GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 23 Oct 2019, 21:32

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

If a > b > 0, then √a^2 - b^2

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

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

Show Tags

New post Updated on: 23 Jul 2019, 07:33
4
12
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

85% (01:01) correct 15% (01:34) wrong based on 301 sessions

HideShow timer Statistics

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?

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.
Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58467
Re: If a > b > 0, then √a^2 - b^2  [#permalink]

Show Tags

New post 08 May 2015, 05:29
5
3
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.
_________________
General Discussion
Intern
Intern
avatar
Joined: 21 Dec 2014
Posts: 14
Re: If a > b > 0, then √a^2 - b^2  [#permalink]

Show Tags

New post 08 May 2015, 05:48
mmhh ok seem to have overcomplicated my thoughts
Intern
Intern
User avatar
Joined: 08 Sep 2014
Posts: 25
GMAT Date: 07-11-2015
GPA: 2.9
WE: Account Management (Computer Software)
Re: If a > b > 0, then √a^2 - b^2  [#permalink]

Show Tags

New post 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.
Director
Director
User avatar
S
Joined: 17 Dec 2012
Posts: 626
Location: India
Re: If a > b > 0, then √a^2 - b^2  [#permalink]

Show Tags

New post 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.
_________________
Srinivasan Vaidyaraman
Sravna Test Prep
http://www.sravnatestprep.com

Holistic and Systematic Approach
Target Test Prep Representative
User avatar
G
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2815
Re: If a > b > 0, then √a^2 - b^2  [#permalink]

Show Tags

New post 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)

Answer: D
_________________

Jeffrey Miller

Head of GMAT Instruction

Jeff@TargetTestPrep.com
TTP - Target Test Prep Logo
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Intern
Intern
avatar
B
Joined: 27 Oct 2018
Posts: 2
Re: If a > b > 0, then √a^2 - b^2  [#permalink]

Show Tags

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?
GMAT Club Bot
Re: If a > b > 0, then √a^2 - b^2   [#permalink] 24 Jun 2019, 16:07
Display posts from previous: Sort by

If a > b > 0, then √a^2 - b^2

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne