<p>If the roots of equation \((a-1)(x^2 + x + 1)^2 = (a+1)(x^4 + x^2 + 1)\) are real and distinct, then the value of \(a\) \(\in\)</p>
Step-by-Step Solution
Key Concept: Recognize that x⁴ + x² + 1 = (x² + x + 1)(x² - x + 1), and use substitution y = x² + x + 1 to transform into a quadratic inequality in y that ensures real, distinct roots in x.
<p><strong>Step 1:</strong> Factor the RHS: x⁴ + x² + 1 = (x² + x + 1)(x² - x + 1)</p><p><strong>Step 2:</strong> Note that x² + x + 1 > 0 for all real x (discriminant = -3 < 0). Divide both sides by (x² + x + 1) (valid when x² + x + 1 ≠ 0):</p><p>(a-1)(x² + x + 1) = (a+1)(x² - x + 1)</p><p><strong>Step 3:</strong> Let y = x² + x + 1. Then x² - x + 1 = y - 2x. Rearranging:</p><p>(a-1)y = (a+1)(y - 2x)</p><p><strong>Step 4:</strong> Alternatively, expand directly: (a-1)x² + (a-1)x + (a-1) = (a+1)x² - (a+1)x + (a+1)</p><p>Simplify: -2x² + 2(a)x + (a-2) = 0, or x² - ax - (a-2)/2 = 0</p><p><strong>Step 5:</strong> For real distinct roots: Δ = a² + 2(a-2) > 0</p><p>a² + 2a - 4 > 0</p><p>a ∈ (-∞, -1-√5) ∪ (-1+√5, ∞)</p><p><strong>Step 6:</strong> Additional constraint: a ≠ 1 (from original equation denominator consideration)</p><p>∴ <strong>Answer: B</strong> is a∈(-∞, -1-√5) ∪ (-1+√5, ∞) with a ≠ 1</p>
Correct Answer: B