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m19#20

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m19#20 [#permalink] New post 28 Jun 2009, 22:06
Which of the following always equals \sqrt{9 + x^2 - 6x} ?

(C) 2008 GMAT Club - m19#20

* x - 3
* 3 + x
* |3 - x|
* |3 + x|
* 3 - x


can someone please give me some insight on why the answer is not A? I think im confused on some theory...
thanks!
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Re: m19#20 [#permalink] New post 29 Jun 2009, 08:12
what is OA?


I pick 3-x.
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Re: m19#20 [#permalink] New post 29 Jun 2009, 11:14
It's C as |3-x| = | -(x-3)|

its like |-5| or |5| = 5
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Re: m19#20 [#permalink] New post 30 Jun 2009, 03:43
millhouse wrote:
Which of the following always equals \sqrt{9 + x^2 - 6x} ?

(C) 2008 GMAT Club - m19#20

* x - 3
* 3 + x
* |3 - x|
* |3 + x|
* 3 - x


can someone please give me some insight on why the answer is not A? I think im confused on some theory...
thanks!



\sqrt{9 + x^2 - 6x} = x - 3 or 3 - x

x - 3 or 3 - x = |3 - x| or |x - 3|

So \sqrt{9 + x^2 - 6x} = |3 - x| or |x - 3|

Thats C.
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Re: m19#20 [#permalink] New post 26 May 2011, 12:30
GMAT TIGER wrote:
millhouse wrote:
Which of the following always equals \sqrt{9 + x^2 - 6x} ?

(C) 2008 GMAT Club - m19#20

* x - 3
* 3 + x
* |3 - x|
* |3 + x|
* 3 - x


can someone please give me some insight on why the answer is not A? I think im confused on some theory...
thanks!



\sqrt{9 + x^2 - 6x} = x - 3 or 3 - x

x - 3 or 3 - x = |3 - x| or |x - 3|

So \sqrt{9 + x^2 - 6x} = |3 - x| or |x - 3|

Thats C.



What's confusing me is that the expression can be factored into (x-3)^2 and (3-x)^2. I did the former, but couldn't find |x-3| as an answer choice. Is it typical to be able to factor a quadratic equation multiple ways? Thanks.
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Re: m19#20 [#permalink] New post 25 Sep 2013, 00:15
Sorry , but i still dont get it \sqrt{9+x^2-6x} can we rewritten as \sqrt{x^2-6x+9} which can be simplied to \sqrt{(x-3)^2} .. Thus giving |x-3| which can be x-3 or 3-x
Thats my understanding.. could anyone plz help
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Re: m19#20 [#permalink] New post 25 Sep 2013, 00:29
Expert's post
thinktank wrote:
Sorry , but i still dont get it \sqrt{9+x^2-6x} can we rewritten as \sqrt{x^2-6x+9} which can be simplied to \sqrt{(x-3)^2} .. Thus giving |x-3| which can be x-3 or 3-x
Thats my understanding.. could anyone plz help


You are absolutely right. The result is |x-3|, which is the same as |3-x|.

Answer: C.
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Re: m19#20   [#permalink] 25 Sep 2013, 00:29
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