An expression involving the roots of a quadratic polynomial I have the equation $$ 9x^2 - 11x + 1 = 0 $$ whose two roots are $ \alpha $ and $ \beta $ . 
I need to evaluate $$ \frac 1 {(9\alpha-11)^2} + \frac{11\beta - 1} 9$$
What I've tried


*

*Expanded the denominator and add them, but nothing simplifies and I get even a complex expression.

*I found that $ \alpha + \beta = \frac{11}{9} 
\implies \alpha = \frac{11}{9} - \beta $


How to evaluate the value of the expression in an easier way?
 A: Since $9\alpha^2-11\alpha+1=0=9\beta^2-11\beta+1$, by Vieta's theorem we have 
$$ \frac{1}{(9\alpha-11)^2}+\frac{(11\beta-1)}{9} = \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=\left(\frac{11}{9}\right)^2-2\cdot\frac{1}{9}=\color{red}{\frac{103}{81}}$$
A: Just to show another way: 
$$
\eqalign{
  & 9\alpha ^{\,2}  - 11\alpha  + 1 = 0\quad  \Rightarrow \quad \alpha \left( {9\alpha  - 11} \right) =  - 1\quad  \Rightarrow \quad \left( {9\alpha  - 11} \right) =  - 1/\alpha   \cr 
  & 9\beta ^{\,2}  - 11\beta  + 1 = 0\quad  \Rightarrow \quad 11\beta  - 1 = 9\beta ^{\,2}   \cr 
  & 9\left( {\alpha ^{\,2}  + \beta ^{\,2} } \right) - 11\left( {\alpha  + \beta } \right) + 2 = 0\quad  \Rightarrow \quad 9\left( {\alpha ^{\,2}  + \beta ^{\,2} } \right) = 11\left( {\alpha  + \beta } \right) - 2 = 11\left( {{{11} \over 9}} \right) - 2 = {{103} \over 9}  \cr 
  & {1 \over {\left( {9\alpha  - 11} \right)^{\,2} }} + {{11\beta  - 1} \over 9} = \left( { - \alpha } \right)^{\,2}  + \beta ^{\,2}  = 103/81 \cr} 
$$
