Question

Find two consecutive odd positive integers, sum of whose squares is 290.

Answer 1

let first odd positive integer be`x`
let the next integer be`x+2`
acc to question
`x^2 + (x+2)^2 = 290`
`x^2 + x^2 + 4 +4x=290`
`2x^2 + 4x -286=0`
`x^2 +2x-143=0`
`x^2 +13x-11x -143=0`
`x(x+13) -11(x`+13)=0
`(x-11)(x+13)=0`
`x=11,-13`
as x cant be negative, so x=11
`x+2 = 13`
answer