How to find constant term in two quadratic equations

Question

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??

Answer 1

Use your $r$ and your definitions for $\alpha(a,r)$ and $\beta(a,r)$ in your first equation, and solve for $a$. Once you have $a$ and $r$, you've determined $\alpha, \beta, \gamma, \delta$, and so you can use those values to determine $p,q$.

Answer 2

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?

Answer 3

$\alpha + \beta = 1$

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.