GMAT Club Forumhttps://gmatclub.com:443/forum/ If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 =https://gmatclub.com/forum/if-p-1-14-13-and-q-1-14-13-then-p-2-2pq-q-279624.html Page 1 of 1

 Author: Bunuel [ 22 Oct 2018, 01:33 ] Post subject: If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 = If $$p=\frac{1}{√14 − √13}$$ and $$q = \frac{1}{√14 + √13}$$ then $$p^2 + 2pq + q^2 =$$A. 26B. 28C. 52D. 56E. 112

 Author: Chethan92 [ 22 Oct 2018, 02:37 ] Post subject: If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 = 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.

 Author: pushpitkc [ 22 Oct 2018, 02:39 ] Post subject: If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 = Bunuel wrote:If $$p=\frac{1}{√14 − √13}$$ and $$q = \frac{1}{√14 + √13}$$ then $$p^2 + 2pq + q^2 =$$A. 26B. 28C. 52D. 56E. 112First 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)

 Author: PKN [ 22 Oct 2018, 02:40 ] Post subject: If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 = Bunuel wrote:If $$p=\frac{1}{√14 − √13}$$ and $$q = \frac{1}{√14 + √13}$$ then $$p^2 + 2pq + q^2 =$$A. 26B. 28C. 52D. 56E. 112We 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=56Ans. (D)

Author:  dollytaneja51 [ 22 Oct 2018, 05:04 ]
Post subject:  Re: If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 =

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 Author: dave13 [ 22 Oct 2018, 09:46 ] Post subject: If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 = pushpitkc wrote:Bunuel wrote:If $$p=\frac{1}{√14 − √13}$$ and $$q = \frac{1}{√14 + √13}$$ then $$p^2 + 2pq + q^2 =$$A. 26B. 28C. 52D. 56E. 112First 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 ?

 Author: Chethan92 [ 22 Oct 2018, 18:34 ] Post subject: Re: If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 = 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. 26B. 28C. 52D. 56E. 112First 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 = 56Posted from my mobile device

 Author: dave13 [ 23 Oct 2018, 02:49 ] Post subject: 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) [/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 = 56Posted 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 ?

 Author: pushpitkc [ 23 Oct 2018, 03:43 ] Post subject: Re: If p = 1/(√14 − √13) and q = 1/(√14 + √13) then p^2 + 2pq + q^2 = Hey dave13You 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!

 Author: BrushMyQuant [ 21 Feb 2023, 07:50 ] Post subject: Re: If p = 1/(14 13) and q = 1/(14 + 13) then p^2 + 2pq + q^2 = ↧↧↧ 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 = 56So, Answer will be DHope it helps!Watch the following video to learn the Properties of Roots

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