Question

If the sum of two roots of the cubic equation $x^3 - 5x^2 - 2x + 24 = 0$ is $2$, then the roots of the equation are:

• A. $-2, 3, 4$
• B. $2, 3, 0$
• C. $-1, 3, 3$
• D. $-2, -3, -4$

Hint:

Use options to save time! Only option (A) contains the roots $-2, 3, 4$, which sum up to $5$ and have a pairwise product sum of $-2(3) + 3(4) + 4(-2) = -6 + 12 - 8 = -2$.

Answer 1

The Correct Option is A

Step 1: Concept
Let the roots of the cubic equation be $\alpha, \beta, \gamma$. According to Vieta's formulas: \[ \alpha + \beta + \gamma = -\frac{b}{a} \] \[ \alpha\beta\gamma = -\frac{d}{a} \]

Step 2: Meaning
We are given the sum of two roots, say $\alpha + \beta = 2$. We can find the third root $\gamma$ directly from the sum formula.

Step 3: Analysis
From the given cubic equation $x^3 - 5x^2 - 2x + 24 = 0$, we have: \[ \alpha + \beta + \gamma = 5 \] Since $\alpha + \beta = 2$: \[ 2 + \gamma = 5 \implies \gamma = 3 \] Thus, $3$ is one of the roots. Now, use the product of the roots formula: \[ \alpha\beta\gamma = -24 \implies \alpha\beta(3) = -24 \implies \alpha\beta = -8 \] We also have $\alpha + \beta = 2$. Solving these quadratic relations: \[ t^2 - 2t - 8 = 0 \implies (t - 4)(t + 2) = 0 \implies t = 4 \text{ or } t = -2 \] Hence, the roots are $-2, 3, 4$.

Step 4: Conclusion
The roots of the cubic equation are $-2, 3, 4$. Final Answer: (A)