Question

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.

Answer 1

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