Question
Let $$\alpha ,\beta $$ be the roots of the equation $${x^2} - px + r = 0\,\,{\text{and }}\frac{\alpha }{2},2\beta $$ be the roots of the equation $${x^2} - qx + r = 0.\,$$ Then the value of $$r$$ is
A.
$$\frac{2}{9}\left( {p - q} \right)\left( {2q - p} \right)$$
B.
$$\frac{2}{9}\left( {q - p} \right)\left( {2p - q} \right)$$
C.
$$\frac{2}{9}\left( {q - 2p} \right)\left( {2q - p} \right)$$
D.
$$\frac{2}{9}\left( {2p - q} \right)\left( {2q - p} \right)$$
Answer :
$$\frac{2}{9}\left( {2p - q} \right)\left( {2q - p} \right)$$
Solution :
As $$\alpha ,\beta $$ are the roots of $${x^2} - px + r = 0$$
$$\eqalign{
& \therefore \alpha + \beta = p\,\,\,\,\,\,\,\,\,\,\,.....\left( 1 \right) \cr
& {\text{and }}\alpha \beta = r\,\,\,\,\,\,\,.....\left( 2 \right) \cr} $$
Also $$\frac{\alpha }{2},2\beta $$ are the roots of $${x^2} - qx + r = 0$$
$$\therefore \,\,\frac{\alpha }{2} + 2\beta = q\,\,{\text{or }}\alpha + 4\beta = 2q\,\,\,\,\,\,\,.....\left( 3 \right)$$
Solving (1) and (3) for $$\alpha $$ and $$\beta $$ , we get
$$\beta = \frac{1}{3}\left( {2q - p} \right)\,\,{\text{and }}\alpha = \frac{2}{3}\left( {2q - q} \right)$$
Substituting values of $$\alpha $$ and $$\beta $$ , in equation (2),
we get $$\frac{2}{9}\left( {2p - q} \right)\left( {2q - p} \right) = r.$$