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MathGroup Archive 2006

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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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