Quadratic Inequalities

You know how to solve x^2 - 5x + 6 = 0. The roots are x = 2 and x = 3. But suppose the question is not "where does x^2 - 5x + 6 equal zero?" but instead:

Where is this expression positive?

Try a few values. At x = 0: 0 - 0 + 6 = 6 > 0. At x = 2.5: 6.25 - 12.5 + 6 = -0.25 < 0. At x = 5: 25 - 25 + 6 = 6 > 0.

A pattern is forming. The expression is positive at x = 0, negative at x = 2.5, and positive again at x = 5. The sign changes exactly at the roots, x = 2 and x = 3.

The answer to x^2 - 5x + 6 > 0 is not a number — it is a set of intervals: x \in (-\infty, 2) \cup (3, \infty). Every x in that set makes the expression positive; every x outside it makes it zero or negative.

This is a quadratic inequality, and solving one always follows the same three steps: find the roots, determine the sign in each region, and read off the answer. The parabola does the heavy lifting.

The sign-analysis method

Here is the general procedure for solving ax^2 + bx + c > 0 (or < 0, \geq 0, \leq 0).

Step 1. Find the roots. Solve ax^2 + bx + c = 0 to get the roots r_1 and r_2 (with r_1 \leq r_2). Use factoring, the quadratic formula, or completing the square — whatever works.

Step 2. Determine the sign pattern. The roots divide the number line into at most three intervals: (-\infty, r_1), (r_1, r_2), and (r_2, \infty). The sign of f(x) = ax^2 + bx + c is constant on each interval (it can only change at the roots). To find the sign in each interval, you can either:

• Test one value in each interval and compute f at that value, or
• Use the rule: when a > 0, f(x) > 0 outside the roots and f(x) < 0 between them; when a < 0, the reverse.

Step 3. Read off the solution. Select the intervals where f(x) has the sign you want. If the inequality is strict (> or <), exclude the roots. If it is non-strict (\geq or \leq), include them.

That is the entire method. Every quadratic inequality reduces to it.

Why the rule works: the factored form

The rule about the sign of f(x) outside and between the roots is not arbitrary. It comes directly from the factored form.

If f(x) = a(x - r_1)(x - r_2) with r_1 < r_2, then the sign of f(x) is the product of three signs: the sign of a, the sign of (x - r_1), and the sign of (x - r_2).

Region	\text{sign}(x - r_1)	\text{sign}(x - r_2)	Product of two factors	\text{sign}(f) if a > 0
x < r_1	-	-	+	+
r_1 < x < r_2	+	-	-	-
x > r_2	+	+	+	+

When a > 0: positive–negative–positive. When a < 0: negative–positive–negative. The pattern is completely determined by the leading coefficient and the roots.

The wavy curve method

The sign table above is clean for a quadratic, but it gets tedious for products of many factors. The wavy curve method (also called the method of intervals) is a visual shortcut that handles any product of linear factors at once.

Here is how it works. Suppose you need to solve

Step 1. Mark the roots on the number line: 1, 3, 5.

Step 2. Start from the far right. For large positive x, every factor (x - k) is positive, so the product is positive. Mark the rightmost region with a +.

Step 3. Cross each root from right to left. At each root, the sign of exactly one factor changes, so the overall sign flips. Moving left from the + region past x = 5: sign flips to -. Past x = 3: flips to +. Past x = 1: flips to -.

Step 4. Read off the answer. You want the product to be positive, so select the + regions: x \in (1, 3) \cup (5, \infty).

For a quadratic a(x - r_1)(x - r_2) > 0 with a > 0, the wavy curve gives exactly the same answer as the sign table: positive outside the roots, negative between them. The wavy curve just makes it faster and more visual — especially when you move to products of three or more factors.

Repeated roots in the wavy curve

If a factor appears twice — say (x - 2)^2(x - 5) > 0 — the sign does not flip at the repeated root. At x = 2, the factor (x - 2)^2 is always non-negative, so it does not change the sign of the product. The wavy curve touches the axis at x = 2 but bounces back without crossing. The sign only flips at simple (odd-multiplicity) roots.

The rule: the sign flips at roots of odd multiplicity and stays the same at roots of even multiplicity. This is the key refinement of the wavy curve method.

The three discriminant cases for quadratic inequalities

The structure of the solution depends on the discriminant D = b^2 - 4ac.

Case 1: D > 0 (two distinct real roots)

This is the standard case. The roots r_1 < r_2 split the line into three intervals, and the sign alternates as described above.

For a > 0: f(x) > 0 when x \in (-\infty, r_1) \cup (r_2, \infty), and f(x) < 0 when x \in (r_1, r_2).

For a < 0: f(x) > 0 when x \in (r_1, r_2), and f(x) < 0 when x \in (-\infty, r_1) \cup (r_2, \infty).

Case 2: D = 0 (one repeated root)

The parabola touches the axis at exactly one point r = -b/(2a) but does not cross it.

For a > 0: f(x) \geq 0 for all x, with equality only at x = r. The inequality f(x) > 0 has solution x \in \mathbb{R} \setminus \{r\} — all real numbers except r.

For a < 0: f(x) \leq 0 for all x, with equality only at x = r. The inequality f(x) < 0 has solution x \in \mathbb{R} \setminus \{r\}.

Case 3: D < 0 (no real roots)

The parabola never touches the axis. Its sign is the same everywhere — determined entirely by a.

For a > 0: f(x) > 0 for all x \in \mathbb{R}. The inequality f(x) > 0 holds everywhere; f(x) < 0 has no solution.

For a < 0: f(x) < 0 for all x \in \mathbb{R}. The inequality f(x) < 0 holds everywhere; f(x) > 0 has no solution.

The graphical method

Sometimes the cleanest approach is entirely visual. Sketch the parabola, mark the roots, and shade the region you want.

For f(x) > 0: shade where the parabola is above the x-axis. For f(x) < 0: shade where the parabola is below the x-axis.

The sketch does not need to be precise — you only need the rough shape (which way it opens), the roots, and the vertex location. The sign pattern follows from the shape.

This graphical approach is especially useful when the problem gives you a graph and asks you to read the inequality solution from it, or when you need a quick sanity check on an algebraic answer.

An interactive sign explorer

Drag the red point to move along the x-axis. Watch the sign of f(x) = x^2 - 4x + 3 = (x-1)(x-3) change as you cross each root.

The formal definition

Common confusions

• "x^2 - 5x + 6 > 0 means x > 2 and x > 3." No. The solution is x < 2 or x > 3 — two separate intervals, joined by "or," not "and." Solving a quadratic inequality is not the same as solving two separate linear inequalities and combining with "and."

• "I can divide both sides by (x - r)." Only if you know the sign of (x - r). Dividing an inequality by a negative number reverses the inequality. Since (x - r) can be positive or negative depending on x, you would need to split into cases. It is almost always easier to keep everything on one side and use sign analysis.

• "If a > 0 and D < 0, then f(x) > 0 has no solution." The opposite: f(x) > 0 for all real x. The parabola never touches the axis and opens upward, so it is always positive. The inequality is satisfied everywhere.

• "Non-strict and strict give the same answer." They differ only at the roots — strict inequalities use open intervals (parentheses), non-strict use closed intervals (brackets). This matters in applications and in exam answers. x \in (2, 5) and x \in [2, 5] are different sets.

• "The wavy curve only works for quadratics." It works for any product of linear factors, including higher-degree polynomials. If you can factor a polynomial into (x - r_1)(x - r_2) \cdots (x - r_n), the wavy curve gives the sign pattern immediately — alternating signs at simple roots, no sign change at repeated (even-multiplicity) roots.

Going deeper

Rational inequalities

The same sign-analysis method works for inequalities involving rational expressions — fractions with polynomials in the numerator and denominator. To solve

treat the numerator and denominator as separate linear factors. The expression changes sign at x = 1 (numerator zero) and is undefined at x = -3 (denominator zero). Mark both on the number line, then apply the wavy curve: start from the right with +, flip at x = 1, flip again at x = -3. Solution: x \in (-\infty, -3) \cup (1, \infty).

The key difference from polynomial inequalities: the expression is undefined at denominator roots, so those points are always excluded from the solution, even for non-strict inequalities.

Systems of quadratic inequalities

Sometimes you need to solve two quadratic inequalities simultaneously — for example, f(x) > 0 and g(x) \leq 0. Solve each inequality separately to get its solution set, then take the intersection (the overlap). The answer is whatever x-values satisfy both conditions at once.

For instance, x^2 - 4 > 0 gives x \in (-\infty, -2) \cup (2, \infty). The inequality x^2 - 9 \leq 0 gives x \in [-3, 3]. The intersection is x \in [-3, -2) \cup (2, 3] — the parts of [-3, 3] that also satisfy x^2 > 4.

Why "union" and "intersection" matter

A quadratic inequality of the form f(x) > 0 with a > 0 and D > 0 gives a union of two intervals. An inequality of the form f(x) < 0 with a > 0 and D > 0 gives a single interval. The shape of the solution — one piece or two — is determined by which side of zero you are asking about and which way the parabola opens. Getting this right is the difference between a correct and incorrect answer.

Where this leads next

• Location of Roots — the sign of f(k) at a specific point k tells you where the roots are, not just where f is positive or negative.
• Solving Inequalities Using Location of Roots — combining sign analysis with root-location conditions to handle parameter-based inequality problems.
• Quadratic Expression and Function — the parabola as a function, with vertex form, range, and extrema.
• Discriminant and Nature of Roots — the discriminant determines whether the sign pattern has two, one, or zero sign changes.
• Absolute Value Inequalities — another family of inequalities where the solution is an interval, handled by a different but related technique.