Multiple choice

The roots of the equation, $(x^{2}+1)^{2}=x(3x^{2}+4x+3)$ are given by

  1. $2-\sqrt {3}$
  2. $\left(\dfrac {-1+i\sqrt {3}}{2}\right)$
  3. $2+\sqrt {3}$
  4. $\dfrac {-1-i\sqrt {3}}{2}$
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A Correct answer
AI explanation

Expand the equation to x^4 + 2x^2 + 1 = 3x^3 + 4x^2 + 3x, which rearranges to x^4 - 3x^3 - 2x^2 - 3x + 1 = 0. Dividing by x^2 gives x^2 - 3x - 2 - 3/x + 1/x^2 = 0, which is a reciprocal equation and can be regrouped as (x^2 + 1/x^2) - 3(x + 1/x) - 2 = 0. Letting y = x + 1/x transforms the equation into y^2 - 2 - 3y - 2 = 0, or y^2 - 3y - 4 = 0, yielding y = 4 or y = -1. For y = 4, the equation x + 1/x = 4 produces x^2 - 4x + 1 = 0, and applying the quadratic formula gives x = (4 +/- sqrt(12))/2 = 2 +/- sqrt(3); therefore, 2 - sqrt(3) is a valid root.

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