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# x and y are positive integers such that x > y. If 2x^1/2 - 2y^1/2 =

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Math Expert
Joined: 02 Sep 2009
Posts: 46129
x and y are positive integers such that x > y. If 2x^1/2 - 2y^1/2 = [#permalink]

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10 Jun 2015, 04:29
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25% (medium)

Question Stats:

79% (01:12) correct 21% (01:46) wrong based on 120 sessions

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x and y are positive integers such that x > y. If $$2\sqrt{x}-2\sqrt{y}=\frac{x-y}{b}$$, which of the following is equivalent to 2b?

A. $$\sqrt{x}-\sqrt{y}$$

B. $$\sqrt{x}+\sqrt{y}$$

C. $$\frac{\sqrt{x}}{\sqrt{y}}$$

D. $$2\sqrt{xy}$$

E. $$2\sqrt{x-y}$$

Kudos for a correct solution.

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x and y are positive integers such that x > y. If 2x^1/2 - 2y^1/2 = [#permalink]

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10 Jun 2015, 08:09
1
Bunuel wrote:
x and y are positive integers such that x > y. If $$2\sqrt{x}-2\sqrt{y}=\frac{x-y}{b}$$, which of the following is equivalent to 2b?

A. $$\sqrt{x}-\sqrt{y}$$

B. $$\sqrt{x}+\sqrt{y}$$

C. $$\frac{\sqrt{x}}{\sqrt{y}}$$

D. $$2\sqrt{xy}$$

E. $$2\sqrt{x-y}$$

Kudos for a correct solution.

Given: $$2\sqrt{x}-2\sqrt{y}=\frac{x-y}{b}$$

i.e. $$2(\sqrt{x}-\sqrt{y})=\frac{\sqrt{x[}{square_root]^2-[square_root]y}^2/b}$$

i.e. $$2(\sqrt{x}-\sqrt{y})=\frac{(\sqrt{x[}{square_root]-[square_root]y})*(\sqrt{x}+\sqrt{y})/b}$$

i.e. $$2=\frac{(\sqrt{x[}{square_root]+[square_root]y})/b}$$

i.e. $$2b=(\sqrt{x}+\sqrt{y})$$

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Re: x and y are positive integers such that x > y. If 2x^1/2 - 2y^1/2 = [#permalink]

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11 Jun 2015, 03:37
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$$2\sqrt{x}-2\sqrt{y}=\frac{(x-y)}{b}$$

1. Factor 2 out:

$$2(\sqrt{x}-\sqrt{y})=\frac{(x-y)}{b}$$

2. Multiply both sides by b:

$$2b(\sqrt{x}-\sqrt{y})=x-y$$

3. Isolate 2b, by dividing both sides by $$(\sqrt{x}-\sqrt{y})$$:

$$2b=\frac{(x-y)}{(\sqrt{x}-\sqrt{y})}$$

4. Multiply right side by conjugate to eliminate square roots from denominator:

$$2b=\frac{(x-y)*}{(\sqrt{x}-\sqrt{y})} * (\sqrt{x}+\sqrt{y})/(\sqrt{x}+\sqrt{y})$$

$$(\sqrt{x}-\sqrt{y})$$ and $$(\sqrt{x}+\sqrt{y})$$ simplify to "x-y" and that cross cancels with numerator, hence leaving us with $$(\sqrt{x}+\sqrt{y})$$

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Math Expert
Joined: 02 Sep 2009
Posts: 46129
Re: x and y are positive integers such that x > y. If 2x^1/2 - 2y^1/2 = [#permalink]

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15 Jun 2015, 01:42
Bunuel wrote:
x and y are positive integers such that x > y. If $$2\sqrt{x}-2\sqrt{y}=\frac{x-y}{b}$$, which of the following is equivalent to 2b?

A. $$\sqrt{x}-\sqrt{y}$$

B. $$\sqrt{x}+\sqrt{y}$$

C. $$\frac{\sqrt{x}}{\sqrt{y}}$$

D. $$2\sqrt{xy}$$

E. $$2\sqrt{x-y}$$

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

To solve this problem we can focus on solving for b as a first step:

$$2b(\sqrt{x}-\sqrt{y})=x-y$$

$$2b=\frac{x-y}{\sqrt{x}-\sqrt{y}}$$

Note that it is OK to divide by $$\sqrt{x}-\sqrt{y}$$, since x > y , which implies that $$\sqrt{x}-\sqrt{y}\neq{0}$$.

We have solved for 2b, but the result does not match any of the answer choices. Most of the choices are not fractions, so we should try to cancel the denominator. Recognizing that x – y is a well disguised “difference of two squares,” we can factor the numerator and denominator:

$$2b=\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}{\sqrt{x}-\sqrt{y}}$$

Cancel $$\sqrt{x}-\sqrt{y}$$ in the numerator and denominator to get $$2b = \sqrt{x}+\sqrt{y}$$.

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Re: x and y are positive integers such that x > y. If 2x^1/2 - 2y^1/2 = [#permalink]

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16 Jun 2015, 06:33
Bunuel wrote:
Bunuel wrote:
x and y are positive integers such that x > y. If $$2\sqrt{x}-2\sqrt{y}=\frac{x-y}{b}$$, which of the following is equivalent to 2b?

A. $$\sqrt{x}-\sqrt{y}$$

B. $$\sqrt{x}+\sqrt{y}$$

C. $$\frac{\sqrt{x}}{\sqrt{y}}$$

D. $$2\sqrt{xy}$$

E. $$2\sqrt{x-y}$$

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

To solve this problem we can focus on solving for b as a first step:

$$2b(\sqrt{x}-\sqrt{y})=x-y$$

$$2b=\frac{x-y}{\sqrt{x}-\sqrt{y}}$$

Note that it is OK to divide by $$\sqrt{x}-\sqrt{y}$$, since x > y , which implies that $$\sqrt{x}-\sqrt{y}\neq{0}$$.

We have solved for 2b, but the result does not match any of the answer choices. Most of the choices are not fractions, so we should try to cancel the denominator. Recognizing that x – y is a well disguised “difference of two squares,” we can factor the numerator and denominator:

$$2b=\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}{\sqrt{x}-\sqrt{y}}$$

Cancel $$\sqrt{x}-\sqrt{y}$$ in the numerator and denominator to get $$2b = \sqrt{x}+\sqrt{y}$$.

Hello,

could you show me the difference of 2 squares? I cannot see it...
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Re: x and y are positive integers such that x > y. If 2x^1/2 - 2y^1/2 = [#permalink]

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16 Jun 2015, 06:57
pacifist85 wrote:
Bunuel wrote:
Bunuel wrote:
x and y are positive integers such that x > y. If $$2\sqrt{x}-2\sqrt{y}=\frac{x-y}{b}$$, which of the following is equivalent to 2b?

A. $$\sqrt{x}-\sqrt{y}$$

B. $$\sqrt{x}+\sqrt{y}$$

C. $$\frac{\sqrt{x}}{\sqrt{y}}$$

D. $$2\sqrt{xy}$$

E. $$2\sqrt{x-y}$$

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

To solve this problem we can focus on solving for b as a first step:

$$2b(\sqrt{x}-\sqrt{y})=x-y$$

$$2b=\frac{x-y}{\sqrt{x}-\sqrt{y}}$$

Note that it is OK to divide by $$\sqrt{x}-\sqrt{y}$$, since x > y , which implies that $$\sqrt{x}-\sqrt{y}\neq{0}$$.

We have solved for 2b, but the result does not match any of the answer choices. Most of the choices are not fractions, so we should try to cancel the denominator. Recognizing that x – y is a well disguised “difference of two squares,” we can factor the numerator and denominator:

$$2b=\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}{\sqrt{x}-\sqrt{y}}$$

Cancel $$\sqrt{x}-\sqrt{y}$$ in the numerator and denominator to get $$2b = \sqrt{x}+\sqrt{y}$$.

Hello,

could you show me the difference of 2 squares? I cannot see it...

x-y= ($$\sqrt{x}$$ - $$\sqrt{y}$$ )( $$\sqrt{x}$$ + $$\sqrt{y}$$ )
This is of the same identity as $$a^2 - b^2$$= (a+b)(a-b)
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Status: Math is psycho-logical
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Re: x and y are positive integers such that x > y. If 2x^1/2 - 2y^1/2 = [#permalink]

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16 Jun 2015, 07:42
Well I am still missing sth.

We are at: 2b=(x−y) / (√x−√y)
The direrence of squares is this: (x-y)^2 = (x-y)(x+y). But in our case x - y is not raised to the second power..

Then I though you squared the nominator and the denominator, which could create the above mentioned difference of squares, but in the denominator (√x - √y )( √x + √y), and should also create this in the nominator (x-y) (x+y).

So, I still cannot see the difference of squares...
Math Expert
Joined: 02 Sep 2009
Posts: 46129
Re: x and y are positive integers such that x > y. If 2x^1/2 - 2y^1/2 = [#permalink]

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16 Jun 2015, 08:34
pacifist85 wrote:
Well I am still missing sth.

We are at: 2b=(x−y) / (√x−√y)
The direrence of squares is this: (x-y)^2 = (x-y)(x+y). But in our case x - y is not raised to the second power..

Then I though you squared the nominator and the denominator, which could create the above mentioned difference of squares, but in the denominator (√x - √y )( √x + √y), and should also create this in the nominator (x-y) (x+y).

So, I still cannot see the difference of squares...

$$(a-b)^2=a^2-2ab+b^2$$ is square of the difference.

$$a^2 - b^2=(a-b)(a+b)$$ is difference of two squares.

In our case we have x - y, which is $$(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})=x-y$$.

Hope it's clear.
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Re: x and y are positive integers such that x > y. If 2x^1/2 - 2y^1/2 = [#permalink]

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Re: x and y are positive integers such that x > y. If 2x^1/2 - 2y^1/2 =   [#permalink] 26 Mar 2017, 11:00
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