Apply the distance, rate, and time problem-solving strategy to find an unknown speed on a round trip where both legs cover the same distance, the return speed is slower by a fixed amount, and the time difference between the two legs is known — producing a rational equation that leads to a quadratic.

Problem: Hamilton rode his bike downhill $12$ miles on the river trail from his house to the ocean and then rode uphill to return home. His uphill speed was $8$ miles per hour slower than his downhill speed. It took him $2$ hours longer to get home than it took him to get to the ocean. Find Hamilton's downhill speed.

1. Read and draw: Sketch two parallel paths — one labeled "downhill" ($12$ miles) and one labeled "uphill" ($12$ miles, $8$ mph slower, $2$ hours longer). Create a rate–time–distance table.

2. Identify: Hamilton's downhill speed.

3. Name: Let $r$ = Hamilton's downhill speed in mph. Then his uphill speed is $r - 8$. Because $D = r \cdot t$, solving for time gives $t = \frac{D}{r}$. Divide each distance by its rate to fill in the time column:

	Rate (mph)	Time (hrs)	Distance (miles)
Downhill	$r$	$\frac{12}{r}$	$12$
Uphill	$r - 8$	$\frac{12}{r - 8}$	$12$

4. Translate: The uphill time is $2$ hours more than the downhill time:

$\frac{12}{r - 8} = \frac{12}{r} + 2$

5. Solve: Multiply both sides by the LCD, $r(r - 8)$:

$r(r - 8) \cdot \frac{12}{r - 8} = r(r - 8) \cdot \left(\frac{12}{r} + 2\right)$

Cancel matching factors on the left; distribute on the right:

$12r = 12(r - 8) + 2r(r - 8)$

Expand: $12r = 12r - 96 + 2r^2 - 16r$. Collect all terms on one side:

$0 = 2r^2 - 16r - 96$

Factor out the GCF of $2$: $0 = 2(r^2 - 8r - 48)$. Factor the trinomial:

$0 = 2(r - 12)(r + 4)$

Apply the Zero Product Property:

$r - 12 = 0 \Rightarrow r = 12 \quad \text{or} \quad r + 4 = 0 \Rightarrow r = -4$

A negative speed is not meaningful, so discard $r = -4$.

6. Check: Downhill at $12$ mph: $\frac{12}{12} = 1$ hour. Uphill at $12 - 8 = 4$ mph: $\frac{12}{4} = 3$ hours. The uphill time is $2$ hours more than the downhill time. $\checkmark$

7. Answer: Hamilton's downhill speed is $12$ mph.

This example demonstrates the equal-distance round trip with known time difference scenario. Unlike the Jazmine running/biking example — where two different distances are covered and the total time is known — here both legs of the trip cover the same $12$ miles, and the linking condition is a time difference rather than a time sum. Expressing each time as $t = \frac{D}{r}$ produces two rational expressions, and setting one equal to the other plus $2$ yields a rational equation. Multiplying both sides by the LCD $r(r - 8)$ clears the denominators and produces the quadratic $2r^2 - 16r - 96 = 0$. After factoring out the GCF and factoring the trinomial, the Zero Product Property gives two solutions, and the negative root is discarded because speed must be positive.