<s>[INST] Why is this approach wrong? 
If $\alpha$ and $\beta$ are the roots of $x^{2} - 4 x - 3$, then find $$\frac{1}{(\alpha-4)^{2}}+\frac{1}{(\beta-4)^{2}}$$


Solution
This is the approach that gives wrong answer.
\begin{array}{l}
x^{2}-4 x-3=0 \\
(x-4)^{2}=19 \\
So, {(\alpha-4)^{2}} {\&(\beta-4)^{2}}=19 \\
\text { Hence, } \frac{1}{(\alpha-4)^{2}}+\frac{1}{(\beta-4)^{2}}=\frac{2}{19}
\end{array}
This is how it is solved :
\begin{array}{l}
x^{2}-4 x-3=0 \\
\alpha^{2}-4 \alpha-3=0 \\
\alpha(\alpha-4)=3 \\
\alpha-4=\frac{3}{\alpha} \\
\& \beta-4=\frac{3}{\beta}
\end{array}
So , $$
\frac{1}{(\alpha-4)^{2}}+\frac{1}{(\beta-4)^{2}}
$$
\begin{array}{l}
=\frac{\alpha^{2}}{9}+\frac{\beta^{2}}{9} \\
=\frac{1}{9}\left(\alpha^{2}+\beta^{2}\right) \\
=\frac{22}{9}
\end{array}
Please tell me why the first approach is wrong .
 [/INST] Another approach would be to go ahead and perform the ratio addition (but don't fully multiply out the binomial-squares in the denominator):
$$ \frac{1}{(\alpha-4)^{2}}+\frac{1}{(\beta-4)^{2}} \ \ = \ \ \frac{(\beta-4)^{2} \ + \ (\alpha-4)^{2}}{(\alpha-4)^{2}· (\beta-4)^{2}} \ \ = \ \ \frac{\alpha^2 \ + \ \beta^2 \ - \ 8·(\alpha + \beta) \ + \ 32}{[ \ \alpha · \beta \ - \ 4·(\alpha + \beta) \ + 16 \ ]^2} $$
$ [ \ \alpha^2 + \beta^2 \ = \ (\alpha + \beta)^2 \ - \ 2·\alpha·\beta \ ] $
$$ = \ \ \frac{(\alpha \ + \ \beta)^2 \ - \ 2·\alpha·\beta \ - \ 8·(\alpha + \beta) \ + \ 32}{[ \ \alpha · \beta \ - \ 4·(\alpha + \beta) \ + 16 \ ]^2} $$
[we then apply Viete's relations: $ \ \alpha + \beta \ = \ -(-4) \ = \ 4 \ \ , \ \ \alpha·\beta \ = \ -3 \ ] $
$$ = \ \ \frac{4^2 \ - \ 2·(-3) \ - \ 8·(4) \ + \ 32}{[ \ (-3) \ - \ 4·4 \ + 16 \ ]^2}  \ \ = \ \ \frac{16 \ + \ 6 \ - \ 32 \ + \ 32}{[ \ (-3) \ - \ 16 \ + 16 \ ]^2} \ \  = \ \ \frac{16 \ + \ 6 }{(-3)^2} \ \ = \ \ \frac{22}{9} \ \  . $$
This method also allows us to proceed without determining the zeroes of the polynomial themselves.
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