Inequality Calculator
Solve a linear, quadratic, or polynomial inequality and see the sign chart behind the answer.
for x² − 5x + 6 > 0
- 1Move everything to one side.The inequality becomes x² − 5x + 6 > 0, so the question is where that expression is positive.
- 2Find the critical points.The expression is zero at x = 2 and x = 3, which split the number line into 3 intervals.
- 3Test one value in each interval.at x = 1 the value is 2 (positive); at x = 2.5 the value is -0.25 (negative); at x = 4 the value is 2 (positive).
- 4Keep the intervals that satisfy the inequality.That gives (−∞, 2) ∪ (3, ∞).
To solve an inequality, move everything to one side, find where that expression equals zero, and test a value in each interval the roots create. For x² − 5x + 6 > 0 the roots are 2 and 3, and the solution is (−∞, 2) ∪ (3, ∞).
Why a sign chart, not just algebra
A linear inequality can be solved the way an equation is — undo each operation until x stands alone. Anything of higher degree cannot, because the expression changes sign only at its roots, and between those roots it keeps whatever sign it has. So the reliable method is to get zero on one side, find the roots, and test one convenient value inside each resulting interval.
That is what the sign chart above shows. The roots cut the number line into pieces; each piece is entirely in the solution or entirely out of it, so a single test value settles the whole interval. The method works the same for a quadratic, a cubic, or a quartic — only the number of intervals changes.
The rule that catches everyone
Multiplying or dividing both sides by a negative number flips the inequality sign. −2x > 6 becomes x < −3, not x > −3. This is why moving everything to one side first is safer than dividing partway through: with zero on the right there is nothing to flip, and the sign chart does the rest.
the roots of f are the only places the sign can change
- 1 Move every term to one side. For 3x − 5 > x + 1, subtracting x + 1 from both sides gives 2x − 6 > 0.
- 2 Solve the matching equation to find the roots. For x² − 5x + 6 > 0 set x² − 5x + 6 = 0, which factors to (x − 2)(x − 3) and gives x = 2 and x = 3.
- 3 Split the number line at those roots. Two roots create three intervals: less than 2, between 2 and 3, and greater than 3.
- 4 Test one value inside each interval. At x = 0 the expression is 6 (positive); at x = 2.5 it is −0.25 (negative); at x = 4 it is 2 (positive).
- 5 Collect the intervals that work. Only the outer two are positive, so the solution is (−∞, 2) ∪ (3, ∞) — the roots themselves are excluded because the sign is strict.
Reading interval notation
A round bracket excludes the endpoint; a square bracket includes it. Infinity always takes a round bracket.
| Inequality | Interval | Means |
|---|---|---|
| x < 3 | (−∞, 3) | Everything below 3, not 3 itself |
| x ≤ 3 | (−∞, 3] | Everything below 3, and 3 |
| x > 3 | (3, ∞) | Everything above 3, not 3 itself |
| 2 < x < 5 | (2, 5) | Strictly between 2 and 5 |
| 2 ≤ x ≤ 5 | [2, 5] | Between 2 and 5, both included |
| x < 2 or x > 5 | (−∞, 2) ∪ (5, ∞) | Two separate pieces joined by a union |
The two answers that are not intervals
Sometimes there is nothing to test. x² + 1 > 0 has no real roots and the expression is positive everywhere, so every real number works; the same expression with < 0 has no solution at all. A quadratic with a negative discriminant never crosses zero, so it keeps one sign across the whole line and the answer is all-or-nothing.
Repeated roots behave differently too. At a root of even multiplicity — the x = 2 in (x − 2)²(x − 5) — the expression touches zero and turns back without changing sign, so the intervals on either side share a sign. Checking a test value on both sides catches this without having to remember the rule.