Quadratic Inequality Solver
Enter the coefficients of a quadratic inequality ax² + bx + c OP 0. The solver computes the discriminant, finds the roots, reasons about the sign of the parabola via its sign chart and returns the solution set in interval notation.
Two facts decide everything
A quadratic inequality asks where a parabola sits above or below the axis. Unlike its linear counterpart the answer is rarely a half-line: a parabola can be on the wanted side over a bounded stretch, over everything outside one, over the whole line, or nowhere.
Which of those applies is decided before any interval is written down. This solver computes the discriminant, finds the roots when they exist, reads the sign of the parabola from its leading coefficient, and reports the solution as an interval or a union of two.
Only two things matter: whether the parabola crosses the axis, and which way it opens. The discriminant answers the first and the leading coefficient's sign the second, and between them they select the outcome before the roots are used at all.
That is why the discriminant is reported as its own step. Reading it first tells you what shape of answer to expect, and an answer of the wrong shape is then immediately suspicious.
How to use this calculator
- Enter the three coefficients With everything on the left and zero on the right. An inequality comparing against a non-zero value must be rearranged first.
- Choose the relation Strict or inclusive, above or below zero. Inclusive relations add the roots themselves to the solution, which changes the brackets rather than the intervals.
- Read the discriminant Positive means two roots; zero means a single touching point; negative means the parabola never reaches the axis.
- Read the sign chart line It states where the quadratic is positive and where negative, which is the reasoning the interval answer is built from.
The formula, and where it comes from
Δ = b² − 4ac roots = (−b ± √Δ)/(2a) sign(quadratic) = sign(a) outside the roots, opposite between them
The sign rule is the heart of the method and needs no case analysis once stated. A parabola opening upward is positive far out in both directions and dips negative only between its roots; one opening downward does the reverse. The rest follows from matching that against the relation chosen.
With a positive discriminant there are two roots, and the solution is either the stretch between them or the two unbounded pieces outside — never anything else. Which one depends on whether the relation wants the leading coefficient's sign or its opposite.
A zero discriminant means the parabola touches the axis without crossing. The quadratic keeps one sign everywhere except at that point, so the answer is one of four: everything, everything except that point, that point alone, or nothing.
A negative discriminant means the parabola never reaches the axis, so its sign is the sign of the leading coefficient at every value of x. The inequality is then either always true or always false, and the solver reports which without computing any roots.
What each input means
- a Quadratic coefficient — form field “a (x² coefficient)”
- Its sign decides which way the parabola opens and therefore where the quadratic is positive. A value of zero makes the problem linear, and it is handed to the linear inequality routine.
- b Linear coefficient — form field “b (x coefficient)”
- Enters through the discriminant and the roots, but says nothing on its own about the shape.
- c Constant term — form field “c (constant)”
- The value of the quadratic at zero, and part of the discriminant.
- OP Relation — form field “Inequality”
- Strict or inclusive, above or below. Inclusive relations close the interval endpoints at the roots.
Worked examples
Every number below is produced by the same calculation engine the tool above runs. Nothing here is typed by hand, so the walkthrough cannot drift from what you get when you enter the same values yourself.
Below zero between two roots
An upward parabola with roots at 2 and 3, asked where it is at most zero.
Inputs a (x² coefficient) = 1, b (x coefficient) = -5, c (constant) = 6, Inequality = le
- Inequality 1x² − 5x + 6 ≤ 0
- Discriminant Δ = b² − 4ac = 1
- Roots x₁ = 2, x₂ = 3
- Sign chart The quadratic is positive outside the roots, negative between them.
- Solution [2, 3]
Result [2, 3]
The parabola opens upward, so it is negative only between its roots — and the answer is that closed stretch. The endpoints are included because the relation allows equality, and the quadratic is exactly zero there.
This is the bounded case, and it happens only when the relation wants the sign opposite to the leading coefficient. Any answer running off to infinity here would be the wrong half of the sign chart.
The complementary question
The same quadratic, now asked where it is strictly above zero. Reversing the relation reverses which region is wanted.
Inputs a (x² coefficient) = 1, b (x coefficient) = -5, c (constant) = 6, Inequality = gt
- Inequality 1x² − 5x + 6 > 0
- Discriminant Δ = b² − 4ac = 1
- Roots x₁ = 2, x₂ = 3
- Sign chart The quadratic is positive outside the roots, negative between them.
- Solution (−∞, 2) ∪ (3, +∞)
Result (−∞, 2) ∪ (3, +∞)
The answer is now two unbounded pieces rather than one bounded stretch — everything to the left of the smaller root and everything to the right of the larger. The two solutions together with the roots cover the whole line, which they must.
The endpoints are excluded here because the relation is strict and the quadratic is zero at the roots. Only the bracket style differs from the inclusive version, not the intervals themselves.
A parabola that never reaches the axis
An upward parabola with a negative discriminant, asked where it is at least zero.
Inputs a (x² coefficient) = 1, b (x coefficient) = 0, c (constant) = 1, Inequality = ge
- Inequality 1x² + 0x + 1 ≥ 0
- Discriminant Δ = b² − 4ac = -4
- Sign Δ < 0, so the parabola never crosses the x-axis. The expression is always positive.
Result all real x
No roots exist, so the quadratic keeps the sign of its leading coefficient everywhere — positive throughout. The relation asks for non-negative, which is satisfied at every real number.
Asking the opposite question would give no solution at all. When the discriminant is negative the answer is always one of those two extremes and never an interval.
Reading the result
The shape of the answer is set by the discriminant
Two roots give one bounded interval or two unbounded ones. A repeated root gives a single point or its complement. No roots give everything or nothing. Recognising the case before reading the intervals is the fastest error check available.
Complementary relations partition the line
Reversing the relation and swapping strict for inclusive gives exactly the complement of the original answer. If two such solutions overlap or leave a gap beyond the roots, one of them is wrong.
Endpoints belong to inclusive relations only
The roots are where the quadratic equals zero, so they satisfy an inclusive relation and fail a strict one. That is the whole difference between round and square brackets in the answer.
When you would use this
Finding where a modelled quantity stays acceptable
A quantity described by a quadratic — profit against price, height against time — is above or below a threshold on an interval. Rearranging against zero and solving gives that range.
Establishing a domain restriction
A square root requires its radicand to be non-negative, and when that radicand is a quadratic the condition is this inequality. Solving it gives the parent function's domain.
Assumptions and limitations
What this calculator assumes
- The inequality has been rearranged so that zero is on the right.
- The leading coefficient is non-zero; a zero value is handled as a linear inequality instead.
- The sign of the quadratic is determined by the leading coefficient and the position relative to the roots.
- Endpoints are included exactly when the relation is inclusive.
Where it stops being the right tool
- Quadratics only: a cubic or higher needs a full sign chart across more roots.
- One inequality at a time, so a system of simultaneous conditions must be intersected by hand.
- Roots are shown as decimals rather than exact surds, so an irrational boundary appears rounded.
Common mistakes
Solving the equation and stopping
Why it happens. Finding the roots feels like the hard part, and they are the boundaries of the answer. They are not the answer: the solution is a region, and the roots only mark where it changes.
How to avoid it. Use the roots to build intervals, then test the sign in each. The sign chart line does that reasoning explicitly.
Ignoring the sign of the leading coefficient
Why it happens. The rule that a quadratic is negative between its roots is remembered without the condition attached, and it reverses entirely when the parabola opens downward.
How to avoid it. Check which way it opens before choosing the region. A downward parabola is positive between its roots, which inverts every conclusion.
Comparing against a non-zero value
Why it happens. An inequality such as a quadratic exceeding 5 looks ready to solve, and the roots of the quadratic itself are easy to find. They are not the boundaries of that problem.
How to avoid it. Subtract the right-hand side first so the comparison is against zero. The constant term changes, and so do the roots.
Frequently asked questions
How is the sign chart built?
From the direction the parabola opens. With a positive leading coefficient the quadratic is positive outside its roots and negative between them; with a negative one the pattern reverses. The solver matches that against the relation you chose.
What if the discriminant is negative?
The parabola never crosses the axis, so it keeps the sign of its leading coefficient everywhere. The inequality is then satisfied by every real number or by none, and the solver reports which without finding any roots.
What about a repeated root?
The parabola touches zero at one point and keeps one sign elsewhere. Depending on the relation the answer is that single point, everything except it, all real numbers, or nothing — four outcomes rather than an interval.
Why is the answer sometimes two intervals?
Because the region where a parabola takes the sign of its leading coefficient is everything outside its roots, which is two unbounded pieces. Only the region between the roots is a single bounded interval.