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 $x^2-qx+r=0$. 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

Answer: (D)

Since $\alpha$ and $\beta$ are the roots of equation
$x^2-px+r=0$, we get
$\alpha +\beta =p(1)$
$\alpha \beta =r(2)$
Also, since $\alpha /2$ and $2\beta$ are the roots of equation
$x2-qx+r=0$, we get
$\frac{\alpha }{2}+2\beta =q$
$\Rightarrow \alpha +4\beta =2q(3)$
$\alpha \beta =r$
Solving Eqs. (1) and (3), we get
$\alpha =\frac{2(2p-q)}{3}$ and $\beta =\frac{2q-p}{3}$
Substituting $\alpha$ and $\beta$ in Eq. (3), we get
$r=\frac{2}{9}(2p-q)(2q-p)$