Find two consecutive positive integers, sum of whose square is 613.

Let the two consecutive positive integers be x and x + 1

According to the given condition

x2 + (x + 1)2 = 613

x2 + x2 + 1 + 2x = 613

2x2 + 2x – 612 = 0

x2 + x - 306 = 0

x2 + 18x – 17x - 306 = 0

x(x + 18) - 17(x + 18) = 0

(x + 18)(x - 17) = 0

The roots of the equation are those value for which(x + 18)(x - 17) = 0

x = 17 or - 18

Hence the two positive integers are 17 and 18 or - 17 and - 18

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