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

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)

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

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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!

↧↧↧ 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

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)

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 ?

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 ?