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(B) For a quadratic equation $Ax^2 + Bx + C = 0$,the product of roots is $\alpha \beta = \frac{C}{A}$ and the sum of roots is $\alpha + \beta = -\frac

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$-1$

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(B) For a quadratic equation $Ax^2 + Bx + C = 0$,the product of roots is $\alpha \beta = \frac{C}{A}$ and the sum of roots is $\alpha + \beta = -\frac{B}{A}$.Given the product of roots is $2$,we have $\frac{3a + 4}{a + 1} = 2$.Solving for $a$: $3a + 4 = 2(a + 1)$ $\Rightarrow 3a + 4 = 2a + 2$ $\Rightarrow a = -2$.Now,substitute $a = -2$ into the expression for the sum of roots:Sum of roots $= -\frac{2a + 3}{a + 1} = -\frac{2(-2) + 3}{-2 + 1} = -\frac{-4 + 3}{-1} = -\frac{-1}{-1} = -1$.

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

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