Question 3

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

Solution:

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).
$\therefore p\left(2\right)=0$ and $p(-3)=0$
$\Rightarrow 2^2+(a+1)\left(2\right)+b=0$
$\Rightarrow 4+2a+2+b=0$
$\Rightarrow 2a+b=-6$ ..(i)
and ${(-3)}^2+(a+1)(-3)+b=0$
$\Rightarrow 9-3a-3+b=0$
$\Rightarrow 3a-b=6$ …(ii)
On adding Eqs.(i) and (ii), we get
$5a=0\Rightarrow a=0$
Put the value of a in Eq (i), we get
$2\times 0+b=-6\Rightarrow b=-6$
So, the required values are $a=0$ and $b=-6$