If the zeroes of the quadratic polynomial x2 + (a + 1) x + b are 2 and -3, then :

x2 + (a + 1)x + b∵ x = 2 is a zeroand x = – 3 is another zero∴ (2)2 + (a + 1)2 + 6 = 0and (- 3)2 + (a + 1) (- 3) + b = 0⇒ 4 + 2cr + 2 + & = 0and 9 – 3a – 3 + b = 0⇒ 2a + b = – 6 …(i)and – 3a + b = – 6 …(ii)Solving (i) and (ii), we get 5a = 0⇒ a = 0 and b = – 6.

If the zeroes of the quadratic polynomial x2 + (a + 1) x + b are 2 and -3, then :

x2 + (a + 1)x + b∵ x = 2 is a zeroand x = – 3 is another zero∴ (2)2 + (a + 1)2 + 6 = 0and (- 3)2 + (a + 1) (- 3) + b = 0⇒ 4 + 2cr + 2 + & = 0and 9 – 3a – 3 + b = 0⇒ 2a + b = – 6 …(i)and – 3a + b = – 6 …(ii)Solving (i) and (ii), we get 5a = 0⇒ a = 0 and b = – 6.

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If the zeroes of the quadratic polynomial x² + (a + 1) x + b are 2 and -3, then :

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a = -7, b = -1

a = 5, b = -1

a = 2, b = -6

a = 0, b = -6

answer is D.

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Detailed Solution

x² + (a + 1)x + b
∵ x = 2 is a zero
and x = – 3 is another zero
∴ (2)² + (a + 1)² + 6 = 0
and (- 3)² + (a + 1) (- 3) + b = 0
⇒ 4 + 2cr + 2 + & = 0
and 9 – 3a – 3 + b = 0
⇒ 2a + b = – 6 …(i)
and – 3a + b = – 6 …(ii)
Solving (i) and (ii), we get 5a = 0
⇒ a = 0 and b = – 6.