Q3 of 58 Page 108

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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