Mathematics - Quadratic Equation Question with Solution | TestHub

$\begin{array}{rl}& x^2+\sqrt{3}\mathrm{x}-16=0<\beta \\ & {\mathrm{P}}_{\mathrm{n}}+\sqrt{3}{\mathrm{P}}_{\mathrm{n}-1}-16{\mathrm{P}}_{\mathrm{n}-2}=0\\ & {\mathrm{P}}_{25}+\sqrt{3}{\mathrm{P}}_{24}-16{\mathrm{P}}_{23}=0\\ & \therefore \frac{{\mathrm{P}}_{25}+\sqrt{3}{\mathrm{P}}_{24}}{2{\mathrm{P}}_{23}}=8\end{array}$
Similarly
$x^2+3x-1=0<{\sum }_{\delta }^{\gamma }{Q}_n={\gamma }^n+{\delta }^n$
$\begin{array}{rl}& {\mathrm{Q}}_{25}-{\mathrm{Q}}_{23}={\gamma }^{25}+{\delta }^{25}-{\gamma }^{23}-{\delta }^{23}\\ & ={\gamma }^{23}\left({\gamma }^2-1\right)+{\delta }^{23}\left({\delta }^2-1\right)\\ & ={\gamma }^{23}(-3\gamma )+{\delta }^{23}(-3\gamma )\\ & =-3\left[{\gamma }^{24}+{\delta }^{24}\right]\\ & =-3{\mathrm{Q}}_{24}\\ & \therefore \frac{{\mathrm{Q}}_{25}-{\mathrm{Q}}_{23}}{{\mathrm{Q}}_{24}}=-3\end{array}$
$\frac{{\mathrm{P}}_{25}+\sqrt{3}{\mathrm{P}}_{24}}{2{\mathrm{P}}_{23}}+\frac{{\mathrm{Q}}_{25}-{\mathrm{Q}}_{23}}{{\mathrm{Q}}_{24}}=8-3=5$