Mathematics - Quadratic Equation Question with Solution | TestHub

If $r_1$ and $r_2$ are roots of the quadratic $a{x}^2+bx+c=0$

then ${r_1}^n=31{r_1}^{n-2}-228{r_1}^{n-1}$ and ${r_2}^n=31{r_2}^{n-2}-228{r_2}^{n-1}$

So, ${r_1}^n-{r_2}^n=31\left({r_1}^{n-2}-{r_2}^{n-2}\right)-228\left({r_1}^{n-1}-{r_2}^{n-1}\right)$

Or $a_n-31a_{n-1}+228a_{n-2}=0$ where $a_n={r_1}^n-{r_2}^n$

Since $19 \& 12$ are the roots of the quadratic $x^2-31x+228=0$ 

we know, $a_n-31a_{n-1}+228a_{n-2}=0$

Now, for $n=10, {\alpha }_{10}-31{\alpha }_3+228{\alpha }_8=0$

$\Rightarrow \frac{31{\alpha }_9-{\alpha }_{10}}{57{\alpha }_8}=4$