How do you solve #|x ^ { 2} - 32| > 4#?

1 Answer
Bill K.
Sep 17, 2017

The solution set consists of all those real values of #x# for which #x<-6# or #-2sqrt(7) < x < 2sqrt(7)# or #x>6#. This can be written as a union of open intervals: #(-infty,-6) uu (-2sqrt(7),2sqrt(7)) uu (6,infty)#.

Explanation:

Solutions #x# of this inequality will satisfy #x^2-32>4# or #x^2-32<-4#. In the first case, adding 32 to both sides gives #x^2>36# so that #x>6# or #x<-6#. In the second case, adding 32 to both sides gives #x^2<28# so that #-sqrt(28) < x < sqrt(28)#, which is equivalent to #-2sqrt(7) < x < 2sqrt(7)#.

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