Three consecutive positive integers are such that the sum of the square of the first and the product of other two is 46, find the integers.

Let the three consecutive numbers be a, a + 1, a + 2


Given, there are three consecutive integers such that the sum of square of the first and the product of the other two is 46.


a2 + (a + 1)(a + 2) = 46


2a2 + 3a + 2 = 46


2a2 + 3a – 44 = 0


2a2 + 11a – 8a – 44 = 0


a(2a + 11) – 4(2a + 11) = 0


(a – 4)(2a + 11) = 0


Thus, a = 4


Numbers are 4, 5, 6


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