Mathematics · Quadratic Equations

JEE Main 2025 — 2 April, Morning Shift — Question 24

Let $P_{n}=\alpha^{n}+\beta^{n}, n \in N$ . If $P_{10}=123, P_{9}=76, P_{8}=47$ and $P_{1}=1$ , then the quadratic equation having roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is:

Option A:
$x^{2}+x-1=0$
Correct
Option B:
$x^{2}-x+1=0$
Option C:
$x^{2}+x+1=0$
Option D:
$x^{2}-x-1=0$

Answer: A

Step-by-step solution

$\because P_{10}=P_{9}+P_{8}$

$\Rightarrow P_{10}-P_{9}-P_{8}=0$

By Newton's method

Therefore, the equation is

$x^{2}-x-1=0$ as $P_{1}=1$

$\alpha+\beta=1$ and $\alpha \beta=1$

$\therefore$ Quadratic equation whose roots are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is

$x^{2}-\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) x+\frac{1}{\alpha \beta}=0$

$x^{2}-\left(\frac{\alpha+\beta}{\alpha \beta}\right) x+\frac{1}{\alpha \beta}=0$

$x^{2}-(-1) x-1=0$

$x^{2}+x-1=0$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 2 April, Morning Shift
Subject: Mathematics
Chapter: Quadratic Equations
Topic: Transformation of Equations