algebra precalculus - 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??