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555-605 (Medium)|   Algebra|   Roots|      
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Bunuel
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If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 = 22 Oct 2018, 01:33
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D
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Difficulty:

35% (medium)

Question Stats:

75% (02:04) correct 25% (02:41) wrong based on 686 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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D
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If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 = 22 Oct 2018, 02:37
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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 = 22 Oct 2018, 02:39
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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)
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If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 = 22 Oct 2018, 02:40
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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)
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If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 = 22 Oct 2018, 05:04
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If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 = 22 Oct 2018, 09:46
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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)
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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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If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 = 22 Oct 2018, 18:34
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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

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If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 = 23 Oct 2018, 02:49
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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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If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 = 23 Oct 2018, 03:43
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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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If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 = 21 Feb 2023, 07:50
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↧↧↧ 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

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If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 = 18 Jan 2026, 11:13
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