Question

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

• A. $4$
• B. $2$
• C. $1$
• D. $-2$
• E. $-4$

Answer 1

The correct option is B $2$

Given equation is,
$(a+1)x^2+(2a+3)x+(3a+4)=0;a\ne -1$

and sum of the roots $=-1$

$\Rightarrow -\frac{(2a+3)}{(a+1)}=-1$

$\Rightarrow 2a+3=a+1\Rightarrow a=-2$

Product of the roots $=\frac{3a+4}{a+1}=\frac{3(-2)+4}{-2+1}(\because a=-2)$

$=\frac{-6+4}{-1}=2$