how to prove $\frac{b^2}{b_1^2}=\frac{ac}{a_{1c_1}}$? If the Ratio of the roots of $ax^2+bx+c=0$ be equal to the ratio of the roots of $a_1x^2+b_1x+c_1=0$, then how one prove that $\frac{b^2}{b^2_1}=\frac{ac}{a_1 c_1}$?
 A: Hint:
Note that , if the roots of the equation $ax^2+bx+c=0$ are $x_1$ and $x_2$, than
$$
\frac{b}{ac}=\frac{\left(\frac{x_1}{x_2}+1\right)^2}{\frac{x_1}{x_2}}
$$
A: Hint :
let $\alpha$ and $\beta$ be the roots of $ax^2+bx+c=0$ & let $\gamma$ and $\delta$ be the roots of $a_1 x^2+b_1 x+c_1 =0$.
The ratio of their roots are equal if 
\begin{eqnarray*}
\frac{\alpha}{\beta} = \frac{\gamma}{\delta}.
\end{eqnarray*}
Further hint : $\color{red}{\alpha+\beta=-\frac{b}{a}}$ & $\alpha \beta=\frac{c}{a}$
\begin{eqnarray*}
\frac{b^2}{ac} = \color{red}{\frac{b^2}{a^2}} \frac{a}{c} = \frac{\color{red}{(\alpha+\beta)^2}}{\alpha \beta}=\frac{\alpha}{\beta}+2+\frac{\beta}{\alpha} = \cdots
\end{eqnarray*}
A: Let the roots of the first quadratic be $\alpha$ and $t\alpha$ and the roots of the second be $\beta$ and $t\beta$
Then $$\alpha(t+1)=\frac ca$$ and $$\alpha^2t=\frac ca$$
Then $$\frac {b^2}{a^2}=\alpha(t+1)^2=\frac{c}{at}(t+1)^2$$
In exactly the same way,  $$\frac {b_1^2}{a_1^2}=\beta(t+1)^2=\frac{c_1}{a_1t}(t+1)^2$$
Now eliminate all the $t$ terms by division and the result follows immediately
