Question

Find two consecutive positive integers, sum of whose squares are 365.

Answer 1

let 1st positive integer be `x`
second positive interger be `x+1`
acc to question
`(x)^2 +(x+1)^2 = 365`
`x^2 + x^2 +2x + (1)^2 = 365`
`2x^2 + 2x -364= 0`
`x^2 + x - 182=0`
`x^2 + 14x -13x -182=0`
`x(x+14) - 13(x+14) = 0`
`(x+14)(x-13)= 0`
`x= -14 or 13`
cant take negative value so x=13
and x+1=14