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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
Posts: 52231
If 0 < x < 1 and y > 1, (√(x + y − 2√xy) + √(x + y + 2√xy) =  [#permalink]

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19 Sep 2018, 23:50
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Difficulty:

95% (hard)

Question Stats:

34% (02:00) correct 66% (01:33) wrong based on 111 sessions

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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

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Joined: 21 Feb 2018
Posts: 16
Re: If 0 < x < 1 and y > 1, (√(x + y − 2√xy) + √(x + y + 2√xy) =  [#permalink]

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19 Sep 2018, 23:57
The equation can be written as √x-√y+√x+√y => 2√x
Option B

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CEO
Joined: 11 Sep 2015
Posts: 3331
If 0 < x < 1 and y > 1, (√(x + y − 2√xy) + √(x + y + 2√xy) =  [#permalink]

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Updated on: 20 Sep 2018, 13:29
1
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

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Originally posted by GMATPrepNow on 20 Sep 2018, 08:03.
Last edited by GMATPrepNow on 20 Sep 2018, 13:29, edited 1 time in total.
CEO
Joined: 11 Sep 2015
Posts: 3331
Re: If 0 < x < 1 and y > 1, (√(x + y − 2√xy) + √(x + y + 2√xy) =  [#permalink]

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20 Sep 2018, 13:35
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!!!

Cheers,
Brent
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Joined: 11 Sep 2015
Posts: 3331
Re: If 0 < x < 1 and y > 1, (√(x + y − 2√xy) + √(x + y + 2√xy) =  [#permalink]

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20 Sep 2018, 13:38
Top Contributor
funsogu wrote:

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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Test confidently with gmatprepnow.com

Joined: 18 Jun 2018
Posts: 103
Location: United States (AZ)
Concentration: Finance, Healthcare
Re: If 0 < x < 1 and y > 1, (√(x + y − 2√xy) + √(x + y + 2√xy) =  [#permalink]

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20 Sep 2018, 13:42
Oh wow. I see my mistake now. Thanks GMATPrepNow

GMATPrepNow wrote:
funsogu wrote:

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
Re: If 0 < x < 1 and y > 1, (√(x + y − 2√xy) + √(x + y + 2√xy) = &nbs [#permalink] 20 Sep 2018, 13:42
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