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# m19#20

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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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29 Jun 2009, 08:12
what is OA?

I pick 3-x.
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29 Jun 2009, 11:14
It's C as |3-x| = | -(x-3)|

its like |-5| or |5| = 5
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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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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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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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25 Sep 2013, 00:29
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 $$|3-x|$$.

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

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