圆o的内接四边形ABCD中。AB=2,BC=4,CD=6,AD为直径，求证：圆O的半径是方程x^3-14x-12=0的根

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连接ACAD^2=AC^2+CD^2AC^2=AB^2+BC^2-2AB*BCcos(180-∠ADC) =4+16+16cos∠ADC =20+16*(CD/AD)AD^2=20+16*(6/AD)+36 =56+96/ADAD^3-56AD-96=0AD=2R8R^3-112R-96=0R^3-14R-12=0∴圆O的半径是方程x^3-14x-12=0的根