Question 59405
{{{x^2-9<0}}}
solve the inequality.
Find the numbers that make the equation equal 0.
{{{(x+3)(x-3)=0}}}
x+3=0  and x-3=0
x+3-3=0-3  and x-3+3=0+3
x=-3  and x=3
Now you test the intervals (-infinity,-3), (-3,3), (3,infinity)
By plugging in a test number of your choice in each number, if the result is less than 0 (negative) the interval is part of the solution.
For (-infinity,-3) test -4
{{{(-4)^2-9}}}
{{{16-9}}}
{{{7}}} 7 is greater tahn 0 (positive), so the interval is not included.
For (-3,3) test 0
{{{(0)^2-9}}}
{{{0-9}}}
{{{-9}}}  -9 is less than 0 (negative), so (-3,3) is part of the solution.
For (3,infinity) test 4
{{{(4)^2-9}}}
{{{16-9}}}
{{{7}}} is greater than 0 (positive), so (3,infinity) is not part of the solution.
The soltuion is (-3,3) in interval notation
{{{-3<x<3}}} is also a notation commonly used.
:
Notice that if the inequality was {{{x^2-9>0}}}, the soltuion would have been (-infinity,-3)U(3,infinity).  Think about it and fun with it.
Happy Calculating!!!