Method Two: Solve x2 + 2x - 8 = 0, then test chosen values
Solve x2 + 2x - 8 = 0

x2 + 2x - 8 = 0                       Factor

( x + 4 )( x - 2 ) = 0                Solve

x ∈ { -4, 2 }


Graph of the solution to the quadratic EQUATION

Choose test values

Choose "test" values that do not belong to the solution set of x2 + 2x - 8 = 0, but are representative of each "region" determined by the solution set. Substitute the chosen values into the quadratic inequality to see which values yield a TRUE result.


For example, choose the test values -5, 0 and 3.


Choose representatice test values

Substitute the test values into the original quadratic inequality (x2 + 2x - 8 ≤ 0) to see which values yield a true result.

Test each chosen values

Since -5 and 3 result in FALSE statements, the regions they represent are NOT part of the solution set of the inequality.


Since 0 results in a TRUE statement, the region it represents IS part of the solution set of the inequality.


Solution Set
Solution set to the original inequality