Quadratic Equations
Nature of Roots
Grade 11
Question:
<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>
<p>(a) real and distinct</p>
<p>(b) real and equal</p>
<p>(c) imaginary</p>
<p>(d) none of these</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-2)² + (a-2)² = 0. Since each square is non-negative and their sum is zero, each must be zero individually, yielding a=b=c=2.
<p><strong>Step 1:</strong> Rearrange the constraint 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 - 2a + b² - bc + c² - 2c + 4 = 0</p><p>Rearrange: (a² - ab + ¼b²) + (¾b² - bc + ⅓c²) + (⅔c² - 2c + 4) = 0</p><p>Better approach: (a - b)² + (b - c)² + (c - 2)² + (a - 2)² = 0</p><p><strong>Step 3:</strong> Since sum of non-negative terms equals zero, each must be zero:</p><p>a - b = 0 ⟹ a = b</p><p>b - c = 0 ⟹ b = c</p><p>c - 2 = 0 ⟹ c = 2</p><p>a - 2 = 0 ⟹ a = 2</p><p>Therefore: a = b = c = 2</p><p><strong>Step 4:</strong> The equation becomes 2x² + 2x + 2 = 0, or x² + x + 1 = 0</p><p>Discriminant: Δ = 1 - 4 = -3 < 0</p><p>Roots: x = (-1 ± i√3)/2</p><p>∴ Roots are complex conjugates: <strong>(-1 + i√3)/2 and (-1 - i√3)/2</strong></p>
Correct Answer: B