How do you solve the inequality #2<=abs(x^2-9)<9# and write your answer in interval notation?

1 Answer
Jun 25, 2017

Answer:

#{x | -3sqrt2 < x <= 11, -sqrt7 <= x < 0, 0 < x <= sqrt7, sqrt11 <= x < 3sqrt2}#

Wolfram Alpha

Explanation:

We can split this into two separate equations:

#2 <= abs(x^2-9) < 9#

#2 <= x^2 - 9 < 9 " " or " " -9 < x^2-9 <= -2#

Solving both of these separately will then give us our solution set.

#2 <= x^2 - 9 < 9#

#11 <= x^2 < 18#

#sqrt11 <= x < 3sqrt2 " " or " " -3sqrt2 < x <= -sqrt11#


#-9 < x^2 - 9 <= -2#

#0 < x^2 <= 7#

#0 < x <= sqrt7 " " or " " -sqrt7 <= x < 0#

Therefore, our set of numbers that #x# could be is:

#{x | -3sqrt2 < x <= 11, -sqrt7 <= x < 0, 0 < x <= sqrt7, sqrt11 <= x < 3sqrt2}#

On the number line, the solution set looks like this:

Wolfram Alpha

Final Answer