If the roots of the equation, ${x}^{3} + p{x}^{2}+qx-1=0$ form an increasing G.P where p and q are real, then
- $p\in \left(-\infty,-3\right)$
- $p\in \left( -3,\infty \right) $
- $p\in\left(-\infty,3\right)$
- $p\in\left(3,\infty\right)$
Let roots be a/r, a, ar. Product of roots = a^3 = 1, so a = 1. Roots are 1/r, 1, r. Since they form an increasing GP, r > 1. The equation is (x-1)(x^2 + (1-p)x + 1) = 0. Sum of roots = 1/r + 1 + r = -p. Since r > 0, r + 1/r >= 2. Thus -p >= 3, so p <= -3.
Let the three roots of the cubic equation be k divided by r, k and kr so they form an increasing geometric progression. Using Vieta's relations for the sum and the product of the roots, we have p equals k times the sum of 1 divided by r, 1 and r, and negative k cubed equals 1. From the product equation, k must equal negative 1. Substituting k equals negative 1 into the sum equation gives p equals the sum of 1 divided by r, 1 and r multiplied by negative 1. Since the progression is increasing and k cubed equals 1, r must be greater than 1. Applying the arithmetic mean and geometric mean inequality to the positive terms 1 divided by r, 1 and r yields their sum is greater than 3. Multiplying by negative 1 reverses the inequality, showing p is strictly less than negative 3. Thus p is in the interval from negative infinity to negative 3.