Question

Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:

• A. \( x^2 - x + 1 = 0 \)
• B. \( x^2 + x - 1 = 0 \)
• C. \( x^2 - x - 1 = 0 \)
• D. \( x^2 + x + 1 = 0 \)

Hint:

For such sequences, derive \( \alpha + \beta \) and \( \alpha \beta \) using known terms and build required equation.

Answer 1

The Correct Option is B

Given a sequence \( P_n = \alpha^n + \beta^n \) with \( P_{10} = 123 \), \( P_9 = 76 \), \( P_8 = 47 \), and \( P_1 = 1 \). The objective is to determine the quadratic equation whose roots are \( \alpha \) and \( \frac{1}{\beta} \).

Concept Used:

For a quadratic equation \( x^2 - Sx + P = 0 \) with roots \( \alpha \) and \( \beta \), where \( S = \alpha + \beta \) and \( P = \alpha\beta \), the sequence \( P_n = \alpha^n + \beta^n \) adheres to the linear recurrence relation:

\[ P_n = S \cdot P_{n-1} - P \cdot P_{n-2} \]

The quadratic equation with roots \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \) can be obtained by substituting \( x = \frac{1}{y} \) into the original equation and simplifying.

Step-by-Step Solution:

Step 1: Derive the recurrence relation from the provided data.

Using the given values \( P_{10} = 123 \), \( P_9 = 76 \), and \( P_8 = 47 \), we observe the relationship:

\[ P_9 + P_8 = 76 + 47 = 123 \]

Since \( P_9 + P_8 = P_{10} \), the sequence follows the recurrence relation:

\[ P_n = P_{n-1} + P_{n-2} \]

Step 2: Identify the quadratic equation for which \( \alpha \) and \( \beta \) are roots.

Comparing \( P_n = P_{n-1} + P_{n-2} \) with the general recurrence \( P_n = S \cdot P_{n-1} - P \cdot P_{n-2} \), we deduce:

\[ S = \alpha + \beta = 1 \] \[ P = \alpha\beta = -1 \]

The given \( P_1 = 1 \) is consistent with \( S = \alpha + \beta = 1 \). The quadratic equation with roots \( \alpha \) and \( \beta \) is \( x^2 - Sx + P = 0 \), which becomes:

\[ x^2 - x - 1 = 0 \]

Step 3: Analyze the roots of the target equation.

The problem requires the quadratic equation with roots \( \alpha \) and \( \frac{1}{\beta} \). From \( \alpha\beta = -1 \), it follows that \( \frac{1}{\beta} = -\alpha \). Therefore, the required roots are \( \alpha \) and \( -\alpha \). Their sum is \( \alpha + (-\alpha) = 0 \) and their product is \( \alpha(-\alpha) = -\alpha^2 \).

Step 4: Calculate the sum and product of the reciprocal roots \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \).

Let the sum of the new roots be \( S' \) and the product be \( P' \).

Sum of new roots:

\[ S' = \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{S}{P} \]

Product of new roots:

\[ P' = \left(\frac{1}{\alpha}\right)\left(\frac{1}{\beta}\right) = \frac{1}{\alpha\beta} = \frac{1}{P} \]

Final Computation & Result:

Step 5: Compute \( S' \) and \( P' \) using \( S=1 \) and \( P=-1 \).

\[ S' = \frac{1}{-1} = -1 \] \[ P' = \frac{1}{-1} = -1 \]

The new quadratic equation is \( x^2 - S'x + P' = 0 \).

\[ x^2 - (-1)x + (-1) = 0 \]

\( x^2 + x - 1 = 0 \)

This result corresponds to one of the provided options.