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Question

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
c) 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=0a=0
Substituting the value of a in Eq. (i), we get
x×0+b=6b=6
So, the required values are a=0 and b=6.


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