If α and β are the roots of the quadratic equation x^2 - 2x + 4 = 0, then what is the equation whose roots are α^3 / β^2 and β^3 / α^2?

Introduction / Context:
This is a problem about transforming the roots of a quadratic equation. The original quadratic has roots α and β, and we are asked to form a new quadratic whose roots are α^3 / β^2 and β^3 / α^2. Instead of solving explicitly for α and β, we use relationships between the sum and product of the roots and apply algebraic identities. This kind of question is common in algebra sections of competitive exams and tests fluency with symmetric polynomials and root transformations.

Given Data / Assumptions:

• Original equation: x^2 - 2x + 4 = 0.
• Roots: α and β.
• We need an equation whose roots are α^3 / β^2 and β^3 / α^2.
• Coefficients are real numbers and the quadratic is monic (leading coefficient 1) in the options.

Concept / Approach:
For the original quadratic x^2 - 2x + 4 = 0, we know from Vieta formulas: α + β = 2, αβ = 4. For the new roots r1 = α^3 / β^2 and r2 = β^3 / α^2, we compute their sum S = r1 + r2 and product P = r1 r2 in terms of α and β. Once S and P are known, the monic quadratic with roots r1 and r2 is x^2 - Sx + P = 0. This approach avoids finding α and β explicitly and instead uses algebraic identities for sums and powers of roots.

Step-by-Step Solution:
Step 1: Compute the product of the new roots. P = (α^3 / β^2) * (β^3 / α^2) = α^(3 - 2) * β^(3 - 2) = αβ. Since αβ = 4, we have P = 4. Step 2: Compute the sum of the new roots S = α^3 / β^2 + β^3 / α^2. Write S as (α^5 + β^5) / (α^2 β^2). We know α^2 β^2 = (αβ)^2 = 4^2 = 16. Step 3: Compute α^5 + β^5 using identities. First compute α^2 + β^2 and α^3 + β^3 and α^4 + β^4. α^2 + β^2 = (α + β)^2 - 2αβ = 2^2 - 2 * 4 = 4 - 8 = -4. α^3 + β^3 = (α + β)^3 - 3αβ(α + β) = 2^3 - 3 * 4 * 2 = 8 - 24 = -16. α^4 + β^4 = (α^2 + β^2)^2 - 2(αβ)^2 = (-4)^2 - 2 * 16 = 16 - 32 = -16. Step 4: Use the identity α^5 + β^5 = (α + β)(α^4 + β^4) - αβ(α^3 + β^3). Substitute values: α^5 + β^5 = 2 * (-16) - 4 * (-16) = -32 + 64 = 32. Step 5: Now compute S: S = (α^5 + β^5) / (α^2 β^2) = 32 / 16 = 2. Step 6: The required quadratic with roots r1 and r2 is x^2 - Sx + P = x^2 - 2x + 4 = 0.

Verification / Alternative check:
One could, in principle, solve the original quadratic x^2 - 2x + 4 = 0 using the quadratic formula, obtain explicit complex roots α and β, then compute α^3 / β^2 and β^3 / α^2 and find the quadratic that has those as roots. This would agree with the simpler Vieta based approach and yield the same equation x^2 - 2x + 4 = 0. The fact that the resulting equation is identical to the original is not unusual, as some transformations of roots preserve the same sum and product.

Why Other Options Are Wrong:
Options a, b and d give quadratics with different sums and products for the roots. For example, x^2 - 4x + 8 = 0 has sum 4 and product 8, which does not match S = 2 and P = 4. Option e, x^2 + 2x + 4 = 0, has sum -2 and product 4, which also does not match. Only x^2 - 2x + 4 = 0 has sum 2 and product 4, exactly corresponding to the derived values S and P for the transformed roots.

Common Pitfalls:
Students sometimes try to expand the expressions α^3 / β^2 and β^3 / α^2 directly without using the symmetry in α and β, which leads to very complicated algebra. Another common error is to misapply the identities for sums of powers, such as α^3 + β^3, or to forget that α^2 β^2 = (αβ)^2. Careful use of Vieta formulas and stepwise computation of α^2 + β^2, α^3 + β^3 and so on keeps the work manageable and accurate.

Final Answer:
Thus, the equation whose roots are α^3 / β^2 and β^3 / α^2 is x^2 - 2x + 4 = 0.