<p><strong>163.</strong> If \(a, b, c \in \mathbb{R}\) and \(a^2 + b^2 + c^2 + 4 = ab + bc + 2c + 2a\), then roots of \(ax^2 + bx + c = 0\) are:</p>
Step-by-Step Solution
Key Concept: Rearrange the constraint as a sum of squares equal to zero: (a-b)² + (b-c)² + (c-1)² + (a-1)² = 0. Since each squared term is non-negative and their sum is zero, each must individually equal zero, forcing a=b=c=1.
<p><strong>Step 1:</strong> Rearrange the given equation:</p><p>a² + b² + c² + 4 = ab + bc + 2c + 2a</p><p>a² + b² + c² + 4 - ab - bc - 2c - 2a = 0</p><p><strong>Step 2:</strong> Rewrite as sum of squares by completing the square:</p><p>a² - ab + ¼b² + ¼b² - bc + c² + c² - 2c + 1 + a² - 2a + 1 - ½b² - c² = 0</p><p>Rearranging strategically:</p><p>(a - b)² + (b - c)² + (c - 1)² + (a - 1)² = 0</p><p><strong>Step 3:</strong> Since each term is a perfect square and their sum equals zero, each must individually be zero:</p><p>a - b = 0 ⟹ a = b</p><p>b - c = 0 ⟹ b = c</p><p>c - 1 = 0 ⟹ c = 1</p><p>a - 1 = 0 ⟹ a = 1</p><p><strong>Step 4:</strong> Therefore a = b = c = 1, so the equation becomes:</p><p>x² + x + 1 = 0</p><p>Discriminant: Δ = 1 - 4 = -3 < 0</p><p>Roots: x = (-1 ± i√3)/2 (complex conjugate pair)</p><p>∴ Answer: B (Roots are complex conjugates)</p>
Correct Answer: B