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:

$x = 2+-sqrt(2)$

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