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 April 6th, 2016, 02:32 PM #1 Newbie   Joined: Apr 2016 From: Waterbury, CT Posts: 2 Thanks: 0 Can you explain to me in details ? QUESTION : Two integers k and p check the relationship k + p = 20 and k * p = 91. What is the value of k² + p² ? ANSWER : k and p have the values 13 and 7, hence k² + p² = 13² + 7² = 218. I still do not understand why k = 13 and p = 7. Could you give me a hint or explain to me in details please ? Thank you in advance for your answer ! April 6th, 2016, 02:38 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 3,016 Thanks: 1600 simultaneous equations ... $k+p=20 \implies k = 20-p$ $kp = 91 \implies (20-p)p = 91$ $(20-p)p = 91$ $0 = p^2 - 20p + 91$ factor the quadratic ... $0 = (p - 13)(p - 7)$ $p$ could be 13, which means $k=7$ or $p$ could be 7, which means $k=13$ Thanks from topsquark April 6th, 2016, 02:41 PM #3 Newbie   Joined: Apr 2016 From: Waterbury, CT Posts: 2 Thanks: 0 Thank you for your fast and clear answer ! April 6th, 2016, 02:41 PM   #4
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Quote:
 Originally Posted by Ruben715 QUESTION : Two integers k and p check the relationship k + p = 20 and k * p = 91. What is the value of k² + p² ? ANSWER : k and p have the values 13 and 7, hence k² + p² = 13² + 7² = 218. I still do not understand why k = 13 and p = 7. Could you give me a hint or explain to me in details please ? Thank you in advance for your answer !
$\displaystyle k + p = 20 \implies k = 20 - p$

So
$\displaystyle kp = 91 \implies (20 - p)p = 20p - p^2 = 91$

Solve for p. Then k = 20 - p.

Note: We get twin solutions (p, k) = (13, 7) and (p, k) = (7, 13). Obviously these are equivalent.

-Dan April 6th, 2016, 04:50 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,681 Thanks: 2659 Math Focus: Mainly analysis and algebra There's no need to calculate $k$ or $p$: $k^2+p^2=(k+p)^2-2kp=20^2-2\times 91=400-182=218$ Thanks from 123qwerty and aurel5 Tags details, explain, integer, relationship, values Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post shreddinglicks Calculus 1 December 16th, 2014 05:56 PM zacharyt Advanced Statistics 1 January 11th, 2014 11:31 PM nishbhatt87 Algebra 11 August 6th, 2012 04:18 PM bigli Calculus 0 March 4th, 2012 10:51 PM

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