Key Concepts and Formulas

• Roots of a Quadratic Equation: If $\alpha$ is a root of $ax^2 + bx + c = 0$, then $a\alpha^2 + b\alpha + c = 0$.
• Recurrence Relations: A recurrence relation defines a sequence based on previous terms. In this case, we derive it from the quadratic equation.
• Algebraic Manipulation: Careful expansion, simplification, and substitution are crucial for solving the problem.

Step-by-Step Solution

Step 1: Derive the Recurrence Relation

Why: We need a recurrence relation to connect $P_{n}$, $P_{n-1}$, and $P_{n-2}$ so that we can simplify the expression.

Given the quadratic equation $x^2 - \sqrt{2}x - \sqrt{3} = 0$, let $\alpha$ and $\beta$ be its roots. Then: $\alpha^2 - \sqrt{2}\alpha - \sqrt{3} = 0 \implies \alpha^2 = \sqrt{2}\alpha + \sqrt{3}$ $\beta^2 - \sqrt{2}\beta - \sqrt{3} = 0 \implies \beta^2 = \sqrt{2}\beta + \sqrt{3}$

Multiplying the first equation by $\alpha^{n-2}$ and the second by $\beta^{n-2}$, we get: $\alpha^n = \sqrt{2}\alpha^{n-1} + \sqrt{3}\alpha^{n-2}$ $\beta^n = \sqrt{2}\beta^{n-1} + \sqrt{3}\beta^{n-2}$

Subtracting the second equation from the first: $\alpha^n - \beta^n = \sqrt{2}(\alpha^{n-1} - \beta^{n-1}) + \sqrt{3}(\alpha^{n-2} - \beta^{n-2})$ Since $P_n = \alpha^n - \beta^n$, we have: $P_n = \sqrt{2}P_{n-1} + \sqrt{3}P_{n-2}$ Thus, $P_n - \sqrt{2}P_{n-1} - \sqrt{3}P_{n-2} = 0$. Replacing $n$ with $k+2$, we have: $P_{k+2} = \sqrt{2}P_{k+1} + \sqrt{3}P_k$

Step 2: Express $P_{12}$ in terms of $P_{11}$ and $P_{10}$

Why: We want to reduce the highest power ($P_{12}$) in the given expression using the recurrence relation.

Setting $k = 10$ in the recurrence relation $P_{k+2} = \sqrt{2}P_{k+1} + \sqrt{3}P_k$, we get: $P_{12} = \sqrt{2}P_{11} + \sqrt{3}P_{10} \quad (*)$

Step 3: Express $P_{11}$ in terms of $P_{10}$ and $P_{9}$

Why: We want to relate $P_{11}$ to $P_9$ since $P_9$ appears in the options.

Setting $k = 9$ in the recurrence relation $P_{k+2} = \sqrt{2}P_{k+1} + \sqrt{3}P_k$, we get: $P_{11} = \sqrt{2}P_{10} + \sqrt{3}P_9 \quad (**)$

Step 4: Substitute $P_{12}$ into the given expression

Why: We substitute the expression of $P_{12}$ that we found in Step 2 to reduce the complexity of the given expression.

Let $E = (11 \sqrt{3}-10 \sqrt{2}) P_{10}+(11 \sqrt{2}+10) P_{11}-11 P_{12}$. Substituting $(*)$, we have: $E = (11 \sqrt{3}-10 \sqrt{2}) P_{10}+(11 \sqrt{2}+10) P_{11}-11(\sqrt{2}P_{11} + \sqrt{3}P_{10})$ $E = 11\sqrt{3}P_{10} - 10\sqrt{2}P_{10} + 11\sqrt{2}P_{11} + 10P_{11} - 11\sqrt{2}P_{11} - 11\sqrt{3}P_{10}$ $E = (11\sqrt{3} - 11\sqrt{3})P_{10} + (-10\sqrt{2})P_{10} + (11\sqrt{2} - 11\sqrt{2})P_{11} + 10P_{11}$ $E = -10\sqrt{2}P_{10} + 10P_{11}$ $E = 10(P_{11} - \sqrt{2}P_{10})$

Step 5: Substitute $P_{11}$ into the simplified expression

Why: We substitute the expression of $P_{11}$ that we found in Step 3 to relate the expression to $P_9$.

From $(**)$, we have $P_{11} = \sqrt{2}P_{10} + \sqrt{3}P_9$, so $P_{11} - \sqrt{2}P_{10} = \sqrt{3}P_9$. Substituting this into the expression for $E$: $E = 10(\sqrt{3}P_9) = 10\sqrt{3}P_9$

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when substituting and expanding.
• Index Consistency: Double-check the indices in the recurrence relation to avoid shifting errors.
• Simplify Incrementally: Perform substitutions one at a time to minimize errors.

Summary

By deriving a recurrence relation from the given quadratic equation and strategically substituting expressions for $P_{12}$ and $P_{11}$, we simplified the given expression to $10\sqrt{3}P_9$. This corresponds to option (A).

Final Answer

The final answer is \boxed{10 \sqrt{3} \mathrm{P}_9}, which corresponds to option (A).