Question

Let $\alpha, \beta$ be the roots of the equation $x^{2}-p x+r=0$ and $\frac{\alpha}{2}, 2 \beta$ be the roots of the equation $x^{2}-q x+r=0$. The value of $r$ is

• A. $\frac{2}{9}(p-q)(2 q-p)$
• B. $\frac{2}{9}(q-p)(2 p-q)$
• C. $\frac{2}{9}(q-2 p)(2 q-p)$
• D. $\frac{2}{9}(2 p-q)(2 q-p)$

Answer 1

Answer: (D)

Since $\alpha$ and $\beta$ are the roots of equation
$x^2-p x+r=0$, we get
$\alpha+\beta=p(1)$
$\alpha \beta=r(2)$
Also, since $\alpha / 2$ and $2 \beta$ are the roots of equation
$x 2-q x+r=0$, we get
$\frac{\alpha}{2}+2 \beta=q $
$\Rightarrow \alpha+4 \beta=2 q(3)$
$\alpha \beta=r$
Solving Eqs. (1) and (3), we get
$\alpha=\frac{2(2 p-q)}{3}$ and $\beta=\frac{2 q-p}{3}$
Substituting $\alpha$ and $\beta$ in Eq. (3), we get
$r=\frac{2}{9}(2 p-q)(2 q-p)$