Five workers have been hired to complete a job. If one additional worker is hired, they could complete the job $8$ days earlier. If the job needs to completed $28$ days earlier, how many additional workers should be hired?
Let $m$ be the number of additional workers that should be hired. Since one additional worker could complete the job $8$ days earlier, it would take $n-8$ days if $n+1$ workers completed the job. Since the job needs to be completed $28$ days earlier, we have $n^2 - 8n = n^2 - 28.$ Simplifying the equation gives $28 - 8n = 0,$ or $n = \frac{28}{8} = \boxed{3.5}.$
Let's assume that each worker can complete the job in $x$ days. So, the total work done by $5$ workers in one day is $5/x$.
If one additional worker is hired, then the total work done by $6$ workers in one day is $6/x$. We are given that $6$ workers can complete the job $8$ days earlier.
Therefore, the total work done by $6$ workers in $8$ days is equal to the total work done by $5$ workers in $8+8=16$ days.
So, we have the equation:
$\dfrac{6}{x} \cdot 8 = \dfrac{5}{x} \cdot 16$
Simplifying, we get:
$\dfrac{48}{x} = \dfrac{80}{x}$
Cross-multiplying, we get:
$48x = 80x$
Solving for $x$, we find:
$32x = 0$
Since we cannot divide by zero, we have no solution for $x$.
Therefore, we conclude that it is not possible to determine the number of additional workers needed to complete the job 28 days earlier based on the given information.