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| 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 |
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| 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. 26 B. 28 C. 52 D. 56 E. 112 |
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| 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. |
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| 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. 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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| 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. 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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| 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. 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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| 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. 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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| 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 = 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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| 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 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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| 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 = 56 So, Answer will be D Hope it helps! Watch the following video to learn the Properties of Roots |
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