x² ≧ 25

Get 0 on the right side:

          x² - 25 ≧ 0

Get the critical numberss by solving the
equation gotten by changing the inequality
symbol to 0.

          x² - 25 = 0

   (x - 5)(x + 5) = 0

x - 5 = 0;  x + 5 = 0

    x = 5;      x = -5

Place the critical values on a number line:

--------o---------o---------
       -5         5

Pick a test value to the left of -5, say -6
Substitute -6 into the inequality

   x² - 25 ≧ 0
(-6)² - 25 ≧ 0
   36 - 25 ≧ 0
        11 ≧ 0

That is true so we shade the region of
the number line to the left of -5

<=======o---------o---------
       -5         5

Now pick a test value between -5 and 5, say 0.
Substitute 0 into the inequality
  x² - 25 ≧ 0
(0)² - 25 ≧ 0
   0 - 25 ≧ 0
      -25 ≧ 0

That is false so we DO NOT shade the region
of the number line between -5 and 5. So we
still have:

<=======o---------o---------
       -5         5

Pick a test value to the right of 5, say 6
Substitute 6 into the inequality
  x² - 25 ≧ 0
(6)² - 25 ≧ 0
  36 - 25 ≧ 0
       11 ≧ 0

That is true so we shade the region of
the number line to the right of 5

<=======o---------o========>
       -5         5

Now we test the critical points themselves:

Testing -5:

   x² - 25 ≧ 0
(-5)² - 25 ≧ 0
   25 - 25 ≧ 0
         0 ≧ 0

That's true, so we darken the circle at -5

<=======⚫--------o========>
       -5         5

Testing 5:

  x² - 25 ≧ 0
(5)² - 25 ≧ 0
  25 - 25 ≧ 0
        0 ≧ 0

That's true, so we darken the circle at 5

<=======⚫--------⚫========>
       -5         5
The interval notation is an abbreviation
of that graph.

     (-∞,-5] U [5,∞)  

Edwin