If $\alpha$ satisfies the equation $\sqrt{\frac{x}{2x+1}} + \sqrt{\frac{2x+1}{x}} = 2$,then the roots of the equation $\alpha^2 x^2 + 4\alpha x + 3 = 0$ are

  • A
    $1, 3$
  • B
    $-1, 1$
  • C
    $2, -3$
  • D
    $3, 4$

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