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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
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