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Square root

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VP
Joined: 28 Dec 2005
Posts: 1484

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15 Oct 2008, 00:12
$$sqrt(4x^2 + 4xy + y^2)$$

x=29
y=-15

dont know what the answer choices are, but am interested in seeing the approaches

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VP
Joined: 17 Jun 2008
Posts: 1479

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15 Oct 2008, 00:27
In the absence of complete question it is difficult to take the approach.

Asuuming, it is a DS question, the expression will be sqrt[(2x + y)^2].

In order to get the value of expression, both x and y are required. Hence, C.
Intern
Joined: 09 Oct 2008
Posts: 18

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15 Oct 2008, 02:43
Is it a data sufficiency question? Please specify.
As scthakur posted before, if DS type question then answer is (c).
VP
Joined: 28 Dec 2005
Posts: 1484

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30 Oct 2008, 13:29
sorry, its a problem solving question
Senior Manager
Joined: 29 Mar 2008
Posts: 341

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Updated on: 30 Oct 2008, 13:57
(2x+y)^2 = 43

Edit:(2x+y) = 43

Forgot to remove the square.
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Originally posted by leonidas on 30 Oct 2008, 13:36.
Last edited by leonidas on 30 Oct 2008, 13:57, edited 1 time in total.
SVP
Joined: 07 Nov 2007
Posts: 1738
Location: New York

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30 Oct 2008, 13:36
pmenon wrote:
$$sqrt(4x^2 + 4xy + y^2)$$

x=29
y=-15

dont know what the answer choices are, but am interested in seeing the approaches

"]$$sqrt(4x^2 + 4xy + y^2)$$
= 2x+y = 2*29-15 =43
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VP
Joined: 28 Dec 2005
Posts: 1484

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31 Oct 2008, 14:30
i must be missing something really simple. can one of you bright folks break it down step by step ?
SVP
Joined: 29 Aug 2007
Posts: 2427

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31 Oct 2008, 14:46
pmenon wrote:
i must be missing something really simple. can one of you bright folks break it down step by step ?

I mean is the given question a complete one? Source?
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Intern
Joined: 03 Mar 2008
Posts: 44

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31 Oct 2008, 14:47
pmenon wrote:
i must be missing something really simple. can one of you bright folks break it down step by step ?

4x^2 + 4xy + y^2 = (2x + y)^2

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

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Re: Square root   [#permalink] 31 Oct 2008, 14:47
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Square root

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