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# If a > b > 0, then √a^2 - b^2

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Intern
Joined: 21 Dec 2014
Posts: 14
If a > b > 0, then √a^2 - b^2 [#permalink]

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08 May 2015, 04:07
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84% (00:30) correct 16% (00:54) wrong based on 208 sessions

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If a > b > 0, then $$\sqrt{a^2 - b^2}$$

A. $$a + b - \sqrt{2ab}$$
B. $$a - b + \sqrt{2ab}$$
C. $$\sqrt{(a-b)^2 - 2ab}$$
D. $$(\sqrt{a+b}) (\sqrt{a-b})$$
E. $$(\sqrt{a} + \sqrt{b}) (\sqrt{a}-\sqrt{b})$$

What is the difference between D and E

I thought because of the numbers being all under the square root e.g. in D $$(\sqrt{a+b})$$
that you will have to subtract a from b first giving e.g. answer$$\sqrt{c}$$ and for the other one e.g.$$\sqrt{d}$$

And therefore $$(\sqrt{c})(\sqrt{d})$$ a totally different value?
[Reveal] Spoiler: OA

Last edited by Bunuel on 08 May 2015, 04:25, edited 1 time in total.
Edited the question.
Math Expert
Joined: 02 Sep 2009
Posts: 43862
Re: If a > b > 0, then √a^2 - b^2 [#permalink]

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08 May 2015, 04:29
Expert's post
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Jeli13 wrote:
If a > b > 0, then $$\sqrt{a^2 - b^2}$$

A. $$a + b - \sqrt{2ab}$$
B. $$a - b + \sqrt{2ab}$$
C. $$\sqrt{(a-b)^2 - 2ab}$$
D. $$(\sqrt{a+b}) (\sqrt{a-b})$$
E. $$(\sqrt{a} + \sqrt{b}) (\sqrt{a}-\sqrt{b})$$

What is the difference between D and E

I thought because of the numbers being all under the square root e.g. in D $$(\sqrt{a+b})$$
that you will have to subtract a from b first giving e.g. answer$$\sqrt{c}$$ and for the other one e.g.$$\sqrt{d}$$

And therefore $$(\sqrt{c})(\sqrt{d})$$ a totally different value?

$$a^2 - b^2 = (a + b)(a - b)$$, thus $$\sqrt{a^2 - b^2}=\sqrt{(a + b)(a - b)}$$, which is the same as $$\sqrt{(a + b)}\sqrt{(a - b)}$$.

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Intern
Joined: 21 Dec 2014
Posts: 14
Re: If a > b > 0, then √a^2 - b^2 [#permalink]

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08 May 2015, 04:48
mmhh ok seem to have overcomplicated my thoughts
Intern
Joined: 08 Sep 2014
Posts: 25
GMAT Date: 07-11-2015
GPA: 2.9
WE: Account Management (Computer Software)
If a > b > 0, then √a^2 - b^2 [#permalink]

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21 May 2015, 03:50
I picked numbers for this question, a = 2 and b = 1 yields $$\sqrt{3}$$ and the only answer choice that yields $$\sqrt{3}$$ when substituting a for 2 and b for 1 is D.
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Posts: 13800
Re: If a > b > 0, then √a^2 - b^2 [#permalink]

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25 Jan 2018, 18:43
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: If a > b > 0, then √a^2 - b^2   [#permalink] 25 Jan 2018, 18:43
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