Question

If the roots of the equation \( x^3 - 13x^2 + Kx - 27 = 0 \) are in geometric progression, then \( K = \):

• A. \(-30\)
• B. \(30\)
• C. \(39\)
• D. \(-39\)

Hint:

For equations with roots in geometric progression, express the roots as powers of a common ratio and use Vieta’s formulas to solve for the unknown coefficients.

Answer 1

The Correct Option is C

We are given the cubic equation: \[ x^3 - 13x^2 + Kx - 27 = 0 \] Let the roots be in geometric progression (G.P.). Assume the roots are: \[ a, ar, ar^2 \] Step 1: Using Vieta's Formulas
From Vieta's formulas: - Sum of roots: \[ a + ar + ar^2 = 13 \] Factoring out \( a \), \[ a(1 + r + r^2) = 13 \quad \text{(Equation 1)} \] - Product of roots: \[ a \cdot ar \cdot ar^2 = 27 \] \[ a^3 r^3 = 27 \] Taking cube roots, \[ ar = 3 \quad \Rightarrow \quad a = \frac{3}{r} \] Step 2: Substituting \( a = \frac{3}{r} \) into the Sum Equation
From Equation 1, \[ \frac{3}{r} (1 + r + r^2) = 13 \] Multiplying through by \( r \), \[ 3(1 + r + r^2) = 13r \] \[ 3 + 3r + 3r^2 = 13r \] \[ 3r^2 - 10r + 3 = 0 \] Step 3: Solving the Quadratic Equation
Using the quadratic formula: \[ r = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(3)(3)}}{2(3)} \] \[ r = \frac{10 \pm \sqrt{100 - 36}}{6} \] \[ r = \frac{10 \pm \sqrt{64}}{6} \] \[ r = \frac{10 \pm 8}{6} \] \[ r = \frac{18}{6} = 3 \quad \text{or} \quad r = \frac{2}{6} = \frac{1}{3} \] Step 4: Finding \( K \)
From Vieta's relation for the sum of the product of roots taken two at a time: \[ K = a \cdot ar + ar \cdot ar^2 + ar^2 \cdot a \] \[ K = a^2 r + a^2 r^3 + a^2 r^3 \] From \( a = \frac{3}{r} \), substituting back: \[ K = a^2 r(1 + r + r^2) = \left(\frac{3}{r}\right)^2 r (1 + r + r^2) = \frac{9}{r^2} \cdot r \cdot (1 + r + r^2) \] When \( r = 3 \), \[ K = \frac{9}{9} \cdot 3 \cdot (1 + 3 + 9) = 1 \times 3 \times 13 = 39 \] Final Answer: (3) \( 39 \)