Difficulty: Medium
Correct Answer: 1
Explanation:
Introduction / Context: This equation mixes a radical expression and its reciprocal on the left with a quadratic on the right. An inequality insight about t + 1/t helps bound the left-hand side and identify feasible x-values without brute force algebra.
Given Data / Assumptions:
Concept / Approach: Let t = √A > 0. Then LHS = t + 1/t ≥ 2 by AM ≥ GM, with equality iff t = 1. Also, A = x^2 − x + 1 has minimum 3/4 (at x = 1/2), so t ≥ √(3/4) = √3/2 > 0, ensuring the expression is defined for all real x.
Step-by-Step Solution:
Since t + 1/t ≥ 2, LHS ≥ 2, with equality when t = 1 → A = 1.A = 1 ⇒ x^2 − x + 1 = 1 ⇒ x(x − 1) = 0 ⇒ x = 0 or x = 1.Check RHS at these x: for x = 0, RHS = 2 − 0 = 2; LHS = 1 + 1 = 2 → valid.For x = 1, RHS = 2 − 1 = 1; LHS = 2 → not valid.For any other x, LHS > 2 while RHS ≤ 2, so no equality.Verification / Alternative check: Numerical sampling near x = 0 and x = 1 confirms only x = 0 satisfies the equality.
Why Other Options Are Wrong: 0, 2, 3, 4 solutions contradict the inequality structure; only one real solution exists.
Common Pitfalls: Forgetting the equality condition t = 1 for t + 1/t = 2; mishandling the domain by allowing A ≤ 0 (which never occurs here).
Final Answer: 1
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