Question

Three consecutive positive integers are such that the sum of the square of the first and the product of the other two is 46. Find the integers. [CBSE 2010]

Answer 1

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

According to the given condition,
$$\begin{gathered} x^2+\left(x+1\right)\left(x+2\right)=46 \\ \Rightarrow x^2+x^2+3x+2=46 \\ \Rightarrow 2x^2+3x-44=0 \\ \Rightarrow 2x^2+11x-8x-44=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow x\left(2x+11\right)-4\left(2x+11\right)=0 \\ \Rightarrow \left(2x+11\right)\left(x-4\right)=0 \\ \Rightarrow 2x+11=0\mathrm{or}x-4=0 \\ \Rightarrow x=-\frac{11}{2}\mathrm{or}x=4 \end{gathered}$$
∴ x = 4 (x is a positive integer)

When x = 4,
x + 1 = 4 + 1 = 5
x + 2 = 4 + 2 = 6

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