Question

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

Answer 1

Given $(a+1)x^2+(2a+3)x+(3a+4)=0$

We have that if equation is $a_1{x}^2+b_{1}x+c_1=0$

then sum of roots $=\frac{-b_1}{a_1}$ and product of roots $=\frac{+a}{a_1}$

Now Here in eq (1) $a_1=(a+1)$

$b_1=(2a+3)$

$c_1=(3a+4)$

then product of roots $=\frac{-b_1}{a_1}=\frac{-(2a+3)}{(a+1)}=2$ ..... given

$\Rightarrow -2a-3=2(a+1)$ [ cross multiply]

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

$\Rightarrow -4a=5$

$\Rightarrow a=\frac{-5}{4}$

Now sum of roots = $\frac{c_1}{a_1}=\frac{(3a+4)}{(a+1)}$

substituting value of $a$ in above we get

sum of roots $=\frac{3\times \left(\frac{-5}{4}\right)+4}{\frac{-5}{4}+1}=\frac{-\frac{15}{4}+\frac{16}{4}}{\frac{-5+4}{4}}$

$=\frac{\frac{1}{4}}{\frac{-1}{4}}=-1$

so sum of roots $=-1$