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

As $\alpha ,\beta$ are the roots of $x^2-px+r=0$
$\therefore \alpha +\beta =p$ .......(1)
and $\alpha \beta =r$ .......(2)
Also $\frac{\alpha }{2},2\beta$ are the roots of $x^2-qx+r=0$
$\therefore \frac{\alpha }{2}+2\beta =q$ or $\alpha +4\beta =2q$ .....(3)
Solving (1) and (3) for $\alpha and\beta$,we get
$\beta =\frac{1}{3}(2q-p)$ and $\alpha =\frac{2}{3}(2q-q)$
Substituting values of $\alpha$ and $\beta$, in equation (2),
we get $\frac{2}{9}(2p-q)(2q-p)=r.$