$If r is the ratio of the roots of a{x}^2+bx+c=0 then \frac{{\left(r+1\right)}^2}{r} is$ 

$\begin{array}{l}\text{Let }\alpha \text{,}\beta \text{\hspace{0.17em} are the roots of }a{x}^2+bx+c=0\\ \text{Given they are in the ratio }r:1\\ Let \alpha =\beta r sum of roots=\beta +\beta r=\frac{-b}{a}\\ \beta \left(1+r\right)=\frac{-b}{a}\\ {\beta }^2{(1+r)}^2=\frac{b^2}{a^2}---\left(1\right)\\ product of roots=\beta ·\beta r=\frac{c}{a} \\ \Rightarrow {\beta }^{2}r=\frac{c}{a}-----\left(2\right)\\ \frac{Eq1}{Eq2}\\ \frac{{\left(r+1\right)}^2}{r}=\frac{\left(\frac{b^2}{a^2}\right)}{\frac{c}{a}}\\ =\boxed{\frac{b^2}{ac}}\end{array}$