find 2 consecutive positive integers , sum of whose squares is 925 ?
Let the two consecutive positive ingers be = x , (x + 1)
According to the question,
x2 + (x + 1)2 = 925
x2 + x2 + 1 + 2x = 925
2x2 + 2x - 924 = 0
x2 + x - 462 = 0
x2 - 21x + 22x - 462 = 0
x (x - 21) + 22 (x - 21) = 0
(x - 21) (x + 22) = 0
x = 21
... the two consecutive positive integers are 21 and 22.