solve x^2-2x+1=0 (completing the square)?

[2]
x^2-2x+1=0
(x)^2-2*x*1+(1)^2=0
(x-1)^2=0
x-1=0 [square-rooting both sides]
x=1

x^2 - 2x + 1 = 0
(x - 1)^2 = 0
x - 1 = 0

x - 1 = 0
x = 1

∴ x = 1

(x - 1)(x - 1) = 0
x = 1 (twice)

use the quadratic formula x=(2+/-sqrt(4-4))/2
(2+/-0)/2
x=1 d.r.

(x-1)^2...
ax^2+bx+c
Find factors of a*c that form b.. -1 and -1 forms the -2x in the middle...
x-1 and x-1 or (x-1)^2

(x-1)^2

This square has already been completed

(x-1)^2=0
x=1