radical equation

[dimitris]

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

[Andrzej Kozlowski]

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

[dh]

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