If a and b are the roots of the equation x² - 7x - 1 = 0, then the value of (a²¹ + b²¹ + a¹⁷ + b¹⁷)/

Q: Wait, why can't I just use the quadratic formula to find $a$ and $b$?

You absolutely could, but you would be walking into a trap! The roots are $x = \frac{7 \pm \sqrt{53}}{2}$. Trying to raise these to the power of $21$ or $19$ would lead to an algebraic nightmare of binomial expansions. In JEE Advanced, when you see high powers, it is almost always a signal to use the properties of the equation itself rather than brute-forcing the roots.

Q: How did you know to group $a^{21}$ with $a^{17}$?

That is the core of the strategy! We look for a common factor that reveals the relationship we derived. Since we found that $a^4 + 1 = 51a^2$, we want to create an $a^4$ term. By grouping $a^{21} + a^{17}$, we can factor out $a^{17}$ to get $a^{17}(a^4 + 1)$. It is all about pattern recognition.

Q: What if the denominator $a^{19} + b^{19}$ is zero?

Great observation! We must ensure the denominator is non-zero. The discriminant of our quadratic is $D = (-7)^2 - 4(1)(-1) = 53$. Since $D > 0$, the roots are real and distinct. Furthermore, since the roots are $\frac{7 \pm \sqrt{53}}{2}$, both roots are positive (or at least one is positive and the other is negative, but their magnitudes are such that the sum of their powers won't vanish). You are safe to divide!

Question

If $a$ and $b$ are the roots of the equation $x^2 - 7x - 1 = 0$, then the value of $\frac{a^{21} + b^{21} + a^{17} + b^{17}}{a^{19} + b^{19}}$ is equal to

Sigma Percentile · Accepted Answer

### The Illusion of Complexity: A Journey into Algebraic Elegance

Imagine you are staring at the expression:

$$\frac{a^{21} + b^{21} + a^{17} + b^{17}}{a^{19} + b^{19}}$$

Your first instinct might be panic. How on earth are we supposed to calculate $a^{21}$?

If you try to find the roots of $x^2 - 7x - 1 = 0$ using the quadratic formula, you will find yourself drowning in a sea of square roots and binomial expansions. But here is the secret: in the world of JEE Advanced, the most intimidating problems often have the most beautiful, hidden exits. We are not here to calculate; we are here to manipulate.

### The Golden Key: The Root Property

The most fundamental truth about a root is that it belongs to the equation. If $a$ is a root of $x^2 - 7x - 1 = 0$, then it must satisfy the equation perfectly:

$$a^2 - 7a - 1 = 0$$

This is our golden key. We do not need to know what $a$ is; we only need to know how it behaves.

Let us rearrange this into a more useful form: $a^2 - 1 = 7a$. This simple shift is the first step in our tactical maneuver.

### The Tactical Maneuver: Squaring the Reality

Look closely at the target expression. The powers are $21, 17,$ and $19$. Notice the gap between $21$ and $17$ is exactly $4$.

This is a massive hint! We need a relationship that connects $a^4$ to lower powers. How do we get from $a^2$ to $a^4$? We square it.

Let us take our rearranged equation $a^2 - 1 = 7a$ and square both sides:

$$(a^2 - 1)^2 = (7a)^2$$

Expanding the left side gives us $a^4 - 2a^2 + 1 = 49a^2$. With a quick move of the $2a^2$ term to the right, we arrive at the elegant identity:

$$a^4 + 1 = 51a^2$$

This is the engine that will drive our solution.

### The Strategic Grouping

Now, let us return to our "monstrous" fraction:

$$\frac{a^{21} + b^{21} + a^{17} + b^{17}}{a^{19} + b^{19}}$$

Let us group the terms by their base:

$$\frac{(a^{21} + a^{17}) + (b^{21} + b^{17})}{a^{19} + b^{19}}$$

Now, factor out the smallest power from each group:

$$\frac{a^{17}(a^4 + 1) + b^{17}(b^4 + 1)}{a^{19} + b^{19}}$$

Do you see it? The expression $a^4 + 1$ has appeared, and we know exactly what it equals!

### The Elegant Cancellation

We substitute our identity $a^4 + 1 = 51a^2$ and $b^4 + 1 = 51b^2$ into the numerator:

$$\frac{a^{17}(51a^2) + b^{17}(51b^2)}{a^{19} + b^{19}}$$

When we multiply $a^{17}$ by $a^2$, the powers add up to $19$. The numerator becomes $51a^{19} + 51b^{19}$.

Factoring out the $51$, we get $51(a^{19} + b^{19})$. The denominator is $a^{19} + b^{19}$. They cancel out perfectly, leaving us with the final answer:

**51**

This is the beauty of mathematics. We started with a problem that seemed to require infinite calculation, and through logical structure and algebraic symmetry, we reduced it to a single, clean number. Never fear the high powers; look for the pattern, trust the identity, and let the algebra do the work for you.

Animated Solution for Mathematics - Quadratic Equations: If $a$ and $b$ are the roots of the equation $x^{2} - 7 x - 1 = 0$ , then the value of $\frac{a ^{21} + b ^{21} + a ^{17} + b ^{17}}{a ^{19} + b ^{19}}$ is equal to

Enter Numerical Value:

Visualized Solution

Analyze the Given Equation

Apply the Root Property

Isolate the Linear Term

Square Both Sides

Expand and Simplify

Find the Value of $a^{4} + 1$

Rearrange the Target Expression

Factor the Numerator

Substitute the Derived Values

Simplify the Expression

Final Calculation

The Sigma Insight: Relation Between Roots and Coefficients

The Illusion of Complexity

The Golden Key

The Tactical Maneuver

The Strategic Grouping

The Elegant Cancellation

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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; } If a and b are the roots of the equation x2−7x−1=0, then the value of a19+b19a21+b21+a17+b17​ is equal to