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

Q: Wait, why can't I just calculate $\alpha$ and $\beta$ using the quadratic formula and plug them into $a_n = \alpha^n - \beta^n$?

I love the enthusiasm, but that is the classic 'trap' of JEE Advanced! If you calculate $\alpha = \frac{6 + \sqrt{36 + 8}}{2} = 3 + \sqrt{11}$, then calculating $(3 + \sqrt{11})^{10}$ is a nightmare of binomial expansion. In competitive exams, we look for the *structure* of the equation, not the brute-force calculation. The recurrence relation method is elegant, fast, and error-proof.

Q: How did you know to multiply by $\alpha^8$? Could it have been $\alpha^7$ or $\alpha^9$?

Great question! We chose $\alpha^8$ because we wanted to create the term $\alpha^{10}$ to match the highest power in our target expression ($a_{10}$). If we had multiplied by $\alpha^7$, we would have ended up with $\alpha^9$ as the highest power. Always look at the highest power in your target expression and align your multiplication to reach it.

Q: Is this method applicable to every quadratic equation?

Absolutely! This is a specific case of a powerful tool called 'Newton's Sums.' For any quadratic equation $ax^2 + bx + c = 0$, the sequence $S_n = p\alpha^n + q\beta^n$ will always satisfy the recurrence relation $a(S_n) + b(S_{n-1}) + c(S_{n-2}) = 0$. Once you master this, you can solve problems involving high powers of roots in seconds.

Question

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

Options:

(A) 1

(B) 2

(C) 3

(D) 4

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a mountain. The peak is the value of the expression:

$$\frac{a_{10} - 2a_8}{2a_9}$$

You have a map, which is the equation $x^2 - 6x - 2 = 0$, and you have the definition of your sequence $a_n = \alpha^n - \beta^n$.

The amateur climber tries to scale this mountain by calculating the exact values of $\alpha$ and $\beta$. They reach for the quadratic formula, find the surds, and immediately get buried in an avalanche of binomial expansions.

But you are an elite JEE aspirant. You know that the mountain is not meant to be climbed by force; it is meant to be navigated by structure.

### The Root Philosophy

Let us begin by respecting the definition of a root. When we say $\alpha$ is a root of $x^2 - 6x - 2 = 0$, we are saying that $\alpha$ is a key that unlocks this specific lock.

It satisfies the equation perfectly: $\alpha^2 - 6\alpha - 2 = 0$. This is not just an equation; it is a transformation rule.

It tells us that any power of $\alpha$ can be expressed in terms of lower powers. Specifically, $\alpha^2 = 6\alpha + 2$. This is the seed of our solution.

### The Power Scaling

Now, look at our target: $\frac{a_{10} - 2a_8}{2a_9}$. We need the tenth power, the ninth power, and the eighth power.

Our current equation only gives us the second power. How do we bridge this gap? We use the power of multiplication.

If we multiply our foundational equation $\alpha^2 - 6\alpha - 2 = 0$ by $\alpha^8$, we get:

$$\alpha^8(\alpha^2 - 6\alpha - 2) = 0 \Rightarrow \alpha^{10} - 6\alpha^9 - 2\alpha^8 = 0$$

Suddenly, the tenth power is within our grasp! We have created a bridge between the tenth, ninth, and eighth powers.

### The Birth of the Sequence

We must do the same for $\beta$, because $\beta$ is also a root. We get $\beta^{10} - 6\beta^9 - 2\beta^8 = 0$.

Now, we have two beautiful, parallel equations. The definition of our sequence is $a_n = \alpha^n - \beta^n$.

This minus sign is a massive hint. It tells us to subtract the $\beta$ equation from the $\alpha$ equation:

$$(\alpha^{10} - 6\alpha^9 - 2\alpha^8) - (\beta^{10} - 6\beta^9 - 2\beta^8) = 0$$

When we group the terms by their powers, we obtain:

$$(\alpha^{10} - \beta^{10}) - 6(\alpha^9 - \beta^9) - 2(\alpha^8 - \beta^8) = 0$$

Look closely. This is exactly $a_{10} - 6a_9 - 2a_8 = 0$.

### The Final Calculation

We are almost at the summit. We need to find the value of $\frac{a_{10} - 2a_8}{2a_9}$.

From our derived equation $a_{10} - 6a_9 - 2a_8 = 0$, we can isolate the terms we need:

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

Now, substitute this into our target fraction:

$$\frac{6a_9}{2a_9} = 3$$

The $a_9$ terms cancel out, leaving us with the final result of **3**. The mountain is conquered.

You did not need to calculate a single root. You used the internal logic of the equation to solve the problem. This technique is a specific application of **Newton's Sums**, a powerful tool that will serve you well throughout your JEE journey.

Animated Solution for Mathematics - Quadratic Equations: Let $α$ and $β$ be the roots of $x^{2} - 6 x - 2 = 0$ , with $α > β$ . 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

Select Answer:

Visualized Solution

Understanding the Given Equation

Applying Root Property for $α$

Generating Higher Powers for $α$

Applying the Same Logic for $β$

Combining Equations to Form $a_{n}$

Grouping Similar Powers

Substituting $a_{n}$

Rearranging the Expression

Final Calculation

Key Takeaway: Newton's Sums

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Root Philosophy

The Power Scaling

The Birth of the Sequence

The Final Calculation

Similar Questions

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:

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Comprehension Passage

$a_{12} =$

If $a_{4} = 28$ , then $p + 2 q =$

Let $α$ and $β$ be the roots of $x^{2} + 3 x - 16 = 0$ , and $γ$ and $δ$ be the roots of $x^{2} + 3 x - 1 = 0$ . If $P_{n} = α^{n} + β^{n}$ and $Q_{n} = γ^{n} + δ^{n}$ , then $\frac{P _{25} + 3 P _{24}}{2 P _{23}} + \frac{Q _{25} - Q _{23}}{Q _{24}}$ is equal to

Select Answer:

Visualized Solution

Understanding the Given Equation

Applying Root Property for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Generating Higher Powers for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Applying the Same Logic for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Combining Equations to Form .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } an​