How do you solve #abs(x - 6)> 2#?

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Answer 1 · 2015-04-06T21:31:45

If #|x - 6|>2#
it implies that #(|x - 6|)^2>2^2# is also true

#=> (x - 6)^2>2^2#

#=> (x - 6)^2 - 2^2>0#

The left hand side is a difference of two squares that behaves like,
#(a - b)(a + b) = a^2 - b^2#

#=> (x - 6 - 2)(x - 6 + 2)>0#

#=> (x - 8)(x - 4)>0#

The answer is #x<4# # uu # #x>8#