NIMCET 2018 — Mathematics PYQ

We know that, if ${\alpha }_1,{\alpha }_2$ be the roots of the quadratic equation ax${}^2+$bx+c=0 then

$${\alpha }_1+{\alpha }_2=\frac{-\mathrm{b}}{\mathrm{a}}\mathrm{and}{\alpha }_1.{\alpha }_2=\frac{\mathrm{c}}{\mathrm{a}}$$

Calculations:
We know that, if ${\alpha }_1$, ${\alpha }_2$ be the roots of the quadratic equation $a{x}^2+bx+c=0$ then ${\alpha }_1+{\alpha }_2=\frac{-b}{a}$ and ${\alpha }_1.{\alpha }_2=\frac{c}{a}$
Given, $\alpha$, $\beta$ be the roots of the equation $x^2-px+r=0$
$\Rightarrow \alpha \beta =r$ and $\alpha +\beta =p$ .... (1)
Also, $\frac{\alpha }{2}$, $2\beta$ be the roots of the equation $x^2-qx+r=0$
$\Rightarrow \alpha \beta =r$ and $\frac{\alpha }{2}+2\beta =q$

$$\Rightarrow \alpha +4\beta =4q....(2)$$
$$<br>\Rightarrow \beta =\frac{2q-p}{3}$$
$$\Rightarrow \alpha =p-\frac{2q-p}{3}$$
$$<br>\Rightarrow \alpha =\frac{2p-2q}{3}$$
$$\Rightarrow r=\alpha \beta$$
$$<br>\Rightarrow r=\frac{2p-2q}{3}\frac{2q-p}{3}$$
$$\Rightarrow r=\frac{2}{9}(p-q)(2q-p)$$
Hence, if $\alpha$, $\beta$ be the roots of the equation $x^2-px+r=0$ and $\frac{\alpha }{2}$, $\beta$ be the roots of the equation $x^2-qx+r=0$.
Then the value of $r$ is $\frac{2}{9}(p-q)(2q-p)$