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Q4: Find two consecutive positive integers, sum of whose squares is 365.

Tuesday, 2 April 2013

Q4: Find two consecutive positive integers, sum of whose squares is 365.
Answer:
 Let first number = x
Then second number will one more so that next number wil x+1
Given that sum of whose squares is 365.
      x2+ (  x + 1)2 = 365
use formula of (a +b)2 = a2 + 2ab +b2
    x2+   x2+ 2x + 1 – 365 = 0
  2x2+ 2x – 364 = 0
Divide by 2 to simplify it
    x2+   x – 182 = 0
factorize it now


Therefore required numbers are 13 and 14

3 comments:

  1. To understand the solution of this problem, I visited so many sites, guides, texts etc. But nowhere I could find the solution of my doubt.
    My doubt was how 2X is coming. Here you clearly solved that doubt and problem (see your explanation line no. 5. Use the formula (a+b)^2 = a^2+2ab+b^2.

    Yes, you're a teacher thinks from students side..

    Thanks a lot. God bless you.

    ReplyDelete
  2. Why we divided it by 2 to simplify it..can u tell this?

    ReplyDelete
    Replies
    1. to take 2 common from all, which is then multiplied to 0 on other side

      Delete