Apply the distance, rate, and time problem-solving strategy to find an unknown speed when a single traveler covers two consecutive segments at different uniform rates, with the individual segment distances and the total time all known, producing a rational equation that leads to a quadratic.

Problem: Jazmine trained for $3$ hours on Saturday. She ran $8$ miles and then biked $24$ miles. Her biking speed is $4$ mph faster than her running speed. What is her running speed?

1. Read and draw: Sketch the route with two arrows — one labeled "run" ($8$ miles) and one labeled "bike" ($24$ miles). The total time for both segments is $3$ hours. Create a rate–time–distance table.

2. Identify: Jazmine's running speed.

3. Name: Let $r$ = Jazmine's running speed in mph. Then her biking speed is $r + 4$. 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)
Run	$r$	$\frac{8}{r}$	$8$
Bike	$r + 4$	$\frac{24}{r + 4}$	$24$

4. Translate: The running time plus the biking time equals $3$ hours:

$\frac{8}{r} + \frac{24}{r + 4} = 3$

5. Solve: Multiply both sides by the LCD, $r(r + 4)$:

$r(r + 4) \cdot \frac{8}{r} + r(r + 4) \cdot \frac{24}{r + 4} = 3 \cdot r(r + 4)$

Cancel matching factors: $8(r + 4) + 24r = 3r(r + 4)$. Distribute: $8r + 32 + 24r = 3r^2 + 12r$. Combine like terms on the left: $32r + 32 = 3r^2 + 12r$. Rearrange to standard form:

$0 = 3r^2 - 20r - 32$

Factor: $0 = (3r + 4)(r - 8)$. Apply the Zero Product Property:

$(3r + 4) = 0 \Rightarrow r = -\frac{4}{3}$ or $(r - 8) = 0 \Rightarrow r = 8$

A negative speed is not meaningful, so discard $r = -\frac{4}{3}$.

6. Check: Running at $8$ mph: $\frac{8}{8} = 1$ hour. Biking at $12$ mph: $\frac{24}{12} = 2$ hours. Total: $1 + 2 = 3$ hours. $\checkmark$

7. Answer: Jazmine's running speed is $8$ mph.

This example demonstrates the known-distances-with-total-time scenario. Because the individual distances are given but the individual times are unknown, each time is expressed as a rational expression using $t = \frac{D}{r}$. Adding these rational expressions and setting their sum equal to the known total time produces a rational equation. Unlike the airplane headwind/tailwind example — where clearing the LCD yielded a linear equation — here the LCD is $r(r + 4)$ and clearing it produces a quadratic equation, $3r^2 - 20r - 32 = 0$. The quadratic is solved by factoring and applying the Zero Product Property, and the negative root is discarded because speed must be positive.