JEE MAIN - Mathematics (2025 - 3rd April Morning Shift - No. 9)

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 $Q_n=\gamma^n+\hat{o}^n$, then $\frac{P_{25}+\sqrt{3} P_{24}}{2 P_{23}}+\frac{Q_{25}-Q_{23}}{Q_{24}}$ is equal to
4
3
5
7

Explanation

JEE Main 2025 (Online) 3rd April Morning Shift Mathematics - Quadratic Equation and Inequalities Question 3 English Explanation 1

$$\begin{aligned} & \Rightarrow \quad P^n=\gamma^n+\delta^n \\ & P_{25}-P_{23}=\left(\gamma^{25}-\gamma^{23}\right)+\left(\delta^{25}-\delta^{23}\right) \\ & =\gamma^{23}\left(\gamma^2-1\right)+\delta^{23}\left(\delta^2-1\right) \\ & =\gamma^{23}(-3 \gamma)+\delta^{23}(-3 \delta)=-3\left[\gamma^{24}+\delta^{24}\right] \end{aligned}$$

$$\begin{aligned} &\Rightarrow \frac{P_{25}-P_{23}}{P_{24}}=(-3)\\ &\text { Similarly, } \end{aligned}$$

JEE Main 2025 (Online) 3rd April Morning Shift Mathematics - Quadratic Equation and Inequalities Question 3 English Explanation 2

$$\begin{aligned} & \Rightarrow Q_{25}+\sqrt{3} Q_{24}=\left(\alpha^{25}+\sqrt{3} \alpha^{24}\right)+\left(\beta^{25}+\sqrt{3} \beta^{24}\right) \\ & =\alpha^{23}\left(\alpha^2+\sqrt{3} \alpha\right)+\beta^{23}\left(\beta^2+\sqrt{3} \beta\right) \\ & =\alpha^{23}(16)+16 \beta^{23} \\ & \Rightarrow \frac{Q_{25}+\sqrt{3} Q_{24}}{2 \cdot Q_{23}}=\frac{16\left(\alpha^{23}+\beta^{23}\right)}{2\left(\alpha^{23}+\beta^{23}\right)}=8 \\ & \Rightarrow \frac{Q_{25}+\sqrt{3} Q_{24}}{2 Q_{23}}+\frac{\left(P_{25}-P_{23}\right)}{P_{24}}=8+(-3)=5 \end{aligned}$$

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