Question

find the two consecutive positive integer, sum of their square is 365.​

Answer 1

let the no. be  x and x+1

 then we have x²+(x+1)²=365

                    2x²+2x-364=0

                          x²+x-182=0

               x²+14x-13x-182=0

                      (x+14)(x-13)=0

                That gives x=13 Avoiding negative value

                one number is 13 and other one 14.

Answer 2

Let first number be x
Let second number be (x+1)


According to given condition, we have

x2+(x+1)2=365 {(a+b)2=a2+b2+2ab}



⇒x2+x2+1+2x=365




⇒2x2+2x−364=0


Dividing equation by 2, we get

x2+x−182=0


⇒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.


Mark it as brainliest answer