Question

Let $\alpha$ and $\beta$ 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 ${}^{\prime }{r}^{\prime }$ 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 A $\frac{2}{9}(p-q)(2q-p)$

For quadratic equation $x^2-px+r=0$

,product of roots $=\alpha \beta =r$ and sum of roots$=\alpha +\beta =p.....\left(1\right)$

For quadratic equation $x^2-qx+r=0$

,product of roots $=\alpha \beta =r$ and sum of roots$=\frac{\alpha }{2}+2\beta =q\Rightarrow \alpha +4\beta =2q....\left(2\right)$

Subtracting equation (1) from (2) ,

$\alpha +4\beta =2q$

$\underset{–}{-\alpha -\beta =-p}$

$3\beta =2q-p\Rightarrow \beta =\frac{2q-p}{3}$

Now putting the value of $\beta$ in equation (1),

$\alpha +\frac{2q-p}{3}=p$

$\Rightarrow \alpha =p-\frac{2q-p}{3}\Rightarrow \alpha =\frac{2(p-q)}{3}$

Thus, $r=\alpha \beta =\frac{2(p-q)}{3}\times \frac{(2q-p)}{3}=\frac{2}{9}(p-q)(2q-p)$