Key Concepts and Formulas

• Vieta's Formulas: For a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$, we have $\alpha + \beta = -\frac{b}{a}$ and $\alpha \beta = \frac{c}{a}$. For $x^2 - sx + p = 0$, $\alpha + \beta = s$ and $\alpha \beta = p$.
• Recurrence Relation for Sums of Powers: If $\alpha$ and $\beta$ are roots of $x^2 - sx + p = 0$, then $P_n = \alpha^n + \beta^n$ satisfies the recurrence $P_n = sP_{n-1} - pP_{n-2}$ for $n \ge 2$.
• Quadratic Equation from Roots: A quadratic equation with roots $r_1$ and $r_2$ is given by $x^2 - (r_1 + r_2)x + r_1r_2 = 0$.

Step-by-Step Solution

Step 1: Identify the Recurrence Relation

We are given $P_{10} = 123$, $P_9 = 76$, and $P_8 = 47$. We want to find a recurrence relation of the form $P_n = A P_{n-1} + B P_{n-2}$. We test if the simplest relation $P_n = P_{n-1} + P_{n-2}$ holds.

For $n = 10$, we check if $P_{10} = P_9 + P_8$: $123 = 76 + 47$ $123 = 123$ Since the equation holds true, the recurrence relation is $P_n = P_{n-1} + P_{n-2}$. This step is crucial because it links the given values of $P_n$ to the coefficients of the original quadratic equation.

Step 2: Determine the Sum and Product of the Original Roots

We have the recurrence relation $P_n = P_{n-1} + P_{n-2}$. Comparing this to the general recurrence relation $P_n = s P_{n-1} - p P_{n-2}$, where $s = \alpha + \beta$ and $p = \alpha \beta$, we can equate the coefficients.

• Coefficient of $P_{n-1}$: $s = 1$, so $\alpha + \beta = 1$.
• Coefficient of $P_{n-2}$: $-p = 1$, so $p = -1$, and $\alpha \beta = -1$.

We also have $P_1 = 1$, which is consistent since $P_1 = \alpha + \beta = 1$.

Step 3: Find the Sum and Product of the Reciprocal Roots

We want to find the quadratic equation with roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$. First, we find the sum of the new roots: $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta}$ Substituting the values we found in Step 2: $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{1}{-1} = -1$

Next, we find the product of the new roots: $\frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha \beta}$ Substituting the value we found in Step 2: $\frac{1}{\alpha \beta} = \frac{1}{-1} = -1$

Step 4: Construct the New Quadratic Equation

The quadratic equation with roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is given by: $x^2 - \left(\frac{1}{\alpha} + \frac{1}{\beta}\right) x + \frac{1}{\alpha \beta} = 0$ Substituting the sum and product we found in Step 3: $x^2 - (-1)x + (-1) = 0$ $x^2 + x - 1 = 0$

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when comparing recurrence relations and applying Vieta's formulas. A common mistake is to misinterpret the sign of the product of roots.
• Recurrence Relation Verification: Always verify the recurrence relation with the given values before proceeding. This ensures that you are using the correct relation.
• Vieta's Formulas: Remember that Vieta's formulas provide a direct relationship between the coefficients of a polynomial and the sums and products of its roots.

Summary

By recognizing and applying the recurrence relation $P_n = P_{n-1} + P_{n-2}$, we determined that $\alpha + \beta = 1$ and $\alpha \beta = -1$. Then we calculated the sum and product of the reciprocals of the roots, $\frac{1}{\alpha}$ and $\frac{1}{\beta}$, which are both $-1$. Using these values, we constructed the quadratic equation $x^2 + x - 1 = 0$.

Final Answer

The final answer is $\boxed{x^2 + x - 1 = 0}$, which corresponds to option (A).