Multiple choice

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:

  1. $\displaystyle \left ( 0, \frac{1}{2} \right )$
  2. $\displaystyle \left (- \frac{1}{2}, 0 \right ) \cup \left ( 0, \frac{1}{2} \right )$
  3. $(- \infty, - 2) \cup (2, \infty)$
  4. $\displaystyle \left ( -\frac{1}{2}, 0 \right )$
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B Correct answer
Explanation

The equation can be rewritten by dividing by (x^2+x+1)^2. Let y = (x^2+x+1)/(x^2-x+1) or similar substitutions. The condition for real and distinct roots leads to the interval (-1/2, 0) union (0, 1/2).

AI explanation

Factoring x^4 + x^2 + 1 as (x^2 + x + 1)(x^2 - x + 1), the original equation becomes (x^2 + x + 1)[(a - 1)(x^2 - x + 1) + (a + 1)(x^2 + x + 1)] = 0. The first factor yields no real roots, so we examine the quadratic part: (2a)(x^2) + (2)(x) + (2a) = 0, which simplifies to a(x^2) + x + a = 0. For this equation to have two distinct real roots, its discriminant must be greater than zero, so 1 - 4(a)^2 > 0. Solving this inequality yields a^2 < 1/4, meaning a is between -1/2 and 1/2. Since a cannot be 0 without reducing the degree of the equation, the set of values is the open interval from -1/2 to 0 combined with the open interval from 0 to 1/2.

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