Solution to the equation $\sqrt{x^2 - 2x + 1} - 5 = 0$ I had this equation on my exam :
$$\sqrt{x^2-2x+1} - 5 = 0$$
My friends said the the solution could be :
$$|x-1| = -5$$
 So the solution is nothing!
But I say the solution is:
$$x^2-2x+1 = 25  $$
so $$x = 6\ |\ x = -4$$  
My Question here is which solution is right, and why.
Thanks in advance.
 A: It is $$\sqrt{x^2-2x+1} - 5 = 0 \\ \Rightarrow \sqrt{(x-1)^2} - 5 = 0 \\ \Rightarrow |x-1|-5=0 \\ \Rightarrow |x-1|=5 \\ \Rightarrow x-1=5 \text{ or } x-1=-5 \\ \Rightarrow x=6 \text{ or } x=-4$$ So, your solution is correct!! 
$$$$ 
If the equation is $\sqrt{x^2-2x+1} + 5 =0$ then we have the following: 
$$\sqrt{x^2-2x+1} + 5 = 0 \\ \Rightarrow \sqrt{(x-1)^2} + 5 = 0 \\ \Rightarrow |x-1|+5=0 \\ \Rightarrow |x-1|=-5$$ 
So, there is no solution. 
In this case the solution of your friend is correct. 
A: Your friends added an extraneous negative sign. Start with

$\sqrt{x^2 - 2x + 1} - 5 = 0$

We add 5 to both sides, giving

$\sqrt{x^2 - 2x + 1} = 5$

Taking the square root gives us

$|x - 1| = 5$

Solving this gives 

$x = 6 | x = -4$

as you said.
A: Your solution is definitely correct:
\begin{align}
&\sqrt{x^2 - 2x +1} - 5 = 0    \\
\Rightarrow \; &x^2 -2x+1 = 25 \\
\Rightarrow \; &(x-6)(x+4) = 0  \\
\Rightarrow \; &x = 6 \text{  or  } x=-4
\end{align}
Your friend's approach is also correct, but he (or she) has a typo. After correcting this, you get:
\begin{align}
&\sqrt{x^2 - 2x +1} - 5 = 0    \\
\Rightarrow \; &|x-1| = 5 \; \text{  (not  }-5)\\
\Rightarrow \; &x = 6 \text{  or  } x=-4
\end{align}
