Question

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. $\frac{2}{9}(2p-q)(2q-p)$
• B. $\frac{2}{9}(q-p)(2p-q)$
• C. $\frac{2}{9}(q-2p)(2q-p)$
• D. $\frac{2}{9}(2p-q)(2q-p)$

Answer 1

The correct option is D

$\frac{2}{9}(2p-q)(2q-p)$

The equation $x^2-px+r=0$ has roots $\alpha ,\beta$ and the equation $x^2-qx+r=0$ has roots $\frac{\alpha }{2}and2\beta \Rightarrow r=\alpha \beta$ and $\alpha +\beta =p,$
and $\frac{\alpha }{2}+2\beta =q\Rightarrow \beta =\frac{2q-p}{3}$ and $\alpha =\frac{2(2p-q)}{3}$
$\Rightarrow \alpha \beta =r=\frac{2}{9}(2q-p)(2p-q)$