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

 It is currently 25 Mar 2019, 11:26

### 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

# If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 =

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

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 53831
If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 =  [#permalink]

### Show Tags

22 Oct 2018, 01:33
00:00

Difficulty:

35% (medium)

Question Stats:

71% (02:06) correct 29% (02:29) wrong based on 102 sessions

### HideShow timer Statistics

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

_________________
Director
Joined: 18 Jul 2018
Posts: 762
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)
If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 =  [#permalink]

### Show Tags

22 Oct 2018, 02:37
1
1
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.
_________________
Press +1 Kudo If my post helps!
Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3365
Location: India
GPA: 3.12
If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 =  [#permalink]

### Show Tags

22 Oct 2018, 02:39
1
Bunuel wrote:
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)
_________________
You've got what it takes, but it will take everything you've got
Director
Status: Learning stage
Joined: 01 Oct 2017
Posts: 965
WE: Supply Chain Management (Energy and Utilities)
If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 =  [#permalink]

### Show Tags

22 Oct 2018, 02:40
Bunuel wrote:
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

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)
_________________
Regards,

PKN

Rise above the storm, you will find the sunshine
Manager
Joined: 26 Feb 2017
Posts: 89
Location: India
GPA: 3.99
Re: If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 =  [#permalink]

### Show Tags

22 Oct 2018, 05:04
2
Posted from my mobile device
Attachments

A534BC51-AC08-437E-A099-78A249E75CBB.jpeg [ 96.38 KiB | Viewed 789 times ]

VP
Joined: 09 Mar 2016
Posts: 1279
If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 =  [#permalink]

### Show Tags

22 Oct 2018, 09:46
pushpitkc wrote:
Bunuel wrote:
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 ?
Director
Joined: 18 Jul 2018
Posts: 762
Location: India
Concentration: Finance, Marketing
WE: Engineering (Energy and Utilities)
Re: If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 =  [#permalink]

### Show Tags

22 Oct 2018, 18:34
1
dave13 wrote:
pushpitkc wrote:
Bunuel wrote:
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 ?

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

Posted from my mobile device
_________________
Press +1 Kudo If my post helps!
VP
Joined: 09 Mar 2016
Posts: 1279
If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 =  [#permalink]

### Show Tags

23 Oct 2018, 02:49
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 ?
Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3365
Location: India
GPA: 3.12
Re: If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 =  [#permalink]

### Show Tags

23 Oct 2018, 03:43
1
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!
_________________
You've got what it takes, but it will take everything you've got
Re: If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 =   [#permalink] 23 Oct 2018, 03:43
Display posts from previous: Sort by

# If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 =

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

 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®.