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

a = - 7, b = - 1

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a = 5, b = - 1

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a = 2, b = - 6

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a = 0, b = - 6

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Solution

The correct option is D $a = 0, b = - 6$
$x^{2} + (a + 1) x + b$ is the quadratic polynomial.

$2$ and $- 3$ are the zeros of the quadratic polynomial.
Thus, $2 + (- 3) = \frac{- (a + 1)}{1}$
$=> \frac{(a + 1)}{1} = 1$
$=> a + 1 = 1$
$=> a = 0$
Also, $2 \times (- 3) = b$
$=> b = - 6$

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Similar questions

x^{2} + (a + 1) x + b

- 3

a = - 7, b = - 1

a = 5, b = - 1

a = 2, b = - 6

a = 0, b = - 6

The zero 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) $a = 2, b = - 6$
c) $a = 0, b = - 6$

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