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MathGroup Archive 2006

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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

> 



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