The solution to this is ugly, right?

[Jack Brennen]

Note that the quartic can be reduced from using two parameters
"a" and "b" to using a single parameter "c":

Substitute:

   c = (a-b)/2
   u = (b-x)+c

Solve:

     (u+c)^(-1/2) + (u-c)^(-1/2) = 1 for u.


In words, you are looking for two numbers with difference 2*c,
such that the harmonic mean of their square roots is 2.  The
solution u is the arithmetic mean of the two numbers.

For instance, if c=27/8, the two numbers are 9/4 and 9; the
harmonic mean of their square roots is 2, so the solution u
is the arithmetic mean of 9/4 and 9; u = 45/8.

The solution for u is still a quartic depending on c, but
it will at least appear a bit less ugly.



Mitch Harris wrote:
> On Fri, May 30, 2008 at 12:04 PM, David W. Wilson <wilson.d at anseri.com> wrote:
>> Solve (a-x)^(-1/2) + (b-x)^(-1/2) = 1 for x.
>>
>> The solution is an ugly quartic?
> 
> Mma helped me reduce this to:
> 
> a^2 + 6 a b - 2 a^2 b + b^2 - 2 a b^2 +  a^2 b^2 +
> (-8 a + 2 a^2 - 8 b + 8 a b - 2 a^2 b + 2 b^2 - 2 a b^2) x +
> (8 - 6 a + a^2 - 6 b + 4 a b + b^2) x^2 +
> (4 - 2 a - 2 b) x^3 +
> x^4
> = 0
> 
> I'd have to say yes, ugly and quartic.
> 
> Mitch
> 
>