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

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Joined: 02 Sep 2009
Posts: 50729
If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 =  [#permalink]

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22 Oct 2018, 00:33
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Difficulty:

35% (medium)

Question Stats:

73% (01:40) correct 27% (02:01) wrong based on 66 sessions

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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

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

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22 Oct 2018, 01:37
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.

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

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22 Oct 2018, 01:39
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)
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If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 =  [#permalink]

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22 Oct 2018, 01:40
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)
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Joined: 26 Feb 2017
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Re: If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 =  [#permalink]

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22 Oct 2018, 04:04
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Joined: 09 Mar 2016
Posts: 1111
If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 =  [#permalink]

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22 Oct 2018, 08:46
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 ?
Senior Manager
Joined: 18 Jul 2018
Posts: 380
Location: India
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Re: If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 =  [#permalink]

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22 Oct 2018, 17:34
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

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VP
Joined: 09 Mar 2016
Posts: 1111
If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 =  [#permalink]

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23 Oct 2018, 01:49
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 ?
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Location: India
GPA: 3.12
Re: If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 =  [#permalink]

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23 Oct 2018, 02:43
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 $$(2√14)^2 = 4*14 = 56$$

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Re: If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 = &nbs [#permalink] 23 Oct 2018, 02:43
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