Question

The zero of the quadratic polynomial $x^2+(a+1)x+b$ 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

Answer 1

Answer: (D) a = 0, b = – 6

Let $p\left(x\right)=x^2+(a+1)x+b$
Given that, 2 and -3 are the zeroes of the quadratic polynomial p (x)
$\therefore p\left(2\right)=0andp(-3)=0,$

Solving for $p\left(2\right)$

$\Rightarrow$ $2^2+(a+1)\left(2\right)+b=0$
$\Rightarrow$ $4+2a+2+b=0$
$\Rightarrow$ $2a+b=-6$ ....(i)

Solving for $p(-3),$
$(-3{)}^2+(a+1\left)\right(-3)+b=0$
$\Rightarrow$ $9–3a–3+b=0$
$\Rightarrow$ $3a–b=6$ ....(ii)

On adding $Eqs.\left(i\right)and\left(ii\right)$, we get
$5a=0\Rightarrow a=0$
Put the value of a in Eq. (i), we get
$x\times 0+b=-6\Rightarrow b=-6$
so, the required values are $a=0$ and $b=-6$
[1 mark]