(a) The roots of the equation \(2x^{2} + (p + 1)x + 9 = 0\), are 1 and 3, where p and q are constants. Find the values of p and q.
(b) The weight of an object varies inversely as the square of its distance from the centre of the earth. A small satellite weighs 80kg on the earth’s surface. Calculate, correct to the nearest whole number, the weight of the satellite when it is 800km above the surface of the earth. [Take the radius of the earth as 6,400km].
Explanation
(a) \(2x^{2} + (p + 1)x + q = 0\)
The roots of the equation are 1 and 3, therefore (x - 1)(x - 3) = 0.
Equation : \(x^{2} - 3x - x + 3 = 0\)
\(x^{2} - 4x + 3 = 0\)
\(\therefore 2x^{2} + (p + 1)x + q \equiv x^{2} - 4x + 3\)
\(\frac{2x^{2} + (p + 1)x + q}{2} = x^{2} + \frac{(p + 1)}{2} x + \frac{q}{2}\)
\(\implies \frac{q}{2} = 3\)
\(q = 6\)
\(\frac{p + 1}{2} = -4 \implies p + 1 = -8\)
\(p = -9\)
(p, q) = (-9, 6).
(b) \(W \propto \frac{1}{d^{2}} \)
\(W = \frac{k}{d^{2}}\)
\(k = Wd^{2}\)
\(W = \frac{80 \times (6400)^{2}}{d^{2}}\) (equation of variation).
\(W = \frac{80 \times (6400)^{2}}{(6400 + 800)^{2}}\)
\(W = \frac{80 \times (6400)^{2}}{(7200)^{2}}\)
= 63.21 kg