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If a > b > 0, then √a^2 - b^2

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Jeli13
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If a > b > 0, then √a^2 - b^2 [#permalink] New post   Updated on: 23 Jul 2019, 07:33
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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. \((\sqrt{a} + \sqrt{b}) (\sqrt{a}-\sqrt{b})\)



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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?
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D

Originally posted by Jeli13 on 08 May 2015, 05:07.
Last edited by Bunuel on 23 Jul 2019, 07:33, edited 2 times in total.
Edited the question.
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Re: If a > b > 0, then √a^2 - b^2 [#permalink] New post   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 \(\sqrt{(a + b)}\sqrt{(a - b)}\).

Answer: D.
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Jeli13
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Re: If a > b > 0, then √a^2 - b^2 [#permalink] New post   08 May 2015, 05:48
mmhh ok seem to have overcomplicated my thoughts
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Re: If a > b > 0, then √a^2 - b^2 [#permalink] New post   21 May 2015, 04:50
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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] New post   08 Mar 2018, 15:04
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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. \((\sqrt{a} + \sqrt{b}) (\sqrt{a}-\sqrt{b})\)



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


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)

Answer: D
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Re: If a > b > 0, then √a^2 - b^2 [#permalink] New post   24 Jun 2019, 16:07
BuggerinOn wrote:
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.



You clearly got it right, but aren't "a" and "b" supposed to be negative?
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Re: If a > b > 0, then √a^2 - b^2 [#permalink] New post   05 Jul 2020, 03:16
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\(\sqrt{a^2-b^2}\)
=\(\sqrt{(a+b)(a-b)}\)
=\(\sqrt{a+b} \sqrt{a-b}\)

Answer: D
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Re: If a > b > 0, then √a^2 - b^2 [#permalink] New post   14 Aug 2021, 21:24
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\(\sqrt{6} = \sqrt{2*3} = \sqrt{2} * \sqrt{3}\)

Similarly, \(\sqrt{a^2 - b^2} = \sqrt{(a - b)(a + b)} = \sqrt{a - b}\sqrt{a + b}\)

Answer D
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Re: If a > b > 0, then √a^2 - b^2 [#permalink] New post   15 Aug 2021, 06:32
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. \((\sqrt{a} + \sqrt{b}) (\sqrt{a}-\sqrt{b})\)



a^2 -b^2= a+b*a-b
if square root is taken '

Then we get \/a+b*a-b

if there is any confusion then plugging in should do the trick
let a=2 and b=1
then we get only D as the right option therefore IMO D
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Re: If a > b > 0, then a^2 - b^2 [#permalink] New post   14 Jan 2024, 17:47
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