Quadratic Equations
Nature of Roots
Grade 11
Question:
<p>If the two roots of the equation, \((a-1)(x^4 + x^2 + 1) + (a+1)(x^2 + x + 1)^2 = 0\) are real and distinct, then the set of all values of \(a\) is</p>
<p>\(\left(-\dfrac{1}{2}, 0\right)\)</p>
<p>\((-\infty, -2) \cup (2, \infty)\)</p>
<p>\(\left(-\dfrac{1}{2}, 0\right) \cup \left(0, \dfrac{1}{2}\right)\)</p>
<p>\(\left(0, \dfrac{1}{2}\right)\)</p>
Step-by-Step Solution
Key Concept: Recognize that x⁴ + x² + 1 = (x² + x + 1)(x² - x + 1), transforming the equation into a quadratic in (x² + x + 1). The constraint for two real distinct roots depends on when the substituted quadratic has positive discriminant.
<p><strong>Step 1:</strong> Factor x⁴ + x² + 1. Note that x⁴ + x² + 1 = (x² + x + 1)(x² - x + 1).</p><p><strong>Step 2:</strong> Let y = x² + x + 1. Then the equation becomes:<br/>(a-1)(y)(y - 2x) + (a+1)y² = 0<br/>Actually, factor correctly: x⁴ + x² + 1 = (x² + x + 1)(x² - x + 1)</p><p><strong>Step 2 (Corrected):</strong> The equation is (a-1)(x² + x + 1)(x² - x + 1) + (a+1)(x² + x + 1)² = 0<br/>Factor out (x² + x + 1): (x² + x + 1)[(a-1)(x² - x + 1) + (a+1)(x² + x + 1)] = 0</p><p><strong>Step 3:</strong> Since x² + x + 1 > 0 for all real x (discriminant = 1 - 4 = -3 < 0), we need:<br/>(a-1)(x² - x + 1) + (a+1)(x² + x + 1) = 0</p><p><strong>Step 4:</strong> Expand: (a-1)x² - (a-1)x + (a-1) + (a+1)x² + (a+1)x + (a+1) = 0<br/>= 2ax² + 2x + 2a = 0<br/>= ax² + x + a = 0</p><p><strong>Step 5:</strong> For two real distinct roots: Δ = 1 - 4a² > 0<br/>4a² < 1<br/>a² < 1/4<br/>|a| < 1/2<br/>-1/2 < a < 1/2</p><p><strong>Step 6:</strong> Also need a ≠ 0 (for quadratic), so a ∈ (-1/2, 0) ∪ (0, 1/2)</p><p>∴ Answer: C</p>
Correct Answer: C