Question

If the sum of two of the roots of \(x^3 + px^2 + qx + r = 0\) is zero, then \(pq =\)

• A. \(-r\)
• B. \(2r\)
• C. \(-2r\)
• D. \(r\)

Hint:

Use Vieta's formulas and substitute the root directly into the cubic.

Answer 1

The Correct Option is D

Step 1: Formula / Definition}
\[ \alpha + \beta + \gamma = -p, \quad \alpha\beta + \beta\gamma + \gamma\alpha = q, \quad \alpha\beta\gamma = -r \]
Step 2: Calculation / Simplification}
Given \(\alpha + \beta = 0 \Rightarrow \beta = -\alpha\)
\(\alpha + \beta + \gamma = -p \Rightarrow 0 + \gamma = -p \Rightarrow \gamma = -p\)
Substitute \(\gamma = -p\) into equation:
\((-p)^3 + p(-p)^2 + q(-p) + r = 0\)
\(-p^3 + p^3 - pq + r = 0\)
\(-pq + r = 0 \Rightarrow pq = r\)
Step 3: Final Answer
\[ pq = r \]