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

Options

A. `a =- 7, b =- 1`
B. `a = 5, b=-1`
C. `a =2, b =- 6`
D. `a =0, b=-6`

Answer

D

Explanation

Let `p(x) =x^(2) +(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`
`rArr 2^(2)+(a+1) (2)+b =0`
`rArr 4+2a +2 +b = 0`
`rArr 2a +b =- 6` ..(i)
and `(-3)^(2) +(a+1) (-3)+b = 0`
`rArr 9 - 3a - 3 +b = 0`
`rArr 3a - b = 6` ...(ii)
On adding Eqs.(i) and (ii), we get
`5a = 0 rArr a = 0`
Put the valye of a in Eq (i), we get
`2 xx 0 +b =- 6 rArr b =- 6`
So, the required values are `a = 0` and `b =- 6`