How to find constant term in two quadratic equations Let $\alpha$ and $\beta$ be the roots of the equation $x^2 - x + p=0$ and let $\gamma$ and $\delta$ be the roots of the equation $x^2 -4x+q=0$. If $\alpha , \beta , \gamma , \delta$ are in Geometric progression then what is the value of $p$ and $q$?
My approach:
From the two equations,
$$\alpha + \beta = 1$$,
$$\alpha \beta = p$$,
$$\gamma + \delta = 4$$, and,
$$\gamma \delta = q$$.
Since $\alpha , \beta , \gamma , \delta$ are in G. P., let  $\alpha = \frac{a}{r^3}$, $\beta = \frac{a}{r^1}$, $\gamma = ar$, 
$\delta = ar^3$.
$$\therefore \alpha \beta \gamma \delta = a^4 = pq$$
Now,
$$\frac{\alpha + \beta}{\gamma + \delta} = \frac{1}{r^4}$$
$$\frac{1}{4} = \frac{1}{r^4}$$
$$\therefore r = \sqrt(2)$$
From here I don't know how to proceed. Am I unnecessarily complicating the problem??

$\alpha + \beta = 1$
Also,
$\alpha + \beta = \frac{a}{r^3} + \frac{a}{r^1}$
So we have,
$\frac{a}{r^3} + \frac{a}{r^1} = 1$
Put value of $r$,
$\frac{a}{2\sqrt2} + \frac{a}{\sqrt 2}= 1$
$a = \frac{2\sqrt2}{3}$
Now you can find roots.
