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

Vedclass · Accepted Answer

$-1$

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

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 is:

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