Question

find two consecutive positive integers, sum of whose square is 365

Answer 1

 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

factorizing the above , we will get..

Now , ⇒x2+14x−13x−182=0

 

⇒x(x+14)−13(x+14)=0

 

⇒(x+14)(x−13)=0

 

⇒x=13,−14

 

Therefore first number = 13   {We discard -14 because it is given that number is positive).

Second number = x+1=13+1=14

Therefore two consecutive positive integers are 13 and 14 whose sum of squares is equal to 365.


the #RIHAAN
hope this will help you.

Answer 2

Let the numbers be x+1and x
Then
x^2+(x+1)^2=365
x^2+x^2+1+2x=365
2x^2+2x-364=0
Dividing whole by2
x^2+x-182=0
From quadratic formula we find zeros
So we get
13 and or -14
So answer us 13 and14