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# If a > b > 0, then √a^2 - b^2

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

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Updated on: 08 May 2015, 05:25
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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

Originally posted by Jeli13 on 08 May 2015, 05:07.
Last edited by Bunuel on 08 May 2015, 05:25, edited 1 time in total.
Edited the question.
Math Expert
Joined: 02 Sep 2009
Posts: 44567
Re: If a > b > 0, then √a^2 - b^2 [#permalink]

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08 May 2015, 05:29
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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, 05:48
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, 04:50
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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Re: If a > b > 0, then √a^2 - b^2 [#permalink]

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08 Mar 2018, 15:04
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. $$(\sqrt{a} + \sqrt{b}) (\sqrt{a}-\sqrt{b})$$

Main Idea: Identify that a formula can be used

Details We see that root(a^2 - b^2) is nothing by root((a+b) (a-b))= root(a+b)*root(a-b)

Hence D.
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Re: If a > b > 0, then √a^2 - b^2 [#permalink]

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12 Mar 2018, 16:35
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. $$(\sqrt{a} + \sqrt{b}) (\sqrt{a}-\sqrt{b})$$

Notice that (a^2 - b^2) is a difference of squares, which we can factor. Simplifying the given expression, we have:

√(a^2 - b^2) = √(a +b)(a - b) = √(a + b) x √(a - b)

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Re: If a > b > 0, then √a^2 - b^2   [#permalink] 12 Mar 2018, 16:35
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