(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 obje...
Question 1 Report
(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].
Answer Details
(a) The equation is \(2x^{2}+(p+1)x+q=0\) with roots \(1\) and \(3\) (the constant term is the unknown \(q\), not a fixed number).
Sum of roots \(=1+3=4=-\dfrac{p+1}{2}\Rightarrow p+1=-8\Rightarrow p=-9\).
Product of roots \(=1\times 3=3=\dfrac{q}{2}\Rightarrow q=6\).
\(p=-9,\ q=6\).
(b) Weight \(W\propto\dfrac{1}{d^{2}}\), so \(W=\dfrac{k}{d^{2}}\). On the surface, \(d=6400\text{ km}\), \(W=80\text{ kg}\):
\(k=80\times 6400^{2}\).
At \(800\text{ km}\) above the surface, \(d=6400+800=7200\text{ km}\):
\(W=80\times\left(\dfrac{6400}{7200}\right)^{2}=80\times\dfrac{64}{81}=\dfrac{5120}{81}=63.2\).
Weight \(\approx \mathbf{63\ kg}\).
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