Solve $\sqrt{2x-5} - \sqrt{x-1} = 1$ Although this is a simple question I for the life of me can not figure out why one would get a 2 in front of the second square root when expanding. Can someone please explain that to me? 
Example: solve $\sqrt{(2x-5)} - \sqrt{(x-1)} = 1$
Isolate one of the square roots: $\sqrt{(2x-5)} = 1 + \sqrt{(x-1)}$
Square both sides: $2x-5 = (1 + \sqrt{(x-1)})^{2}$
We have removed one square root.
Expand right hand side: $2x-5 = 1 + 2\sqrt{(x-1)} + (x-1)$-- I don't understand?
Simplify: $2x-5 = 2\sqrt{(x-1)} + x$
Simplify more: $x-5 = 2\sqrt{(x-1)}$
Now do the "square root" thing again:
Isolate the square root: $\sqrt{(x-1)} = \frac{(x-5)}{2}$
Square both sides: $x-1 = (\frac{(x-5)}{2})^{2}$
Square root removed
Thank you in advance for your help

To get rid of the square root, denote: $\sqrt{x-1}=t\Rightarrow x=t^2+1$. Then:
$$\sqrt{2x-5} - \sqrt{x-1} = 1 \Rightarrow \\
\sqrt{2t^2-3}=t+1\Rightarrow \\
2t^2-3=t^2+2t+1\Rightarrow \\
t^2-2t-4=0 \Rightarrow \\
t_1=1-\sqrt{5} \text{ (ignored, because $t>0$)},t_2=1+\sqrt{5}.$$
Now we can return to $x$:
$$x=t^2+1=(1+\sqrt5)^2+1=7+2\sqrt5.$$
