About the Inequality Solver
This inequality solver handles linear inequalities like 3x − 7 > 2x + 5, quadratic and higher-degree polynomial inequalities like x² − 5x + 6 ≤ 0, and compound (double) inequalities like −3 < 2x + 1 ≤ 7. It returns the solution set in interval notation — for example (−∞, 2) ∪ [3, ∞) — plus set-builder notation and the critical points where the expression changes sign.
Behind the answer is a sign chart: the calculator moves everything to one side, finds the real roots, tests a value in every interval between them and checks the roots themselves (which belong to the solution only for ≤ or ≥). The table shows each test so you can follow the reasoning, and roots of quadratics are shown exactly, including square roots such as (5 − √13)/2.
Type < and > or <= and >= (≤ and ≥ also work), using one letter as the variable, with brackets and fractions as needed. It is meant for algebra and precalculus students checking homework, and for anyone who needs to know for which values a condition holds.
How to use the inequality solver
- 1Type the inequality using <, >, <= or >= — for example x^2 - 4 > 0.
- 2For a double inequality, write it as a chain such as -2 < 3x + 1 <= 10.
- 3Read the solution in interval notation and set-builder form.
- 4Check the sign chart to see which regions pass the test.
Formula and method
Any polynomial inequality can be rewritten as f(x) compared with 0 by subtracting the right side from the left. A polynomial can only change sign at its real roots, so those roots (the critical points) split the number line into intervals on which f has a constant sign. Testing one value in each interval tells you which intervals satisfy the inequality; the roots themselves are included only for ≤ and ≥.
For a linear inequality this is the same as isolating x, remembering that multiplying or dividing by a negative number reverses the inequality sign. A compound inequality a < g(x) < b is two conditions that must both hold, so the calculator collects the critical points of both, tests every region against both conditions, and keeps the regions where both are true.
- f(x)
- Left side minus right side
- ( )
- Open interval: endpoint not included (< or >)
- [ ]
- Closed interval: endpoint included (≤ or ≥)
- ∪
- Union: the solution is in either interval
Worked examples
Quadratic inequality
x² − 5x + 6 = (x − 2)(x − 3) is zero at 2 and 3. Testing x = 2.5 gives −0.25 (negative), while x = 1 and x = 4 give positive values. Because the sign is ≤, the endpoints are included: [2, 3].
Linear inequality
Subtract 2x from both sides: x − 7 > 5. Add 7: x > 12. In interval notation that is (12, ∞), with a round bracket because 12 itself is not included.
Dividing by a negative flips the sign
Subtract 4: −2x ≥ 6. Dividing by −2 reverses the sign: x ≤ −3, which is (−∞, −3].
Compound inequality
Subtract 1 from all three parts: −4 < 2x ≤ 6. Divide all three by 2: −2 < x ≤ 3.
Quadratic with irrational roots, outside the roots
The roots are (5 ± √13)/2 ≈ 0.697 and 4.303. The parabola opens upward, so it is positive outside the roots, giving two intervals joined by ∪.
Frequently asked questions
How do you write the answer to an inequality in interval notation?+
Use round brackets for endpoints that are not included (< or >) and square brackets for endpoints that are (≤ or ≥). Infinity always gets a round bracket, and separate pieces are joined with ∪, as in (−∞, 1) ∪ [4, ∞).
Why does the inequality sign flip when dividing by a negative?+
Multiplying or dividing by a negative number reverses the order of numbers on the number line: 2 < 5 but −2 > −5. So the inequality sign must flip to stay true.
How do you solve a quadratic inequality?+
Move everything to one side, find the roots of the quadratic, and test a point in each interval they create. For ax² + bx + c with a > 0, the expression is negative between the roots and positive outside them.
What is a compound inequality?+
It combines two inequalities. "And" inequalities such as −3 < 2x + 1 ≤ 7 require both to hold and usually give one bounded interval. Solve by doing the same operation to all three parts.
What if a quadratic inequality has no real roots?+
Then the quadratic never crosses zero and has the same sign for every x. Depending on the direction, the solution is either all real numbers (−∞, ∞) or no solution (∅).