$Iftherootsofa{x}^2+bx+c=0are\frac{k+1}{k},\frac{k+2}{k+1},then{(a+b+c)}^2=$

$\begin{array}{l}If\frac{k+1}{k},\frac{k+2}{k+1}arerootsa{x}^2+bx+c=0\\ sumofroots\Rightarrow \frac{k+1}{k}+\frac{k+2}{k+1}=\frac{-b}{a}--\left(1\right)\\ productofroots\Rightarrow \left(\frac{k+1}{k}\right)\left(\frac{k+2}{k+1}\right)=\frac{c}{a}\\ \Rightarrow \frac{k+2}{k}=\frac{c}{a}\\ \Rightarrow 1+\frac{2}{k}=\frac{c}{a}\\ \therefore k=\frac{2a}{c-a}\\ from\left(1\right)\frac{a+c}{2a}+\frac{2c}{a+c}=\frac{-b}{a}\\ \Rightarrow -2b(a+c)={(a+c)}^2+2ac\\ \Rightarrow -2ab-2bc=a^2+c^2+2ac+2ac\\ \Rightarrow b^2-2ac=a^2+b^2+2ac+2bc+2ab\\ \Rightarrow b^2-2ac={(a+b+c)}^2\end{array}$