The solution to this is ugly, right?

[Robert Israel]

Quartic, yes, but maybe this make it a bit less ugly:
You can write it as x = ((b-r)^2 - r^2)/(1-r)^2
where r satisfies r^4 - 2 r^3 + (b-a) (1-r)^2 = 0.
Which roots produce solutions depend on what branches
of the square roots you use.  If b-a = 2 c^3, one of the
values of r has the series expansion

r = 2-1/4*c^3+1/32*c^6+1/256*c^9-3/2048*c^12-1/4096*c^15
+15/131072*c^18+11/524288*c^21-91/8388608*c^24-35/16777216*c^27
+153/134217728*c^30+969/4294967296*c^33-4389/34359738368*c^36
-1771/68719476736*c^39+...

while the other three are

r = c-1/2*c^2+1/12*c^3+1/24*c^4-1/72*c^5-1/96*c^6+5/1296*c^7
+35/10368*c^8-1/768*c^9-77/62208*c^10+91/186624*c^11+1/2048*c^12
-1309/6718464*c^13-2717/13436928*c^14+1/12288*c^15
+55913/644972544*c^16-67925/1934917632*c^17-5/131072*c^18
+1621477/104485552128*c^19+...

(one for each of the three values of c).

Cheers,
Robert Israel

On Fri, 30 May 2008, 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
>