Quadratic Inequalities in One Variable

Solve quadratic inequalities in one variable algebraically and graphically. Grade 11.

Lesson 2.8 of Linear Functions & Systems of Equations in Math 20-1 — Alberta curriculum lessons.

What you'll learn in this lesson

• Understanding Quadratic Inequalities. Discover what quadratic inequalities in one variable look like and build visual intuition using the parabola y = x² - 4 to see how above-and-below the x-axis translates into a solution set.
• Solving Graphically — Worked Examples. Apply a four-step graphical procedure to solve quadratic inequalities: move everything to one side, find the zeros, sketch the parabola, then identify the correct region and state the solution.

Inside the lesson — a free preview

What Do They Look Like?

Definition — Quadratic Inequality in One Variable. A quadratic inequality in one variable takes one of these four forms:

ax^2 + bx + c < 0

ax^2 + bx + c > 0

ax^2 + bx + c ≤ 0

ax^2 + bx + c ≥ 0

The solution is the set of all x-values that make the inequality true.

Key Idea — Think Graphically. When you graph y = ax^2 + bx + c, the inequality is asking:

For which x-values is the parabola above or below the x-axis?

The x-intercepts (zeros) of the parabola are the boundary points of the solution. Everything else is just deciding which side of the boundary satisfies the inequality.

Building Intuition with y = x^2 - 4

This parabola has zeros at x = -2 and x = 2 (where it crosses the x-axis) and a vertex at (0, -4).

Now let's solve four different inequalities using this same parabola. For each one, we highlight the part of the parabola that satisfies the inequality and translate that into a number line.

Part a) — Solve x^2 - 4 < 0. This asks: where is the parabola below the x-axis? That's the dip between the two zeros.

Inequality is strict (<), so the boundary points are NOT included — use open dots.

Part a) — Solution. Solution: -2 < x < 2

Part b) — Solve x^2 - 4 ≤ 0. Same as part a), but now we include the boundary points because of the "or equal to" (≤).

The parabola is at or below the x-axis from x = -2 to x = 2, including the zeros.

Part b) — Solution. Solution: -2 ≤ x ≤ 2

Part c) — Solve x^2 - 4 > 0. This asks: where is the parabola above the x-axis? That's the two outer arms of the parabola.

Boundary points are NOT included — use open dots.

Part c) — Solution. Solution: x < -2 or x > 2

Part d) — Solve x^2 - 4 ≥ 0. Same as part c), but now we include the zeros because of ≥.

The parabola is at or above the x-axis everywhere except the dip between x = -2 and x = 2.

Part d) — Solution. Solution: x ≤ -2 or x ≥ 2

Worked example: Solve x^2 + x > 6

Example 1 — x^2 + x > 6. Step 1: Move everything to one side:

x^2 + x > 6

x^2 + x - 6 > 0

Step 2: Find the zeros by factoring x^2 + x - 6 = 0:

(x + 3)(x - 2) = 0

x = -3 or x = 2

Step 3: Sketch the parabola. It opens upward (since a = 1 > 0), with zeros at x = -3 and x = 2.

Step 4: We want where the parabola is above the x-axis (because x^2 + x - 6 > 0). That's the two outer arms.

Solution: x < -3 or x > 2

Create a free Studyio account to take the full lesson with voice-over teaching, practice questions and instant feedback.