The Law of Trichotomy Applied to Quadratics in One Variable

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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 x2 + x – 6 < 0.
Discussion and Solution: When considering the inequality x2 + x – 6 < 0 the Law of Trichotomy dictates that we be aware of the equality x2 + x – 6 = 0 as well as the inequality x2 + x – 6 > 0. As in the previous discussion, the equation x2 + 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 x2 + x – 6 = 0 to be {2, –3}. We can now sketch the graph of the equation x2 + x – 6 = 0 as shown in Fig. 4.

The graph of the equation x2 + 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 x2 + x – 6 < 0 or x2 + x – 6 > 0.
  • The blue ray (–∞, –3) is part of the solution set for x2 + x – 6 < 0 or x2 + x – 6 > 0.
  • The red ray (2, ∞) is part of the solution set for x2 + x – 6 < 0 or x2 + 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 x2 + x – 6 < 0. When 0 is substituted into x2 + x – 6 < 0 we obtain – 6 < 0; a true statement. Therefore, the interval (–3, 2) is part of the solution set for x2 + x – 6 < 0.
Test – 4 from the ray (–∞, –3) in the inequality x2 + x – 6 < 0. When – 4 is substituted into x2 + 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 x2 + 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 x2 + x – 6 > 0. Therefore, the ray (–∞, –3) is part of the solution set for x2 + x – 6 > 0.
Test 3 from the ray (2, ∞) in the inequality x2 + x – 6 > 0. (observe that I switched inequalities). When 3 is substituted into x2 + x – 6 > 0 we obtain 9 + 3 – 6 > a true statement. Consequently, the ray (2, ∞) is part of the solution set for x2 + x – 6 > 0.
A summary of the test results shows that the solution set for the inequality x2 + x – 6 > 0 is
(–∞, –3) ∪ (2, ∞) and the solution set for x2 + 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 ∅.

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