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).
-
A2b/a , − b/a
-
B2b/a , b/a
-
C−2b/a , − b/a
-
DNone of these
Answer
Correct Answer: 2b/a , b/a
Explanation
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:
Discriminant: B^2 − 4AC = (−3ab)^2 − 4*a^2*2b^2 = 9a^2b^2 − 8a^2b^2 = a^2b^2.√Discriminant = |ab|; algebraically we take ±ab in the formula.x = [3ab ± ab]/(2a^2) = (ab(3 ± 1))/(2a^2) = (b/a) * (3 ± 1)/2.Thus roots: x = 2b/a and x = b/a.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:
2b/a , b/a