Roots of a parameterized quadratic: Find the roots of a^2 x^2 − 3ab x + 2b^2 = 0 in terms of a and b (assume a ≠ 0).

Introduction / Context:We are to solve a^2 x^2 − 3ab x + 2b^2 = 0 for x in terms of parameters a and b. Treat this as a standard quadratic with respect to x and apply the quadratic formula. Parameterized quadratics commonly appear in algebraic identities and comparisons.Given Data / Assumptions:

• Quadratic: a^2 x^2 − 3ab x + 2b^2 = 0.
• a ≠ 0 (to ensure a^2 is invertible); b can be any real.

Concept / Approach:Identify coefficients: A = a^2, B = −3ab, C = 2b^2. Use x = [−B ± √(B^2 − 4AC)] / (2A) and simplify by factoring ab where possible.

Step-by-Step Solution:

Verification / Alternative check:Plug x = 2b/a: a^2*(4b^2/a^2) − 3ab*(2b/a) + 2b^2 = 4b^2 − 6b^2 + 2b^2 = 0. Similarly for x = b/a: a^2*(b^2/a^2) − 3ab*(b/a) + 2b^2 = b^2 − 3b^2 + 2b^2 = 0.

Why Other Options Are Wrong:

• Signs with negatives do not satisfy the original quadratic.
• “None of these” is incorrect since the explicit pair 2b/a and b/a works.

Common Pitfalls:Dropping parameters during simplification or mishandling the square root of a^2b^2. Keep factors symbolic and simplify carefully.

Final Answer: