(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