How many roots lie in the interval $(0,1)$? 
If $p$th , $q$th , $r$th term of a Geometric Progression be a $27 , 8$ and
  $12$ respectively, then how many root of the quadratic equation $ px^2
 + 2qx -2r = 0  $ lie in the interval $(0,1)$ ?

Does this problem requiring taking logarithms or can be solved without them.
Thanks in advance.
 A: This began as a comment to Foool's answer, but grew.
The ratios given lead to a set of solutions $r = q + k$, $p = q + 3k$, where the term ratio of the geometric progression is $\left(\frac{3}{2}\right)^\frac{1}{k}$. (Foool's answer is therefore the special case $q=k=1$). Note that $k$ is integral and non-zero, but may be negative; $q$ can be any integer.
Therefore you have $$px^2 + 2qx - 2r = (q+3k)x^2 + 2qx - 2(q+k)$$
Jumping straight to the quadratic equation, $$\begin{eqnarray}x & = & \frac{-q \pm \sqrt{q^2 + 2(q+3k)(q+k)} }{q+3k} \\
& = & \frac{-q \pm \sqrt{3q^2 + 8qk + 6k^2} }{q+3k}\end{eqnarray}$$
So it looks quite complicated.
We can derive a Sturm sequence:
$$\begin{eqnarray}
P_0 & = & (q+3k)x^2 + 2qx - 2(q+k) \\
P_1 & = & (q+3k)x + q \\
P_2 & = & -qx + 2(q+k) \\
P_3 & = & \frac{q^2+8qk+6k^2}{q}
\end{eqnarray}$$
where I've scaled P_1 by a positive scalar as a computational optimisation. If we consider sign changes in
$$\begin{eqnarray}
P_0(0) & = & - 2(q+k) \\
P_1(0) & = & q \\
P_2(0) & = & 2(q+k) \\
P_3(0) & = & \frac{q^2+8qk+6k^2}{q}
\end{eqnarray}$$
we have one between $P_0(0)$ and $P_2(0)$, and a second if $$ \frac{(q+k)(q^2+8qk+6k^2)}{q} < 0$$
If we consider sign changes in
$$\begin{eqnarray}
P_0(1) & = & q+5k \\
P_1(1) & = & 2q+3k \\
P_2(1) & = & q+2k \\
P_3(1) & = & \frac{q^2+8qk+6k^2}{q}
\end{eqnarray}$$
And again we see that it's quite complicated. With some patience one can draw a graph showing the regions where the number of sign changes is the same, and hence derive the number of roots, but I think there may be some criterion missing from the question.
A: Let the first term of the G.P be  $a$ and common ratio is $d$.
So, $a \times d^{p-1} = 27, a \times d^{q-1} = 8 $ and $ a \times d^{r-1} = 12$
Now lets take the ratio two at a time, it will give three equations $$d^{p-q}=\left (\frac32\right)^3 \tag{1}$$ $$ d^{r-q}=\left (\frac32\right) \tag{2}$$ $$ d^{p-r}=\left (\frac32\right)^2 \tag{3}$$
Clearly, $d=\frac32$  and $p=4,q=1,r=2$ so now our equation becomes $$ 4x^2+2x-4=0$$
We can use quadratic formula for the rest, if my algebra is correct the roots are $\frac{1}{4} \left(-1-\sqrt{17}\right)$  and $ \frac{1}{4} \left(-1+\sqrt{17}\right)$ of which only the later lies in $(0,1)$.
