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Question

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

A
a = -7 and b = -1
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B
a = 5 and b = -1
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C
a = 2 and b = -6
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D
a = 0 and b = -6
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Solution

The correct option is D a = 0 and b = -6
Given, f(x)=x2+(a+1)x+b.
And, 2 and -3 are zeroes of f(x)

f(2)=0
(2)2+(a+1)×2+b=0
4+2a+2+b=0
6+2a+b=0(i)

and f(3)=0
(3)2+(a+1)×(3)+b=0
93a3+b=0
63a+b=0(ii)

On subtracting Eq. (ii) from Eq. (i), we get
5a=0a=0

6+b=0 [from Eq. (i)]
b=6

Hence,a=0 and b=6

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