M01-19

Question

If  x > y > 0 , what is the value of  \frac{\sqrt{2x}+\sqrt{2y}}{x-y} ?

(1)  x+y=4+2\sqrt{xy} 
 
(2)  x-y=9

Bunuel · Accepted Answer

Official Solution:

If  x > y > 0 , what is the value of  \frac{\sqrt{2x}+\sqrt{2y}}{x-y} ? 
 
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}} . 
 
So, we should find the value of  \sqrt{x}-\sqrt{y} .
 
(1)  x+y=4+2\sqrt{xy} 
 
 x-2\sqrt{xy}+y=4 ;
 
 (\sqrt{x}-\sqrt{y})^2=4 ;
 
 \sqrt{x}-\sqrt{y}=2  (note that since  x-y > 0 , the second solution  \sqrt{x}-\sqrt{y}=-2  is not valid). Sufficient.
 
(2)  x-y=9 . This information is clearly not sufficient on its own.

Answer: A

JackSparr0w · Answer

Does stmt 2, x-y=9, become:

((√x)^2) - ((√y)^2)= 9 :then

(√x+√y)(√x-√y) = 9

so the two factors above are either both 3 or (-3), which doesn't allow us to determine the sign of our rephrased question?

If so, is there a way to notice this without completing the work?

Bunuel · Answer

That's not correct. How did you get that from (√x+√y)(√x-√y) = 9 both factors are either 3 or -3?

sabineodf · Answer

Forgive my ignorance but I am a little bit confused about what happens here
(1) x+y=4+2\sqrt{xy}

x-2\sqrt{xy}+y=4;

(\sqrt{x}-\sqrt{y})^2=4; <----HERE where did the 2 go?

And also I am new to this forum so I am unfamiliar with how to format, and this is just what it looked like when I pressed paste. Sorry !

Thanks for any help, I am very grateful

Sabine

Bunuel · Answer

Apply a^2-2ab+b^2=(a-b)^2 : x-2\sqrt{xy}+y=(\sqrt{x})^2-2\sqrt{x}\sqrt{y}+(\sqrt{y})^2=(\sqrt{x}-\sqrt{y})^2 . P.S. Writing Mathematical Formulas on the Forum: rules-for-posting-please-read-this-before-posting-133935.html#p1096628

ankuradhikary · Answer

I think this is a  high-quality  question and I agree with explanation.

ENEM · Answer

Hi  Bunuel ,

if we apply

(a-b)^2  to the denominator, where does the how does it lead to  ( \sqrt{x} - \sqrt{y} ) ( \sqrt{x} + \sqrt{y} ) :

why not simply (x+y) (x-y)

Bunuel · Answer

First of all we are applying  a^2-b^2=(a-b)(a+b)  to the expression in the denominator, not  (a-b)^2=a^2-2ab+b^2 .

In the denominator we have x - y, which can be written as  x-y=(\sqrt{x})^2-(\sqrt{y})^2=(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}) .

Hope it's clear.

shahul. · Answer

Hi  Bunuel  and  chetan2u   VeritasPrepKarishma

I could'nt understand why the second solution √x−√y= - 2 is not valid when x−y>0. kindly explain if it has to solved algebraically

Bunuel · Answer

We are given that x > y > 0, so x > y, so √x > √y, so √x − √y > 0 and cannot equal to -2.

KarishmaB · Answer

Another way to look at it (if that helps): 
√x and √y have to be positive (principal square roots). So the only way √x−√y= - 2, if √x is 2 less than √y.
But if x - y > 0 i.e. x > y, then that is not possible.

sanjay1810 · Answer

Hi  Bunuel ,

Can you please elaborate on how st 2 is not sufficient ?

x - y = 9 
( \sqrt{x} + \sqrt{y} ) ( \sqrt{x} - \sqrt{y} ) = 9 = 9 . 1 OR 3. 3
Since it's given the x> y >0, sum of two numbers cannot be equal to difference of the same two numbers. So 3,3 is out. So 9, 1 is still left.
Both numbers positive, one greater than the other, 
so
\sqrt{x} + \sqrt{y} = 9
\sqrt{x} - \sqrt{y} = 1
this can be solved and we get the ans which obviously i wrong as per OA.

What's wrong here and why is 2 not sufficient?

Thanks.

Bunuel · Answer

You are assuming that  \sqrt{x} + \sqrt{y}  and  \sqrt{x} - \sqrt{y}  are integers, which is not given. Why cannot  \sqrt{x} + \sqrt{y}  be 81 and  \sqrt{x} - \sqrt{y}  be 1/9? x - y = 9 has infinitely many solutions, so you cannot get the single numerical value of  \sqrt{x} - \sqrt{y}

sanjay1810 · Answer

Fell for the trap. Thanks  Bunuel  as always for the quick response - eye opener!

Pari28 · Answer

I think this is a  high-quality  question and I agree with explanation.

vikastgmat · Answer

Hi @VeritasPrepKarishma,
Can you please explain why 
√x and √y have to be positive (principal square roots).

Bunuel · Answer

\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.

https://gmatclub.com/forum/download/file.php?id=39502 
 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 {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 .

CrackverbalGMAT · Answer

From the question data, x and y are both positive numbers and x>y
The value of the expression needs to be evaluated.

From statement I alone, x + y = 4 + 2√xy.

Transposing the variable terms onto the LHS, we have x + y - 2√xy. = 4
The expression on the LHS represents the simplified form of  (√x - √y)^2 . Therefore, the equation can be rewritten as  (√x - √y)^2  = 4.

Taking the square root on both sides, we have (√x - √y) = ± 2; since the question says that both x and y are positive, (√x - √y) cannot be equal to – 2, therefore, (√x - √y) = 2.

In the expression given in the question stem, factoring out common terms, we have,
 \frac{√2 (√x + √y )}{ (x-y)}

The denominator can be expressed as the product of (√x - √y) (√x + √y ).

Therefore,  \frac{√2 (√x + √y )}{ (x-y)}  =  \frac{√2 (√x + √y )}{ (√x - √y) (√x + √y )}

Cancelling off (√x + √y ) and substituting the value of (√x - √y), value of the expression =  \frac{√2 }{ 2}  =  \frac{1}{ √2} 
Statement I alone is sufficient to answer the question. Answer options B, C and E can be eliminated.

From statement II alone, x – y = 9.

If x = 25 and y = 16, the value of the expression equal to √2
If x = 26 and y = 17, the value of the expression is not equal to √2
Statement II alone is insufficient to find a unique value for the given expression. Answer option D can be eliminated.

The correct answer option is A.

Hope that helps!
Aravind BT

KarishmaB · Answer

vikastgmat 
√x is by definition called the principal square root or the positive square root. The negative square root is given by -√x.

So if we are given  \sqrt{4} , it means we are given the positive square root which will be 2 only.

So a question such as "what is the value of 
 \sqrt{4} + 3 " would be answered with 5.

If the negative square root has to be written, we need to write: "what is the value of 
 -\sqrt{4} + 3 " and this would be answered with 1.

PS - Tag me at VeritasKarishma

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