Modulus Equations and Inequalities

Definition and Properties of the Modulus

The modulus of a number |x| is the distance from the point x to 0 on the number line. |x| = x (if x ≥ 0), |x| = −x (if x < 0).

• The modulus is never negative: |x| ≥ 0.
• |−x| = |x|, |ab| = |a| · |b|, |x|² = x², √(x²) = |x|.
• The expression |x − a| denotes the distance between x and a.

Basic Forms

The following rules are the basis of all problems with a modulus (a > 0).

• If |f(x)| = a, then f = a or f = −a.
• If |f(x)| < a, then −a < f < a.
• If |f(x)| > a, then f < −a or f > a.
• If |f(x)| = |g(x)|, then f = g or f = −g.
• If |f(x)| = g(x), first set the condition g ≥ 0, then solve f = g or f = −g. Check the root against g ≥ 0.
• If a < 0, then |f| = a has no solution, and the solution of |f| > a is the whole number line.

Several Moduli

1. Find the zero of each expression inside a modulus
2. The zeros divide the number line into intervals
3. On each interval, remove the modulus according to the sign
4. Solve the resulting equation
5. Check that the root belongs to that interval

The root must lie in the interval. Otherwise it is an extraneous root.

Examples

• |2x − 3| = 5: 2x − 3 = 5 or 2x − 3 = −5. So x = 4 or x = −1.
• |x − 2| < 3: −3 < x − 2 < 3, that is, −1 < x < 5. The integer solutions are 0, 1, 2, 3, 4.
• |x − 1| = 2x − 4: the condition is 2x − 4 ≥ 0, x ≥ 2. From x − 1 = 2x − 4 we get x = 3 (it works). From x − 1 = −2x + 4 we get x = 5/3, which is less than 2, an extraneous root. Answer: x = 3.