The solution to this is ugly, right?

[Mitch Harris]

On Fri, May 30, 2008 at 1:40 PM, Jack Brennen <jb at brennen.net> wrote:
> 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.

namely:
-4 c^2 + c^4 +
4 c^2 u +
(8 - 2 c^2) u^2 +
-4 u^3 +
u^4
==0

>
>
>
> 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
>>
>>
>
>



-- 
Mitch Harris