Let $\alpha$ and $\beta$ be the roots of $x^2+\sqrt{3 x}-16=0$, and $\gamma$ and $\delta$ be the roots of…

Let $\alpha$ and $\beta$ be the roots of $x^2+\sqrt{3 x}-16=0$, and $\gamma$ and $\delta$ be the roots of $x^2+3 x-1=0$. If $P_n=\alpha^n+\beta^n$ and $Q_n=\gamma^n+\delta^n$, then $\frac{\mathrm{P}_{25}+\sqrt{3 \mathrm{P}_{24}}}{2 \mathrm{P}_{23}}+\frac{\mathrm{Q}_{25}-\mathrm{Q}_{23}}{\mathrm{Q}_{24}}$ is equal to
  1. $3$
  2. $4$
  3. $5$
  4. $7$

Solution

$\begin{aligned} & x^2+\sqrt{3} \mathrm{x}-16=0 \lt \beta \\ & \mathrm{P}_{\mathrm{n}}+\sqrt{3} \mathrm{P}_{\mathrm{n}-1}-16 \mathrm{P}_{\mathrm{n}-2}=0 \\ & \mathrm{P}_{25}+\sqrt{3} \mathrm{P}_{24}-16 \mathrm{P}_{23}=0 \\ & \therefore \frac{\mathrm{P}_{25}+\sqrt{3} \mathrm{P}_{24}}{2 \mathrm{P}_{23}}=8\end{aligned}$
Similarly
$x^2+3 x-1=0 \lt \sum_\delta^\gamma \quad Q_n=\gamma^n+\delta^n$
$\begin{aligned}
& \mathrm{Q}_{25}-\mathrm{Q}_{23}=\gamma^{25}+\delta^{25}-\gamma^{23}-\delta^{23} \\ &=\gamma^{23}\left(\gamma^2-1\right)+\delta^{23}\left(\delta^2-1\right) \\ &=\gamma^{23}(-3 \gamma)+\delta^{23}(-3 \gamma) \\ &=-3\left[\gamma^{24}+\delta^{24}\right] \\ &=-3 \mathrm{Q}_{24} \\ & \therefore \frac{\mathrm{Q}_{25}-\mathrm{Q}_{23}}{\mathrm{Q}_{24}}=-3
\end{aligned}$
$\frac{\mathrm{P}_{25}+\sqrt{3} \mathrm{P}_{24}}{2 \mathrm{P}_{23}}+\frac{\mathrm{Q}_{25}-\mathrm{Q}_{23}}{\mathrm{Q}_{24}}=8-3=5$ /

Asked in: JEE Main 2025 (03 Apr Shift 1)

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