Question 364929
{{{sqrt(2k-1) - sqrt(k-1) = 1}}}
Transpose to get {{{sqrt(2k-1) = sqrt(k-1) + 1}}}.
Square both sides: {{{2k-1 = 1+2sqrt(k-1) + k-1}}}.
Simplify:  {{{k-1 = 2sqrt(k-1)}}}.
Square both sides again: {{{k^2-2k+1 = 4(k-1)}}},
{{{k^2-2k+1 = 4k-4)}}},
or {{{k^2 - 6k+5 = 0}}},
then (k-5)(k-1) = 0, or k = 5 or k = 1.
Both satisfy the original equation, so the solution set is {1,5}.