It the roots of the equation  x3 + px2 + qx − 1  =  0   form an increasing G.P. where p and q are real, then

To determine the conditions under which the roots of the polynomial $x^3 + px^2 + qx - 1 = 0$ form an increasing geometric progression (G.P.), we need to analyze the properties of the roots in terms of G.P.

1. ### Break Down the Problem

1. Define the roots of the polynomial as $a, ar, ar^2$ where $a > 0$ (to ensure the G.P. is increasing) and $r > 1$.
2. Use Vieta's formulas to relate the coefficients of the polynomial to the sums and products of its roots.

2. ### Relevant Concepts

1. Vieta's relations for a cubic equation give us:

   • $a + ar + ar^2 = -p$
   • $a \cdot ar + a \cdot ar^2 + ar \cdot ar^2 = q$
   • $a \cdot ar \cdot ar^2 = 1$ (since the constant term is -1)

   From this we get the following equations:

   • $a(1 + r + r^2) = -p$
   • $a^2r(1 + r) = q$
   • $a^3r^3 = 1$

3. ### Analysis and Detail

1. From the equation $a^3r^3 = 1$:

   • $a = \frac{1}{r^{3}}$

2. Substituting $a$ in the first Vieta's relation:

   • $\frac{1}{r^{3}}(1 + r + r^2) = -p$ leads to: $-p = \frac{1 + r + r^2}{r^3}$

3. Substituting $a$ in the second Vieta's relation:

   • $\left(\frac{1}{r^3}\right)^2 r(1 + r) = q$ leads to: $q = \frac{(1 + r)}{r^5}$

4. ### Verify and Summarize

1. We have established the relationships for $p$ and $q$:

   • $p = -\frac{1 + r + r^2}{r^3}$
   • $q = \frac{(1 + r)}{r^5}$

2. These equations can be rearranged to express $p$ and $q$ in terms of the common ratio $r$ while ensuring $r > 1$ for the roots to be indeed an increasing G.P.

Final Answer

The relations for $p$ and $q$ when the roots of the polynomial form an increasing geometric progression are given by: $p = -\frac{1 + r + r^2}{r^3}, \quad q = \frac{1 + r}{r^5}, \quad \text{where } r > 1$