If 0

Question

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

BrentGMATPrepNow · Accepted Answer

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

prudhvi0410 · Answer

The equation can be written as √x-√y+√x+√y => 2√x
Option B

Sent from my ONE A2003 using  GMAT Club Forum mobile app

BrentGMATPrepNow · Answer

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

allan89 · Answer

Would solve using the property root x^2 = |x|

above expression is equal to root ((root x - root y)^2) + root ((root x + root y)^2)

= | root x - root y| + | root x + root y|

since root x < root y

= root y- root x + root + root y
= 2* root y

Cheers

Kinshook · Answer

Asked: If  0 < x < 1  and  y > 1 ,  \sqrt{x + y − 2\sqrt{xy}} + \sqrt{x + y + 2\sqrt{xy}} =

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

IMO E

brt · Answer

according to the property
when x<=0, then root x2=-x
when x>=0, then root x2=x

since y>x......
we get 2root y

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions