How do you solve #x^2-2x+1=0# by completing the square?

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Answer 1 · 2017-09-28T22:13:12

#(x-1)^2=0#

#x^2-2x+1=0#

Let's start by subtracting #1# from both sides so that the #x# terms are isolated:

#x^2-2x=-1#

Now, let's compute #(b/2)^2# and add it to both sides of the equation:

#(b/2)^2=(-2/2)^2=1#

#x^2-2x+1=-1+1#

Let's factorize the left hand side and simplify:

#(x-1)(x-1)=0#

#(x-1)^2=0#