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

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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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New post 22 Oct 2018, 00:33
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A
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E

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  35% (medium)

Question Stats:

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

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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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New post 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.

D is the answer.
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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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New post 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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New post 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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Re: If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 =  [#permalink]

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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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New post 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 ? :?
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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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New post 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

Posted from my mobile device
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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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New post 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 ? :grin:
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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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New post 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 \((2√14)^2 = 4*14 = 56\)

Hope that clears your confusion!
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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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If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 =

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