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

It is an identity:
$$(1+x)^2=1+2x+x^2$$
How? Well, consider this:
$$\begin{equation*}
\begin{split}
(1+x)^2&=(\color{blue}{1}+\color{red}{x})(1+x)\\
&=\color{blue}{1}(1+x)+\color{red}{x}(1+x)\\
&=1+x+x+x^2\\
&=1+2x+x^2\\
\end{split}
\end{equation*}$$
In general, $(a+b)^2=a^2+2ab+b^2$, which you can try to deduce yourself.
Hope this helps. Ask anything if not clear :)
