Question

Three consecutive positive integers are such that the sum of the square of the smallest and the product of the other two is 67. Find the numbers, using the quadratic equation.

Answer 1

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

According to the question,

(x)² + (x + 1) (x + 2) = 67

⇒ x² + x² + 3x + 2 = 67

⇒ 2x² + 3x + 2 – 67 = 0

⇒ 2x² + 3x – 65 = 0

⇒ 2x² + 13x – 10х – 65 = 0

⇒ x(2x + 13) – 5(2x + 13) = 0

⇒ (x – 5) (2x + 13) = 0

⇒ x – 5 = 0 or 2x + 13 = 0

⇒ x = 5 or 2x = –13

x = `(-13)/2`

∵ `(-13)/2` is not a positive integer.

So x = 5,

x + 1

= 5 + 1

= 6

And x + 2

= 5 + 2

= 7

Therefore, the three consecutive positive integers are 5, 6 and 7.