Condition for reciprocal roots (standard form): If the quadratic equation (x^2)/a + (x)/b + 1/c = 0 has one root equal to the reciprocal of the other (both nonzero), which condition among a, b, c must hold?

Introduction / Context:A quadratic px^2 + qx + r = 0 has roots that are reciprocals of each other (both nonzero) if and only if p = r. This follows from the Vieta relations: product of roots = r / p, which must equal 1 for reciprocal roots, hence r = p. We rewrite the given equation in standard form to identify p and r and then apply the criterion.Given Data / Assumptions:

• Equation: (x^2)/a + (x)/b + 1/c = 0.
• a, b, c ≠ 0; roots are finite and nonzero.
• We seek the relation among a, b, c ensuring reciprocal roots.

Concept / Approach:Express in standard quadratic form px^2 + qx + r = 0. Here p = 1/a, q = 1/b, r = 1/c. For reciprocal roots, require r = p ⇒ 1/c = 1/a ⇒ a = c.

Step-by-Step Solution:

Verification / Alternative check:If a = c, then product of roots = r/p = (1/c) / (1/a) = a/c = 1, so roots are reciprocals (assuming both nonzero). The condition is both necessary and sufficient.

Why Other Options Are Wrong:

• a = b or b = c: These do not force r = p.
• ac = 1: Irrelevant to r/p; it constrains magnitudes but not the required equality 1/c = 1/a.

Common Pitfalls:Confusing the “sum of roots” condition with “reciprocal roots.” Only the product matters here. Ensure the quadratic is in standard form before applying criteria.

Final Answer: