OG Pg 186 Problem 245 : Quant Question Archive [LOCKED]
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# OG Pg 186 Problem 245

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Current Student
Joined: 03 Dec 2007
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Schools: NYU STERN
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OG Pg 186 Problem 245 [#permalink]

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18 Jan 2009, 12:07
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Ok so I feel stupid asking this but i am having a problem at a certain point in the problem:

If x+y=a and x-y=b then 2xy= ?

The answer is: 2xy= (a^2 - b^2)/2

I got it to this:

y= [2a/2] - [(a+b)/2]

why is the numerator then in the next step: 2a-a-b and not 2a-a+b?

Thanks!
Director
Joined: 12 Jul 2008
Posts: 518
Schools: Wharton
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Re: OG Pg 186 Problem 245 [#permalink]

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18 Jan 2009, 14:00
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KUDOS
hardaway7 wrote:
Ok so I feel stupid asking this but i am having a problem at a certain point in the problem:

If x+y=a and x-y=b then 2xy= ?

The answer is: 2xy= (a^2 - b^2)/2

I got it to this:

y= [2a/2] - [(a+b)/2]

why is the numerator then in the next step: 2a-a-b and not 2a-a+b?

Thanks!

hardaway, when you post the question, you don't need to post page number and problem number from OG. Just copy the question w/ all the choices.

$$(x+y)^2 = x^2 + 2xy + y^2$$
$$(x-y)^2 = x^2 - 2xy + y^2$$

we know that:
$$a^2 = (x+y)^2$$
$$b^2 = (x-y)^2$$

so,
$$a^2 - b^2 = (x^2 + 2xy + y^2) - (x^2 - 2xy + y^2)$$
$$= x^2 - x^2 + 2xy + 2xy + y^2 - y^2$$
$$= 4xy$$

$$4xy = a^2 - b^2$$
$$2xy = (a^2 - b^2)/2$$
Re: OG Pg 186 Problem 245   [#permalink] 18 Jan 2009, 14:00
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