If the zeroes of the quadratic polynomial $x^{2} + (a + 1) x + b$ . are 2 and $- 3$ , then

(a) $a = - 7, b = - 1$ (b) $a = 5, b = - 1$ (c) $a = 2, b = - 6$ (d) $a = 0, b = - 6$

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Solution

The given quadratic equation is $x^{2} + (a + 1) x + b = 0$ .
Since the zeroes of the given equation are 2 and –3.
So,
$α = 2$ and $β = - 3$
Now,
$Sum of zeroes = - \frac{Coefficient of x}{Coefficient of x^{2}} \Rightarrow 2 + (- 3) = - \frac{(a + 1)}{1} \Rightarrow - 1 = - a - 1 \Rightarrow a = 0$
$Product of zeroes = \frac{Constant term}{Coefficient of x^{2}} \Rightarrow 2 \times (- 3) = \frac{b}{1} \Rightarrow b = - 6$
So, a = 0 and b = −6

Hence, the correct answer is option D.

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