The quadratic inequality calculator above solves expressions of the form ax² + bx + c compared against zero and returns the answer as an interval, a union of intervals, a single point or the empty set. It finds the roots, works out the sign of the expression on each region the roots cut the number line into, and assembles the regions that satisfy your comparison. The sign chart it shows is not decoration — it is the actual method, and reading it makes the answer self-evident instead of something you have to trust.
Arb Digital publishes this page because quadratic inequalities are where a lot of otherwise confident algebra falls apart. Students who can solve ax² + bx + c = 0 without hesitation freeze when the equals sign becomes a less-than, and the usual response is to memorise two rules — "inside the roots" and "outside the roots" — that are correct only when a is positive and that quietly fail the moment it is not. Working from the sign of the expression rather than from a remembered rule never fails, and it generalises to cubics and rational inequalities without any new ideas.
What This Quadratic Inequality Calculator Does
It handles all four comparisons, both signs of the leading coefficient, and every discriminant case, including the ones with no real roots at all. It reports the discriminant, both roots when they exist, and the vertex of the parabola, then presents the sign chart and the resulting solution set. Where the answer is two separate pieces it returns them as a union; where the answer is a single point, as with a non-strict inequality on a repeated root, it returns that point as a set.
That is a different question from the one our quadratic equation solver answers. That page finds the values of x where the expression equals zero — normally two numbers. This page finds the values of x where the expression is positive or negative — normally a whole range of numbers, and sometimes two ranges. The roots are an ingredient here rather than the output. If you need help reading the answer once you have it, the interval notation calculator converts between interval notation, inequality notation and set-builder form.
How to Use It
- Get everything on one side. An inequality like x² + 3 > 4x becomes x² − 4x + 3 > 0. The calculator assumes the right-hand side is zero.
- Enter a, b and c. Include the sign. A missing x term means b is 0; a missing constant means c is 0.
- Pick the comparison. Choose carefully between strict and non-strict — it changes whether the roots themselves belong to the answer.
- Read the sign chart. Each row is an interval or a boundary point, with the sign of the expression there. The filled rows are the ones in your solution.
- Check the headline interval. Round brackets exclude an endpoint, square brackets include it, and infinity always takes a round bracket.
The Method and How It's Calculated
A quadratic expression can only change sign where it equals zero, because it is continuous. So the roots are the only places anything interesting happens, and between consecutive roots the sign is constant. That single fact is the whole method: find the roots with x = (−b ± √(b² − 4ac)) ÷ 2a, mark them on the number line, pick any convenient test value inside each resulting region, and evaluate the expression there. Its sign at that one point is its sign across the whole region.
Take the defaults, x² − 5x + 6 < 0. The discriminant is 25 − 24 = 1, which is positive, so there are two distinct real roots: (5 ± 1) ÷ 2, giving 2 and 3. Those cut the line into three regions. Test x = 0 in the first: 0 − 0 + 6 = 6, positive. Test x = 2.5 in the middle: 6.25 − 12.5 + 6 = −0.25, negative. Test x = 4 in the last: 16 − 20 + 6 = 2, positive. We asked for where the expression is negative, so the answer is the middle region alone, written (2, 3). The brackets are round because the inequality is strict and at x = 2 and x = 3 the expression is exactly zero, not less than zero.
Change nothing but the comparison and the answer changes shape. With ≤ 0 the endpoints join the solution and it becomes [2, 3]. With > 0 the answer is the two outer regions, (−∞, 2) ∪ (3, ∞). With ≥ 0 it becomes (−∞, 2] ∪ [3, ∞). Paul's Online Notes covers the same test-point procedure in its section on Polynomial Inequalities, extended to higher degrees.
Why the Sign Chart Beats a Memorised Rule
The rule most people carry is "less than means between the roots, greater than means outside them". It works, until a is negative. Consider −x² + 5x − 6 < 0, which is the default multiplied through by −1. The roots are still 2 and 3, but the parabola now opens downwards, so the expression is negative outside the roots and positive between them. The answer is (−∞, 2) ∪ (3, ∞) — precisely the opposite of what the remembered rule predicts. Nobody who tested a point would have got this wrong; everybody who relied on the rule does.
The sign chart also survives the cases where the rule has nothing to say. If the discriminant is negative there are no roots to be inside or outside of, and the expression keeps one sign across the entire real line. If the discriminant is exactly zero the parabola touches the axis at one point and never crosses, so it has the sign of a everywhere except at that single point. In both cases the answer is either everything or nothing or one isolated point, and the test-point method finds it without any special-case reasoning.
Finally, the method scales. A cubic with three roots gives four regions and the same procedure applies unchanged. A rational inequality adds the points where the denominator is zero to the list of places the sign can flip, with the extra rule that those points are never included in the solution. Learning to think in sign charts once removes the need to memorise a separate rule for each family.
The Three Discriminant Cases
When the discriminant is positive there are two distinct real roots and three regions, with the expression taking the sign of a in the outer two and the opposite sign in the middle. This is the standard case and the one every worked example uses.
When the discriminant is exactly zero the two roots coincide. The parabola is tangent to the x-axis and never crosses it, so the expression has the sign of a everywhere else and equals zero only at that repeated root. The consequences are sharp. For a > 0, the inequality > 0 has the solution "all real numbers except the root", ≥ 0 has the solution "all real numbers", < 0 is empty, and ≤ 0 has the solution consisting of exactly one point. Those last two catch people out constantly, because a solution set containing a single number looks like a mistake and is not.
When the discriminant is negative there are no real roots, the parabola sits entirely above or entirely below the axis, and the expression never changes sign. The answer is either the whole real line or the empty set, decided purely by the sign of a and the direction of the inequality. Checking the discriminant before anything else tells you which of the three worlds you are in, and our factoring trinomials calculator is a quick way to confirm the roots when they turn out to be rational.
Strict Versus Non-Strict, and What the Endpoints Mean
The difference between < and ≤ is exactly two points, and it is the detail most often lost between the working and the final line. At each root the expression equals zero. Zero is not less than zero, so a strict inequality excludes the roots and gets round brackets. Zero is less than or equal to zero, so a non-strict inequality includes them and gets square brackets.
Infinity never gets a square bracket, because it is not a number and cannot be attained. Writing (−∞, 2] is correct; writing [−∞, 2] is not, whatever the inequality was. When the answer comes in two pieces, the union symbol joins them and the pieces are written in increasing order — (−∞, 2) ∪ (3, ∞), never the reverse. The absolute value inequality calculator produces answers in the same two-piece shape for a related reason, since an absolute value condition also splits the line into an inside and an outside.
When a Is Zero, and Other Edge Cases
If you enter a = 0 the expression is not quadratic at all — it is the linear inequality bx + c compared with zero, and the calculator says so rather than dividing by zero and returning nonsense. A linear inequality has at most one boundary point and its solution is a single ray. If b is also zero, the expression is the constant c, and the answer is either the entire real line or the empty set depending on whether c satisfies the comparison. These sound like pedantic cases, but they arise all the time when the coefficients come out of a larger problem and the x² term happens to cancel. The wider chapter on Solving Equations and Inequalities in Paul's Online Notes works through the linear, quadratic, rational and absolute-value families side by side.
Two further practicalities. First, very large or very small coefficients can push the discriminant into floating-point trouble, where b² and 4ac are nearly equal and their difference loses precision — if the discriminant comes back as a tiny non-zero number, treat it as a possible repeated root. Second, the roots displayed here are decimals. When the discriminant is a perfect square the true roots are rational and worth writing exactly; when it is not, the exact answer involves a surd and the decimal is a rounding of it. Our square root calculator will simplify the radical if you want the exact form, and the polynomial graph plotter shows the shape the sign chart is describing.
Arb Digital builds educational tools and content that show the working, because a page that explains its answer earns trust and links that a bare result never will.
Browse All Free Tools Talk To Our TeamCommon Mistakes to Avoid
- Forgetting to flip the sign when multiplying by a negative — turning −x² + 5x − 6 < 0 into x² − 5x + 6 < 0 is wrong. Multiplying both sides by −1 makes it x² − 5x + 6 > 0.
- Applying the "between the roots" rule with a negative leading coefficient — the parabola opens downwards, so the interior and exterior regions swap signs and the rule gives the exact opposite of the right answer.
- Using round brackets on a non-strict inequality — with ≤ or ≥ the roots satisfy the condition and belong in the solution set, so they need square brackets.
- Writing a bracket on infinity — infinity is never attained, so it always takes a round bracket regardless of the comparison.
- Assuming there is always an answer — a positive-definite quadratic is never negative, so a strict less-than inequality on it has a genuinely empty solution set. Empty is a valid result, not an error.
Related Free Tools From Arb Digital
Find the roots on their own with the quadratic equation solver, extend the same sign-chart reasoning to degree three with the cubic equation calculator, combine and expand expressions using the polynomial arithmetic calculator, or divide one polynomial by another with the polynomial division calculator. The full free online tools hub lists every algebra calculator we publish.
Frequently Asked Questions
Move everything to one side so the other side is zero, find the roots of the resulting quadratic, mark them on a number line, and test one value inside each region. The regions whose test value satisfies your comparison form the solution.
It depends on the sign of the leading coefficient. With a positive, the expression is negative between the roots and positive outside them. With a negative, that is reversed. Testing a point is safer than remembering which case you are in.
There are no real roots, so the expression never changes sign. It is positive everywhere if the leading coefficient is positive and negative everywhere if it is negative, which makes the solution either all real numbers or the empty set.
Yes. If the discriminant is zero and the inequality is non-strict in the direction opposite to the parabola's opening, the only value satisfying it is the repeated root, so the solution set contains exactly one point.
A round bracket excludes the endpoint and a square bracket includes it. A strict inequality excludes the roots because the expression equals zero there, while a non-strict inequality includes them.
Then it is not a quadratic. It reduces to a linear inequality with at most one boundary point, and the calculator solves that case separately rather than attempting to divide by zero.
Yes. Find every real root, use them to divide the number line into regions, and test one point in each. The only extra care needed is at repeated roots, where the sign may touch zero without crossing.
This page explains an algebraic method for educational purposes only, and results should be checked against your own working before being used in assessed coursework.
