Last visit was: 01 May 2026, 18:54 It is currently 01 May 2026, 18:54
Home
Messages
Tests
Schools
Events
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
555-605 (Medium)|   Algebra|   Roots|      
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 01 May 2026
Posts: 110,001
Own Kudos:
812,342
 [23]
Given Kudos: 105,976
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 110,001
Kudos: 812,342
 [23]
If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 = 22 Oct 2018, 01:33
Kudos
Add Kudos
23
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

35% (medium)

Question Stats:

75% (02:04) correct 25% (02:41) wrong based on 675 sessions

History

Date
Time
Result
Not Attempted Yet
If \(p=\frac{1}{√14 − √13}\) and \(q = \frac{1}{√14 + √13}\) then \(p^2 + 2pq + q^2 =\)

A. 26
B. 28
C. 52
D. 56
E. 112
ShowHide Answer
Official Answer
D
Most Helpful Reply
User avatar
Chethan92
User avatar
Joined: 18 Jul 2018
Last visit: 21 Apr 2022
Posts: 899
Own Kudos:
1,511
 [6]
Given Kudos: 95
Location: India
Concentration: Operations, General Management
Schools: IIMB EPGP'22 (M)
GMAT 1: 590 Q46 V25
GMAT 2: 690 Q49 V34
WE:Engineering (Energy)
Products:
My Reviews
Schools: IIMB EPGP'22 (M)
GMAT 2: 690 Q49 V34
Posts: 899
Kudos: 1,511
 [6]
If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 = 22 Oct 2018, 02:37
4
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
Factorizing both P and Q gives,

P = \(\sqrt{14} + \sqrt{13}\)

Q = \(\sqrt{14} - \sqrt{13}\)

Question reduced to \((p+q)^2\) = \((2\sqrt{14})^2\) = 4*14 = 56.

D is the answer.
User avatar
pushpitkc
User avatar
Joined: 26 Feb 2016
Last visit: 19 Feb 2025
Posts: 2,800
Own Kudos:
6,239
 [5]
Given Kudos: 47
Location: India
GPA: 3.12
Posts: 2,800
Kudos: 6,239
 [5]
If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 = 22 Oct 2018, 02:39
2
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
Bunuel
If \(p=\frac{1}{√14 − √13}\) and \(q = \frac{1}{√14 + √13}\) then \(p^2 + 2pq + q^2 =\)

A. 26
B. 28
C. 52
D. 56
E. 112
Show more

First step is to simplify p and q by multiplying and dividing with it's conjugate. More on that here.

Now the values of p and q are \(√14 + √13\) and \(√14 - √13\) respectively.

\(p^2 + 2pq + q^2\) = \((p+q)^2\) Now, p+q = \(√14 + √13 + √14 - √13 = 2√14\).

Therefore, the value of \((p+q)^2\) is \((2√14)^2 = 4*14 = 56\)(Option D)
General Discussion
User avatar
PKN
User avatar
Joined: 01 Oct 2017
Last visit: 11 Oct 2025
Posts: 809
Own Kudos:
1,641
 [1]
Given Kudos: 41
Status:Learning stage
WE:Supply Chain Management (Energy)
Posts: 809
Kudos: 1,641
 [1]
If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 = 22 Oct 2018, 02:40
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Bunuel
If \(p=\frac{1}{√14 − √13}\) and \(q = \frac{1}{√14 + √13}\) then \(p^2 + 2pq + q^2 =\)

A. 26
B. 28
C. 52
D. 56
E. 112
Show more

We know, \(p^2 + 2pq + q^2 =(p+q)^2\)=\((\frac{1}{√14 − √13}+\frac{1}{√14 + √13})^2\)=\((\frac{√14 + √13+√14 - √13}{(√14 + √13)(√14 - √13)})^2\)=\((\frac{2√14}{(√14)^2-(√13)^2})^2\)=4*14=56

Ans. (D)
User avatar
dollytaneja51
User avatar
Joined: 26 Feb 2017
Last visit: 02 Apr 2019
Posts: 69
Own Kudos:
60
 [4]
Given Kudos: 14
Location: India
GPA: 3.99
Posts: 69
Kudos: 60
 [4]
If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 = 22 Oct 2018, 05:04
4
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Posted from my mobile device
Attachments

A534BC51-AC08-437E-A099-78A249E75CBB.jpeg
A534BC51-AC08-437E-A099-78A249E75CBB.jpeg [ 96.38 KiB | Viewed 12764 times ]

User avatar
dave13
User avatar
Joined: 09 Mar 2016
Last visit: 15 Mar 2026
Posts: 1,086
Own Kudos:
1,138
 [1]
Given Kudos: 3,851
Posts: 1,086
Kudos: 1,138
 [1]
If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 = 22 Oct 2018, 09:46
Kudos
Add Kudos
Bookmarks
Bookmark this Post
pushpitkc
Bunuel
If \(p=\frac{1}{√14 − √13}\) and \(q = \frac{1}{√14 + √13}\) then \(p^2 + 2pq + q^2 =\)

A. 26
B. 28
C. 52
D. 56
E. 112

First step is to simplify p and q by multiplying and dividing with it's conjugate. More on that here.

Now the values of p and q are \(√14 + √13\) and \(√14 - √13\) respectively.

\(p^2 + 2pq + q^2\) = \((p+q)^2\) Now, p+q = \(√14 + √13 + √14 - √13 = 2√14\).

Therefore, the value of \((p+q)^2\) is \((2√14)^2 = 4*14 = 56\)(Option D)
Show more

pushpitkc if i follow this formula \((p+q)^2\) i get this:

\(((√14 + √13) + (√14 - √13))^2 = (√14 )^2 + √14 *( - √13 ) + √13 *√14 +(√13)^2\) why ? :?
User avatar
Chethan92
User avatar
Joined: 18 Jul 2018
Last visit: 21 Apr 2022
Posts: 899
Own Kudos:
1,511
 [1]
Given Kudos: 95
Location: India
Concentration: Operations, General Management
Schools: IIMB EPGP'22 (M)
GMAT 1: 590 Q46 V25
GMAT 2: 690 Q49 V34
WE:Engineering (Energy)
Products:
My Reviews
Schools: IIMB EPGP'22 (M)
GMAT 2: 690 Q49 V34
Posts: 899
Kudos: 1,511
 [1]
If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 = 22 Oct 2018, 18:34
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
dave13
pushpitkc
Bunuel
If \(p=\frac{1}{√14 − √13}\) and \(q = \frac{1}{√14 + √13}\) then \(p^2 + 2pq + q^2 =\)

A. 26
B. 28
C. 52
D. 56
E. 112

First step is to simplify p and q by multiplying and dividing with it's conjugate. More on that here.

Now the values of p and q are \(√14 + √13\) and \(√14 - √13\) respectively.

\(p^2 + 2pq + q^2\) = \((p+q)^2\) Now, p+q = \(√14 + √13 + √14 - √13 = 2√14\).

Therefore, the value of \((p+q)^2\) is \((2√14)^2 = 4*14 = 56\)(Option D)

pushpitkc if i follow this formula \((p+q)^2\) i get this:

\(((√14 + √13) + (√14 - √13))^2 = (√14 )^2 + √14 *( - √13 ) + √13 *√14 +(√13)^2\) why ? :?
Show more

Cancel out the terms sqrt(13) inside and then solve. You'll get (2sqrt(14))^2 = 4*14 = 56

Posted from my mobile device
User avatar
dave13
User avatar
Joined: 09 Mar 2016
Last visit: 15 Mar 2026
Posts: 1,086
Own Kudos:
1,138
Given Kudos: 3,851
Posts: 1,086
Kudos: 1,138
If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 = 23 Oct 2018, 02:49
Kudos
Add Kudos
Bookmarks
Bookmark this Post
First step is to simplify p and q by multiplying and dividing with it's conjugate. More on that here.

Now the values of p and q are \(√14 + √13\) and \(√14 - √13\) respectively.

\(p^2 + 2pq + q^2\) = \((p+q)^2\) Now, p+q = \(√14 + √13 + √14 - √13 = 2√14\).

Therefore, the value of \((p+q)^2\) is \((2√14)^2 = 4*14 = 56\)(Option D) [/quote]

pushpitkc if i follow this formula \((p+q)^2\) i get this:

\(((√14 + √13) + (√14 - √13))^2 = (√14 )^2 + √14 *( - √13 ) + √13 *√14 +(√13)^2\) why ? :?[/quote]

Cancel out the terms sqrt(13) inside and then solve. You'll get (2sqrt(14))^2 = 4*14 = 56

Posted from my mobile device[/quote]





thanks but there are three radicals with term term 13

if i cancel \(( - √13 ) + √13\) i will still be left with \((√13)^2\) same question is with radical 14 :?


and will get (√14 )^2 + √14 + √14 +(√13)^2 :?

hey pushpitkc gmat mathmaster :) are you there ? :grin:
User avatar
pushpitkc
User avatar
Joined: 26 Feb 2016
Last visit: 19 Feb 2025
Posts: 2,800
Own Kudos:
6,239
 [1]
Given Kudos: 47
Location: India
GPA: 3.12
Posts: 2,800
Kudos: 6,239
 [1]
If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 = 23 Oct 2018, 03:43
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hey dave13

You have unnecessarily confused yourself here.

We have p = √14 + √13, q = √14 - √13 and need to find \((p+q)^2\)

First step is to find the value of p+q which is √14 + √13 + √14 - √13 = 2√14
(Here the √13 cancels each other and we are left with two √14)

The second step is to find the square of the value of (p+q) which is \((2√14)^2 = 4*14 = 56\)

Hope that clears your confusion!
User avatar
BrushMyQuant
User avatar
Joined: 05 Apr 2011
Last visit: 01 May 2026
Posts: 2,287
Own Kudos:
2,684
 [1]
Given Kudos: 100
Status:Tutor - BrushMyQuant
Location: India
Concentration: Finance, Marketing
Schools: XLRI (A)
GMAT 1: 700 Q51 V31
GPA: 3
WE:Information Technology (Computer Software)
Expert
Expert reply
Schools: XLRI (A)
GMAT 1: 700 Q51 V31
Posts: 2,287
Kudos: 2,684
 [1]
If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 = 21 Feb 2023, 07:50
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
↧↧↧ Weekly Video Solution to the Problem Series ↧↧↧



Given that \(p=\frac{1}{√14 − √13}\) and \(q = \frac{1}{√14 + √13}\) and we need to find the value of \(p^2 + 2pq + q^2\) =

\(p^2 + 2pq + q^2\) = \((p + q)^2\) = \( ( \frac{1}{√14 − √13} + \frac{1}{√14 + √13} ) ^ 2\) = \((\frac{√14 + √13 + √14 − √13}{ (√14 − √13) * (√14 + √13)} )^2\)

Now, denominator becomes (a-b) * (a+b) and will be equal to \(a^2 - b^2\)

=> \((\frac{√14 + √13 + √14 − √13}{ (√14)^2 − (√13)^2} )^2\) = \((\frac{2√14 }{14 - 13} )^2\) = 4 * 14 = 56

So, Answer will be D
Hope it helps!

Watch the following video to learn the Properties of Roots

User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 39,014
Own Kudos:
1,122
Posts: 39,014
Kudos: 1,122
If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 = 18 Jan 2026, 11:13
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
Moderators:
Bunuel
Math Expert
110001 posts
Krunaal
Tuck School Moderator
852 posts