Let the original speed of the aircraft = $x$ km/h.
Setting up the equation:
Original time = $\dfrac{2800}{x}$ hours
Reduced speed = $(x - 100)$ km/h, so new time = $\dfrac{2800}{x-100}$ hours
Since time increased by 30 minutes = $\dfrac{1}{2}$ hour:
$$\frac{2800}{x-100} - \frac{2800}{x} = \frac{1}{2}$$
$$2800\left(\frac{x - (x-100)}{x(x-100)}\right) = \frac{1}{2}$$
$$\frac{2800 \times 100}{x(x-100)} = \frac{1}{2}$$
$$x(x - 100) = 560000$$
$$x^2 - 100x - 560000 = 0$$
Solving by factorisation (or quadratic formula):
$$x = \frac{100 \pm \sqrt{10000 + 2240000}}{2} = \frac{100 \pm \sqrt{2250000}}{2} = \frac{100 \pm 1500}{2}$$
Taking positive value: $x = \dfrac{100 + 1500}{2} = 800$ km/h
(Negative value rejected as speed cannot be negative.)
Original duration of flight:
$$t = \frac{2800}{800} = 3.5 \text{ hours}$$
∴ The original duration of the flight is 3.5 hours (3 hours 30 minutes).
Source: Chapter 4, Quadratic Equations
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