Question

If \( \alpha, \beta, \gamma \) are the roots of the equation \[ x^3 + 3x^2 - 10x - 24 = 0. \] If \( \alpha(\beta + \gamma), \beta(\gamma + \alpha) \), and \( \gamma(\alpha + \beta) \) are the roots of the equation \[ x^3 + px^2 + qx + r = 0, \] then find the value of \( q \).

• A. \( -44 \)
• B. \( -28 \)
• C. \( 44 \)
• D. \( 28 \)

Hint:

Use Vieta's formulas and transformations to handle equations where new roots are derived from old ones.

Answer 1

The Correct Option is D

Given that $\alpha$, $\beta$, and $\gamma$ are the roots of the equation $x^3 + 3x^2 - 10x - 24 = 0$. If $\alpha(\beta+\gamma)$,  $\beta(\gamma+\alpha)$, and $\gamma(\alpha+\beta)$ are the roots of the equation $x^3 + px^2 + qx + r = 0$, then find the value of $q$.

Let the given cubic equation be

$$x^3 + 3x^2 - 10x - 24 = 0$$

Since $\alpha$, $\beta$, and $\gamma$ are the roots of this equation, by Vieta's formulas, we have:

\[ \begin{aligned} \alpha + \beta + \gamma &= -3 \\ \alpha\beta + \beta\gamma + \gamma\alpha &= -10 \\ \alpha\beta\gamma &= 24 \end{aligned} \]

We are given that $\alpha(\beta+\gamma)$, $\beta(\gamma+\alpha)$, and $\gamma(\alpha+\beta)$ are the roots of the equation $x^3 + px^2 + qx + r = 0$. We want to find $q$, which is the sum of the products of the roots taken two at a time.

\[ q = \alpha(\beta+\gamma)\beta(\gamma+\alpha) + \beta(\gamma+\alpha)\gamma(\alpha+\beta) + \gamma(\alpha+\beta)\alpha(\beta+\gamma) \]

After expanding and substituting the values obtained using Vieta's formulas, we find that

\[ q = 28 \]

Final Answer: The final answer is $\boldsymbol{28}$.