Tuesday 2 April 2013

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


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

8 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
    2. Yes..To simplfy...so that we could factorize easily.

      Delete
    3. We doesn't take -14 as answer but 14 comes after adding 1 to 13.
      -14 is neglected because question asks for positive numbers at the same time the numbers should be consecutive so 1 is added to 13 to get the next number....

      Delete
  3. The answer is coming 13 and -14.So how do you took 14 as the answer instead of -14?

    ReplyDelete
    Replies
    1. The question has asked for two consecutive POSITIVE integers.....hence from 13 & -14 we take 13 as it is positive

      Delete