If one root of the quadratic equation ax² + bx + c = 0 is equal to the n-th power of the other, then

Q: Wait, why do we reject the negative root here? Or do we?

Great question! In this specific proof, we are working with the relationship between the roots and coefficients. We aren't necessarily solving for a specific numerical value of $x$, but rather establishing an identity. The algebraic manipulation holds regardless of the sign of the root, provided the roots exist. We are essentially treating $\alpha$ as a variable that satisfies the quadratic equation.

Q: Why can't we just use the quadratic formula to find the roots?

You absolutely could, but it would be a nightmare! The quadratic formula gives you roots involving $\sqrt{b^2 - 4ac}$. Trying to raise that to the power of $n$ and then equating it to the other root would lead to an algebraic mess. Vieta's formulas are the 'shortcut' that bypasses the need to explicitly calculate the roots, allowing us to work directly with the coefficients.

Q: How do I know when to multiply by $a$ in the final steps?

Always keep your eyes on the target expression: $(ac^n)^{1/(n+1)} + (a^nc)^{1/(n+1)} + b = 0$. Notice that the target has $a$ inside the parentheses. In our derivation, we had terms like $a \cdot (c/a)^{1/(n+1)}$. Multiplying by $a$ is the standard technique to 'absorb' the coefficient into the radical, transforming $a$ into $a^{(n+1)/(n+1)}$ so it can combine with the other terms.

Question

If one root of the quadratic equation $ax^2 + bx + c = 0$ is equal to the $n$-th power of the other, then show that $(ac^n)^{1/(n+1)} + (a^nc)^{1/(n+1)} + b = 0$

Sigma Percentile · Accepted Answer

### The Symphony of Roots: A Quadratic Journey

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving an equation; we are uncovering a hidden symmetry within the quadratic form $ax^2 + bx + c = 0$.

This problem is a classic, a beautiful dance between the roots and the coefficients that define the parabola's very existence.

### Phase 1: The Root Relationship

Imagine the parabola $y = ax^2 + bx + c$. It cuts the x-axis at two points, which we call $\alpha$ and $\beta$.

The problem gives us a fascinating constraint: one root is the $n$-th power of the other. So, we define our roots as $\alpha$ and $\beta = \alpha^n$.

This isn't just a label; it's a geometric lock waiting for the right key.

### Phase 2: The Power of Vieta

We reach into our toolkit and pull out the most powerful weapon in the quadratic arsenal: Vieta's Formulas. We know that the product of the roots is $\alpha \cdot \beta = \frac{c}{a}$.

Substituting our relationship, we get $\alpha \cdot \alpha^n = \frac{c}{a}$, which simplifies elegantly to $\alpha^{n+1} = \frac{c}{a}$.

By taking the $(n+1)$-th root, we isolate our first root:

$$\alpha = \left( \frac{c}{a} \right)^{\frac{1}{n+1}}$$

This is our anchor.

### Phase 3: The Summation Bridge

Now, we turn to the second Vieta relation: the sum of the roots, $\alpha + \beta = -\frac{b}{a}$. Substituting our roots, we get $\alpha + \alpha^n = -\frac{b}{a}$.

This is the bridge connecting our isolated $\alpha$ to the coefficient $b$. We substitute our expression for $\alpha$ into this sum:

$$\left( \frac{c}{a} \right)^{\frac{1}{n+1}} + \left( \left( \frac{c}{a} \right)^{\frac{1}{n+1}} \right)^n = -\frac{b}{a}$$

### Phase 4: The Algebraic Dance

This is where the magic happens. We have:

$$\left( \frac{c}{a} \right)^{\frac{1}{n+1}} + \left( \frac{c}{a} \right)^{\frac{n}{n+1}} = -\frac{b}{a}$$

To clear the denominator and align with our target, we multiply the entire equation by $a$. This gives us:

$$a \cdot \left( \frac{c}{a} \right)^{\frac{1}{n+1}} + a \cdot \left( \frac{c}{a} \right)^{\frac{n}{n+1}} = -b$$

Now, we bring $b$ to the left side to get:

$$a \cdot \left( \frac{c}{a} \right)^{\frac{1}{n+1}} + a \cdot \left( \frac{c}{a} \right)^{\frac{n}{n+1}} + b = 0$$

### Absorbing the Coefficients

To finish, we must absorb the $a$ into the radicals. We rewrite $a$ as $(a^{n+1})^{\frac{1}{n+1}}$.

For the first term, this becomes:

$$(a^{n+1} \cdot \frac{c}{a})^{\frac{1}{n+1}} = (a^n c)^{\frac{1}{n+1}}$$

For the second term, we have:

$$(a^{n+1} \cdot \frac{c^n}{a^n})^{\frac{1}{n+1}} = (ac^n)^{\frac{1}{n+1}}$$

### The Grand Finale

Substituting these back, we arrive at the beautiful identity:

$$(a^n c)^{\frac{1}{n+1}} + (ac^n)^{\frac{1}{n+1}} + b = 0$$

We have successfully navigated the complexity and arrived at the elegant truth. Remember, in JEE Advanced, it is rarely about brute force; it is about finding the symmetry and letting the algebra flow.

**You have mastered this proof!**

Animated Solution for Mathematics - Quadratic Equations: If one root of the quadratic equation $a x^{2} + b x + c = 0$ is equal to the $n$ -th power of the other, then show that $(a c^{n})^{1/ (n + 1)} + (a^{n} c)^{1/ (n + 1)} + b = 0$

Visualized Solution

Defining the Roots

Product of Roots Formula

Simplifying the Product

Solving for $α$

Sum of Roots Formula

Substitution of $α$

Rearranging the Equation

Manipulating the First Term

Manipulating the Second Term

Final Result

The Sigma Insight: Relation Between Roots and Coefficients

The Symphony of Roots

Phase 1

Phase 2

Phase 3

Phase 4

Absorbing the Coefficients

The Grand Finale

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