Let α and β be the roots of x 2 +√(3 x)-16=0 , and γ and δ be the roots of x 2 +3 x-1=0 . If P n =α n +β n and Q n =γ n +δ n , then frac P 25 +sqrt 3 P 24 2 P 23 +frac Q 25 -Q 23 Q 24 is equal to

Mathematics · Quadratic Equations

JEE Main 2025 — 3 April, Morning Shift — Question 22

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

$\mathrm{P}_{\mathrm{n}}=\alpha^{\mathrm{n}}+\beta^{\mathrm{n}}$ and $\mathrm{Q}_{\mathrm{n}}=\gamma^{\mathrm{n}}+\delta^{\mathrm{n}}$ , then $\frac{P_{25}+\sqrt{3 P_{24}}}{2 P_{23}}+\frac{Q_{25}-Q_{23}}{Q_{24}}$ is equal to

Option A:
3
Option B:
4
Option C:
5
Correct
Option D:
7

Answer: C

Step-by-step solution

$x^{2}+\sqrt{3} x-16=0<\underbrace{\alpha}_{\beta} \quad P_{n}=\alpha^{n}+\beta^{n}$

$P_{n}+\sqrt{3} P_{n-1}-16 P_{n-2}=0$

$\mathrm{P}_{25}+\sqrt{3} \mathrm{P}_{24}-16 \mathrm{P}_{23}=0$

$\therefore \frac{\mathrm{P}_{25}+\sqrt{3} \mathrm{P}_{24}}{2 \mathrm{P}_{23}}=8$

Similarly $\mathrm{x}^{2}+3 \mathrm{x}-1=0$

$\sum_{\delta}^{\gamma} \quad \mathrm{Q}_{\mathrm{n}}=\gamma^{\mathrm{n}}+\delta^{\mathrm{n}}$

$\mathrm{Q}_{25}-\mathrm{Q}_{23}=\gamma^{25}+\delta^{25}-\gamma^{23}-\delta^{23}$

$=\gamma^{23}\left(\gamma^{2}-1\right)+\delta^{23}\left(\delta^{2}-1\right)$

$=\gamma^{23}(-3 \gamma)+\delta^{23}(-3 \gamma)$

$=-3\left[\gamma^{24}+\delta^{24}\right]$

$=-3 \mathrm{Q}_{24}$

$\therefore \frac{\mathrm{Q}_{25}-\mathrm{Q}_{23}}{\mathrm{Q}_{24}}=-3$

$\frac{\mathrm{P}_{25}+\sqrt{3} \mathrm{P}_{24}}{2 \mathrm{P}_{23}}+\frac{\mathrm{Q}_{25}-\mathrm{Q}_{23}}{\mathrm{Q}_{24}}=8-3=5$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 3 April, Morning Shift
Subject: Mathematics
Chapter: Quadratic Equations
Topic: Transformation of Equations