Question

If the roots of the equation \((a-1)(x^2 + x + 1)^2 = (a+1)(x^4 + x^2 + 1)\) are real and distinct then the value of \(a \in\)

• A. \((-\infty, -2) \cup (2, \infty)\)
• B. \((-\infty, -2] \cup [2, \infty)\)
• C. \([-2, 2]\)
• D. \((-2, 2)\)

Hint:

Use \(\Delta \ge 0\) for real roots and \(\Delta>0\) for distinct roots.

Answer 1

The Correct Option is B

Step 1: Formula / Definition}
\[ x^4 + x^2 + 1 = (x^2 + x + 1)(x^2 - x + 1) \]
Step 2: Calculation / Simplification}
\[ (a-1)(x^2+x+1)^2 = (a+1)(x^2+x+1)(x^2-x+1) \]
\[ (a-1)(x^2+x+1) = (a+1)(x^2-x+1) \quad [\because x^2+x+1 \neq 0] \]
\[ x^2 - ax + 1 = 0 \]
For real roots: \(\Delta = a^2 - 4 \ge 0\)
\[ a^2 \ge 4 \Rightarrow a \in (-\infty, -2] \cup [2, \infty) \]
Step 3: Final Answer
\[ a \in (-\infty, -2] \cup [2, \infty) \]