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42% (01:09) correct 58% (01:53) wrong based on 222 sessions
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15 Sep 2014, 23:15



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23 Nov 2014, 06:35
Does stmt 2, xy=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?



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26 Nov 2014, 08:48
I think I was thinking that the factors must be integers, but even then some combinaton of +/1 and 9 would work as well. Thanks.



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17 Feb 2015, 01:10
Forgive my ignorance but I am a little bit confused about what happens here (1) x+y=4+2\sqrt{xy}
x2\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



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25 Feb 2015, 14:56
Bunuel wrote: Official Solution:
Question: \(\frac{\sqrt{2x}+\sqrt{2y}}{xy}=?\) Factor out \(\sqrt{2}\) from the numerator and apply \(a^2b^2=(ab)(a+b)\) 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) \(x+y=4+2\sqrt{xy}\) \(x2\sqrt{xy}+y=4\); \((\sqrt{x}\sqrt{y})^2=4\); \(\sqrt{x}\sqrt{y}=2\) (note that since \(xy \gt 0\) then the second solution \(\sqrt{x}\sqrt{y}=2\) is not valid). Sufficient. (2) \(xy=9\). Not sufficient.
Answer: A Beautiful.
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08 Jul 2016, 15:13
i am little bit confused about this explanation. please explain in this way. as per above explanation, [square_root(x)  [square_root(y)= 2 , since xy>0 [square_root(x) [square_root(y)= 2 is not valid.
what if [square_root(x)= 5 and [square_root(y) = 3 x=25 and y=9 (x>y>0) xy= 16 >0 but [square_root(x) [square_root(y) = 2 question doesn't mention explicitly that both sqrt (x) and sqrt (y) are positive. if both sqrt(x) and sqrt(y) are positive, then sqrt(x) sqrt(y)= 2 or both sqrt(x) and sqrt(y) are negative ,then sqrt(x) sqrt(y)= 2



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15 Sep 2016, 11:37
I think this is a highquality question and I agree with explanation.



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28 Nov 2016, 00:04
Overlaps with m1609
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01 Jan 2017, 22:34
i am quite confused about that question.



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18 Jul 2017, 04:22
Hi Bunuel, if we apply \((ab)^2\) to the denominator, where does the how does it lead to \(( \sqrt{x}  \sqrt{y} ) ( \sqrt{x} + \sqrt{y} ) :\) why not simply (x+y) (xy)
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09 May 2018, 06:06
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



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11 May 2018, 01:19
shahul. wrote: 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 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.
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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.







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