Result from solve() command shows c_0

[parisse]

c_0 is any real (solve algorithm introduces auxiliary unknowns to solve equations with sqrt, c_0 is one of them)
Here it would means that any real x that can be written as c_0^2+1 is solution, i.e. x>=1 is solution since c_0^2+1>=1 if c_0 is real.
This is incorrect for all c, but valid for c in [2,3], i.e. x in [5,10]
Try this

Code:

f:=sqrt(x+3-4*sqrt(x-1))+sqrt(x+8-6*sqrt(x-1))

then

Code:


assume(c>0)
simplify(f(x=c^2+1))

You can confirm this with plot(f,x=4..11)