There are three consecutive positive integers such that the sum of the square of the first and the product of the other two is 154. What are the integers ?

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

Given, sum of the square of the first and the product of the other two is = 154

$\Rightarrow x^2 + (x + 1)(x + 2) = 154 \\[1em] \Rightarrow x^2 + (x^2 + 2x + x + 2) = 154 \\[1em] \Rightarrow 2x^2 + 3x + 2 - 154 = 0 \\[1em] \Rightarrow 2x^2 + 3x - 152 = 0 \\[1em] \Rightarrow 2x^2 + 19x - 16x - 152 = 0 \\[1em] \Rightarrow x(2x + 19) - 8(2x + 19) = 0 \\[1em] \Rightarrow (x - 8)(2x + 19) = 0 \\[1em] \Rightarrow x - 8 = 0 \text{ or } 2x + 19 = 0 \\[1em] x = 8 \text{ or } x = -\dfrac{19}{2}$

Since the integers are positive hence , x ≠ $-\dfrac{19}{2}$

∴ x = 8 , x + 1 = 9 , x + 2 = 10.

Hence , the required numbers are 8, 9, 10.