Question

Find two consecutive positive even integers whose product is 288.

Answer 1

Let the two consecutive positive even integers be x and (x + 2).

According to the given condition,
$$\begin{gathered} x\left(x+2\right)=288 \\ \Rightarrow x^2+2x-288=0 \\ \Rightarrow x^2+18x-16x-288=0 \\ \Rightarrow x\left(x+18\right)-16\left(x+18\right)=0 \end{gathered}$$
$$\begin{gathered} \Rightarrow \left(x+18\right)\left(x-16\right)=0 \\ \Rightarrow x+18=0\mathrm{or}x-16=0 \\ \Rightarrow x=-18\mathrm{or}x=16 \end{gathered}$$
∴ x = 16 (x is a positive even integer)

When x = 16,
x + 2 = 16 + 2 = 18

Hence, the required integers are 16 and 18.