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.
• B.
• C.
• D.

Answer 1

The correct option is D

$CONVENTIONALAPPROACH$:
Since $\alpha ,\beta$ are roots of $x^2-px+r=0$
$\therefore \alpha +\beta =p,\alpha \beta =r....\left(1\right)$
Since $\frac{\alpha }{2},2\beta$ are roots of $x^2-qx+r=0$
$\therefore \frac{a}{2}+2\beta =q,\alpha \beta =r....\left(2\right)$
from (1), $\alpha +\beta =p$
From(2), $\alpha +4\beta =2q$
$\Rightarrow p+3\beta =2q\Rightarrow \beta =\frac{2q-p}{3}$
$\therefore \alpha =p-\frac{2q-p}{3}=\frac{4p-2q}{3}=\frac{2(2p-q)}{3}$
Hence $r=\alpha \beta$
=$\frac{2}{3}(2p-q)\frac{1}{3}(2q-p)=\frac{2}{9}(2p-q)(2q-p)$