If α, β are the roots of ax² + bx + c = 0, (a ≠ 0) and α + δ, β + δ are the roots of Ax² + Bx + C =

Q: Why can't we just use the quadratic formula directly to find the roots and then subtract them?

You absolutely could, but it would be a nightmare of algebra! The quadratic formula gives you $\alpha, \beta = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$. Subtracting these leads to $\frac{\sqrt{b^2-4ac}}{a}$. Squaring this gives $\frac{b^2-4ac}{a^2}$. While this works, it is much more elegant to use the properties of the roots directly. In JEE, time is your most precious resource, and recognizing the invariance of the root difference saves you from unnecessary calculations.

Q: Does the value of $\delta$ matter at all?

That is the beauty of this problem! The value of $\delta$ is completely irrelevant to the final relationship. Because the shift is applied to both roots equally, the distance between them remains constant. Whether you shift the parabola by 1 unit or 100 units, the 'spread' of the roots remains the same, which is why $\delta$ cancels out entirely.

Q: Is this property true for cubic equations too?

Great intuition! This is a specific case of a more general property. For any polynomial, if you shift all roots by the same constant $\delta$, the differences between the roots remain invariant. This means that any symmetric function of the differences of the roots will also remain invariant. It is a powerful tool in higher-level polynomial theory!

Question

If $\alpha, \beta$ are the roots of $ax^2 + bx + c = 0, (a 
eq 0)$ and $\alpha + \delta, \beta + \delta$ are the roots of $Ax^2 + Bx + C = 0, (A 
eq 0)$ for some constant $\delta$, then prove that $(b^2-4ac)/a^2 = (B^2-4AC)/A^2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a coordinate plane, watching a blue parabola defined by $ax^2 + bx + c = 0$. The roots, $\alpha$ and $\beta$, are the points where this curve intersects the $x$-axis.

Now, imagine we take this entire shape and slide it horizontally by a distance $\delta$. The new parabola, $Ax^2 + Bx + C = 0$, has roots $\alpha + \delta$ and $\beta + \delta$.

Geometrically, the shape of the parabola remains unchanged; it has simply moved as a rigid body. Consequently, the distance between the roots must be identical to the distance between the original roots. This is the core insight: the distance between the roots is **invariant under horizontal translation**.

### The Algebraic Toolkit

To prove this, we translate our geometric intuition into the language of algebra. We know the distance between the roots of the first equation is $|\alpha - \beta|$.

For the second equation, the distance is $|(\alpha + \delta) - (\beta + \delta)|$. As expected, the $\delta$ terms cancel out, leaving us with $|\alpha - \beta|$.

Since the distances are equal, their squares must also be equal:
$$(\alpha - \beta)^2 = ((\alpha + \delta) - (\beta + \delta))^2$$

To connect this to the coefficients $a, b, c$, we use the classic identity:
$$(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta$$
This identity serves as our bridge between the roots and the coefficients.

### The Calculation

Let us focus on the first equation, $ax^2 + bx + c = 0$. From Vieta's formulas, we know the sum of the roots is $\alpha + \beta = -\frac{b}{a}$ and the product is $\alpha\beta = \frac{c}{a}$.

Substituting these into our identity, we obtain:
$$(\alpha - \beta)^2 = \left(-\frac{b}{a}\right)^2 - 4\left(\frac{c}{a}\right)$$

Simplifying this, we have $\frac{b^2}{a^2} - \frac{4c}{a}$. Finding a common denominator of $a^2$, we yield:
$$\frac{b^2 - 4ac}{a^2}$$
This expression represents the square of the difference of the roots.

### The Grand Conclusion

We apply this exact same logic to the second equation, $Ax^2 + Bx + C = 0$. The square of the difference of its roots is:
$$\frac{B^2 - 4AC}{A^2}$$

Since we established that the square of the difference of the roots is invariant under the shift $\delta$, these two expressions must be equal. Therefore:
$$\frac{b^2 - 4ac}{a^2} = \frac{B^2 - 4AC}{A^2}$$

We have arrived at the result! This is a beautiful demonstration of how geometry and algebra interact. By understanding the physical meaning of the shift, we bypassed complex calculations and arrived at the truth through symmetry.

Keep this perspective in your toolkit; it will serve you well in your journey through **JEE Advanced**.

Animated Solution for Mathematics - Quadratic Equations: 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 $(b^{2} - 4 a c) / a^{2} = (B^{2} - 4 A C) / A^{2}$

Visualized Solution

The First Quadratic

The Shifted Roots

Invariance of Root Difference

Expressing the Difference

Coefficients of First Equation

Difference Square for Eq 1

Simplifying the First Difference

Difference Square for Second Equation

Equating and Concluding

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Algebraic Toolkit

The Calculation

The Grand Conclusion

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