Lesson · Quadratics and Complex Numbers · Parabolas

Quadratic Inequalities

To solve ax² + bx + c > 0, first solve ax² + bx + c = 0. The roots are the only places the sign can change, so testing one value in each interval they cut settles the whole line. The answer is a set, and the endpoints belong to it only when the inequality is not strict.

Open the interactive lesson Practice

get 0 on one side first: ax² + bx + c > 0 solve = 0 for the roots, then test one value per interval a > 0 with roots r < s: positive outside [r, s], negative between strict > and < exclude the roots; ≥ and ≤ include them

The method in four lines

Core ideas

What this lesson establishes

An inequality asks about a sign, and signs are exactly what a factored quadratic reports.

The sign can only change at a root

Between consecutive roots the expression keeps one sign, so a single test value speaks for a whole interval. That is what makes the method finite rather than a search.

Never divide by an expression containing the variable

Dividing x² > 3x by x assumes x is positive — and if x is negative the inequality reverses, while x = 0 is deleted altogether. Move everything to one side and factor instead.

The answer is a set

It is one interval, two intervals, all of the real line, or nothing at all. Deciding which endpoints belong is a separate question from finding them, and it is settled by whether the sign is strict.

The material

Solving a quadratic inequality

Roots first, signs second, endpoints last.

Definition · Quadratic inequality

An inequality that can be written ax² + bx + c > 0 with a ≠ 0, or with <, ≥ or ≤ in place of >. A solution is any real number that makes the statement true, and solving means describing all of them.

Getting 0 on one side is not cosmetic. "Greater than zero" is a question about a sign, and a factored expression reports signs one factor at a time. "Greater than 5" reports nothing the factors can answer: knowing that x - 1 and x + 4 are both positive does not settle whether their product is above 5 or below it. Rewrite (x - 1)(x + 4) > 5 as x² + 3x - 9 > 0 first, and the factors of that expression do answer the question.

Theorem · Sign changes happen only at roots

Suppose ax² + bx + c has real roots r ≤ s, so that it equals a(x - r)(x - s). On each of x < r, r < x < s and x > s the expression keeps a single sign, so one test value chosen from an interval determines the sign on all of it.

The reason is the factored form. (x - r) is negative to the left of r and positive to the right, and likewise for (x - s); neither factor changes sign anywhere else. A product changes sign only when one of its factors does, so the only candidates are r and s.

Method 1: a sign table

x < -2 x = -2 -2 < x < 3 x = 3 x > 3 x + 2 - 0 + + + x - 3 - - - 0 + ------------------------------------------------------------------ (x + 2)(x - 3) + 0 - 0 + so (x + 2)(x - 3) ≤ 0 on -2 ≤ x ≤ 3, which is [-2, 3] and (x + 2)(x - 3) > 0 on x < -2 or x > 3

Method 2: read it off the parabola

The graph says the same thing in one picture. y > 0 is where the curve is above the x-axis. An upward parabola with two roots dips below the axis exactly between them, so > 0 gives the two outer intervals and < 0 gives the middle one. A downward parabola swaps those two answers, which is why the sign of a has to be checked before anything is written down.

Example · 2x² + 5x > 3

Move everything to one side: 2x² + 5x - 3 > 0. Factor: (2x - 1)(x + 3) > 0, so the roots are 1/2 and -3. a = 2 > 0, so the expression is positive outside the roots: x < -3 or x > 1/2.

Test x = 0, which lies between the roots: 2(0)² + 5(0) - 3 = -3, and -3 > 0 is false. The middle interval is correctly excluded.

Example · x² - 4x + 1 < 0

This does not factor over the integers, and it does not need to. D = 16 - 4 = 12, so the roots are (4 ± √12)/2 = 2 ± √3. a = 1 > 0, so the expression is negative between them: 2 - √3 < x < 2 + √3.

Check with the easiest interior value, x = 2: 4 - 8 + 1 = -3, which is indeed negative. The roots are all the method ever wanted; how they were found is not part of the method.

When there are fewer than two roots

D < 0. The expression never reaches 0, so it keeps one sign everywhere — the sign of a. x² + x + 1 > 0 is true for every real x; x² + x + 1 < 0 has no solutions at all. Both answers are complete answers.
D = 0. There is one root and the curve touches the axis without crossing, so the sign is the same on both sides. (x - 3)² > 0 holds for every x except 3; (x - 3)² ≥ 0 holds for every x; (x - 3)² ≤ 0 holds only at x = 3; (x - 3)² < 0 never holds.
Endpoints. The roots make the expression 0. With ≥ or ≤ they belong to the solution set, and with > or < they do not. Nothing else about the answer changes.

The one manipulation to refuse is division by an expression in x. To solve x² > 3x, do not cancel an x: that would assume x > 0, and it would lose x = 0 besides. Instead x² - 3x > 0, so x(x - 3) > 0, giving x < 0 or x > 3. Cancelling would have returned x > 3 alone, and x = -1 — where 1 > -3 is true — shows what that answer is missing.

Interactive

Quadratic inequalities

Sign analysis on the factored form

Method

Solving ax² + bx + c > 0

Four steps. The first one is the one that gets skipped when the inequality already looks tidy.

Move everything to one side

The comparison must be with 0. Do it by adding and subtracting only — never by dividing by something containing x, whose sign you do not know.

Solve the corresponding equation

Factor it, or use the formula, or note that D < 0 and there are no roots at all. These roots are the only places the sign can change.

Test one value in each interval

Mark the roots on a line and pick a convenient number from each piece — 0 whenever it is available. The sign of the expression there is its sign on the whole piece.

Write the set, then settle the endpoints

Use interval notation from “Interval and Set-Builder Notation” in Linear Equations and Inequalities. A square bracket for a root that satisfies the inequality, a round one for a root that does not, and "or" between two separate intervals.

Solving inequalities

Score 0 / 0

Choose the answer you can justify.

x² - x - 6 ≤ 0

Zero is already on the right. The roots come first and the inequality sign is dealt with last, which keeps the two decisions apart.

x² - x - 6 = (x - 3)(x + 2) roots x = 3 and x = -2 three intervals, one test value in each: x = -3 (-6)(-1) = 6 > 0 NOT in the solution set x = 0 (-3)( 2) = -6 < 0 in the solution set x = 4 ( 1)( 6) = 6 > 0 NOT in the solution set so the expression is negative exactly on -2 < x < 3 endpoints: the inequality is ≤ , and both roots give 0, and 0 ≤ 0 is true, so both are included solution set: -2 ≤ x ≤ 3 which is [-2, 3]

The three test values are the evidence, not decoration: they are what establishes the sign on each interval. One value per interval is enough because the expression cannot change sign without passing through a root, and both roots are already accounted for.

Had the inequality been < 0 rather than ≤ 0, the same three lines of work would give (-2, 3). The roots are found once; the brackets are chosen afterwards.

For a faded sign pass, solve 2x² - x - 6 > 0. The supplied factorisation is (2x + 3)(x - 2) and the critical values are -3/2 and 2. Test one number in each interval, write the result as a union and decide the endpoints only after the signs are known. Compare with (-∞, -3/2) ∪ (2, ∞); square brackets would include points where the expression is 0, not greater than 0.

Common mistakes

Dividing both sides by x, or by a bracketThe sign of the divisor is unknown, so you cannot know whether to reverse the inequality, and if the divisor can be 0 you have deleted a value as well. x² > 3x gives x < 0 or x > 3; cancelling the x returns half of that.

Writing a two-interval answer as one chain3 < x < -2 says x is above 3 and below -2 at the same time, which nothing is: the chain notation always means an intersection. Two separate intervals need the word "or".

Assuming the sign alternates across every rootIt alternates at a simple root and does not at a repeated one. (x - 2)² is positive on both sides of 2, so (x - 2)² > 0 has solution "every x except 2" — a set with a hole in it, which no amount of alternating would produce.

Check your understanding

✓

Finish when the answer is a set and you can say why each endpoint is in or out.

Solve 2x² - 5x - 3 ≥ 0 and write the result in interval notation, then say what changes if the ≥ becomes >. If the endpoint decision needs re-solving rather than one look at the sign, the two halves of the method have run together.

Note: Sign analysis is the same tool Rational and Radical Expressions used on rational expressions, with one addition there: the sign can also change where a denominator is zero, and that value is excluded rather than included. Here the expression is defined everywhere, so the roots are the only critical values.