Solve |x| > 4, when x ϵ R.

|x| > 4


Square


x2 > 16


x2 – 16 > 0


x2 – 42 > 0


(x + 4)(x – 4) > 0


Observe that when x is greater than 4, (x + 4)(x – 4) is positive


And for each root the sign changes hence



We want greater than 0 that is positive part


Hence x should be less than -4 and greater than 4 for (x + 4)(x – 4) to be positive


x less than -4 means x is from negative infinity to -4 and x greater than 4 means x is from 4 to infinity


Hence x (-∞, -4) and x (4, ∞)


Hence the solution set of |x| > 4 is (-∞, -4) U (4, ∞)


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