radical equation

*To*: mathgroup at smc.vnet.net*Subject*: [mg71965] radical equation*From*: "dimitris" <dimmechan at yahoo.com>*Date*: Wed, 6 Dec 2006 06:03:52 -0500 (EST)

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

**Follow-Ups**:**Re: radical equation***From:*Andrzej Kozlowski <akoz@mimuw.edu.pl>