In the quadratic equation ax² + bx + c = 0, Δ = b² - 4ac and α + β, α² + β², α³ + β³, are in G.P. wh

Q: Wait, why can't we just solve for $\alpha$ and $\beta$ using the quadratic formula and plug them in?

You *could*, but that is the path of infinite pain! The quadratic formula introduces square roots that make the algebra messy and prone to error. In JEE Advanced, we value symmetry. By using Vieta's formulas, we keep the expressions clean and elegant, allowing the structure of the equation to reveal itself naturally.

Q: Why does the equation end up as $c\Delta = 0$? Does this mean $c=0$ or $\Delta=0$?

Exactly! It means either the constant term $c$ is zero, or the discriminant $\Delta$ is zero. If $c=0$, one root is zero. If $\Delta=0$, the roots are equal. Both scenarios satisfy the condition that the terms form a GP. It's a beautiful, unified result.

Q: I got stuck at the expansion step. Is there a trick to avoid expanding everything?

Great question. While you can use Newton's Sums for higher powers, for this specific problem, direct expansion is actually quite fast because the higher-order terms ($\alpha^4$ and $\beta^4$) cancel out. Trust the algebra; sometimes the most direct path is the one that looks the most intimidating at first glance.

Question

In the quadratic equation $ax^2 + bx + c = 0$, $\Delta = b^2 - 4ac$ and $\alpha + \beta, \alpha^2 + \beta^2, \alpha^3 + \beta^3$, are in G.P. where $\alpha, \beta$ are the root of $ax^2 + bx + c = 0$, then

Options:

(A) $\Delta 
eq 0$

(B) $b \Delta = 0$

(C) $c \Delta = 0$

(D) $\Delta = 0$

Sigma Percentile · Accepted Answer

### The Symphony of Symmetry: Unlocking the GP Condition

Welcome, future engineer. Today, we are not just solving a quadratic equation; we are exploring the hidden architecture of roots.

When you look at a quadratic equation like $ax^2 + bx + c = 0$, most students see a parabola. But I want you to see something deeper: a system of symmetric relationships.

We are given that the roots $\alpha$ and $\beta$ generate a sequence: $T_1 = \alpha + \beta$, $T_2 = \alpha^2 + \beta^2$, and $T_3 = \alpha^3 + \beta^3$. We are told these three terms form a Geometric Progression (G.P.). This is our starting point, our anchor in the storm.

### Phase 1: The Geometric Key

What does it mean for three numbers to be in a G.P.? It means the ratio between consecutive terms is constant.

Mathematically, this translates to the elegant condition:
$$T_2^2 = T_1 \cdot T_3$$

This is the master key. It allows us to bypass the messy business of finding the roots individually. We don't need to know what $\alpha$ or $\beta$ are; we only need to know how they relate to each other.

By substituting our terms, we get the equation:
$$(\alpha^2 + \beta^2)^2 = (\alpha + \beta)(\alpha^3 + \beta^3)$$

Take a moment to look at this. It is a statement of pure symmetry.

### Phase 2: The Algebraic Expansion

Now, I know what you are thinking. 'Do I really have to expand all of this?' The answer is a resounding yes, but with a twist.

Let's expand the left-hand side (LHS) using the identity $(x+y)^2 = x^2 + y^2 + 2xy$. We get $\alpha^4 + \beta^4 + 2\alpha^2\beta^2$.

Now, look at the right-hand side (RHS). Multiplying $(\alpha + \beta)(\alpha^3 + \beta^3)$ gives us:
$$\alpha^4 + \alpha\beta^3 + \beta\alpha^3 + \beta^4$$

When we set them equal, something magical happens. The $\alpha^4$ and $\beta^4$ terms appear on both sides. They are like ghosts—they vanish instantly!

We are left with:
$$2\alpha^2\beta^2 = \alpha\beta^3 + \beta\alpha^3$$

This is the moment where the problem begins to yield. We can factor out $\alpha\beta$ from the right side, leaving us with:
$$2\alpha^2\beta^2 = \alpha\beta(\alpha^2 + \beta^2)$$

### Phase 3: The Elegant Collapse

Let's bring everything to one side:
$$2\alpha^2\beta^2 - \alpha\beta(\alpha^2 + \beta^2) = 0$$

If we factor out $-\alpha\beta$, we get:
$$-\alpha\beta(\alpha^2 + \beta^2 - 2\alpha\beta) = 0$$

Look closely at that bracket. It is the perfect square identity: $(\alpha - \beta)^2$. So, our entire complex expression has collapsed into the beautiful, compact form:
$$-\alpha\beta(\alpha - \beta)^2 = 0$$

This is the power of algebraic manipulation. We have reduced a high-degree polynomial relationship into a simple product of two fundamental quantities: the product of the roots and the square of their difference.

### Phase 4: The Final Connection

Now, we bring in the heavy artillery: Vieta's formulas. We know that for $ax^2 + bx + c = 0$, the product of the roots is $\alpha\beta = \frac{c}{a}$.

And the difference of the roots? That is tied directly to the discriminant $\Delta = b^2 - 4ac$. Specifically:
$$(\alpha - \beta)^2 = \frac{\Delta}{a^2}$$

Substituting these into our collapsed equation, we get:
$$-\left(\frac{c}{a}ight) \left(\frac{\Delta}{a^2}ight) = 0$$

This simplifies to:
$$-\frac{c\Delta}{a^3} = 0$$

Since $a 
eq 0$ (otherwise it wouldn't be a quadratic equation), the only way for this expression to be zero is if the numerator is zero. Thus, **$c\Delta = 0$**.

### Conclusion

We have arrived at the destination. The condition for these roots to form a G.P. is simply **$c\Delta = 0$**.

It is a result that is both simple and profound. It tells us that either the constant term is zero (meaning one root is zero) or the discriminant is zero (meaning the roots are equal).

You have successfully navigated the algebra and uncovered the underlying truth. Keep this mindset—look for the symmetry, trust the algebra, and the answer will always reveal itself.

Animated Solution for Mathematics - Quadratic Equations: In the quadratic equation $a x^{2} + b x + c = 0$ , $Δ = b^{2} - 4 a c$ and $α + β, α^{2} + β^{2}, α^{3} + β^{3}$ , are in G.P. where $α, β$ are the root of $a x^{2} + b x + c = 0$ , then

Select Answer:

Visualized Solution

Defining the Roots and G.P. Condition

Applying the G.P. Property

Expanding Both Sides

Simplifying the Equation

Rearranging and Factoring

Relating to Coefficients and Discriminant

Substituting into the Equation

Final Conclusion

The Sigma Insight: Relation Between Roots and Coefficients

The Symphony of Symmetry

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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