Q. Let $\alpha,\beta$ be the roots of the equation $x^{2} - px + r = 0$ and $\frac{\alpha}{2},\ 2\beta$ be the roots of the equation $x^{2} - qx + r = 0$. Then the value of $r$ is
A) $\frac{2}{9}(p - q)(2q - p)$
B) $\frac{2}{9}(q - p)(2p - q)$
C) $\frac{2}{9}(q - 2p)(2q - p)$
D) $\frac{2}{9}(2p - q)(2q - p)$
Explanation
(d)
Since, the equation $x^{2} - px + r = 0$ has roots $(\alpha,\ \beta)$ and the equation $x^{2} - qx + r = 0$ has roots $\left( \frac{\alpha}{2},\ 2\beta \right)$
$\therefore\ \ \alpha + \beta = p$ and $r = \alpha\beta$ and $\frac{\alpha}{2} + 2\beta = q$
$\Rightarrow \ \beta = \frac{2q - p}{3}$ and $\alpha = \frac{2(2p - q)}{3}$
$\therefore\ \alpha\beta = r = \frac{2}{9}(2q - p)(2p - q)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.