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Absolute Value

Absolute value strips the sign off a number and asks a simpler question: how far is it from zero? That distance idea is all you need to evaluate the modulus and to solve equations that hide two answers inside one.

MYP 4Extended MathsNumberCriteria A · C~9 min read

Written by Fayek · MYP student

Distance from zero

The absolute value or modulus of a number is its distance from zero on the number line. Distance is never negative, so the modulus is never negative either, whatever sign you feed in.

Absolute value \(|x|\)
The distance of \(x\) from 0. It equals \(x\) itself when \(x\) is positive or zero, and equals \(-x\) (which flips it positive) when \(x\) is negative.
\[ |x| = \begin{cases} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{cases} \]

So \(|5| = 5\) because 5 is five units from zero, and \(|-5| = 5\) because \(-5\) is also five units from zero, just on the other side. The two points are mirror images across zero.

Two ideas, same answer

You can think "distance from zero" or "make it positive". Both give \(|-8| = 8\). The distance picture is the one that explains why modulus equations usually have two solutions.

Evaluating the modulus

When a modulus wraps a calculation, work out what is inside the bars first, then take the absolute value of that single result. Treat the bars like brackets that also strip the sign.

Worked example

Evaluate \(|-7| + |3 - 10|\).

1
First bar: \(|-7| = 7\).
2
Second bar: work out the inside first, \(3 - 10 = -7\), then \(|-7| = 7\).
3
Add the two results: \(7 + 7 = 14\).
\(|-7| + |3 - 10| = 14\)

The bars are not brackets you can skip

\(|3 - 10|\) is \(7\), not \(-7\). Do the subtraction inside, then apply the modulus to the whole thing. Never drop the bars until the inside is a single number.

Solving modulus equations

An equation like \(|x - 3| = 5\) says: "the distance from \(x\) to 3 is 5." A distance of 5 can be reached by going 5 above or 5 below, so the expression inside the bars could equal \(+5\) or \(-5\). That gives you two cases to solve.

Worked example

Solve \(|x - 3| = 5\).

1
The inside can be positive or negative, so split into two cases: \(x - 3 = 5\) or \(x - 3 = -5\).
2
Case 1: \(x - 3 = 5\) gives \(x = 8\).
3
Case 2: \(x - 3 = -5\) gives \(x = -2\).
4
Check both: \(|8 - 3| = |5| = 5\) and \(|-2 - 3| = |-5| = 5\). Both work.
\(x = 8\) or \(x = -2\)

On the number line, 8 and \(-2\) are each exactly 5 units from 3, one to the right and one to the left. That is the two-case idea made visible.

Reading the answers

One case never covers it: writing only \(x = 8\) throws away half the solution. Equally, a modulus can never equal a negative, so \(|x - 3| = -5\) has no solution at all, because a distance cannot be negative.

Where this is assessed

Setting up the two cases and solving each is Criterion A knowledge and understanding, while stating both solutions clearly and checking them is Criterion C communicating. Marks are often split across both cases, so show each one.

Check yourself

1. Evaluate \(|-12|\). +

\(-12\) is 12 units from zero, so \(|-12| = \mathbf{12}\).

2. Solve \(|x + 2| = 6\). +

Two cases: \(x + 2 = 6\) gives \(x = 4\); \(x + 2 = -6\) gives \(x = -8\). Check: \(|4 + 2| = 6\) and \(|-8 + 2| = |-6| = 6\). So \(x = \mathbf{4}\) or \(x = \mathbf{-8}\).

3. Evaluate \(|5 - 9| - |-2|\). +

Inside first: \(5 - 9 = -4\), so \(|5 - 9| = 4\). Also \(|-2| = 2\). Then \(4 - 2 = \mathbf{2}\).

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