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# M01-19

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Math Expert
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15 Sep 2014, 23:15
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Difficulty:

95% (hard)

Question Stats:

42% (01:09) correct 58% (01:53) wrong based on 222 sessions

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If $$x&gt;y&gt;0$$ then what is the value of $$\frac{\sqrt{2x}+\sqrt{2y}}{x-y}$$?

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

(2) $$x-y=9$$

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15 Sep 2014, 23:15
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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.

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23 Nov 2014, 06:35
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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25 Nov 2014, 06:23
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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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}

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
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17 Feb 2015, 01:48
1
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 $$a^2-2ab+b^2=(a-b)^2$$:

$$x-2\sqrt{xy}+y=(\sqrt{x})^2-2\sqrt{x}\sqrt{y}+(\sqrt{y})^2=(\sqrt{x}-\sqrt{y})^2$$.

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25 Feb 2015, 14:56
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.

Beautiful.
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08 Jul 2016, 15:13
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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08 Jul 2016, 23:09
mrinmoyc987 wrote:
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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15 Sep 2016, 11:37
I think this is a high-quality question and I agree with explanation.
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28 Nov 2016, 00:04
1
Overlaps with m16-09
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28 Nov 2016, 01:17
Akela wrote:
Overlaps with m16-09

Thank you. Removed that question.
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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

$$(a-b)^2$$ 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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18 Jul 2017, 04:29
1
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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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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09 May 2018, 08:38
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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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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25 Jun 2018, 01:26
1
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, 01:26

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# M01-19

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