A quadratic inequality might look like this:

$$ax^2 + bx + c \ge 0$$

Or with a different symbol: less than, greater than, or less/greater than or equal to. 

The symbol changes which solutions you keep, but the first step is the same: find where $ax^2+bx+c$ is positive, negative, or zero, then keep the parts that match the symbol.

Graphically, we’re asking: When is the parabola above the x-axis? Or below it? 

Get Everything on One Side

Before you do anything else, you want to rewrite the inequality so one side is zero. That way, you’re working with a single expression:

$$ax^2 + bx + c\ [>,<,\ge,\le]\ 0$$

Example:

$$x^2 < 3x + 2 \Rightarrow x^2 – 3x – 2 < 0$$

Now you’re ready to figure out when this expression is less than zero.

Solve the Related Equation First

Ignore the inequality sign for a moment and solve the related equation:

$$ax^2 + bx + c = 0$$

These solutions, if they exist, are the points where the parabola touches or crosses the x-axis. Think of them as boundaries. They divide the number line into regions where the expression either stays positive or stays negative.

How many of these boundary points you get depends on the discriminant. If it’s positive, you get two real roots. If it’s zero, you get one root. If it’s negative, there are no real roots. For more details on solving regular quadratic equations, check out my post on it!

Use the Roots to Break Up the Number Line

If you have real roots, they split the number line into intervals. Let’s say your roots are r₁ and r₂, with r₁ < r₂. Then the number line splits into three parts: to the left of r₁, between r₁ and r₂, and to the right of r₂.

If there’s only one root, you have two intervals, one on each side of that point.

No real roots means the entire number line stays either positive or negative the whole time. There’s no switching of sign because the graph never crosses zero.

Test Each Region

Pick a number from each interval, any number, and plug it into the original expression:

$$f(x)=a{x}^2+bx+c$$

What matters is whether the result is positive or negative. If one value in the interval makes the expression negative, then the whole interval is negative. Same for positive. That’s because quadratic functions are continuous. They don’t suddenly jump from one sign to another without passing through zero.

Once you know which region has the sign you want, you can answer the inequality.

Putting It All Together

Let’s work through this inequality:

x² − x − 2 < 0

First, solve the related equation:

x² − x − 2 = 0

You can factor this one:

(x − 2)(x + 1) = 0

So the roots are:

x = −1 and x = 2

Now use these roots to split the number line into three regions:
x < −1, −1 < x < 2, and x > 2.

Pick one number from each region and test the sign of the expression x² − x − 2.

Try x = −2:
f(−2) = 4 + 2 − 2 = 4 (positive)

Try x = 0:
f(0) = −2 (negative)

Try x = 3:
f(3) = 9 − 3 − 2 = 4 (positive)

The inequality asks where the expression is less than zero, so we choose the interval where the expression is negative:

(−1, 2)

Since the inequality is strict (<), we don’t include the endpoints.

Examples of Other Tricky Cases

Let’s look at how this works when you don’t have two roots.

One Real Root

Solve:

x² + 4x + 4 ≥ 0

First, factor:

(x + 2)² ≥ 0

This gives one solution: x = −2

Because it’s a perfect square, the graph only touches the x-axis at x = −2 and stays above it everywhere else.

Try x = −3:
f(−3) = 1 (positive)

Try x = 0:
f(0) = 4 (positive)

The expression is always positive, or zero at x = −2. So the inequality f(x) ≥ 0 is true for all real numbers.

(−∞, ∞)

No Real Roots

Solve:

x² + 4x + 5 < 0

Try solving the related equation:

x² + 4x + 5 = 0

Use the discriminant:

D = 4² − 4(1)(5) = 16 − 20 = −4

No real roots, so the parabola never touches the x-axis.

Since the leading coefficient a = 1 > 0, the parabola opens upward and stays above the axis the entire time. That means the expression is always positive.

But our inequality asks when it’s less than zero, which never happens.

No solution.

Writing the Final Answer Clearly

Once you know which intervals work, here’s how to express the solution.

Interval Notation

This uses parentheses and brackets to show where the inequality is true. Use round brackets for strict inequalities like < or >. Use square brackets if the inequality includes equality like ≤ or ≥.

Example:
If the solution is everything between x = −1 and x = 2, and we include both endpoints:

If we exclude both:

(−1, 2)

If it’s two regions, combine them with a union:

(−∞, −3) ∪ (2, ∞)

Set-Builder Notation

This uses a condition to describe the solution. For example:

{ x ∣ x < −3 or x > 2 }

This reads as “the set of all x such that x is less than −3 or greater than 2.”

Graphical Representation

Draw a number line. Mark the boundary points (your roots) and shade the regions where the inequality holds.

Use open circles for strict inequalities and filled circles for inclusive ones. Then shade the correct parts of the line.

This makes it easy to see the solution, especially when comparing intervals.