Key Concepts and Formulas

• Root Property: If $\alpha$ is a root of the polynomial equation $P(x) = 0$, then $P(\alpha) = 0$.
• Recurrence Relation: A recurrence relation is an equation that defines a sequence recursively: each term is defined as a function of the preceding terms.
• Sum of Powers of Roots: Given a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$, we define $S_n = \alpha^n + \beta^n$.

Step-by-Step Solution

1. Applying the Root Property to the Given Equation

Since $\alpha$ is a root of the equation $5x^2 + 6x - 2 = 0$, substituting $\alpha$ into the equation must yield a true statement: $5\alpha^2 + 6\alpha - 2 = 0 \quad (\text{Eq 1})$ Explanation: This step utilizes the definition of a root. Any root of an equation satisfies that equation. This is the foundation for constructing our recurrence relation.

2. Generalizing the Relation for Higher Powers of $\alpha$

To introduce terms related to $S_n$, $S_{n+1}$, and $S_{n+2}$, we multiply every term in (Eq 1) by $\alpha^{n+4}$. This is a crucial step to "shift" the powers of $\alpha$ to align with the indices of $S_n$ that we want. Multiplying (Eq 1) by $\alpha^{n+4}$ (for $n \ge 0$): $\alpha^{n+4}(5\alpha^2 + 6\alpha - 2) = \alpha^{n+4}(0)$ $5\alpha^{n+6} + 6\alpha^{n+5} - 2\alpha^{n+4} = 0 \quad (\text{Eq 2})$ Explanation: This operation preserves the equality and creates terms with exponents $n+6$, $n+5$, and $n+4$, which are exactly what we need when considering $S_{n+6}$, $S_{n+5}$, and $S_{n+4}$. However, to directly obtain a relation between $S_6, S_5, S_4$, we substitute $n=0$. This gives $5\alpha^6 + 6\alpha^5 - 2\alpha^4 = 0 \quad (\text{Eq 2'})$

3. Repeating for the Other Root, $\beta$

Similarly, since $\beta$ is also a root of $5x^2 + 6x - 2 = 0$, it satisfies the equation: $5\beta^2 + 6\beta - 2 = 0 \quad (\text{Eq 3})$ And multiplying (Eq 3) by $\beta^{n+4}$: $\beta^{n+4}(5\beta^2 + 6\beta - 2) = \beta^{n+4}(0)$ $5\beta^{n+6} + 6\beta^{n+5} - 2\beta^{n+4} = 0 \quad (\text{Eq 4})$ Explanation: The exact same logic applies to $\beta$. We derive an identical power relation for $\beta$ as we did for $\alpha$. Again, to directly obtain a relation between $S_6, S_5, S_4$, we substitute $n=0$. This gives $5\beta^6 + 6\beta^5 - 2\beta^4 = 0 \quad (\text{Eq 4'})$

4. Summing the Relations to Involve $S_n$

Now, we add (Eq 2') and (Eq 4') together. This step is where the definition of $S_n = \alpha^n + \beta^n$ comes into play. $(5\alpha^6 + 6\alpha^5 - 2\alpha^4) + (5\beta^6 + 6\beta^5 - 2\beta^4) = 0 + 0$ Group terms with common coefficients: $5(\alpha^6 + \beta^6) + 6(\alpha^5 + \beta^5) - 2(\alpha^4 + \beta^4) = 0$ Explanation: By summing the relations for $\alpha$ and $\beta$ and grouping terms, we naturally form the expressions $\alpha^k + \beta^k$, which are precisely $S_k$.

5. Substituting with $S_n$ Notation

Using the definition $S_k = \alpha^k + \beta^k$, we can rewrite the equation as: $5S_6 + 6S_5 - 2S_4 = 0$ Explanation: This is the relation for $S_6, S_5, S_4$ for this specific quadratic equation.

6. Rearranging to Match the Options

The derived relation is $5S_6 + 6S_5 - 2S_4 = 0$. We can rearrange this to match the format of the given options: $5S_6 + 6S_5 = 2S_4$ This matches option (A).

Common Mistakes & Tips

• Don't start with Vieta's formulas for $S_n$ directly: While Vieta's formulas relate $\alpha+\beta$ and $\alpha\beta$ to coefficients, they don't directly give this type of recurrence for general $S_n$. The method of using the root property is more efficient and general.
• Index Errors: Be very careful when multiplying by $\alpha^n$ and correctly identifying the new exponents.
• The Power of Starting with n=0: In this specific problem, since we want a direct relationship between $S_6, S_5, S_4$, it is easier to substitute $n=0$ after obtaining equations (Eq 2) and (Eq 4).

Summary

This problem illustrates a standard method for deriving relationships between sums of powers of roots of a polynomial equation. The core idea is that each root satisfies the original polynomial equation. By using this fact, we can directly obtain the desired relationship. In this case, we found that $5S_6 + 6S_5 = 2S_4$.

Final Answer The final answer is \boxed{5S_6 + 6S_5 = 2S_4}, which corresponds to option (A).