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

 It is currently 23 Oct 2019, 21:32

### 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

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

# If a > b > 0, then √a^2 - b^2

Author Message
TAGS:

### Hide Tags

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

### Show Tags

Updated on: 23 Jul 2019, 07:33
4
12
00:00

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

### Show Tags

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 $$\sqrt{(a + b)}\sqrt{(a - b)}$$.

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

### Show Tags

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

### Show Tags

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
Joined: 17 Dec 2012
Posts: 626
Location: India
Re: If a > b > 0, then √a^2 - b^2  [#permalink]

### Show Tags

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
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2815
Re: If a > b > 0, then √a^2 - b^2  [#permalink]

### Show Tags

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)

_________________

# Jeffrey Miller

Jeff@TargetTestPrep.com
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
Joined: 27 Oct 2018
Posts: 2
Re: If a > b > 0, then √a^2 - b^2  [#permalink]

### Show Tags

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