Question

Let \( P_n = \alpha^n + \beta^n \), \( P_{10} = 123 \), \( P_9 = 76 \), \( P_8 = 47 \), and \( P_1 = 1 \). The quadratic equation whose roots are \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \) is:

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

Hint:

For recursive sequences like this, identify patterns or relationships between consecutive terms to solve for unknowns.

Answer 1

The Correct Option is D

We are given the sequence \( P_n = \alpha^n + \beta^n \), and the values of \( P_{10}, P_9, P_8, P_1 \). We are tasked with finding the quadratic equation whose roots are \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \). First, we need to find a relationship between \( \alpha \) and \( \beta \) using the given values of \( P_n \). From the recursive sequence, we can calculate the general form for \( P_n \) and use the known values of \( P_8 \), \( P_9 \), and \( P_{10} \) to solve for \( \alpha \) and \( \beta \). Using these, the sum and product of the roots \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \) can be determined as follows: The sum of the roots is: \[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} \] The product of the roots is: \[ \frac{1}{\alpha \beta} \] From the sequence relations, the quadratic equation with these roots is: \[ x^2 - (\frac{1}{\alpha} + \frac{1}{\beta})x + \frac{1}{\alpha \beta} = 0 \] By substituting the values, we find that the quadratic equation is: \[ x^2 - x - 2 = 0 \] Thus, the correct quadratic equation is \( x^2 - x - 2 = 0 \).