If $\alpha$ and $\beta$ are the zeroes of the polynomial $p(x) = x^2 - ax - b$, then the value of $(\alpha + \beta + \alpha\beta)$ is equal to:
Generated by claude-sonnet-4-6 · 2026-06-15 10:34 · grounding rag
Model Answer
For $p(x) = x^2 - ax - b$, comparing with $Ax^2 + Bx + C$: $A=1,\ B=-a,\ C=-b$.
$$\alpha + \beta = \frac{-B}{A} = a, \qquad \alpha\beta = \frac{C}{A} = -b$$
$$\therefore\ \alpha + \beta + \alpha\beta = a + (-b) = a - b$$
Answer: (B) $a - b$
Source: Chapter 2, Section 2.3
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Explanation
- Identify coefficients carefully: the polynomial is $x^2 - ax - b$, so the coefficient of $x$ is $-a$ (not $a$) and the constant term is $-b$.
- Apply the standard formulae: sum of zeroes $= \frac{-(\text{coeff of }x)}{\text{coeff of }x^2}$ and product of zeroes $= \frac{\text{constant term}}{\text{coeff of }x^2}$.
- The most common mistake is using $a$ and $b$ directly without accounting for the minus signs. Always write out the comparison step explicitly to avoid sign errors.