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
 A: I suppose you know this relation: $(a+b)^2=a^2+2ab+b^2$. In the step that you don't understand exactly this relation is used with $a:=1$ and $b:= \sqrt{1-x}$.
A: $$2x-5 = (1 + \sqrt{x-1})^2$$
to expand RHS use this formula or simple mulipty it with itself(to do square).
formula is:
$(a+b)^2=a^2+b^2+2\times a\times b$
so your expansion will be
$$2x-5 = (1^2 + (\sqrt{x-1})^2+2\times1\times \sqrt{x-1})$$
$$2x-5 = (1 + {x-1}+2\times \sqrt{x-1})$$
$$2x-5 = x+2\sqrt{x-1}$$
$$x-5 = 2\sqrt{x-1}$$
now you have your way. 
A: $\sqrt{2x-5} - \sqrt{x-1} = 1$
Let $\sqrt{2x-5} + \sqrt{x-1} = y$
Multiplying, we get
$(2x-5) - (x-1) = y$
$y = x - 4$
\begin{align}
   \sqrt{2x-5} + \sqrt{x-1} &= x - 4 \\
   \sqrt{2x-5} - \sqrt{x-1} &= 1 & \text{subtract}\\
\hline
   2\sqrt{x-1} &= x-5 \\
   4x-4 &= x^2 - 10x + 25 \\
   x^2 -14x + 29 &= 0 \\
   x &= 7 \pm 2 \sqrt 5
\end{align}
A: You can solve the square of a sum by writing out the square as a product of two sums and writing out the multiplication for each pair of terms.
$$(1+\sqrt{x-1})^2 =\\
(1+\sqrt{x-1})(1+\sqrt{x-1})=\\
1\times1 + 1\times\sqrt{x-1}+\sqrt{x-1}\times1+\sqrt{x-1}\times\sqrt{x-1}=\\
1+\sqrt{x-1}+\sqrt{x-1}+x-1=\\
2\sqrt{x-1}+x$$
A: 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 :)
A: 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 $x$:
$$x=t^2+1=(1+\sqrt5)^2+1=7+2\sqrt5.$$
