Question

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

Answer 1

let the first no. be x
then second no. = x+1
ATQ
x^2 + (x+1)^2 = 365
x^2 + x^2 + 1 + 2x = 365
2x^2 + 2x = 364
x^2 + x = 182
x^2 + x - 182 = 0
x^2 + 14x - 13x - 182 = 0
x ( x + 14 ) - 13 ( x + 14) = 0
(x-13)(x+14) = 0
x = 13
x = -14
x cannot be = -14 because it is mentioned in the question that Integers are positive.
So x = 13
No.s are 13 , (13+1)
= 13 and 14

Answer 2

Let Integers be x and x+1

  We are given,
                        x² + (x+1)² = 365
                        x² + x² + 2x + 1 = 365       [(a+b) = a² + b² + 2ab]
                      2x² + 2x - 364 = 0
                         x² + x - 182 = 0
                       x² +14x - 13x - 182 = 0     [By Splitting Middle Term ]
                      x(x+14)  -13(x-14)
                 So, (x-13) (x+14) = 0
 
              so, x = 13, -14
           We are given Integers are positive 

So, Integers are 13 and 14 =(13+1)