Last visit was: 10 Jul 2025, 11:08 It is currently 10 Jul 2025, 11:08
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
User avatar
kook44
Joined: 23 Jan 2006
Last visit: 30 Jul 2006
Posts: 90
Own Kudos:
290
 [132]
Posts: 90
Kudos: 290
 [132]
9
Kudos
Add Kudos
123
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 10 July 2025
Posts: 102,624
Own Kudos:
Given Kudos: 98,170
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,624
Kudos: 740,136
 [40]
19
Kudos
Add Kudos
21
Bookmarks
Bookmark this Post
User avatar
KarishmaB
Joined: 16 Oct 2010
Last visit: 10 Jul 2025
Posts: 16,101
Own Kudos:
74,245
 [20]
Given Kudos: 475
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,101
Kudos: 74,245
 [20]
11
Kudos
Add Kudos
9
Bookmarks
Bookmark this Post
General Discussion
User avatar
kalita
Joined: 06 Apr 2012
Last visit: 30 Sep 2016
Posts: 27
Own Kudos:
153
 [3]
Given Kudos: 48
Posts: 27
Kudos: 153
 [3]
2
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Quote:
If n is positive, which of the following is equal to \(\frac{1}{\sqrt{n+1}-\sqrt{n}}\)

A. 1

B. \(\sqrt{2n+1}\)

C. \(\frac{\sqrt{n+1}}{\sqrt{n}}\)

D. \(\sqrt{n+1}-\sqrt{n}\)

E. \(\sqrt{n+1}+\sqrt{n}\)

This question is dealing with rationalisation of a fraction. Rationalisation is performed to eliminate irrational expression in the denominator. For this particular case we can do this by applying the following rule: \((a-b)(a+b)=a^2-b^2\).

Multiple both numerator and denominator by \(\sqrt{n+1}+\sqrt{n}\): \(\frac{\sqrt{n+1}+\sqrt{n}}{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}=\frac{\sqrt{n+1}+\sqrt{n}}{(\sqrt{n+1})^2-(\sqrt{n})^2)}=\frac{\sqrt{n+1}+\sqrt{n}}{n+1-n}=\sqrt{n+1}+\sqrt{n}\).

Answer: E.

Bunuel - just wanted to clarify an aspect of the roots - the final answer of this problem is E and it is perfectly understood. However, if I want to simplify the \(\sqrt{n+1} + \sqrt{n}\) even more... theoretically I could "unroot" these expressions, so that I get \(2n+1\), however, as the answer B is clearly wrong (and I can see why), I want to but I struggle to understand how to "put the roots back" in the \(2n+1\) to get an equivalent of \(\sqrt{n+1} + \sqrt{n}\). Any thoughts on this matter?

Thanks!
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 10 July 2025
Posts: 102,624
Own Kudos:
740,136
 [1]
Given Kudos: 98,170
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,624
Kudos: 740,136
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
ikokurin
Bu nuel
If n is positive, which of the following is equal to \(\frac{1}{\sqrt{n+1}-\sqrt{n}}\)

A. 1

B. \(\sqrt{2n+1}\)

C. \(\frac{\sqrt{n+1}}{\sqrt{n}}\)

D. \(\sqrt{n+1}-\sqrt{n}\)

E. \(\sqrt{n+1}+\sqrt{n}\)

This question is dealing with rationalisation of a fraction. Rationalisation is performed to eliminate irrational expression in the denominator. For this particular case we can do this by applying the following rule: \((a-b)(a+b)=a^2-b^2\).

Multiple both numerator and denominator by \(\sqrt{n+1}+\sqrt{n}\): \(\frac{\sqrt{n+1}+\sqrt{n}}{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}=\frac{\sqrt{n+1}+\sqrt{n}}{(\sqrt{n+1})^2-(\sqrt{n})^2)}=\frac{\sqrt{n+1}+\sqrt{n}}{n+1-n}=\sqrt{n+1}+\sqrt{n}\).

Answer: E.

Bunuel - just wanted to clarify an aspect of the roots - the final answer of this problem is E and it is perfectly understood. However, if I want to simplify the SQRT(n+1) + SQRT(n) even more... theoretically I could "unsquare" these expressions, so that I get 2n+1, however, as the answer B is clearly wrong (and I can see why), I struggle to understand how to "square back" the 2n+1 to get an equivalent of SQRT(n+1) + SQRT(n). Can you help me out or share your thoughts on the matter? Thanks!

I don't understand what you mean: how can you get \(2n+1\) from \(\sqrt{n+1}+\sqrt{n}\)?
User avatar
kalita
Joined: 06 Apr 2012
Last visit: 30 Sep 2016
Posts: 27
Own Kudos:
Given Kudos: 48
Posts: 27
Kudos: 153
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Quote:
I don't understand what you mean: how can you get \(2n+1\) from \(\sqrt{n+1}+\sqrt{n}\)?

I meant some people might get \(\sqrt{2n+1}\) which is the answer B. However, I can see why \(\sqrt{n+1}+\sqrt{n}\) is NOT equal \(\sqrt{2n+1}\) even though it might be tempting to simplify it to this form (and pick the wrong answer). But my question is can we simplify \(\sqrt{n+1}+\sqrt{n}\) further by "squaring" both terms and then "unsquaring" them/the expression back somehow... or what could be an equivalent of \(\sqrt{n+1}+\sqrt{n}\)?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 10 July 2025
Posts: 102,624
Own Kudos:
740,136
 [2]
Given Kudos: 98,170
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,624
Kudos: 740,136
 [2]
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
ikokurin
I meant some people might get SQRT(2n+1) which is the answer B. However, I can see why SQRT(n+1) + SQRT(n) is NOT equal SQRT(2n+1) even though it might be tempting to simplify it to this form (and pick the wrong answer). But my question is can we simplify SQRT(n+1) + SQRT(n) further by "squaring" both terms and then "squarerooting" them again somehow... or what could be an equivalent of SQRT(n+1) + SQRT(n)?

\(\sqrt{n+1}+\sqrt{n}\) is the simplest form. If you square it you'll get \((\sqrt{n+1}+\sqrt{n})^2=(\sqrt{n+1})^2+2\sqrt{n+1}*\sqrt{n}+\sqrt{n}^2=2n+1+2\sqrt{(n+1)n}\). You cannot take square root from this expression to get anything better than \(\sqrt{n+1}+\sqrt{n}\).

Hope it's clear.
User avatar
kalita
Joined: 06 Apr 2012
Last visit: 30 Sep 2016
Posts: 27
Own Kudos:
Given Kudos: 48
Posts: 27
Kudos: 153
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Quote:
\(\sqrt{n+1}+\sqrt{n}\) is the simplest form. If you square it you'll get \((\sqrt{n+1}+\sqrt{n})^2=(\sqrt{n+1})^2+2\sqrt{n+1}*\sqrt{n}+\sqrt{n}^2=2n+1+2\sqrt{(n+1)n}\). You cannot take square root from this expression to get anything better than \(\sqrt{n+1}+\sqrt{n}\).

Hope it's clear.

I see. What you are saying is clear but your answer does not exactly address what I am after. I can see that \((\sqrt{n+1}+\sqrt{n})^2\) only complicates it further. Sorry to be pertinacious on this - if we do \((\sqrt{n+1})^2+(\sqrt{n})^2\) => we will get \(n+1 + n = 2n + 1\) => can we "undo" the expression \(2n + 1\) somehow to get the equivalent of \(\sqrt{n+1}+\sqrt{n}\)? I promise this is the last one:)

P.S. Also, please let me know if it would be better to send a PM on related "clarifying" questions...
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 10 July 2025
Posts: 102,624
Own Kudos:
Given Kudos: 98,170
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,624
Kudos: 740,136
Kudos
Add Kudos
Bookmarks
Bookmark this Post
ikokurin
\(\sqrt{n+1}+\sqrt{n}\) is the simplest form. If you square it you'll get \((\sqrt{n+1}+\sqrt{n})^2=(\sqrt{n+1})^2+2\sqrt{n+1}*\sqrt{n}+\sqrt{n}^2=2n+1+2\sqrt{(n+1)n}\). You cannot take square root from this expression to get anything better than \(\sqrt{n+1}+\sqrt{n}\).

Hope it's clear.

I see. What you are saying is clear but your answer does not exactly address what I am after. Sorry to be pertinacious on this but I was wondering if we can do (SQRT(n+1))^2 + (SQRT(n))^2 => we will get n+1 + n = 2n + 1 => can we "undo" the expression 2n + 1 somehow to get the equivalent of SQRT(n+1) + SQRT(n)? I promise this is the last one:) Also, please let me know if it would be better to send a PM on related "clarifying" questions...[/quote]

The answer is no, these expressions are not equal.

P.S. Please use formatting, check here: rules-for-posting-please-read-this-before-posting-133935.html#p1096628
User avatar
kalita
Joined: 06 Apr 2012
Last visit: 30 Sep 2016
Posts: 27
Own Kudos:
Given Kudos: 48
Posts: 27
Kudos: 153
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Quote:
The answer is no, these expressions are not equal.

P.S. Please use formatting, check here: rules-for-posting-please-read-this-before-posting-133935.html#p1096628


I understand they are not equal, thanks for help. So I take away there is no way to go from \(2n+1\) (obtained after squaring both terms (\((\sqrt{n+1})^2 + (\sqrt{n})^2\)) into something else that could be an equivalent of\(\sqrt{n+1} + \sqrt{n}\). As mentioned above, for those having issues with exponents/roots, it is possible to make a mistake of simplifying \((\sqrt{n+1})^2 + (\sqrt{n})^2\) into \(\sqrt{2n+1}\) (which is incorrect); nevertheless I wanted to see if there was a way to do something about \(2n+1\) to make it equal to \(\sqrt{n+1} + \sqrt{n}\). For some reason, having inner desire to combine those \(n\) terms to make it all look nicer, it bugs me that leaving the answer as \(\sqrt{n+1} + \sqrt{n}\) is all we can do about this equation; especially after I saw some tricks/solutions relating to the tricky exponent problems and how one can do "wonders" with squaring and unsquaring things :) I was thinking about simplifying this thing into something like, obviously grossly exaggerated, \(^4\sqrt{2n+1}\) or \(\sqrt{2n}+\sqrt{1}\), etc., by "squarerooting" \(2n+1\) back somehow. But again I know the previous examples are plain wrong, just giving an example of what one can go through working through possibilities. Anyhow, enough of this rumble, let me know if you have anything to add...and thanks much for patience.

Regards,
User avatar
mbaiseasy
Joined: 13 Aug 2012
Last visit: 29 Dec 2013
Posts: 323
Own Kudos:
1,989
 [5]
Given Kudos: 11
Concentration: Marketing, Finance
GPA: 3.23
Posts: 323
Kudos: 1,989
 [5]
2
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
\(\frac{1}{\sqrt{n+1}-\sqrt{n}}\)

\(\frac{1}{\sqrt{n+1}-\sqrt{n}} * \frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}\)


\(\frac{\sqrt{n+1}+\sqrt{n}}{n+1-n}\)

\(\frac{\sqrt{n+1}+\sqrt{n}}{1}\)




Answer: E
User avatar
pharm
Joined: 03 Jan 2013
Last visit: 06 Jul 2015
Posts: 15
Given Kudos: 50
Posts: 15
Kudos: 0
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I have a quick question on this ..when the initial fraction was rationalized you used:

\(\sqrt{n+1}+ \sqrt{n} / \sqrt{n+1}+ \sqrt{n}\)

did you change the sign from negative to positive since the question stated "n" is a positive number. Wouldn't you have to use the same denominator when Rationalizing a fraction?
User avatar
pharm
Joined: 03 Jan 2013
Last visit: 06 Jul 2015
Posts: 15
Given Kudos: 50
Posts: 15
Kudos: 0
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Thanks Karishma that cleared things up
User avatar
jlgdr
Joined: 06 Sep 2013
Last visit: 24 Jul 2015
Posts: 1,316
Own Kudos:
Given Kudos: 355
Concentration: Finance
Posts: 1,316
Kudos: 2,717
Kudos
Add Kudos
Bookmarks
Bookmark this Post
kook44
If n is positive, which of the following is equal to \(\frac{1}{\sqrt{n+1}-\sqrt{n}}\)

A. 1

B. \(\sqrt{2n+1}\)

C. \(\frac{\sqrt{n+1}}{\sqrt{n}}\)

D. \(\sqrt{n+1}-\sqrt{n}\)

E. \(\sqrt{n+1}+\sqrt{n}\)

Isn't it much easier to just pick n=1 and then look for target in answer choices?

Cheers!
J :)
User avatar
gmatprav
Joined: 25 Oct 2013
Last visit: 19 Nov 2015
Posts: 111
Own Kudos:
Given Kudos: 55
Posts: 111
Kudos: 180
Kudos
Add Kudos
Bookmarks
Bookmark this Post
jlgdr
kook44
If n is positive, which of the following is equal to \(\frac{1}{\sqrt{n+1}-\sqrt{n}}\)

A. 1

B. \(\sqrt{2n+1}\)

C. \(\frac{\sqrt{n+1}}{\sqrt{n}}\)

D. \(\sqrt{n+1}-\sqrt{n}\)

E. \(\sqrt{n+1}+\sqrt{n}\)

Isn't it much easier to just pick n=1 and then look for target in answer choices?

Cheers!
J :)

What if more than one answer choice gives you same value? first, we have to try original expression with 1 and try each of the choices with 1. If we are lucky we have only one choice matching. but what if there are 2 or even 3 answer choices? we would then have to pick another number. Personally I feel solving it is faster in this case.

Sometimes number picking works faster. knowing when to use number picking is the difficult part.
User avatar
jlgdr
Joined: 06 Sep 2013
Last visit: 24 Jul 2015
Posts: 1,316
Own Kudos:
Given Kudos: 355
Concentration: Finance
Posts: 1,316
Kudos: 2,717
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Ya I guess your right after solving the way Bunuel did it took less than 20 secs

Posted from my mobile device
User avatar
KarishmaB
Joined: 16 Oct 2010
Last visit: 10 Jul 2025
Posts: 16,101
Own Kudos:
74,245
 [2]
Given Kudos: 475
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,101
Kudos: 74,245
 [2]
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
jlgdr
kook44
If n is positive, which of the following is equal to \(\frac{1}{\sqrt{n+1}-\sqrt{n}}\)

A. 1

B. \(\sqrt{2n+1}\)

C. \(\frac{\sqrt{n+1}}{\sqrt{n}}\)

D. \(\sqrt{n+1}-\sqrt{n}\)

E. \(\sqrt{n+1}+\sqrt{n}\)

Isn't it much easier to just pick n=1 and then look for target in answer choices?

Cheers!
J :)

Yes, absolutely it is. I would answer this question by plugging in the values but you have to be careful of two things. When pluggin in values in the options, two or more options might seem to satisfy. If this happens, you need to plug in a different number in those two to get the actual correct answer.
Also, you need to ensure that the value given by option actually does not match the required value before discarding it.
e.g. here if I put n = 1, \(\frac{1}{\sqrt{n+1}-\sqrt{n}}\) = \(\frac{1}{\sqrt{2}-1}\)

while option (E) gives \(\sqrt{n+1}+\sqrt{n}\) = \(\sqrt{2}+1\)

You cannot discard option (E) because it doesn't look the same. You must rationalize the value obtained from the expression and then compare it with what you get from option (E). So you must be careful.
avatar
Jannnn04
Joined: 22 Jan 2018
Last visit: 05 Feb 2018
Posts: 4
Given Kudos: 115
Posts: 4
Kudos: 0
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
If n is positive, which of the following is equal to \(\frac{1}{\sqrt{n+1}-\sqrt{n}}\)

A. 1

B. \(\sqrt{2n+1}\)

C. \(\frac{\sqrt{n+1}}{\sqrt{n}}\)

D. \(\sqrt{n+1}-\sqrt{n}\)

E. \(\sqrt{n+1}+\sqrt{n}\)

This question is dealing with rationalisation of a fraction. Rationalisation is performed to eliminate irrational expression in the denominator. For this particular case we can do this by applying the following rule: \((a-b)(a+b)=a^2-b^2\).

Multiple both numerator and denominator by \(\sqrt{n+1}+\sqrt{n}\): \(\frac{\sqrt{n+1}+\sqrt{n}}{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}=\frac{\sqrt{n+1}+\sqrt{n}}{(\sqrt{n+1})^2-(\sqrt{n})^2)}=\frac{\sqrt{n+1}+\sqrt{n}}{n+1-n}=\sqrt{n+1}+\sqrt{n}\).

Answer: E.


Hi! I do not understand, how you came up with the last part of the solution where you just simplify and take the expression from the denominator away.
thank you!
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 10 July 2025
Posts: 102,624
Own Kudos:
740,136
 [1]
Given Kudos: 98,170
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 102,624
Kudos: 740,136
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Jannnn04
Bunuel
If n is positive, which of the following is equal to \(\frac{1}{\sqrt{n+1}-\sqrt{n}}\)

A. 1

B. \(\sqrt{2n+1}\)

C. \(\frac{\sqrt{n+1}}{\sqrt{n}}\)

D. \(\sqrt{n+1}-\sqrt{n}\)

E. \(\sqrt{n+1}+\sqrt{n}\)

This question is dealing with rationalisation of a fraction. Rationalisation is performed to eliminate irrational expression in the denominator. For this particular case we can do this by applying the following rule: \((a-b)(a+b)=a^2-b^2\).

Multiple both numerator and denominator by \(\sqrt{n+1}+\sqrt{n}\): \(\frac{\sqrt{n+1}+\sqrt{n}}{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}=\frac{\sqrt{n+1}+\sqrt{n}}{(\sqrt{n+1})^2-(\sqrt{n})^2)}=\frac{\sqrt{n+1}+\sqrt{n}}{n+1-n}=\sqrt{n+1}+\sqrt{n}\).

Answer: E.


Hi! I do not understand, how you came up with the last part of the solution where you just simplify and take the expression from the denominator away.
thank you!

The denominator is n+1-n, which is 1:

n + 1 - n= 1.
User avatar
dave13
Joined: 09 Mar 2016
Last visit: 23 Nov 2024
Posts: 1,114
Own Kudos:
Given Kudos: 3,851
Posts: 1,114
Kudos: 1,087
Kudos
Add Kudos
Bookmarks
Bookmark this Post
kook44
If n is positive, which of the following is equal to \(\frac{1}{\sqrt{n+1}-\sqrt{n}}\)

A. 1

B. \(\sqrt{2n+1}\)

C. \(\frac{\sqrt{n+1}}{\sqrt{n}}\)

D. \(\sqrt{n+1}-\sqrt{n}\)

E. \(\sqrt{n+1}+\sqrt{n}\)


this link is great source about rationalizing denominator :) https://www.wtamu.edu/academic/anns/mps/ ... nalize.htm
 1   2   
Moderators:
Math Expert
102624 posts
PS Forum Moderator
685 posts