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MathGroup Archive 2006

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Re: radical equation

  • To: mathgroup at smc.vnet.net
  • Subject: [mg71979] Re: [mg71965] radical equation
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Thu, 7 Dec 2006 06:25:21 -0500 (EST)
  • References: <200612061103.GAA02757@smc.vnet.net>

On 6 Dec 2006, at 20:03, 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
>

The solution seems rather obvious so perhaps I have misunderstood  
your question, but:


Reduce[{req, 0 < x < 1, 0 < c < 1}, {x}, Reals]


0 < c < 1 && x == Root[#1^3 - 8*#1^2 - 16*c*#1 + 24*#1 +
       16*c - 16 & , 1]

gives the single real solution in a very short time. To convince  
yourself that it is correct you can look at the plots:

f[c_] = x /. ToRules[Last[%]];

Plot[f[c], {c, 0, 1}]

Plot[Chop[First[req] /. x -> f[c]], {c, 0, 1}]

Andrzej Kozlowski
Tokyo, Japan



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