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M01-19

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M01-19  [#permalink]

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New post 16 Sep 2014, 00:15
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A
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E

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New post 16 Sep 2014, 00:15
6
3
Official Solution:


Question: \(\frac{\sqrt{2x}+\sqrt{2y}}{x-y}=?\) Factor out \(\sqrt{2}\) from the numerator and apply \(a^2-b^2=(a-b)(a+b)\) 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 \(x-y \gt 0\) then the second solution \(\sqrt{x}-\sqrt{y}=-2\) is not valid). Sufficient.

(2) \(x-y=9\). Not sufficient.


Answer: A
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Re: M01-19  [#permalink]

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New post 23 Nov 2014, 07:35
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?
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New post 25 Nov 2014, 07:23
JackSparr0w wrote:
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?


That's not correct. How did you get that from (√x+√y)(√x-√y) = 9 both factors are either 3 or -3?
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Re: M01-19  [#permalink]

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New post 26 Nov 2014, 09:48
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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New post 17 Feb 2015, 02:10
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

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New post 17 Feb 2015, 02:48
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sabineodf wrote:
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


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
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Re: M01-19  [#permalink]

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New post 25 Feb 2015, 15:56
Bunuel wrote:
Official Solution:


Question: \(\frac{\sqrt{2x}+\sqrt{2y}}{x-y}=?\) Factor out \(\sqrt{2}\) from the numerator and apply \(a^2-b^2=(a-b)(a+b)\) 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 \(x-y \gt 0\) then the second solution \(\sqrt{x}-\sqrt{y}=-2\) is not valid). Sufficient.

(2) \(x-y=9\). Not sufficient.


Answer: A


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Re: M01-19  [#permalink]

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New post 08 Jul 2016, 16:13
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 x-y>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)
x-y= 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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New post 09 Jul 2016, 00:09
mrinmoyc987 wrote:
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 x-y>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)
x-y= 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


The square root function cannot give negative result.
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Re M01-19  [#permalink]

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New post 15 Sep 2016, 12:37
I think this is a high-quality question and I agree with explanation.
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New post 28 Nov 2016, 01:04
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Overlaps with m16-09
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New post 01 Jan 2017, 23:34
i am quite confused about that question.
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New post 18 Jul 2017, 05:22
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)
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New post 18 Jul 2017, 05:29
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Kaal wrote:
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)


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.
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Re: M01-19  [#permalink]

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New post 09 May 2018, 07:06
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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New post 09 May 2018, 09:38
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


We are given that x > y > 0, so x > y, so √x > √y, so √x − √y > 0 and cannot equal to -2.
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Re: M01-19  [#permalink]

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New post 11 May 2018, 02:19
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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New post 25 Jun 2018, 02:26
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.
M01-19 &nbs [#permalink] 25 Jun 2018, 02:26

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