What is one of the possible values of x-\frac{1}{x}?

Consider the following equation:<br><br>6x^{2}-25x+\frac{6}{x^{2}}+\frac{25}{x}+12=0

  1. A. \frac{1}{2}
  2. B. \frac{3}{2}
  3. C. 2
  4. D. \frac{5}{2}

Correct Answer: B. \frac{3}{2}

Explanation

Group terms: 6(x^2 + \frac{1}{x^2}) - 25(x - \frac{1}{x}) + 12 = 0. Let y = x - \frac{1}{x}, then x^2 + \frac{1}{x^2} = y^2 + 2. The equation becomes 6(y^2 + 2) - 25y + 12 = 0 \implies 6y^2 - 25y + 24 = 0. Factoring gives (3y - 8)(2y - 3) = 0, so y = \frac{8}{3} or y = \frac{3}{2}.

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