Roots of the equation $\sqrt {\dfrac {x}{1-x}}+\sqrt {\dfrac {1-x}{x}}=2\dfrac {1}{6}$ are
- $\dfrac {3}{2},\dfrac {2}{3}$
- $\dfrac {9}{13},\dfrac {4}{13}$
- $\dfrac {1}{3},\dfrac {3}{1}$
- $\dfrac {1}{13},\dfrac {13}{1}$
Let y = sqrt(x/(1-x)). The equation becomes y + 1/y = 13/6. Solving y^2 - (13/6)y + 1 = 0 gives y = 3/2 or 2/3. If sqrt(x/(1-x)) = 3/2, then x/(1-x) = 9/4, so 4x = 9 - 9x, 13x = 9, x = 9/13. If y = 2/3, x = 4/13.
Let t equal the square root of x divided by (1 minus x), so the equation becomes t plus 1 divided by t equals 13 divided by 6. Multiplying by 6t gives 6t squared minus 13t plus 6 equals 0, which factors into (2t minus 3) times (3t minus 2) equals 0. For t equal to 3 divided by 2, solving the square root of x divided by (1 minus x) equals 3 divided by 2 gives x equal to 9 divided by 13. For t equal to 2 divided by 3, solving the square root of x divided by (1 minus x) equals 2 divided by 3 gives x equal to 4 divided by 13. The roots are 9 divided by 13 and 4 divided by 13.