My primary recommendation is that we always simultaneously consider an equation and its two inequality siblings as a group and that we graph all three on the same number line.

Example 2: Suppose it is required to solve the inequality x^{2} + x – 6 < 0.

Discussion and Solution: When considering the inequality x^{2} + x – 6 < 0 the Law of Trichotomy dictates that we be aware of the equality x^{2} + x – 6 = 0 as well as the inequality x^{2} + x – 6 > 0. As in the previous discussion, the equation x^{2} + x – 6 = 0 is sometimes called the boundary equation because its graph is the boundary between the graphs of the two inequalities. Our strategy will be to graph the equation and test numbers from the various resulting rays and intervals formed by that graph.

Factoring (Distributive Property) and The Zero Factor Property show the solution set of the equation x^{2} + x – 6 = 0 to be {2, –3}. We can now sketch the graph of the equation x^{2} + x – 6 = 0 as shown in Fig. 4.

The graph of the equation x^{2} + x – 6 = 0 divides the real line into an interval (–3, 2) (green) and two rays (blue (–∞, –3) and red (2, ∞)).

- The interval (–3, 2) is part of the solution set for either x
^{2}+ x – 6 < 0 or x^{2}+ x – 6 > 0. - The blue ray (–∞, –3) is part of the solution set for x
^{2}+ x – 6 < 0 or x^{2}+ x – 6 > 0. - The red ray (2, ∞) is part of the solution set for x
^{2}+ x – 6 < 0 or x^{2}+ x – 6 > 0.

We need to test one number from each of the rays and the interval in either of the inequalities.

**Test 0 from the interval (–3, 2)** in the inequality x^{2} + x – 6 < 0. When 0 is substituted into x^{2} + x – 6 < 0 we obtain – 6 < 0; a true statement. Therefore, the interval (–3, 2) is part of the solution set for x^{2} + x – 6 < 0.

**Test – 4 from the ray (–∞, –3)** in the inequality x^{2} + x – 6 < 0. When – 4 is substituted into x^{2} + x – 6 < 0 we obtain 16 – 4 – 6 < 0; a false statement. Therefore – 4, and consequently no number in the ray (–∞, –3), is a solution of the inequality x^{2} + x – 6 < 0. It now follows from the Law of Trichotomy that every number in the ray (–∞, –3) is a solution of the other inequality x^{2} + x – 6 > 0. Therefore, the ray (–∞, –3) is part of the solution set for x^{2} + x – 6 > 0.

**Test 3 from the ray (2, ∞)** in the inequality x^{2} + x – 6 > 0. (observe that I switched inequalities). When 3 is substituted into x^{2} + x – 6 > 0 we obtain 9 + 3 – 6 > a true statement. Consequently, the ray (2, ∞) is part of the solution set for x^{2} + x – 6 > 0.

A summary of the test results shows that the solution set for the inequality x^{2} + x – 6 > 0 is

(–∞, –3) ∪ (2, ∞) and the solution set for x^{2} + x – 6 > 0 is the interval (–3, 2).

Illustrated in Fig. 5. Graphing the three siblings on the same number line illustrates the relation between the three.

Rule for Quadratics: When working with quadratic equations and inequalities in one variable;

- If the equation has two real solutions, the resulting interval will be the solution set for one of the inequalities and the union of the rays will be the solution set for the other inequality.
- If the equation has one real solution the union of the resulting two rays will be the solution set for one of the inequalities and the solution set for the other inequality is the empty set ∅.
- If the equation has no real solution, the solution set for one of the inequalities is the set of real numbers R and the solution set for the other inequality is the empty set ∅.