x} = \sqrt{y} + 1\).

So, the wording of the question is not precise overall.

Hope it's clear.­
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