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??

let $$\beta=\alpha y$$,$$\gamma=\alpha y^2$$,$$\delta=\alpha y^3$$ then we get from the first equation
$$\alpha^2-\alpha=\alpha^2y^2-\alpha y$$
from here we get $$\alpha=\frac{1}{1+y}$$
analogously we get from the second equation:$$\alpha=\frac{4}{y^2(1+y)}$$ combining these equations we have $$y^2=4$$ can you finish now?
