Let α and β be the roots of x² - 6x - 2 = 0. If aₙ = αⁿ - βⁿ for n ≥ 1, then the value of (a₁₀ - 2a₈

Q: Wait, why can't we just use the quadratic formula to find $\alpha$ and $\beta$?

You absolutely could, but it would lead to a nightmare of irrational numbers and complex arithmetic! The beauty of this problem lies in its structure. By using the property that roots satisfy their parent equation, we bypass the need for the actual values of $\alpha$ and $\beta$ entirely. In JEE Advanced, recognizing this structural shortcut is often the difference between finishing the paper and running out of time.

Q: How did you know to group $\alpha^{10}$ with $2\alpha^8$?

That is a great observation! We looked at the target expression $a_{10} - 2a_8$. By definition, $a_{10} = \alpha^{10} - \beta^{10}$ and $a_8 = \alpha^8 - \beta^8$. When you subtract $2a_8$ from $a_{10}$, you naturally get $(\alpha^{10} - \beta^{10}) - 2(\alpha^8 - \beta^8)$. Rearranging this to $(\alpha^{10} - 2\alpha^8) - (\beta^{10} - 2\beta^8)$ allows us to factor out $\alpha^8$ and $\beta^8$, which perfectly aligns with the relation $\alpha^2 - 2 = 6\alpha$ that we derived earlier. Always let the structure of the target expression guide your algebraic steps.

Q: Is there a faster way to solve this?

Yes! This is a classic application of 'Newton's Sums.' For any quadratic equation $x^2 - px - q = 0$, the sequence $a_n = \alpha^n - \beta^n$ satisfies the recurrence relation $a_n - pa_{n-1} - qa_{n-2} = 0$. In our case, $p=6$ and $q=-2$, so $a_n - 6a_{n-1} - 2a_{n-2} = 0$. This means $a_{10} - 6a_9 - 2a_8 = 0$, which immediately gives $a_{10} - 2a_8 = 6a_9$. It is a powerful tool to keep in your arsenal for competitive exams.

Question

Let $\alpha$ and $\beta$ be the roots of $x^2 - 6x - 2 = 0$. If $a_n = \alpha^n - \beta^n$ for $n \ge 1$, then the value of $\frac{a_{10} - 2a_8}{3a_9}$ is:

Options:

(A) 4

(B) 1

(C) 2

(D) 3

Sigma Percentile · Accepted Answer

### The Hidden Symmetry of Roots

Imagine you are standing before a locked gate. You have a key, but it looks like a complex, tangled mess of high-power polynomials.

In many JEE Advanced problems, the initial sight of terms like $\alpha^{10}$ and $\beta^8$ is designed to make you panic. But today, we are going to learn how to look past the complexity and find the elegant, hidden symmetry that makes this problem collapse in seconds.

### The DNA of the Equation

We start with the quadratic equation $x^2 - 6x - 2 = 0$. This equation is the 'DNA' of our problem.

The roots, $\alpha$ and $\beta$, are not just random numbers; they are bound by this specific constraint. The most fundamental property of a root is that it satisfies its parent equation.

Therefore, we know with absolute certainty that $\alpha^2 - 6\alpha - 2 = 0$. By rearranging this, we get the key to our puzzle:

$$\alpha^2 - 2 = 6\alpha$$

This simple relation is the bridge between the high powers we fear and the manageable terms we need.

### The Algebraic Dance

Our objective is to evaluate the expression $\frac{a_{10} - 2a_8}{3a_9}$, where $a_n = \alpha^n - \beta^n$.

Let's focus on the numerator: $a_{10} - 2a_8$. Substituting the definition of our sequence, we get $(\alpha^{10} - \beta^{10}) - 2(\alpha^8 - \beta^8)$.

Now, let's perform an algebraic dance. We group the $\alpha$ terms and the $\beta$ terms:

$$(\alpha^{10} - 2\alpha^8) - (\beta^{10} - 2\beta^8)$$

Look closely at the first group, $\alpha^{10} - 2\alpha^8$. We can factor out $\alpha^8$ to get $\alpha^8(\alpha^2 - 2)$.

And what do we know about $\alpha^2 - 2$? From our DNA equation, we know it is exactly $6\alpha$!

So, the first group becomes $\alpha^8(6\alpha) = 6\alpha^9$. By the exact same logic, the second group, $\beta^{10} - 2\beta^8$, becomes $6\beta^9$.

The entire numerator has now simplified to $6\alpha^9 - 6\beta^9$, or $6(\alpha^9 - \beta^9)$.

### The Grand Finale

We have arrived at the final step. The numerator is $6a_9$, and the denominator is $3a_9$.

The expression becomes:

$$\frac{6a_9}{3a_9}$$

The $a_9$ terms cancel out beautifully, leaving us with:

$$\frac{6}{3} = 2$$

It is a moment of pure mathematical satisfaction—the high powers vanished, the complexity dissolved, and we are left with a simple, elegant integer.

Remember, in the heat of the JEE exam, don't rush to calculate. Look for the structure, trust the symmetry, and let the algebra do the heavy lifting for you. The final answer is **2**.

Animated Solution for Mathematics - Quadratic Equations: Let $α$ and $β$ be the roots of $x^{2} - 6 x - 2 = 0$ . If $a_{n} = α^{n} - β^{n}$ for $n \geq 1$ , then the value of $\frac{a _{10} - 2 a _{8}}{3 a _{9}}$ is:

Select Answer:

Visualized Solution

The Problem Setup

Roots Satisfy the Equation

Rearranging for $α$

Rearranging for $β$

Expanding the Target Numerator

Grouping Like Terms

Factoring Out Common Powers

The Magic Substitution

Simplifying the Numerator

Recognizing the Sequence Term

Final Evaluation

Newton's Sums (Pro Tip)

The Sigma Insight: Relation Between Roots and Coefficients

The Hidden Symmetry of Roots

The DNA of the Equation

The Algebraic Dance

The Grand Finale

Similar Questions

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Comprehension Passage

$a_{12} =$

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