Absolute value measures how far a number is from zero, ignoring direction. That simple idea powers distance formulas, error tolerances, and a whole family of equations and inequalities in algebra. This calculator does three jobs: it finds |x| for any number, solves equations and inequalities of the form |ax + b| ? c with a number-line picture, and computes the modulus of a complex number.
How to use the absolute value calculator
- Pick a mode at the top.
- |x| of a number: type any number, including decimals and fractions such as −3/4.
- Solve |ax + b| ? c: enter a and b for the expression inside the bars, choose =, <, ≤, > or ≥, and enter c. The tape gives the solution in inequality and interval notation, and the number line shades it.
- |a + bi|: enter the real and imaginary parts to get the modulus, its exact radical form when one exists, the argument and the conjugate.
Absolute value definitions
| Form | Equivalent statement (c > 0) | Solution shape |
|---|---|---|
| |u| = c | u = c or u = −c | two points |
| |u| < c | −c < u < c | one interval |
| |u| > c | u < −c or u > c | two rays |
| |u| = c with c < 0 | impossible | no solution |
| |u| > c with c < 0 | always true | all real numbers |
Useful properties: |ab| = |a|·|b|, |a/b| = |a|/|b|, and the triangle inequality |a + b| ≤ |a| + |b|.
Worked examples
A number: |−12.5| = −(−12.5) = 12.5. Both 12.5 and −12.5 sit 12.5 units from zero.
An equation: |2x − 3| = 7 splits into 2x − 3 = 7 (x = 5) and 2x − 3 = −7 (x = −2). Both answers are 7 units from 1.5, the point where the inside equals zero.
An inequality: |2x − 3| < 7 becomes −7 < 2x − 3 < 7. Adding 3 gives −4 < 2x < 10, and dividing by 2 gives −2 < x < 5, the open interval (−2, 5).
A complex number: |3 − 4i| = √(3² + (−4)²) = √25 = 5.
Reading absolute value as distance
The expression |x − a| is the distance between x and a. That reading makes inequalities intuitive: |x − 10| ≤ 0.5 describes every x within half a unit of 10, the interval [9.5, 10.5]. Engineers write tolerances this way: a shaft specified as 25.00 mm ± 0.05 mm must satisfy |d − 25| ≤ 0.05.
The equation |2x − 3| < 7 can be rewritten as |x − 1.5| < 3.5 by dividing through by 2, which says “x is within 3.5 of 1.5”, giving the same interval from −2 to 5.
Pitfalls
- Negative right-hand sides. |u| = −4 has no solution, but |u| > −4 is true for every x. Check the sign of c before splitting.
- Flipping signs. When a is negative, dividing by it reverses each inequality sign. The calculator notes when this happens.
- Isolate first. For 3|x + 1| − 2 = 10, first get |x + 1| = 4 on its own, then split.
- “Absolute value makes things positive” is not quite right. |0| = 0, and −|x| is never positive. Think “distance,” not “make positive.”
For the distance between two specific numbers, use the absolute difference calculator. The same idea in two and three dimensions is the distance calculator, and vector length is covered by the dot product calculator.
Frequently asked questions
What is absolute value?
The absolute value of a number is its distance from zero on the number line, written |x|. Distance is never negative, so |−12.5| = 12.5 and |12.5| = 12.5. Only |0| equals 0.
How do you solve an absolute value equation?
Isolate the absolute value, then split it into two equations: |2x − 3| = 7 means 2x − 3 = 7 or 2x − 3 = −7, which gives x = 5 or x = −2. If the right side is negative there is no solution.
How do absolute value inequalities work?
A 'less than' inequality becomes one compound inequality: |2x − 3| < 7 means −7 < 2x − 3 < 7, so −2 < x < 5. A 'greater than' inequality becomes two separate pieces: |2x − 3| > 7 means x < −2 or x > 5.
What is the absolute value of a complex number?
It is the distance from 0 in the complex plane, also called the modulus: |a + bi| = √(a² + b²). For 3 − 4i that is √(9 + 16) = 5.