Question

The product of two consecutive positive even integers is 48. Find the greatest positive integer.

Answer 1

From that statement we can create the following equation,

`n\cdot \left(n+2\right)=48`

solving for n,

`\begin{gathered} n^2+2n=48 \\ n^2+2n-48=0 \\ n_(1,\:2)=(-2\pm √(2^2-4\cdot \:1\cdot \left(-48\right)))/(2\cdot \:1) \\ n_(1,\:2)=(-2\pm \:14)/(2\cdot \:1) \\ n_1=(-2+14)/(2\cdot \:1),\:n_2=(-2-14)/(2\cdot \:1) \\ n=6,\:n=-8 \end{gathered}`

We can only use the positive number for this problem, therefore n = 6

From the above, the set of numbers is 6 and 6+2=8, since 6*8=48.

Answer: the greatest integer is 8