Re: radical equation

*To*: mathgroup at smc.vnet.net*Subject*: [mg71988] Re: radical equation*From*: dh <dh at metrohm.ch>*Date*: Thu, 7 Dec 2006 06:25:37 -0500 (EST)*Organization*: hispeed.ch*References*: <el68mn$2ui$1@smc.vnet.net>

Hi Dimitris, I think you have one real and 2 complex roots. Consider: Reduce[{(2 - x)^2 - 4*Sqrt[1 - x]*Sqrt[1 - c*x] == 0, 0 < c < 1, 0 < x < 1}, x] this give (0 < c < 1 && x ==Root[-16 + 16 c + 24 #1 - 16 c #1 - 8 #1^2 + #1^3 &, 1] this means the first root of the polynomial. You can easily verify that the other 2 roots (e.g. Root[-16 + 16 c + 24 #1 - 16 c #1 - 8 #1^2 + #1^3 &, 2])are complex. Daniel dimitris wrote: > Consider the following equation > > req = (2 - x)^2 - 4*Sqrt[1 - x]*Sqrt[1 - c*x] == 0; > > where {0<x<1 and 0<c<1}. > > Ommiting the trivial zero root, the following roots are provided by > Solve > > sols=DeleteCases[Solve[req, x], {x -> 0}] > {{x -> 8/3 - (8 - 48*c)/(6*(-17 + 45*c + 3*Sqrt[3]*Sqrt[11 - 62*c + > 107*c^2 - 64*c^3])^(1/3)) + > (2/3)*(-17 + 45*c + 3*Sqrt[3]*Sqrt[11 - 62*c + 107*c^2 - > 64*c^3])^(1/3)}, > {x -> 8/3 + ((1 + I*Sqrt[3])*(8 - 48*c))/(12*(-17 + 45*c + > 3*Sqrt[3]*Sqrt[11 - 62*c + 107*c^2 - 64*c^3])^(1/3)) - > (1/3)*(1 - I*Sqrt[3])*(-17 + 45*c + 3*Sqrt[3]*Sqrt[11 - 62*c + > 107*c^2 - 64*c^3])^(1/3)}, > {x -> 8/3 + ((1 - I*Sqrt[3])*(8 - 48*c))/(12*(-17 + 45*c + > 3*Sqrt[3]*Sqrt[11 - 62*c + 107*c^2 - 64*c^3])^(1/3)) - > (1/3)*(1 + I*Sqrt[3])*(-17 + 45*c + 3*Sqrt[3]*Sqrt[11 - 62*c + > 107*c^2 - 64*c^3])^(1/3)}} > > Applying the principle of the argument it can be proved that req has > only one solution in > 0<x<1 for 0<c<1. > > Direct substitution of sols to req does not lead anywhere since Solve > is not capable of doing this kind of verification for extraneous roots. > > Any ideas? > > Dimitris >