How do I solve a quadratic inequality?

In three steps. Move everything to one side so that zero is left on the other. Find the zeros of the quadratic (the discriminant, or a special product). Sketch the parabola — opening upwards for a > 0, downwards for a < 0 — and read off which intervals it lies on the required side of the axis. Write the answer as an interval.

Are quadratic inequalities part of the core syllabus?

Yes. Quadratic inequalities belong to the core level. It is the higher-degree polynomial inequalities and the rational ones — the second half of this lesson — that belong to the extended level.

What if the discriminant is negative?

The quadratic has no zeros, so its sign is the sign of a everywhere. The inequality x² + x + 1 > 0 is therefore satisfied by every real number, while x² + x + 1 < 0 has no solutions at all and its solution set is the empty set.

Why does a double root not change the sign?

Because the factor (x − a)² is a square and is never negative. The graph touches the axis there and bounces off instead of crossing it. On a sign chart that means the same sign stands on both sides of a root of even multiplicity — and that is the only thing multiplicity changes in this method.

Why is the end at a zero of the denominator always open?

Because the expression has no value there — it is neither positive, nor negative, nor zero. The strictness of the inequality only decides the ends that come from the numerator. A zero of the denominator falls outside the domain and is excluded however the inequality is written.

Advanced level

Quadratic and rational inequalities

Q: Are quadratic inequalities part of the core syllabus?

Yes. Quadratic inequalities belong to the core level. It is the higher-degree polynomial inequalities and the rational ones — the second half of this lesson — that belong to the extended level.

Q: What if the discriminant is negative?

The quadratic has no zeros, so its sign is the sign of a everywhere. The inequality x² + x + 1 > 0 is therefore satisfied by every real number, while x² + x + 1 < 0 has no solutions at all and its solution set is the empty set.

Q: Why does a double root not change the sign?

Because the factor (x − a)² is a square and is never negative. The graph touches the axis there and bounces off instead of crossing it. On a sign chart that means the same sign stands on both sides of a root of even multiplicity — and that is the only thing multiplicity changes in this method.

Q: Why is the end at a zero of the denominator always open?

Because the expression has no value there — it is neither positive, nor negative, nor zero. The strictness of the inequality only decides the ends that come from the numerator. A zero of the denominator falls outside the domain and is excluded however the inequality is written.

Solving an inequality means finding a whole set rather than a single number — and it always takes two steps: the zeros first, then the sign between them. The parabola and the sign of a quadratic, the three discriminant cases, the sign chart, and polynomial and rational inequalities with the trap of a zero denominator.

Before you start

This topic builds on earlier ideas. Before you start, it's worth working through the lessons below — they'll make everything click:

Where this is used

Real situations where you count exactly the way this lesson teaches:

A thrown ball and time above a height
A ball thrown straight up at 20 m/s is at height h(t) = 20t − 5t². Asking how long it stays above 15 m is the inequality 20t − 5t² ≥ 15, that is t² − 4t + 3 ≤ 0, whose solution is [1, 3]. The ball spends exactly 2 seconds above that height.
Fencing and the largest enclosure
With 40 m of fencing for a rectangular run the sides satisfy x + y = 20. The condition "an area of at least 96 m²" is x(20 − x) ≥ 96, so x² − 20x + 96 ≤ 0 and x ∈ [8, 12]. A side shorter than 8 m or longer than 12 m no longer reaches the required area, even though it uses the same fencing.
Pharmacy and the therapeutic window
The concentration of a drug in the blood is the rational expression C(t) = 100t / (t² + 4) mg/l. The condition C(t) ≥ 20 reduces to 20t² − 100t + 80 ≤ 0, that is t² − 5t + 4 ≤ 0, giving t ∈ [1, 4]. The drug is effective for 3 hours, and that is the interval after which the next dose is due.
Manufacturing and the break-even run
With a fixed cost of 5,000 and a variable cost of 12 per unit, the unit cost is k(n) = (5,000 + 12n) / n. The condition "no more than 20 per unit" is the rational inequality (5,000 + 12n) / n ≤ 20, which gives n ≥ 625. Below 625 units the run does not pay off, even though each extra unit costs only 12.

All formulas

Factored form
$ax^2 + bx + c = a(x - x_1)(x - x_2), \quad \Delta > 0$
the starting point of every quadratic inequality
Sign of a quadratic
$\Delta > 0 \implies \text{sign of } a \ \text{outside the roots, } -a \ \text{between them}$
the parabola crosses the axis twice
Zero discriminant
$\Delta = 0 \implies \text{sign of } a \ \text{everywhere except } x_0$
the parabola touches the axis at one point
Negative discriminant
$\Delta < 0 \implies \text{sign of } a \ \text{for every } x$
the parabola never meets the axis
Rational inequality
$\frac{P(x)}{Q(x)} \ge 0 \iff P(x) \cdot Q(x) \ge 0 \ \text{and} \ Q(x) \neq 0$
multiply by the SQUARE of the denominator; the domain survives

An equation has solutions you can list. An inequality usually has infinitely many, so the answer is a set — most often an interval or a union of two. The method is the same every time and has two steps: the zeros first, then the sign of the expression between them.

Quadratic inequalities are core material. The higher-degree polynomial and rational ones that close this lesson are extended material — but the method is identical, there are simply more points on the axis.

The parabola says where a quadratic is negative

The inequality $x^2 - x - 6 < 0$ asks: for which $x$ does the graph lie below the axis. The zeros come out exactly as in a quadratic equation: $\Delta = 1 + 24 = 25$ , so $x_1 = -2$ and $x_2 = 3$ .

Opening upwards, because a = 1 > 0. The parabola lies below the axis exactly between its zeros.

The shaded stretch is the answer: $x \in (-2, 3)$ . The ends are open, because at $-2$ and $3$ the expression equals zero and the inequality is strict.

Had the sign been the other way — $x^2 - x - 6 > 0$ — the answer would be the two pieces outside the roots:

x \in (-\infty, -2) \cup (3, +\infty)

That is the general rule: for $\Delta > 0$ a quadratic has the sign of $a$ outside its roots and the opposite sign between them.

The three discriminant cases

Discriminant	Parabola	Sign of the quadratic
$\Delta > 0$	crosses the axis twice	sign of $a$ outside the roots, opposite between them
$\Delta = 0$	touches the axis once	sign of $a$ everywhere except $x_0$ , zero at $x_0$
$\Delta < 0$	never meets the axis	sign of $a$ for every $x$

The last two rows give answers that surprise on first meeting. The inequality $x^2 - 4x + 4 > 0$ has $\Delta = 0$ and the root $x_0 = 2$ , so it is satisfied by every number except two: $x \in \mathbb{R} \setminus \{2\}$ . And $x^2 + x + 1 < 0$ has $\Delta = -3 < 0$ and opens upwards, so no number satisfies it — its solution set is $\varnothing$ .

Solve −2x² + 8x ≥ 0.

The sign chart

For a polynomial of higher degree, sketching a graph stops being practical. What replaces it is a sign chart: the roots on the axis, and the sign of the expression between them. Start on the right, where the sign is the sign of the leading coefficient, and moving left flip the sign at every root of odd multiplicity.

Multiplicity is the only thing to think about here. The factor $(x - a)^2$ is a square, so it is never negative — the graph touches the axis and bounces off, and the sign does not change.

The sign chart of (x + 2)(x − 1)(x − 4). The two bands above it are the two pieces of one solution set of the inequality > 0.

Reading it off: $(x + 2)(x - 1)(x - 4) > 0$ holds wherever the chart shows a plus, that is

x \in (-2, 1) \cup (4, +\infty)

A union of two intervals is the norm here, not the exception — the more roots, the more pieces.

Solve (x + 2)(x − 1)² ≥ 0.

Rational inequalities

An inequality $\dfrac{P(x)}{Q(x)} \ge 0$ may not be multiplied through by $Q(x)$ : the sign of that denominator is unknown, and multiplying by a negative number reverses the inequality. What you may multiply by is $Q^2(x)$ , which is positive whenever it is not zero:

\frac{P(x)}{Q(x)} \ge 0 \iff P(x) \cdot Q(x) \ge 0 \ \text{and} \ Q(x) \neq 0

The fraction becomes a product and goes back to the sign chart. One difference remains, and it is the heart of this half of the lesson: a zero of the denominator is excluded regardless of the inequality sign. Even in a non-strict inequality that end stays open, because the expression has no value there at all.

The hollow dot at −2 is the zero of the denominator. The end at 1 is filled, because the inequality is not strict and 1 is inside the domain.

The answer: $x \in (-2, 1]$ . The two ends of one interval carry different brackets, and that is not an oversight — one comes from the numerator, the other from the domain.

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0

How many integers: x² − 8x + 12 < 0

Common mistakes

Multiplying the inequality by the denominator — $Q(x)$ may be negative, in which case the inequality has to be reversed. The safe move is to put everything over one fraction and switch to the product $P \cdot Q$ .
Closing the end at a zero of the denominator — in $\dfrac{x - 1}{x + 2} \le 0$ the end at $-2$ stays open, even though the inequality is not strict.
Flipping the sign at a double root — at an even multiplicity the graph bounces off the axis, so the sign is the same on both sides.
Quoting one interval instead of a union — $x^2 > 4$ has the solution $(-\infty, -2) \cup (2, +\infty)$ , not " $x > 2$ ".
Dropping the roots in a non-strict inequality — with $\ge$ and $\le$ the zeros of the numerator belong to the solution set.
Ignoring the sign of $a$ — for $a < 0$ the parabola opens downwards and the whole reading is reversed; finding the roots alone is not enough.

Formula card

Topic: Quadratic and rational inequalities

Factored form
$ax^2 + bx + c = a(x - x_1)(x - x_2), \quad \Delta > 0$
the starting point of every quadratic inequality
Sign of a quadratic
$\Delta > 0 \implies \text{sign of } a \ \text{outside the roots, } -a \ \text{between them}$
the parabola crosses the axis twice
Zero discriminant
$\Delta = 0 \implies \text{sign of } a \ \text{everywhere except } x_0$
the parabola touches the axis at one point
Negative discriminant
$\Delta < 0 \implies \text{sign of } a \ \text{for every } x$
the parabola never meets the axis
Rational inequality
$\frac{P(x)}{Q(x)} \ge 0 \iff P(x) \cdot Q(x) \ge 0 \ \text{and} \ Q(x) \neq 0$
multiply by the SQUARE of the denominator; the domain survives

The parabola y = x² − x − 6. It lies below the axis exactly over the interval (−2, 3) — the solution set of x² − x − 6 < 0.

The inequality (x − 1) / (x + 2) ≤ 0. The end at 1 is closed, while the end at −2 stays open because the denominator vanishes there.

Frequently asked questions

A balance scale in perfect equilibrium rendered as a geometric form, with a straight line crossing an axis like an orbital path, against a cosmic nebula in violet and magenta

Algebra

What x really means — unknown, variable and parameter

The letter x does not always mean the same thing. Sometimes it is a number you are hunting for, sometimes a quantity that varies continuously, sometimes a blank waiting for any number at all. Here is where algebraic notation came from, why it took two thousand years to arrive, what exactly separates an unknown from a variable and a parameter, and why an inequality flips when you multiply by a negative number.

11 min