How do you solve #|x^2 - 4x + 4| =2#?

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Answer 1 · 2016-09-02T22:39:55

#x = 2+-sqrt(2)#

Note that #x^2-4x+4 = (x-2)^2# which will be non-negative for any Real value of #x#

Hence we find #abs(x^2-4x+4) = (x-2)^2# and our equation becomes:

#(x-2)^2 = 2#

Hence:

#x-2 = +-sqrt(2)#

Add #2# to both sides to get:

#x = 2+-sqrt(2)#