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Question 3
The zero of the quadratic polynomial x2+(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

Let p(x)=x2+(a+1)x+b
Given that, 2 and -3 are the zeroes of the quadratic polynomial p (x)
p(2)=0 and p(3)=0,

Solving for p(2)

22+(a+1)(2)+b=0
4+2a+2+b=0
2a+b=6 ....(i)

Solving for p(3),
(3)2+(a+1)(3)+b=0
93a3+b=0
3ab=6 ....(ii)

On adding Eqs.(i) and (ii), we get
5a=0 a=0
Put the value of a in Eq. (i), we get
x×0+b=6 b=6
so, the required values are a=0 and b=6

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