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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?

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Since $9\alpha^2-11\alpha+1=0=9\beta^2-11\beta+1$, by Vieta's theorem we have $$ \frac{1}{(9\alpha-11)^2}+\frac{(11\beta-1)}{9} = \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=\left(\frac{11}{9}\right)^2-2\cdot\frac{1}{9}=\color{red}{\frac{103}{81}}$$

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  • $\begingroup$ How the expression is equated to $ \alpha^2 + \beta^2 $ ? $\endgroup$ – Ravi Prakash Jul 13 '17 at 18:13
  • $\begingroup$ @RaviPrakash: since $\alpha(9\alpha-11)=-1$ and $\beta^2=\frac{11\beta-1}{9}$... $\endgroup$ – Jack D'Aurizio Jul 13 '17 at 18:14
  • $\begingroup$ How $ \alpha(9\alpha-11) = -1 $ ? $\endgroup$ – Ravi Prakash Jul 13 '17 at 18:18
  • $\begingroup$ @RaviPrakash: Since $9\alpha^2-11\alpha+1=0$, then $\alpha(9\alpha-11)=-1$. $\endgroup$ – Jack D'Aurizio Jul 13 '17 at 18:20
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    $\begingroup$ Of course. :) Well done! $\endgroup$ – Ravi Prakash Jul 13 '17 at 18:21
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Just to show another way:

$$ \eqalign{ & 9\alpha ^{\,2} - 11\alpha + 1 = 0\quad \Rightarrow \quad \alpha \left( {9\alpha - 11} \right) = - 1\quad \Rightarrow \quad \left( {9\alpha - 11} \right) = - 1/\alpha \cr & 9\beta ^{\,2} - 11\beta + 1 = 0\quad \Rightarrow \quad 11\beta - 1 = 9\beta ^{\,2} \cr & 9\left( {\alpha ^{\,2} + \beta ^{\,2} } \right) - 11\left( {\alpha + \beta } \right) + 2 = 0\quad \Rightarrow \quad 9\left( {\alpha ^{\,2} + \beta ^{\,2} } \right) = 11\left( {\alpha + \beta } \right) - 2 = 11\left( {{{11} \over 9}} \right) - 2 = {{103} \over 9} \cr & {1 \over {\left( {9\alpha - 11} \right)^{\,2} }} + {{11\beta - 1} \over 9} = \left( { - \alpha } \right)^{\,2} + \beta ^{\,2} = 103/81 \cr} $$

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