Let $α,β$ be the roots of $x^2-x+p=0$ and $γ,δ$ be the roots of $x^2-4x+q = 0$ where $p$ and $q$ are integers. If $α,β,γ,δ$ are in GP then $p + q$ is? 
Let $α,β$ be the roots of $x^2-x+p=0$ and $γ,δ$ be the roots of
$x^2-4x+q = 0$ where $p$ and $q$ are integers. If $α,β,γ,δ$ are in GP
then $p + q$ is ?

My solution approach :-
I have assumed $α,β,γ,δ$  to be as follows ;
$α = \frac{a}{r^3}$
$β = \frac{a}{r}$
$γ = ar$
$δ = ar^3$
Here the common ratio is $r^2$.
Now $α+β = 1$ and $γ+δ = 4$ and solving these two equations in terms of $a$ and $r$ gives the value of $r= \sqrt{2} \text{ and } \sqrt{-2}$ which leads to the values of $a$ to be $\frac{2\sqrt{2}}{3}$ and $\frac{-2\sqrt{2}}{3}$ respectively.
In both the above cases, $α,β,γ,δ$ turns out to be the same i.e. $\frac{1}{3},\frac{2}{3},\frac{4}{3},\frac{8}{3}$ which gives the value of $p+q = \frac{34}{9}$.
Now if I assume $α,β,γ,δ$  to be as follows ;
$α = a$
$β = ar$
$γ = ar^2$
$δ = ar^3$
Here the common ratio is just $r$.
But in this case $r$ values turn out to be $\pm2$ and for $r=2$ we get $α,β,γ,δ$ to be $\frac{1}{3},\frac{2}{3},\frac{4}{3},\frac{8}{3}$ whereas for $r=-2$ we get $α,β,γ,δ$ to be $-1,2,-4,8$ and hence $p+q = \frac{34}{9} \text{ or }-34 $.
Now I am getting confused why there is this difference in both the approaches. What am I doing wrong? Please help me on this !!!
Thanks in advance !!!

The question can also be approached from the direction of working with the zeroes of the polynomials themselves.  For $ \ x^2 - x + p \ = \ 0 \ $ and $ \ x^2 - 4x + q \ = \ 0 \ \ , $ we have
$$ \alpha \ , \ \beta \ \ = \ \ \frac{1 \pm  \sqrt{1 - 4p}}{2} \ = \ \frac12 \ \pm \ \frac12 \sqrt{1-4p} \ = \ \frac12 \ \pm \ \frac12\sqrt{D_1} \ \ , $$
$$ \gamma \ , \ \delta \ \ = \ \ \frac{4 \pm  \sqrt{16 - 4q}}{2} \ = \ 2 \ \pm \ \sqrt{4-q} \ = \ 2 \ \pm \ \sqrt{D_2} \ \ . $$
We wish for the zeroes to lie in a geometric progression $ \ \alpha \ , \ \beta \ , \ \gamma \ , \ \delta \ \ , $ which for real zeroes requires:
•  either a ratio $ \ r > 0 \ $ with all of the zeroes positive and in the indicated sequence, or
• a ratio $ \ r < 0 \ $ with the zeroes alternating in sign and falling on the real-number line in the sequence $ \ \gamma \ , \ \alpha \ , \ (0) \ , \ \beta \  , \ \delta \ \   . $
[Note:  we are using $ \ r \ $ as the ratio between zeroes here, rather than $ \ r^2 \ . $ ]
We can see already what is necessary for the first case:  if $ \ \alpha \ > 0 \ $ (and likewise for the other zeroes) and the zeroes are all to be distinct, we must have $ \ 0 < 1-4p < 1 \ \Rightarrow \ 0 < p < \frac14 \ \ . $ Thus, if $ \ p \ $ and $ \ q \ $ are to be integers, the first case above has no solution.  (We will relax this condition later to see what occurs.)  On the other hand, if we permit $ \ \alpha < 0 \ $ with the other zeroes having alternating signs, we simply have $ \ 1 < 1-4p   \ \Rightarrow \ p < 0 \ \ . $  (We can also find constraints for $ \ q \ $ here, but we'll shortly see that a quite rigid requirement will be set.)
If we first consider the required ratio for each pair  of zeroes, we find
$$ \frac12 · ( \ 1 +  \sqrt{D_1} \ ) \ = \ r · \frac12 · ( \ 1 -  \sqrt{D_1} \ ) \ \ \Rightarrow \ \ D_1 \ = \ \left( \frac{r \ - \ 1}{r \ + \ 1} \right)^2 \ \ , $$
$$ 2 +  \sqrt{D_2}   \ = \ r ·   ( \ 2 -  \sqrt{D_2} \ ) \ \ \Rightarrow \ \ D_2 \ = \ \left( \frac{2r \ - \ 2}{r \ + \ 1} \right)^2 \ = \ 4 · \left( \frac{r \ - \ 1}{r \ + \ 1} \right)^2 \ = \ 4D_1 \ \ . $$
From this, we obtain $ \ 4 - q \ = \ 4 · (1-4p) \ \Rightarrow \ q \ = \ 16p \ \ , $ and consequently that $ \ p + q \ = \ 17p \ \ . $
We now need to "couple" these pairs of zeroes through the condition $ \ \gamma \ = \ r · \beta \ $ or
$$ 2 -  \sqrt{D_2}   \ = \ r ·  \frac12 · ( \ 1 +  \sqrt{D_1} \ ) \ \ \Rightarrow \ \ 2·2 \ -  \ 2·2\sqrt{D_1}   \ = \ r · ( \ 1 +  \sqrt{D_1} \ ) $$
$$ \Rightarrow \ \  D_1 \ = \ \left( \frac{4 \ - \ r}{r \ + \ 4} \right)^2  \ \ . $$
Equating the two ratio expressions for $ \ D_1 \ $ yields either
$$ \left( \frac{r \ - \ 1}{r \ + \ 1} \right) \ = \ \left( \frac{4 \ - \ r}{r \ + \ 4} \right) \ \ \ \text{or} \ \ \ \left( \frac{r \ - \ 1}{r \ + \ 1} \right) \ = \  \left( \frac{r \ - \ 4}{r \ + \ 4} \right) \ \ ; $$
the first of these leads to $ \ r^2 + 3r - 4 \ = \ -r^2 + 3r + 4 \ \Rightarrow \ 2r^2 \ = \ 8 \ \Rightarrow \ r^2 \ = \ 4 \ \ , $ while the second equation produces $ \ r^2 + 3r - 4 \ = \ r^2 - 3r - 4 \ \Rightarrow \ 6r  \ = \ 0 \   , $ which is a "trivial" result.
The first of the cases (for which we already know $ \ p \ $ and $ \ q \ $ cannot be integral) gives us
$ \mathbf{ r = 2 \ \ : }  \quad D_1 \ = \ \left( \frac{2 \ - \ 1}{2 \ + \ 1} \right)^2 \ = \ \frac19 \ = \ 1 - 4p \ \ \Rightarrow \ \ p \ = \ \frac29 \ \ \Rightarrow \ \ p+q \ = \ \frac{34}{9} \ \  $
[we do have a rational-number solution] with the zeroes being $ \ \alpha \ , \ \beta   \ =   \   \frac12 \ \pm \ \frac12\sqrt{\frac19} \ = \ \frac13 \ , \ \frac23 \ $ and $ \ \gamma \ , \ \delta \  = \ 2 \ \pm \ \sqrt{4·\frac19} \ = \ \frac43 \ , \ \frac83 \ \ . $
The alternating-sign case is
$ \mathbf{ r = -2 \ \ : }  \quad D_1 \ = \ \left( \frac{[-2] \ - \ 1}{[-2] \ + \ 1} \right)^2 \ = \ 9 \ = \ 1 - 4p \ \ \Rightarrow \ \ p \ = \ -2 \ \ \Rightarrow \ \ p+q \ = \ -34 \ \  $
with the zeroes being $ \ \alpha \ , \ \beta   \ =   \  \frac12 \ \pm \  \frac12·\sqrt{9} \ = \ -1 \ , \ 2 \ $ and $ \ \gamma \ , \ \delta \  = \ 2 \ \pm \ \sqrt{4·9} \ = \ -4 \ , \ 8 \ \ , $ arranged on the $ \ x-$axis in the order $ \ \gamma \ , \ \alpha \ ,  \ \beta \  , \ \delta \ \   . $
Graphs for the two cases are presented below.
$ \ \ $
We won't need to be concerned about the situation for complex zeroes.  We would then have
$$ \alpha  \ , \ \beta = \overline{\alpha} \ \ = \ \  \frac12 \ \pm \ i · \frac12 \sqrt{4p-1}   \ \ , \ \  \gamma \ , \ \delta = \overline{\gamma} \ \ = \ \   2 \ \pm \ i · \sqrt{q-4}   \ \ , \ \ p > \frac14 \ , \ q > 4 \ \ . $$  Since these are conjugate-pair zeroes which may be expressed as $ \ \zeta · e^{\pm i \phi} \ $ and $ \ \xi · e^{\pm i \psi} \ \ , $ respectively, there is no single complex ratio $ \ r \ = \ \rho · e^{i \theta} \ \ , $ with $ \ \rho \ , \ \zeta \ $ and $ \ \xi \ $ being real numbers, that will put all four zeroes into a geometric progression.


