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

Q: Wait, why can't we just calculate $\alpha$ and $\beta$ using the quadratic formula and plug them into the expression?

You absolutely could, but that is the path of maximum resistance! The roots of $x^2 - 6x - 2 = 0$ are $3 \pm \sqrt{11}$. Raising $(3 + \sqrt{11})^{10}$ and $(3 - \sqrt{11})^{10}$ to such high powers is a computational nightmare and a recipe for algebraic errors. In JEE Advanced, when you see high powers of roots, it is almost always a signal to use the recurrence relation derived from the quadratic equation itself.

Q: How do I know to multiply by $\alpha^8$ specifically?

Look at the target expression: $\frac{a_{10} - 2a_8}{2a_9}$. The highest power is $10$ and the lowest is $8$. By multiplying the base equation $\alpha^2 - 6\alpha - 2 = 0$ by $\alpha^8$, we generate the term $\alpha^{10}$ and keep the other terms in the sequence ($a_9$ and $a_8$). It is all about aligning the powers to match the terms in your target expression.

Q: Why does the subtraction $(\alpha^{10} - \beta^{10}) - 6(\alpha^9 - \beta^9) - 2(\alpha^8 - \beta^8) = 0$ work so perfectly?

This works because of the linearity of the expression. Since both $\alpha$ and $\beta$ satisfy the same quadratic relation, the operator $L(x) = x^2 - 6x - 2$ acts on both roots to produce zero. When you subtract the two equations, you are essentially applying the operator to the difference of the powers, which preserves the structure of the sequence $a_n$.

Question

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

Options:

(A) 3

(B) $-3$

(C) 6

(D) $-6$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

In the landscape of JEE Advanced mathematics, we often encounter problems that seem designed to exhaust our patience. You see an equation like $x^2 - 6x - 2 = 0$, and your first instinct might be to reach for the quadratic formula.

You calculate the roots, $\alpha$ and $\beta$, and find they are $3 \pm \sqrt{11}$. Then, you look at the target expression: $\frac{a_{10} - 2a_8}{2a_9}$, where $a_n = \alpha^n - \beta^n$.

If you try to calculate $\alpha^{10}$ and $\beta^{10}$ directly, you are walking into a trap. The beauty of this problem lies not in brute force, but in recognizing the hidden structure of the quadratic equation.

### The Power of Satisfaction

Let us pause and look at the quadratic equation $x^2 - 6x - 2 = 0$. By definition, if $\alpha$ is a root, it must satisfy the equation. This is a fundamental truth that students often overlook.

It means that $\alpha^2 - 6\alpha - 2 = 0$, or more usefully, $\alpha^2 = 6\alpha + 2$. This simple rearrangement is our key. It tells us that any power of $\alpha$ can be expressed in terms of its lower powers, serving as the seed of a recurrence relation.

### Scaling the Mountain

We need to reach the tenth power, $a_{10}$. We scale up our relation $\alpha^2 = 6\alpha + 2$ by multiplying by $\alpha^8$:

$$\alpha^{10} = 6\alpha^9 + 2\alpha^8$$

This is the bridge we needed. We have successfully linked the tenth power to the ninth and eighth powers. Because $\beta$ is also a root of the same equation, it must obey the exact same law:

$$\beta^{10} = 6\beta^9 + 2\beta^8$$

### The Art of Subtraction

Now, we have two parallel realities: one for $\alpha$ and one for $\beta$. We want to find $a_n = \alpha^n - \beta^n$. If we subtract the $\beta$ equation from the $\alpha$ equation, the structure of the sequence $a_n$ emerges naturally:

$$(\alpha^{10} - \beta^{10}) = 6(\alpha^9 - \beta^9) + 2(\alpha^8 - \beta^8)$$

Substituting our sequence definition $a_n$, this becomes:

$$a_{10} = 6a_9 + 2a_8$$

This is a beautiful, linear recurrence relation. It tells us that the tenth term of our sequence is entirely determined by the ninth and eighth terms, bypassing the need to calculate the actual values of $\alpha$ or $\beta$.

### The Final Cancellation

Look back at our target: $\frac{a_{10} - 2a_8}{2a_9}$. We have our relation $a_{10} = 6a_9 + 2a_8$. If we rearrange this to isolate the numerator, we get:

$$a_{10} - 2a_8 = 6a_9$$

Now, the final step is almost poetic. We substitute this into our target fraction:

$$\frac{6a_9}{2a_9}$$

The $a_9$ terms, which seemed so intimidating at the start, simply cancel out. We are left with $\frac{6}{2}$, which results in the final answer of **3**.

### Reflection

This problem is a masterclass in mathematical efficiency. It teaches us that in the JEE, the most complex-looking expressions often have the simplest underlying structures.

When you see high powers of roots, do not calculate; relate. Build the recurrence, observe the symmetry, and let the algebra do the heavy lifting for you. You have just turned a terrifying exponent problem into a simple arithmetic one.

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

Select Answer:

Visualized Solution

The Quadratic Equation

Locating $α$ and $β$

Defining the Goal

Roots Satisfy the Equation

Generating Powers of 10

Expanding the $α$ Equation

The $β$ Equation

Combining the Equations

Grouping Like Powers

Introducing $a_{n}$

The Simplified Relation

Matching the Target Numerator

Final Substitution

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Power of Satisfaction

Scaling the Mountain

The Art of Subtraction

The Final Cancellation

Reflection

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