Question

two consecutive positive integers that the sum of their squares is 365

Answer 1

let the first no be =x
let the second no be =x+1
ATQ
x^2+(x+1)^2=365

x^2+x^2+2x+1=365

2x^2+2x+1=365

2x^2+2x+1-365=0

2x^2+2x-364=0

divide the equation by 2

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=0. x-13=0


x=14 x=13

the no are 13,14

Answer 2

Step-by-step explanation:

so the two numbers be x , x+1

so given

x² + (x+1)² = 365

x² + x² + 1 + 2x = 365

2x² + 2x - 364 = 0

x² + x - 182 = 0

x² + 14x - 13x - 182 = 0

x(x + 14) - 13(x + 14) = 0

(x - 13)(x + 14) = 0

so x = 13  

so the  

consecutive numbers are 13,14