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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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