Question

If $\alpha ,\beta$ are the roots of the equation $x^2-px+r=0$ and $\frac{\alpha }{2},2\beta$ are the roots of the equation $x^2-qx+r=0$, then the value of $r$ in terms of $p$ and $q$ 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. $\frac{2}{9}(2p-q)(2q-p)$

Answer 1

The correct option is D $\frac{2}{9}(2p-q)(2q-p)$

Since $\alpha ,\beta$ are the roots of $x^2-px+r=0$
$\therefore \alpha +\beta =p$ and $\alpha \beta =r$
It is given that $\frac{\alpha }{2}$ and $2\beta$ are the roots of the equation $x^2-qx+r=0$
$\therefore \frac{\alpha }{2}+2\beta =q$ and
$\frac{\alpha }{2}\times 2\beta =r$
Solving
$\alpha +\beta =p\frac{\alpha }{2}+2\beta =q$
$\Rightarrow \frac{3\alpha }{2}=2p-q\Rightarrow \alpha =\frac{2}{3}(2p-q)\Rightarrow \beta =\frac{1}{3}(2q-p)$

We know that,
$r=\alpha \beta \Rightarrow r=\frac{2}{9}(2p-q)(2q-p)$