Solving 11.11 Quadratic Inequalities- Methods and Examples
What Quadratic Inequalities Actually Are
A quadratic inequality is an equation where a variable is squared and the result is compared using inequality signs like <, >, ≤, or ≥ instead of an equal sign. It looks something like x² + 5x + 6 > 0. You're not finding a single answer—you're finding the range of values that make the statement true.
Most students struggle here because they expect one clean answer. Quadratic inequalities don't work that way. The solution is almost always an interval or a union of intervals on the number line.
The Core Method: Find the Zeros, Test the Regions
This is the most reliable approach. It works every time, no exceptions.
Step 1: Replace the Inequality with an Equality
Turn > or < into =. So x² + 5x + 6 > 0 becomes x² + 5x + 6 = 0. Solve this quadratic equation using factoring, the quadratic formula, or completing the square.
Step 2: Find the Roots
Solving x² + 5x + 6 = 0 gives you x = -2 and x = -3. These are your boundary points. They divide the number line into three regions: left of -3, between -3 and -2, and right of -2.
Step 3: Test Each Region
Pick any value from each region and plug it into the original inequality. If the test works, that entire region is part of your solution.
- Test x = -4 (left region): (-4)² + 5(-4) + 6 = 16 - 20 + 6 = 2 > 0 ✓
- Test x = -2.5 (middle region): (-2.5)² + 5(-2.5) + 6 = 6.25 - 12.5 + 6 = -0.25 < 0 ✗
- Test x = 0 (right region): 0² + 5(0) + 6 = 6 > 0 ✓
Step 4: Check the Inequality Sign
For > or <, the boundary points are not included. For ≥ or ≤, they are included. Our example uses >, so the solution is (-∞, -3) ∪ (-2, ∞).
The Graphical Method: See It to Believe It
Graph the quadratic function y = ax² + bx + c. The inequality tells you which parts of the graph you're interested in.
- y > ax² + bx + c: Look at where the graph is above the x-axis
- y < ax² + bx + c: Look at where the graph is below the x-axis
- y ≥ or ≤: Include the points where the graph touches the x-axis
The x-intercepts are still your boundaries. This method is faster when you need a visual sense of the solution, but the test-point method is more precise for written work.
Comparing Solution Methods
| Method | Best For | Speed | Accuracy |
|---|---|---|---|
| Test Point / Boundary | All problems, especially exams | Medium | High |
| Graphical | Visual learners, quick checks | Fast | Medium |
| Sign Chart / Interval | Multiple inequalities, advanced problems | Fast | High |
| Quadratic Formula | Non-factorable quadratics | Slow | High |
Examples Worked Out
Example 1: x² - 4x - 5 ≤ 0
Factor: (x - 5)(x + 1) ≤ 0
Roots: x = 5, x = -1
Since ≤ includes boundaries, test both regions.
Solution falls between the roots: [-1, 5]
Example 2: 2x² + 3x - 9 > 0
Factor: (2x - 3)(x + 3) > 0
Roots: x = 1.5, x = -3
Test left of -3: works ✓
Test between -3 and 1.5: fails ✗
Test right of 1.5: works ✓
Solution: (-∞, -3) ∪ (1.5, ∞)
Example 3: x² + 4x + 4 ≥ 0
Factor: (x + 2)² ≥ 0
Root: x = -2 (double root)
Since the square of any real number is ≥ 0, this is always true. Solution: all real numbers (-∞, ∞).
Common Mistakes That Blow the Answer
- Dividing by negative numbers: If you multiply or divide both sides by a negative, the inequality sign flips. People forget this constantly.
- Including boundaries incorrectly: Check your sign. > means open interval, ≥ means closed.
- Forgetting to test all regions: Three regions, three tests. Don't skip any.
- Solving instead of testing: Finding x-values where the quadratic equals zero is step one. You still need to test the regions.
Getting Started: Your Quick Checklist
- Move everything to one side so the inequality reads "expression > 0" or "expression < 0"
- Set the expression equal to zero and solve for the roots
- Plot those roots on a number line
- Test one point in each created region
- Write the solution using interval notation
- Double-check: does your answer include or exclude the boundaries?
When the Quadratic Doesn't Factor
Not all quadratics factor nicely. When you hit something like x² + 4x + 7 < 0, reach for the quadratic formula: x = (-b ± √(b² - 4ac)) / 2a.
Calculate the discriminant (b² - 4ac) first. If it's negative, the quadratic never touches the x-axis. That means it's either always positive or always negative—test a single point to find out which.