I am trying to solve the following equation in Mathematica:
Solve[(xc - x) == 1/(xr - (x + x0)^2) + 1/(xr - (x - x0)^2), x]
However, this yields
{{x -> Root[-2 x0^2 - x0^4 xc + 2 xr + 2 x0^2 xc xr - xc xr^2 + (x0^4 - 2 x0^2 xr + xr^2) #1 + (-2 + 2 x0^2 xc + 2 xc xr) #1^2 + (-2 x0^2 - 2 xr) #1^3 - xc #1^4 + #1^5 &, 1]}, {x -> Root[-2 x0^2 - x0^4 xc + 2 xr + 2 x0^2 xc xr - xc xr^2 + (x0^4 - 2 x0^2 xr + xr^2) #1 + (-2 + 2 x0^2 xc + 2 xc xr) #1^2 + (-2 x0^2 - 2 xr) #1^3 - xc #1^4 + #1^5 &, 2]}, {x -> Root[-2 x0^2 - x0^4 xc + 2 xr + 2 x0^2 xc xr - xc xr^2 + (x0^4 - 2 x0^2 xr + xr^2) #1 + (-2 + 2 x0^2 xc + 2 xc xr) #1^2 + (-2 x0^2 - 2 xr) #1^3 - xc #1^4 + #1^5 &, 3]}, {x -> Root[-2 x0^2 - x0^4 xc + 2 xr + 2 x0^2 xc xr - xc xr^2 + (x0^4 - 2 x0^2 xr + xr^2) #1 + (-2 + 2 x0^2 xc + 2 xc xr) #1^2 + (-2 x0^2 - 2 xr) #1^3 - xc #1^4 + #1^5 &, 4]}, {x -> Root[-2 x0^2 - x0^4 xc + 2 xr + 2 x0^2 xc xr - xc xr^2 + (x0^4 - 2 x0^2 xr + xr^2) #1 + (-2 + 2 x0^2 xc + 2 xc xr) #1^2 + (-2 x0^2 - 2 xr) #1^3 - xc #1^4 + #1^5 &, 5]}}
If I remove the second fraction to the right of ==, I get a regular three solutions for x. Is there a reason why the same thing doesn't happen when there are two fractions?
Rootobjects instead (best it can do). – Roman Feb 21 '21 at 20:46Rootobjects are perfectly good explicit formulae, they just don't conform to Bronze Age notions of what an explicit formula is. They are often superior to formulae using radicals, as they tend to be simpler and more stable for purposes of numerical evaluation. – John Doty Feb 21 '21 at 22:23Rootof this sort is a closed form. – John Doty Feb 22 '21 at 18:49