How do solve #x^2>4# and write the answer as a inequality and interval notation?

2 Answers
Jan 6, 2018

Answer:

#x<-2# or #x>2#
Interval notation:
#(-oo,-2)uu(2,oo)#

Explanation:

To solve, all we need to do is take the square root of both sides to get #x# by itself:
#sqrt(x^2)>sqrt(4)#

#|x|>2#

Which is true either when #x>2# or when #x<-2#

To express the answer in interval notation,we think about which interval satisfies this inequality. It's simply all numbers less than #-2#, and all numbers greater than #2#. This is the interval from #-oo# to #-2# and from #2# to #oo#. Note that the interval does not include #2# or #-2# themselves.

We can write the interval like so:
#(-oo,-2)uu(2,oo)#

The #uu# (#U# for union) symbol means to combine the two intervals.

Jan 9, 2018

Answer:

#x < -2 vv x > 2#

#x in (-oo, -2) uu (2, oo)#

Explanation:

Given:

#x^2 > 4#

Subtract #4# from both sides to get:

#x^2-4 > 0#

Factor the left hand side to find:

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

The left hand side is positive if both of the factors are positive or both of the factors are negative.

Hence:

#x < -2" "# or #" "x > 2#

In interval notation:

#x in (-oo, -2) uu (2, oo)#