Question

If the sum of the roots of the equation $\left(a+1\right)x^2+\left(2a+3\right)x+\left(3a+4\right)=0,$ 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

Answer: (B)

Given equation is,
$(a+1)x^2+(2a+3)x+(3a+4)=0$ ;
$a\ne -1$
and sum of the roots $=-1$
$\Rightarrow -\frac{(2a+3)}{(a+1)}=-1$
$\Rightarrow 2a+3=a+1$
$\Rightarrow a=-2$
$\therefore$ Product of the roots $=\frac{3a+4}{a+1}=\frac{3(-2)+4}{-2+1}$
$(\because a=-2)$
$=\frac{-6+4}{-1}=2$