Let p and q be real numbers such that p ≠ 0, p³ ≠ -q. If α and β are nonzero complex numbers satisfy

Q: Wait, why do we use the identity $\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$ instead of trying to solve for $\alpha$ and $\beta$ individually?

That is a fantastic question! Solving for $\alpha$ and $\beta$ individually would involve solving a system of equations that leads to a quadratic in $\alpha$ (or $\beta$), which is often messy and unnecessary. In JEE problems, we look for symmetry. By using the identity, we directly find the value of the product $\alpha\beta$ without ever needing to know the specific values of $\alpha$ or $\beta$. It is the elegant, efficient path.

Q: Why does the product of the roots $\frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha}$ just become 1? Is it always that simple?

Yes, it really is that simple! Because the roots are reciprocals of each other, their product is always 1. Whenever you see roots like $\frac{1}{x}$ and $x$, or $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$, the product is guaranteed to be 1. It is a beautiful simplification that saves you time.

Q: I got stuck on the simplification of the sum of roots $S$. How do I handle the complex fraction $\frac{(\alpha+\beta)^2 - 2\alpha\beta}{\alpha\beta}$?

Don't panic! When you have a fraction like $\frac{A}{B}$, just substitute the expressions for $A$ and $B$ first. Then, multiply the entire numerator and denominator by the common denominator of the internal fractions (in this case, $3p$). This will clear out the 'fraction within a fraction' and leave you with a clean, manageable expression.

Question

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

Options:

(A) $(p^3+q)x^2 - (p^3+2q)x + (p^3+q) = 0$

(B) $(p^3+q)x^2 - (p^3-2q)x + (p^3+q) = 0$

(C) $(p^3-q)x^2 - (5p^3-2q)x + (p^3-q) = 0$

(D) $(p^3-q)x^2 - (5p^3+2q)x + (p^3-q) = 0$

Sigma Percentile · Accepted Answer

### The Elegance of Symmetry: A Journey into Quadratic Construction

Imagine you are standing at the threshold of a classic JEE Advanced problem. You are given two complex numbers, $\alpha$ and $\beta$, bound by the constraints $\alpha + \beta = -p$ and $\alpha^3 + \beta^3 = q$.

Your mission is to construct a new quadratic equation whose roots are the ratios $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$. Let us peel back the layers of this problem together.

### Phase 1: The Quest for the Product

To build any quadratic equation, we rely on the fundamental form: $x^2 - Sx + P = 0$, where $S$ is the sum of the roots and $P$ is their product.

Our first target is the product $P = \frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha}$. As we look at this, the beauty of the problem reveals itself—the $\alpha$ and $\beta$ terms cancel out perfectly, leaving us with $P = 1$.

To find the sum $S = \frac{\alpha}{\beta} + \frac{\beta}{\alpha}$, we first need to determine the value of $\alpha\beta$. We utilize the cubic identity:
$$\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)$$

Substituting our known values, we get $q = (-p)^3 - 3\alpha\beta(-p)$. Simplifying this, we find $q = -p^3 + 3p\alpha\beta$.

With a little algebraic rearrangement, we isolate the product:
$$\alpha\beta = \frac{p^3 + q}{3p}$$

### Phase 2: The Sum of Roots

Now, let us tackle the sum $S = \frac{\alpha}{\beta} + \frac{\beta}{\alpha}$. By taking a common denominator, we get:
$$S = \frac{\alpha^2 + \beta^2}{\alpha\beta}$$

For the numerator, we use the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$. Substituting our known values, we have:
$$\alpha^2 + \beta^2 = (-p)^2 - 2\left(\frac{p^3 + q}{3p}\right)$$

Combining these into the expression for $S$, we have:
$$S = \frac{p^2 - 2\left(\frac{p^3 + q}{3p}\right)}{\frac{p^3 + q}{3p}}$$

To simplify this, we multiply the numerator and the denominator by $3p$. The expression transforms into:
$$S = \frac{3p^3 - 2(p^3 + q)}{p^3 + q}$$

Expanding the numerator, we get $3p^3 - 2p^3 - 2q$, which simplifies beautifully to:
$$S = \frac{p^3 - 2q}{p^3 + q}$$

### Phase 3: The Final Construction

We have arrived at the finish line. Our quadratic equation is $x^2 - Sx + P = 0$. Substituting our values, we get:
$$x^2 - \left(\frac{p^3 - 2q}{p^3 + q}\right)x + 1 = 0$$

To reach the standard polynomial form, we multiply the entire equation by the denominator $(p^3 + q)$. This yields the **final result**:
$$(p^3 + q)x^2 - (p^3 - 2q)x + (p^3 + q) = 0$$

This result is clean, symmetric, and derived entirely from the relationships between the roots. This is the heart of JEE mathematics—not brute force, but the strategic use of identities to reveal the underlying structure of the problem.

Select Answer:

Visualized Solution

Identify the Goal

The Cubic Identity

Substitution for $α β$

Solving for $α β$

Finding the Sum of Roots $S$

Calculating $α^{2} + β^{2}$

Simplifying the Sum of Roots $S$

Finding the Product of Roots $P$

Constructing the Equation

Final Simplification

Key Takeaway

The Sigma Insight: Relation Between Roots and Coefficients

The Elegance of Symmetry

Phase 1

Phase 2

Phase 3

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Select Answer:

Visualized Solution

Identify the Goal

The Cubic Identity

Substitution for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } αβ

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } αβ

Finding the Sum of Roots .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α2+β2

Simplifying the Sum of Roots .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S

Finding the Product of Roots .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P