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How do you solve #X^2>16#?

1 Answer

Aug 8, 2015

#x > 4# or #x < -4#

Explanation:

If #x >=0# and #x^2 > 16#
then #x > sqrt(16) = 4#

If #x > 0# and #x^2 > 16#
then # -x > sqrt(16) = 4#
and (multiplying by #(-1)# which reverses the inequality sign)
#color(white)("XXXX")##x < -4#

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