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If 0 < x < 1 and y > 1, (√(x + y − 2√xy) + √(x + y + 2√xy) =

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Joined: 02 Sep 2009
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If 0 < x < 1 and y > 1, (√(x + y − 2√xy) + √(x + y + 2√xy) =  [#permalink]

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New post 20 Sep 2018, 00:50
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

31% (01:52) correct 69% (01:34) wrong based on 87 sessions

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Re: If 0 < x < 1 and y > 1, (√(x + y − 2√xy) + √(x + y + 2√xy) =  [#permalink]

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New post 20 Sep 2018, 00:57
The equation can be written as √x-√y+√x+√y => 2√x
Option B

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If 0 < x < 1 and y > 1, (√(x + y − 2√xy) + √(x + y + 2√xy) =  [#permalink]

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New post Updated on: 20 Sep 2018, 14:29
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Top Contributor
Bunuel wrote:
If \(0 < x < 1\) and \(y > 1\), \(\sqrt{x + y − 2\sqrt{xy}} + \sqrt{x + y + 2\sqrt{xy}} =\)


(A) √x
(B) 2√x
(C) √y +√x
(D) √y
(E) 2√y


It's probably not immediately apparent, but x - 2√(xy) + y is a form of the SPECIAL PRODUCT x² - 2xy + y²
Notice that (√x - √y)² = (√x - √y)(√x - √y) = x - 2√(xy) + y
Likewise, (√x + √y)² = (√x + √y)(√x + √y) = x + 2√(xy) + y

So...
\(\sqrt{x + y − 2\sqrt{xy}} + \sqrt{x + y + 2\sqrt{xy}} =\) (√x - √y)² + (√x + √y)²

-------------ASIDE-------------------------
IMPORTANT: before we conclude that (√x - √y)² = (√x - √y), we must recognize that, since 0 < x < y, √x - √y is NEGATIVE
Since the square root of a value cannot be NEGATIVE, it must be the case that (√x - √y)² = (√y - √x)
Here's an illustrative example:
(√1 - √9)² = (1 - 3)²
= (-2)²
= 4
= 2
Notice that √1 - √9 DOES NOT equal 2; it equals -2
However, √9 - √1 DOES equal 2
So, we can say that (√1 - √9)² = √9 - √1

Likewise, (√x - √y)² = (√y - √x)
-------------BACK TO THE QUESTION------------------------

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

Answer: E

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Originally posted by GMATPrepNow on 20 Sep 2018, 09:03.
Last edited by GMATPrepNow on 20 Sep 2018, 14:29, edited 1 time in total.
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Re: If 0 < x < 1 and y > 1, (√(x + y − 2√xy) + √(x + y + 2√xy) =  [#permalink]

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New post 20 Sep 2018, 14:35
Top Contributor
Bunuel wrote:
If \(0 < x < 1\) and \(y > 1\), \(\sqrt{x + y − 2\sqrt{xy}} + \sqrt{x + y + 2\sqrt{xy}} =\)


(A) √x
(B) 2√x
(C) √y +√x
(D) √y
(E) 2√y


Another (easier) approach is to just plug in values for x and y
How about x = 0.25 and y = 4

When we plug those values into the expression, we get:
\(\sqrt{x + y − 2\sqrt{xy}} + \sqrt{x + y + 2\sqrt{xy}} =\) \(\sqrt{0.25 + 4 − 2\sqrt{1}} + \sqrt{0.25 + 4 + 2\sqrt{1}}\)
= 3/2 + 5/2
= 8/2
= 4

So, when x = 0.25 and y = 4, the given expression evaluates to 4
Now which of the following answer choices evaluates to 4 when x = 0.25 and y = 4??
(A) √0.25 = 0.5 NO GOOD
(B) 2√0.25 = (2)(0.5) = 1 NO GOOD
(C) √4 +√0.25 = 2 + 0.5 = 2.5 NO GOOD
(D) √4 = 2 NO GOOD
(E) 2√4 = (2)(2) = 4 GREAT!!!

Answer: E

Cheers,
Brent
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Re: If 0 < x < 1 and y > 1, (√(x + y − 2√xy) + √(x + y + 2√xy) =  [#permalink]

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New post 20 Sep 2018, 14:38
Top Contributor
funsogu wrote:
Bunuel I don't understand why the OA is E. Shouldn't it be B? Please help. Thanks.

Bunuel wrote:
If \(0 < x < 1\) and \(y > 1\), \(\sqrt{x + y − 2\sqrt{xy}} + \sqrt{x + y + 2\sqrt{xy}} =\)


(A) √x
(B) 2√x
(C) √y +√x
(D) √y
(E) 2√y


Turns out there's a big problem with B.
See my (edited) post above.

Cheers,
Brent
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Joined: 18 Jun 2018
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Re: If 0 < x < 1 and y > 1, (√(x + y − 2√xy) + √(x + y + 2√xy) =  [#permalink]

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New post 20 Sep 2018, 14:42
Oh wow. I see my mistake now. Thanks GMATPrepNow

GMATPrepNow wrote:
funsogu wrote:
Bunuel I don't understand why the OA is E. Shouldn't it be B? Please help. Thanks.

Bunuel wrote:
If \(0 < x < 1\) and \(y > 1\), \(\sqrt{x + y − 2\sqrt{xy}} + \sqrt{x + y + 2\sqrt{xy}} =\)


(A) √x
(B) 2√x
(C) √y +√x
(D) √y
(E) 2√y


Turns out there's a big problem with B.
See my (edited) post above.

Cheers,
Brent
GMAT Club Bot
Re: If 0 < x < 1 and y > 1, (√(x + y − 2√xy) + √(x + y + 2√xy) = &nbs [#permalink] 20 Sep 2018, 14:42
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