Quadratic roots identity (Vieta's relations): If α and β are the roots of ax^2 + bx + c = 0 (a ≠ 0), find the value of (α/β) + (β/α) in terms of a, b, c.

Introduction / Context:
This problem checks your comfort with Vieta's relations for quadratics and your ability to manipulate symmetric expressions in the roots. Instead of solving for the roots explicitly, we will express (α/β) + (β/α) directly using a, b, and c.

Given Data / Assumptions:

• Quadratic: ax^2 + bx + c = 0 with a ≠ 0.
• Roots: α and β (possibly real or complex).
• We require (α/β) + (β/α).

Concept / Approach:
By Vieta's relations, α + β = −b/a and αβ = c/a. Note that (α/β) + (β/α) = (α^2 + β^2)/(αβ). Also, α^2 + β^2 = (α + β)^2 − 2αβ. Substitute Vieta's expressions and simplify carefully to avoid sign errors.

Step-by-Step Solution:
α + β = −b/aαβ = c/aα^2 + β^2 = (α + β)^2 − 2αβ = (b^2/a^2) − 2(c/a)(α/β) + (β/α) = (α^2 + β^2)/(αβ) = [ (b^2/a^2) − 2(c/a) ] / (c/a)= (b^2/a − 2c)/c = (b^2 − 2ac)/(ac)

Verification / Alternative check:
Plugging a simple example (e.g., a = 1, b = −3, c = 1 with roots 1 and 2) confirms the formula numerically equals (1/2 + 2/1) = 2.5 and (b^2 − 2ac)/(ac) = (9 − 2)/1 = 7; wait, with a = 1, b = −3, c = 2 (roots 1 and 2), (α/β + β/α) = 1/2 + 2 = 2.5 and (b^2 − 2ac)/(ac) = (9 − 4)/(2) = 2.5—consistent.

Why Other Options Are Wrong:
Dividing by 2ac or adding 2ac in the numerator changes the value; these arise from algebra slips in the transformation from α^2 + β^2 to (α/β + β/α).

Common Pitfalls:
Forgetting that α^2 + β^2 = (α + β)^2 − 2αβ, or mishandling complex fractions while clearing denominators.

Final Answer:
(b^2 - 2ac) / (ac)