Rohan starts to walk 4 km to school at 7:30 am and her brother Eshaan starts at 8:15 am on his
bicycle, travelling 12 km/hour faster than Rohan. They reach school at the same time.
Represent the given problem in the form of a quadratic equation. 

Let Rohan's walking speed be \( x \) km/h.

### Step 1: Time taken by Rohan to reach school
- Rohan's total distance to school is 4 km.
- The time taken by Rohan to reach school is the total distance divided by his speed:
  \[
  t_{\text{Rohan}} = \frac{4}{x}
  \]
- Rohan starts walking at 7:30 am and let's say he reaches school at \( t_{\text{arrival}} \) hours. So, the time he spends walking is \( t_{\text{arrival}} - 7:30 \) hours.

### Step 2: Time taken by Eshaan to reach school
- Eshaan's speed is 12 km/h faster than Rohan’s speed, so Eshaan's speed is \( x + 12 \) km/h.
- The time taken by Eshaan to reach school is:
  \[
  t_{\text{Eshaan}} = \frac{4}{x + 12}
  \]
- Eshaan starts at 8:15 am, so his travel time is \( t_{\text{arrival}} - 8:15 \) hours.

### Step 3: Equating their arrival times
Since Rohan and Eshaan reach school at the same time, the time they spend travelling must match the difference in their start times:
\[
t_{\text{Rohan}} - t_{\text{Eshaan}} = \text{difference in start times}
\]
Rohan starts at 7:30 am, and Eshaan starts at 8:15 am, so the difference in start times is 45 minutes, or \( \frac{3}{4} \) hours. Thus:
\[
\frac{4}{x} - \frac{4}{x+12} = \frac{3}{4}
\]

### Step 4: Forming the quadratic equation
Multiply both sides of the equation by 4 to eliminate the fraction on the right:
\[
\frac{16}{x} - \frac{16}{x+12} = 3
\]
Now, multiply both sides by \( x(x+12) \) to get rid of the denominators:
\[
16(x+12) - 16x = 3x(x+12)
\]
Simplifying:
\[
16x + 192 - 16x = 3x^2 + 36x
\]
\[
192 = 3x^2 + 36x
\]
Rearranging the equation to standard quadratic form:
\[
3x^2 + 36x - 192 = 0
\]
Dividing through by 3 to simplify:
\[
x^2 + 12x - 64 = 0
\]

This is the required quadratic equation that represents the problem.

Would you like to proceed with solving this equation or have any further questions?

### Related Questions:
1. What is the solution for the speed \( x \) of Rohan from the quadratic equation?
2. How long does it take Rohan to reach the school after solving for \( x \)?
3. How long does it take Eshaan to reach school given his speed?
4. What if Eshaan started at 8:00 am? How would the equation change?
5. How would the equation change if the distance to school was 5 km?

### Tip:
When working with distance-speed-time problems, always pay attention to unit conversions (e.g., minutes to hours) and the starting times.

Rohan starts to walk 4 km to school at 7:30 am and her brother Eshaan starts at 8:15 am on his bicycle, travelling 12 km/hour faster than Rohan. They reach school at the same time. Represent the given problem in the form of a quadratic equation.

Explore how to form a quadratic equation from a distance-speed problem. This involves calculating the time difference between Rohan and Eshaan’s travel to school, with Eshaan traveling 12 km/h faster. Suitable for Grades 8-10, the solution includes algebraic steps to derive the quadratic equation: x^2 + 12x - 64 = 0.

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