Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 0 down vote accepted

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 $(0,1)$.

share|cite|improve this answer
(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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.