Wolfram|Alpha gives me no solution exists, but I could solve the problem: (x+2(x-1)^0.5)^0.5-(x-2(x-1)^0.5)^0.5=3
https://www.wolframalpha.com/input/?i=%28x%2B2*%28x-1%29%5E0.5%29%5E0.5-%28x-2*%28x-1%29%5E0.5%29%5E0.5%3D3
Radicals are tricky. Equations with radicals, doubly so. I have learnt from your nice example. I hope you have too.
Yep, you're right. I'm so stupid and I feel miserable wasting everybody's time because I wasn't attentive.
I spot careless uses of implications elsewhere in the sheet too. The silent use of the rule Sqrt[a] Sqrt[b] == Sqrt[a*b] is problematic. However, in this case the faulty solution x=13/4 is introduced not in the first step, but in the very last step: x=13/4 does not satisfy the condition pt. x<2.
Sqrt[a] Sqrt[b] == Sqrt[a*b]
x=13/4
pt. x<2
Gianluca nailed it.
Thanks to everyone for your responses.
There is a mistake in your first step. When you take the square of an equation, you cannot write the double implication <==> but only ==>.
<==>
==>
A simple plot reveals that there is no solution for $f(x)=3$:
f[x_] := (x + 2 (x - 1)^0.5)^0.5 - (x - 2 (x - 1)^0.5)^0.5; ReImPlot[f[x], {x, -2, 3}, PlotRange -> All, Frame -> True, GridLines -> Automatic, ImageSize -> Large]
Hi Yes, my solution gave me a false response, but I can't understand this discrepancy. Anyway, have a nice day.
Plug that solution into the equation to see if it actually works.