Question

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

• A. a = -7 and b = -1
• B. a = 5 and b = -1
• C. a = 2 and b = -6
• D. a = 0 and b = -6

Answer 1

The correct option is D a = 0 and b = -6

Given, $f\left(x\right)=x^2+(a+1)x+b$.
And, 2 and -3 are zeroes of f(x)

$\Rightarrow f\left(2\right)=0$
$\Rightarrow (2{)}^2+(a+1)\times 2+b=0$
$\Rightarrow 4+2a+2+b=0$
$\Rightarrow 6+2a+b=0\dots \left(i\right)$

$\text{and}f(-3)=0$
$\Rightarrow (-3{)}^2+(a+1)\times (-3)+b=0$
$\Rightarrow 9-3a-3+b=0$
$\Rightarrow 6-3a+b=0\dots \left(ii\right)$

On subtracting Eq. (ii) from Eq. (i), we get
$5a=0\Rightarrow a=0$

$\Rightarrow 6+b=0\text{[from Eq. (i)]}$
$\Rightarrow b=-6$

Hence,$a=0\text{and}b=-6$