Answer :
Let integers be x and (x+1)
So,
x[tex] x^{2} + (x+1)^{2} = 365
or, x^{2} + x^{2} + 2x + 1 = 365
or, 2 x^{2} + 2x - 364 = 0 [/tex]
or, [tex] x^{2} + 14x - 13x - 182 = 0
or, (x +14)(x-13) = 0[/tex]
or, x = 13
So, 2 positive integers are 13 and (13+1) = 14
So,
x[tex] x^{2} + (x+1)^{2} = 365
or, x^{2} + x^{2} + 2x + 1 = 365
or, 2 x^{2} + 2x - 364 = 0 [/tex]
or, [tex] x^{2} + 14x - 13x - 182 = 0
or, (x +14)(x-13) = 0[/tex]
or, x = 13
So, 2 positive integers are 13 and (13+1) = 14
[tex] {\huge{\sf{\bold{\boxed{\color{Pink}{answer}}}}}} [/tex]
Let us say, the two consecutive positive integers be x and x + 1.
Therefore, as per the given statement,
[tex] \tt x^2 + (x + 1)^2 = 365 [/tex]
[tex] \tt ⇒ x^2 + x^2 + 1 + 2x = 365 [/tex]
[tex] \tt ⇒ 2x^2 + 2x – 364 = 0 [/tex]
[tex] \tt ⇒ x^2 + x – 182 = 0 [/tex]
[tex] \tt ⇒ x^2 + 14x – 13x – 182 = 0 [/tex]
[tex] \tt ⇒ x(x + 14) -13(x + 14) = 0 [/tex]
[tex] \tt ⇒ (x + 14)(x – 13) = 0 [/tex]
Thus, either, x + 14 = 0 or x – 13 = 0, [/tex]
[tex] \tt ⇒ x = – 14 or x = 13 [/tex]
since, the integers are positive, so x can be 13, only.
So, x + 1 = 13 + 1 = 14