If  $\alpha ,\beta$ are the roots of  $a{x}^2+bx+c=0$ ${\alpha }_1,-\beta$ are the roots of $a_1{x}^2+b_{1}x+c_1=0$ then $\alpha ,{\alpha }_1$ are the roots of the equation

$\alpha +\beta =\frac{-b}{a}-\left(1\right)$

$\alpha \beta =\frac{c}{a}-\left(2\right);{\alpha }_1-\beta =\frac{-b_1}{a_1}-\left(3\right)$

$-{\alpha }_1\beta =\frac{c_1}{a_1}-\left(4\right)$

$\left(1\right)+\left(2\right)\Rightarrow \alpha +{\alpha }_1=\frac{-b}{a}-\frac{b_1}{a_1}$

Div (1) by (2) and (3) by(4) 

$\Rightarrow \frac{1}{\beta }+\frac{1}{\alpha }=\frac{-b}{c}$ and $\frac{-1}{\beta }+\frac{1}{\alpha }=\frac{-b_1}{c_1}$

Adding  $\Rightarrow \frac{1}{\alpha }+\frac{1}{{\alpha }_1}=\frac{-b}{c}-\frac{b_1}{c_1}$

$\alpha {\alpha }_1=\frac{\frac{-b}{a}-\frac{b_1}{a_1}}{\frac{-b}{c}-\frac{b_1}{c_1}}$

$x^2-\left(-\frac{b}{a}-\frac{b_1}{a_1}\right)x+\left(\frac{-\frac{b}{a}-\frac{b_1}{c_1}}{\frac{-b}{c}-\frac{b_1}{c_1}}\right)=0$

$\Rightarrow {\left(\frac{b}{a}+\frac{b_1}{a_1}\right)}^{-1}{x}^2+x+{\left(\frac{b}{c}+\frac{b_1}{c_1}\right)}^{-1}=0$