Question

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

Answer 1

Let the three consecutive positive integers be x, x + 1 and x + 2.

According to the given condition, 

x² + (x + 1)(x + 2) = 46 

⇒ x² + x² + 3x + 2 = 46 

⇒ 2x² + 3x – 44 = 0 

⇒ 2x² + 11x – 8x – 44 = 0 

⇒ x(2x + 11) – 4(2x + 11) = 0 

⇒ (2x + 11)(x – 4) = 0 

⇒ 2x + 11 = 0 or x – 4 = 0 

⇒ `x = -11/2` or x = 4 

∴ x = 4   ...( x is a positive integer) 

x + 1 = 4 + 1 = 5 

x + 2 = 4 + 2 = 6

Hence, the required integers are 4, 5 and 6. 

Answer 2

Let the numbers be x, x + 1, x + 2.

According to the question,

x² + (x + 1)(x + 2) = 46

2x² + 3x – 44 = 0

⇒ 2x² + 11x – 8x – 44 = 0

⇒ x(2x + 11) – 4(2x + 11) = 0

⇒ (2x + 11)(x – 4) = 0

∴ 2x + 11 = 0 and x – 4 = 0

⇒ `x = - 11/2` and x = 4

But x can’t be negative

∴ x = 4

So, numbers are 4, 5 and 6.