Let $\alpha, \, \beta$ be the roots of the equation $x^2 - px + r = 0$ and let $\frac{\alpha}{2},\,\, 2\beta$ be the roots of the equation $x^2 - qx + r = 0$. Then the value of $r$ is one of the following.
$$\bbox[yellow, 5px]{(a) \,\frac{2}{9}(p-q)(2q-p) \qquad (b) \,\frac{2}{9}(q-p)(2p-q) }$$ $$\bbox[yellow, 5px]{ (c) \, \frac{2}{9}(q-2p)(2q-p) \qquad (d) \, \frac{2}{9}(2p-q)(2q-p) }$$
Now, I get the following equations for sum and product of the roots.
$$ \bbox[yellow, 5px] { \alpha + \beta = p } $$ $$ \bbox[yellow, 5px] { \frac{\alpha}{2} + 2 \beta = q } $$ $$ \bbox[yellow, 5px] { \alpha \beta = r } $$
Manipulating them, I can see that answer is option (d)
$$ \bbox[yellow, 5px] { r = \, \frac{2}{9}(2p-q)(2q-p) } $$
But I had seen one video on youtube (its not in English), where the author of the video just chose $ \alpha =2 $ and $ \beta = 1$ and arrived at the values $ p = q = 3$ and $ r = 2$ and based on that he reasoned that the answer would be option (d). I want to know the reasoning behind this.
Thanks
If the possible expressions of the solutions are given, it is "allowed" to try them with arbitrary values and if a single of them works, you are done.
This is because you have found counterexamples of the wrong expressions and as the correct expression is guaranteed to be there, you have found it by elimination.
Trivial example: is $8n-6$
If you plug $n=1$, $8n-6=2$ and you can conclude.