The sum of the squares of two consecutive positive numbers is 61.?

Answer 1

Let the two consecutive positive numbers be ( n ) and ( n+1 ). The equation can be set up as ( n^2 + (n+1)^2 = 61 ). Expanding this gives ( n^2 + n^2 + 2n + 1 = 61 ), which simplifies to ( 2n^2 + 2n - 60 = 0 ). Dividing by 2, we have ( n^2 + n - 30 = 0 ), which factors to ( (n-5)(n+6) = 0 ), yielding ( n = 5 ) (since ( n ) must be positive). Thus, the consecutive numbers are 5 and 6.