# 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.

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What does "G.P" stand for? – user2468 Mar 17 '12 at 7:14
@J.D. edited the question. – vikiiii Mar 17 '12 at 7:16
Assuming $p, q, r$ to be positive integers, what can you say immediately about the value of the quadratic at $x=0$ and the general location of the roots? Then you should see that everything depends on the value at $x=1$. What cases are there to consider? – Mark Bennet Mar 17 '12 at 7:29
Are $p$, $q$ and $r$ fixed? If yes, you could just solve the equation and check... Otherwise, what is the precise question? – Johannes Kloos Mar 17 '12 at 7:29
@Johannes Kloos : yes they are fixed.I have to find the number of roots lying in the interval (0,1) .How i can achieve this? – vikiiii Mar 17 '12 at 7:37

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 $(0,1)$.

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(4,1,2) are only a set of values of p,q,r , there will be infinitely many G.P. satisfying the relation given in the question.For example if d= $$\frac{3^(1/2)}{2^(1/2)}$$ – Tomarinator Mar 17 '12 at 9:47
@5T0M, that's roughly where I started with my alternative answer, but actually the greater problem isn't the term ratio (which just scales the indices and doesn't change the roots) but the offset in the progression. – Peter Taylor Mar 17 '12 at 10:09

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.

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