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

The equation $x^2-px+r=0$ has roots $(\alpha ,\beta )$ and the equation $x^2-qx+r=0$ has roots $(\frac{\alpha }{2},2\beta )$ .

using properties of roots of quadratic equation

$\Rightarrow r=\alpha \beta$ and $\alpha +\beta =p$ $\dots \left(1\right)$

and $\frac{\alpha }{2}+2\beta =q$

$\alpha +4\beta =2q\Rightarrow (\alpha +\beta )+3\beta =2q\Rightarrow 3\beta =2q-p$

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

substitute in $\left(1\right)$ we get

$\alpha =p-\frac{2q-p}{3}=\frac{4p-2q}{3}=\frac{2(2p-q)}{3}$

$\Rightarrow \alpha \beta =r=\frac{2}{9}(2q-p)(2p-q)$ .

option $D$ is correct