If the product of the roots of the equation $(a+1)x^2+(2a+3)x+(3a+4)=0$ be $2$, then the sum of roots is

The correct option is C: $-1$

The given quadratic equation is $(a+1)x^2+(2a+3)x+(3a+4)=0$

It is given that $\alpha \beta =2$

We know that the product of the roots is given by

$\alpha \beta =\frac{\text{Constant term}}{\text{Coefficient of}{x}^2}$

$\Rightarrow \frac{3a+4}{a+1}=2$

$\Rightarrow 3a+4$ = $2a+2$

$\Rightarrow a=-2$

Also $\alpha +\beta =\frac{-\text{Coefficient of x}}{\text{Coefficient of}{x}^2}=-\frac{2a+3}{a+1}$

Putting this value of $a=-2$, we get sum of roots

$=-\frac{(2\times -2)+3}{-2+1}=-\frac{(-4+3)}{(-2+1)}=-1$