Question

If \( \alpha, \beta, \gamma \) are the roots of the equation \( x^3 + 3x^2 - 10x - 24 = 0 \). If \( \alpha>\beta>\gamma \) and \( \alpha^3 + 3\beta^2 - 10\gamma - 24 = 11k \), then \( k = \):

• A. 1
• B. 11
• C. 5
• D. 55

Hint:

To solve cubic equations efficiently, apply Vieta's formulas to express the sums and products of the roots.

Answer 1

The Correct Option is C

We are given the cubic equation: \[ x^3 + 3x^2 - 10x - 24 = 0 \] with roots \( \alpha, \beta, \gamma \) such that \( \alpha>\beta>\gamma \). Step 1: Find the Roots Using the Rational Root Theorem, we check possible rational roots among \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24 \). Checking \( x = 3 \): \[ 3^3 + 3(3^2) - 10(3) - 24 = 27 + 27 - 30 - 24 = 0 \] Thus, \( x - 3 \) is a factor of the polynomial. Performing synthetic division on \( x^3 + 3x^2 - 10x - 24 \) by \( (x - 3) \):

\[ \begin{array}{r|rrrr} 3 & 1 & 3 & -10 & -24  \\ & & 3 & 18 & 24  \\ \hline & 1 & 6 & 8 & 0  \\ \end{array} \]

The quotient polynomial is: \[ x^2 + 6x + 8 = 0 \] Factoring: \[ (x + 2)(x + 4) = 0 \] Thus, the roots are: \[ \alpha = 3, \quad \beta = -2, \quad \gamma = -4 \] Step 2: Compute the Given Expression \[ \alpha^3 + 3\beta^2 - 10\gamma - 24 \] Substituting the values: \[ 3^3 + 3(-2)^2 - 10(-4) - 24 \] \[ = 27 + 3(4) + 40 - 24 \] \[ = 27 + 12 + 40 - 24 = 55 \] \[ = 11k \] Solving for \( k \): \[ 11k = 55 \Rightarrow k = 5 \] Final Answer: \(\boxed{5}\)