Let α and β be the roots of the equation, 5x² + 6x - 2 = 0. If Sₙ = αⁿ + βⁿ, n = 1, 2, 3, ..., then:

Q: Why can't we just solve for $\alpha$ and $\beta$ using the quadratic formula and then raise them to the power of 6?

While you *could* technically find the roots using the quadratic formula, it would lead to a nightmare of surds and complex algebra. The beauty of this method is that it bypasses the need to know the exact values of $\alpha$ and $\beta$. It relies on the *property* of the roots, which is a much more powerful and elegant approach in JEE Advanced.

Q: How did you know to multiply by $\alpha^4$ specifically?

Great question! Look at the target expression: $5S_6 + 6S_5 = 2S_4$. The highest power is $S_6$ (which is $\alpha^6 + \beta^6$) and the lowest is $S_4$ (which is $\alpha^4 + \beta^4$). Since we start with $\alpha^2$, we need to multiply by $\alpha^4$ to reach the highest power, $\alpha^6$. It's all about matching the indices!

Q: Does this work for any quadratic equation?

Absolutely! This is a specific case of Newton's Sums. For any quadratic equation $ax^2 + bx + c = 0$, the power sums $S_n$ will always satisfy the recurrence relation $aS_n + bS_{n-1} + cS_{n-2} = 0$. It's a universal tool in your algebraic toolkit.

Question

Let $\alpha$ and $\beta$ be the roots of the equation, $5x^2 + 6x - 2 = 0$. If $S_n = \alpha^n + \beta^n, n = 1, 2, 3, \dots$, then:

Options:

(A) $5S_6 + 6S_5 = 2S_4$

(B) $6S_6 + 5S_5 + 2S_4 = 0$

(C) $6S_6 + 5S_5 = 2S_4$

(D) $5S_6 + 6S_5 + 2S_4 = 0$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing on the precipice of a calculation. You are given a quadratic equation, $5x^2 + 6x - 2 = 0$, and asked to find a relationship between $S_6, S_5,$ and $S_4$, where $S_n = \alpha^n + \beta^n$.

At first glance, this might look like a tedious exercise in exponentiation. But stop. Take a breath. In the world of JEE Advanced, we don't brute-force; we seek the underlying structure.

### The Root Property: The Key to the Kingdom

The most fundamental truth about a root $\alpha$ is that it is not just a number; it is a key that unlocks the equation. If $\alpha$ is a root of $5x^2 + 6x - 2 = 0$, then it must satisfy the equation perfectly.

This gives us the identity:
$$5\alpha^2 + 6\alpha - 2 = 0$$

This is our starting point. It is simple, yet it contains the DNA of the entire problem.

### Scaling Up: The Art of Manipulation

We need to reach $S_6$, which involves $\alpha^6$. We currently have $\alpha^2$. How do we bridge this gap? We scale up.

By multiplying our identity $5\alpha^2 + 6\alpha - 2 = 0$ by $\alpha^4$, we don't just change the numbers; we elevate the entire expression to the power we need. When we multiply, the exponents add up, leading us to:
$$5\alpha^6 + 6\alpha^5 - 2\alpha^4 = 0$$

This is the moment of clarity. We have successfully created a relationship for $\alpha$ that mirrors our target.

### The Power of Symmetry

Now, we repeat this for $\beta$. Because $\beta$ is also a root, it follows the exact same path: $5\beta^2 + 6\beta - 2 = 0$. Multiplying by $\beta^4$ gives us:
$$5\beta^6 + 6\beta^5 - 2\beta^4 = 0$$

Now, we have two beautiful, symmetric equations. What happens when we add them? We get:
$$(5\alpha^6 + 6\alpha^5 - 2\alpha^4) + (5\beta^6 + 6\beta^5 - 2\beta^4) = 0$$

By grouping the terms with the same coefficients, we arrive at:
$$5(\alpha^6 + \beta^6) + 6(\alpha^5 + \beta^5) - 2(\alpha^4 + \beta^4) = 0$$

### The Final Synthesis

Look at what we have created. The brackets are exactly our definitions of $S_6, S_5,$ and $S_4$. The equation simplifies to:
$$5S_6 + 6S_5 - 2S_4 = 0$$

Transposing the term $-2S_4$ to the other side, we get:
$$5S_6 + 6S_5 = 2S_4$$

This is the elegance of mathematics. We didn't need to calculate $\alpha$ or $\beta$. We simply leveraged the structure of the equation itself.

Remember this, for it is a powerful tool: for any quadratic $ax^2 + bx + c = 0$, the power sums $S_n$ will always obey the recurrence:
$$aS_n + bS_{n-1} + cS_{n-2} = 0$$

Keep this in your toolkit, and you will conquer any problem that comes your way.

Animated Solution for Mathematics - Quadratic Equations: Let $α$ and $β$ be the roots of the equation, $5 x^{2} + 6 x - 2 = 0$ . If $S_{n} = α^{n} + β^{n}, n = 1, 2, 3, \dots$ , then:

Select Answer:

Visualized Solution

Analyzing $5 x^{2} + 6 x - 2 = 0$

Root Property for $α$

Multiplying by $α^{4}$

Repeating for $β$

Adding the Equations

Grouping Coefficients

Substituting $S_{n}$

Final Result

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Root Property

Scaling Up

The Power of Symmetry

The Final Synthesis

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Animated Solution for Mathematics - Quadratic Equations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let α and β be the roots of the equation, 5x2+6x−2=0. If Sn​=αn+βn,n=1,2,3,…, then:

Select Answer:

Visualized Solution

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 5x2+6x−2=0

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

Multiplying by .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α4

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