Mr Tan took part in a 42 km marathon run. He used $x$ minutes to run the first 25 km at a constant speed.

1. Convert the time from minutes to hours and form the speed expression.

Time taken for the first 25 km is $x$ minutes.

In hours, this is $\dfrac{x}{60}$ hours.

$\text{Speed}=\frac{\text{distance}}{\text{time}}=\frac{25}{x/60}=\frac{1500}{x}$

1. Form the equation for the last 17 km using speed, distance and time.

Initial speed from part (a) is $\dfrac{1500}{x}$ km/h.

Speed for the last 17 km is therefore
$\frac{1500}{x}-2$
km/h.

Time for the last 17 km is $(x-22.5)$ minutes, so in hours it is
$\frac{x-22.5}{60}.$

Using $\text{distance}=\text{speed}\times\text{time}$,
$17=\left(\frac{1500}{x}-2\right)\left(\frac{x-22.5}{60}\right).$

2. Simplify the equation to the required quadratic form.

$17=\left(\frac{1500}{x}-2\right)\left(\frac{x-22.5}{60}\right)$

Multiply both sides by $60$:
$1020=\left(\frac{1500}{x}-2\right)(x-22.5)$

Multiply both sides by $x$:
$1020x=(1500-2x)(x-22.5)$

Expand:
$1020x=1500x-33750-2x^2+45x$
$1020x=1545x-33750-2x^2$

Rearrange:
$2x^2-525x+33750=0$

3. Solve the quadratic equation.

$2x^2-525x+33750=0$

Using the quadratic formula,
$x=\frac{-(-525)\pm\sqrt{(-525)^2-4(2)(33750)}}{2(2)}$

$x=\frac{525\pm\sqrt{275625-270000}}{4}$

$x=\frac{525\pm\sqrt{5625}}{4}$

$x=\frac{525\pm75}{4}$

Hence,
$x=\frac{600}{4}=150$
or
$x=\frac{450}{4}=112.5.$

1. Find the possible total times using the two possible values of $x$.

From part (c), $x=112.5$ or $150$.

Total time for the marathon
$=x+(x-22.5)=2x-22.5 \text{ minutes}.$

If $x=112.5$,
$2x-22.5=2(112.5)-22.5=202.5\text{ min}.$

If $x=150$,
$2x-22.5=2(150)-22.5=277.5\text{ min}.$

2. Select the time that is more than 4 hours and convert it to hours and minutes.

More than $4$ hours means more than
$4\times 60=240\text{ min}.$

So the valid total time is $277.5$ min.

$277.5\text{ min}=240\text{ min}+37.5\text{ min}=4\text{ h }37.5\text{ min}.$