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

It is currently 21 Jan 2019, 10:02

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
Events & Promotions in January
PrevNext
SuMoTuWeThFrSa
303112345
6789101112
13141516171819
20212223242526
272829303112
Open Detailed Calendar
  • GMAT Club Tests are Free & Open for Martin Luther King Jr.'s Birthday!

     January 21, 2019

     January 21, 2019

     10:00 PM PST

     11:00 PM PST

    Mark your calendars - All GMAT Club Tests are free and open January 21st for celebrate Martin Luther King Jr.'s Birthday.
  • The winners of the GMAT game show

     January 22, 2019

     January 22, 2019

     10:00 PM PST

     11:00 PM PST

    In case you didn’t notice, we recently held the 1st ever GMAT game show and it was awesome! See who won a full GMAT course, and register to the next one.

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

 
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: 08 May 2015, 04:25
4
9
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

85% (01:02) correct 15% (01:35) wrong based on 330 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, 04:07.
Last edited by Bunuel on 08 May 2015, 04:25, edited 1 time in total.
Edited the question.
Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 52343
Re: If a > b > 0, then √a^2 - b^2  [#permalink]

Show Tags

New post 08 May 2015, 04:29
2
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.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

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, 04:48
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)
If a > b > 0, then √a^2 - b^2  [#permalink]

Show Tags

New post 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.
Director
Director
User avatar
S
Joined: 17 Dec 2012
Posts: 625
Location: India
Re: If a > b > 0, then √a^2 - b^2  [#permalink]

Show Tags

New post 08 Mar 2018, 14:04
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 Holistic Solutions
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: 2830
Re: If a > b > 0, then √a^2 - b^2  [#permalink]

Show Tags

New post 12 Mar 2018, 15:35
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
_________________

Jeffery Miller
Head of GMAT Instruction

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

GMAT Club Bot
Re: If a > b > 0, then √a^2 - b^2 &nbs [#permalink] 12 Mar 2018, 15:35
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  


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

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

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.