Question

If -2 and -3 are the zeroes of the quadratic polynomial \( x^2 + (a+1)x + b \), then:

• A. \( a = -2, b = 6 \)
• B. \( a = 2, b = -6 \)
• C. \( a = -2, b = -6 \)
• D. \( a = 4, b = 6 \)

Hint:

For quadratic equations \( ax^2 + bx + c = 0 \): \[ \text{Sum of roots} = -\frac{b}{a}, \quad \text{Product of roots} = \frac{c}{a}. \]

Answer 1

The Correct Option is A

Step 1: Using the sum and product of roots formula:
\[ \alpha + \beta = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2} \] \[ \alpha \beta = \frac{\text{Constant term}}{\text{Coefficient of } x^2}. \] Step 2: Substituting \( \alpha = -2 \), \( \beta = -3 \):
\[ (-2) + (-3) = - (a+1) \quad \Rightarrow \quad -5 = -a -1 \quad \Rightarrow \quad a = -2. \] \[ (-2) \times (-3) = b \quad \Rightarrow \quad b = 6. \]