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. \( a = 6, b = 6 \)

Hint:

For any quadratic polynomial \( ax^2 + bx + c \), the sum and product of the zeroes are given by: - Sum of zeroes: \( -\frac{b}{a} \) - Product of zeroes: \( \frac{c}{a} \)

Answer 1

The Correct Option is C

Step 1: The sum of the zeroes of the quadratic polynomial is \( -a-1 \). Since the zeroes are -2 and -3, the sum of the zeroes is: \[ -2 + (-3) = -5 \] Thus, \[ -a - 1 = -5 \quad \Rightarrow \quad a = -2 \] Step 2: The product of the zeroes is \( b \). The product of -2 and -3 is: \[ (-2) \times (-3) = 6 \] Thus, \( b = 6 \). Thus, the correct answer is \( a = -2 \) and \( b = -6 \).