Let a, b, c be real numbers with a ≠ 0 and let α, β be the roots of the equation ax² + bx + c = 0. E

Q: Why do we divide by $a^3$? Why not just $a$?

That is a brilliant observation. If we divide by $a$, we get $a^2x^2 + bcx + \frac{c^3}{a} = 0$. This doesn't help us isolate the ratios $\frac{b}{a}$ and $\frac{c}{a}$ cleanly. Dividing by $a^3$ is a strategic move to normalize the leading coefficient to $1$ while simultaneously creating the exact powers of $\frac{c}{a}$ and $\frac{b}{a}$ we need to map back to $\alpha$ and $\beta$. It is about aligning the 'DNA' of the new equation with the 'DNA' of the old one.

Q: Can I just use the quadratic formula to find the roots of the new equation?

Technically, yes, but you are walking into a trap. The discriminant will be a massive expression involving $a, b, c$. You will spend 15 minutes simplifying it, only to realize you could have solved it in 30 seconds by recognizing the structure. In JEE Advanced, the method is often more important than the result. We want to be elegant, not just brute-force calculators.

Q: How did we know that the roots would be $\alpha^2\beta$ and $\alpha\beta^2$?

We didn't 'know' it at the start—we deduced it! By forcing the equation into the form $x^2 - (\text{Sum})x + (\text{Product}) = 0$, we identified the sum and product of the new roots. Once you have the sum $S = \alpha^2\beta + \alpha\beta^2$ and the product $P = (\alpha^2\beta)(\alpha\beta^2)$, the roots are simply the solutions to the quadratic equation formed by those values. It is a beautiful game of pattern matching.

Question

Let $a, b, c$ be real numbers with $a 
eq 0$ and let $\alpha, \beta$ be the roots of the equation $ax^2 + bx + c = 0$. Express the roots of $a^3x^2 + abcx + c^3 = 0$ in terms of $\alpha, \beta$.

Sigma Percentile · Accepted Answer

### The Symphony of Symmetry: Unlocking Quadratic Transformations

Welcome, future engineers. Today, we aren't just solving a quadratic equation; we are performing a surgical operation on algebraic structure.

When you look at a problem like this in the JEE Advanced exam, your first instinct might be to panic—to reach for the quadratic formula and start calculating discriminants. But I want you to pause and take a breath.

In the world of competitive mathematics, the most powerful tool you possess is not your speed, but your ability to see the **hidden symmetry**.

### Phase 1: The DNA of the Quadratic

We start with the original equation: $ax^2 + bx + c = 0$. We are told its roots are $\alpha$ and $\beta$.

Immediately, your mind should jump to Vieta's relations. These are the DNA of any quadratic, telling us exactly how the coefficients $a, b,$ and $c$ are woven into the roots:

$$\alpha + \beta = -\frac{b}{a}$$

$$\alpha\beta = \frac{c}{a}$$

These two equations are our master keys. They allow us to translate any expression involving $a, b,$ and $c$ into the language of $\alpha$ and $\beta$. Never underestimate the power of this translation; it is the bridge between the unknown and the known.

### Phase 2: The Art of Normalization

Now, look at our target equation: $a^3x^2 + abcx + c^3 = 0$. It looks intimidating, but it is just a transformation of the original.

To make sense of it, we must normalize it. Since $a 
eq 0$, we divide the entire equation by $a^3$:

$$\frac{a^3x^2}{a^3} + \frac{abcx}{a^3} + \frac{c^3}{a^3} = 0$$

This simplifies beautifully to:

$$x^2 + \left(\frac{bc}{a^2}ight)x + \left(\frac{c}{a}ight)^3 = 0$$

Do you see it now? The structure is emerging. We have successfully isolated the ratios $\frac{b}{a}$ and $\frac{c}{a}$, mapping the equation back to our roots.

### Phase 3: The Pattern Recognition

This is where the magic happens. Let's look at the coefficient of $x$, which is $\frac{bc}{a^2}$.

We can rewrite this as $\left(\frac{b}{a}ight) \cdot \left(\frac{c}{a}ight)$. Using our Vieta's relations, where $\frac{b}{a} = -(\alpha + \beta)$ and $\frac{c}{a} = \alpha\beta$, the coefficient becomes:

$$-(\alpha + \beta)(\alpha\beta) = -(\alpha^2\beta + \alpha\beta^2)$$

Now, look at the constant term: $\left(\frac{c}{a}ight)^3$. Since $\frac{c}{a} = \alpha\beta$, this is simply $(\alpha\beta)^3$, or $\alpha^3\beta^3$.

To make this fit the standard quadratic form $x^2 - (	ext{Sum})x + (	ext{Product}) = 0$, we can express this product as $(\alpha^2\beta)(\alpha\beta^2)$.

### The Grand Finale

We have reconstructed our equation as:

$$x^2 - (\alpha^2\beta + \alpha\beta^2)x + (\alpha^2\beta)(\alpha\beta^2) = 0$$

Compare this to the standard form $x^2 - Sx + P = 0$. The sum of our new roots is $S = \alpha^2\beta + \alpha\beta^2$, and the product is $P = (\alpha^2\beta)(\alpha\beta^2)$.

It is undeniable: the roots of our target equation are **$\alpha^2\beta$ and $\alpha\beta^2$**.

This, my friends, is the beauty of mathematics. We didn't need to calculate a single discriminant or perform complex algebra. We simply looked at the structure, respected the symmetry, and let the logic flow.

Animated Solution for Mathematics - Quadratic Equations: Let $a, b, c$ be real numbers with $a \neq = 0$ and let $α, β$ be the roots of the equation $a x^{2} + b x + c = 0$ . Express the roots of $a^{3} x^{2} + ab c x + c^{3} = 0$ in terms of $α, β$ .

Visualized Solution

The Original Quadratic

Vieta's Relations

The Target Equation

Normalizing the Equation

Transforming the $x$ Coefficient

Transforming the Constant Term

The Reconstructed Equation

Identifying the Roots

The Sigma Insight: Relation Between Roots and Coefficients

The Symphony of Symmetry

Phase 1

Phase 2

Phase 3

The Grand Finale

Similar Questions

Let $α, β$ be the roots of the equation $(x - a) (x - b) = c, c \neq = 0$ . Then the roots of the equation $(x - α) (x - β) + c = 0$ are

Let $p$ and $q$ be real numbers such that $p \neq = 0, p^{3} \neq = - q$ . If $α$ and $β$ are nonzero complex numbers satisfying $α + β = - p$ and $α^{3} + β^{3} = q$ , then a quadratic equation having $α / β$ and $β / α$ as its roots is

If $α, β$ are the roots of $a x^{2} + b x + c = 0, (a \neq = 0)$ and $α + δ, β + δ$ are the roots of $A x^{2} + B x + C = 0, (A \neq = 0)$ for some constant $δ$ , then prove that $(b^{2} - 4 a c) / a^{2} = (B^{2} - 4 A C) / A^{2}$

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 $(a c^{n})^{1/ (n + 1)} + (a^{n} c)^{1/ (n + 1)} + b = 0$

If $α$ and $β$ are the roots of the equation $x^{2} + p x + 2 = 0$ and $\frac{1}{α}$ and $\frac{1}{β}$ are the roots of the equation $2 x^{2} + 2 q x + 1 = 0$ , then $(α - \frac{1}{α}) (β - \frac{1}{β}) (α + \frac{1}{β}) (β + \frac{1}{α})$ is equal to :

If $α, β$ are the roots of the equation $x^{2} - (5 + 3^{l o g_{3} 5} - 5^{l o g_{5} 3}) x + 3 (3^{(l o g_{3} 5)^{1/3}} - 5^{(l o g_{5} 3)^{2/3}} - 1) = 0$ then the equation, whose roots are $α + \frac{1}{β}$ and $β + \frac{1}{α}$ , is

Let $λ \neq = 0$ be a real number. Let $α, β$ be the roots of the equation $14 x^{2} - 31 x + 3 λ = 0$ and $α, γ$ be the roots of the equation $35 x^{2} - 53 x + 4 λ = 0$ . Then $\frac{3 α}{β}$ and $\frac{4 α}{γ}$ are the roots of the equation :

If $α$ and $β$ ( $α < β$ ) are the roots of the equation $x^{2} + b x + c = 0$ , where $c < 0 < b$ , then

If one root is square of the other root of the equation $x^{2} + p x + q = 0$ , then the relation between $p$ and $q$ is

Let $α, β$ be the roots of the equation $x^{2} - 4 λ x + 5 = 0$ and $α, γ$ be the roots of the equation $x^{2} - (32 + 23) x + 7 + 3 λ 3 = 0$ . If $β + γ = 32$ , then $(α + 2 β + γ)^{2}$ is equal to ______.