Absolute Value

What is absolute value?

The absolute value of a real number \(x\), written \( |x| \), is the distance from \(0\) on the number line. Distance is never negative, so \( |x| \ge 0 \) for every real \(x\). Two simple facts follow immediately:

• \( |x| = x \) if \( x \ge 0 \).
• \( |x| = -x \) if \( x < 0 \) (flips the sign to make it positive).

Formally, we write the piecewise definition:

\[ |x|=\begin{cases} x, & x\ge 0, -x, & x<0. \end{cases} \]

Quick intuition

Think “how far from zero?” The number \(7\) is seven units to the right of zero, so \( |7|=7 \). The number \(-7\) is seven units to the left, so its distance is also \(7\), and \( |-7|=7 \). The only number at distance zero from zero is \(0\) itself, so \( |0|=0 \).

Evaluating absolute values

• Pure numbers: \( |-12|=12 \), \( |0|=0 \), \( |3-10|=|-7|=7 \).
• With variables (substitute then evaluate): if \( x=-3 \), then \( |2x|=|2(-3)|=|-6|=6 \).
• Mixed expressions: \( |-2|^2 + |3-7| = 2^2 + 4 = 8 \).

Key properties

• Non-negativity: \( |x| \ge 0 \) and \( |x|=0 \iff x=0 \).
• Evenness (symmetry): \( |x| = |-x| \).
• Product rule: \( |ab| = |a|\,|b| \).
• Quotient rule: \( \left|\dfrac{a}{b}\right| = \dfrac{|a|}{|b|} \) for \( b\neq 0 \).
• Triangle inequality: \( |a+b| \le |a| + |b| \). (Distances add at most.)
• Reverse triangle inequality: \( \bigl||a|-|b|\bigr| \le |a-b| \).

Solving equations with absolute value

Almost every GCSE question comes down to “remove the bars by considering the two sign cases”. Here are the essential patterns.

1) \( |x| = a \)

• If \( a < 0 \): no solution (absolute value cannot be negative).
• If \( a \ge 0 \): \( x=a \) or \( x=-a \).

Example: \( |x|=5 \Rightarrow x=5 \) or \( x=-5 \).

2) \( |x-c| = a \)

Interpret \( |x-c| \) as “distance of \(x\) from \(c\)”. Solutions are all points exactly \(a\) units from \(c\): \( x=c-a \) or \( x=c+a \) (when \( a \ge 0 \)).

Example: \( |x-2|=3 \Rightarrow x=2-3=-1 \) or \( x=2+3=5 \).

3) \( |ax+b| = a_0 \) (with a scale)

First solve \( |ax+b| = a_0 \) in two linear cases, then divide to isolate \(x\): \( ax+b=a_0 \) or \( ax+b=-a_0 \).

Example: \( |2x-1|=7 \Rightarrow 2x-1=7 \) or \( 2x-1=-7 \). So \( x=4 \) or \( x=-3 \).

4) Equations with absolute value on both sides

Sometimes you see \( |\,\cdot\,| \) on both sides. A common approach is to consider sign cases or square both sides (carefully, since squaring may introduce extraneous solutions).

Example: Solve \( |x-3| = |2x-1| \).

• Case 1: \( x-3 = 2x-1 \Rightarrow -3+1 = x \Rightarrow x=-2 \).
• Case 2: \( x-3 = -(2x-1) \Rightarrow x-3 = -2x+1 \Rightarrow 3x = 4 \Rightarrow x=\dfrac{4}{3} \).

Solving inequalities with absolute value

Inequalities with \( |x-a| \) describe points within a certain distance of \(a\), or at least that distance away from \(a\). Two master patterns cover almost all questions.

Pattern A: “within” — \( |x-a| \le b \) (or \( < \))

If \( b \ge 0 \) then \( |x-a| \le b \) means \(x\) lies within \(b\) units of \(a\):

\[ |x-a| \le b \quad \Longleftrightarrow \quad a-b \le x \le a+b. \]

Example: \( |x-4| \le 2 \Rightarrow 2 \le x \le 6 \).

Pattern B: “at least” — \( |x-a| \ge b \) (or \( > \))

If \( b \ge 0 \), then \( |x-a| \ge b \) means \(x\) is at least \(b\) units away from \(a\):

\[ |x-a| \ge b \quad \Longleftrightarrow \quad x \le a-b \ \text{ or }\ x \ge a+b. \]

Example: \( |x+1| \ge 3 \Rightarrow x \le -4 \ \text{or}\ x \ge 2 \).

Working with a coefficient

If the inside is linear but scaled, isolate it first. For instance, \( |2x-5| \le 9 \Rightarrow -9 \le 2x-5 \le 9 \Rightarrow -4 \le 2x \le 14 \Rightarrow -2 \le x \le 7 \).

Graph view

The graph of \( y=|x| \) is a symmetric “V” meeting at the origin \((0,0)\). It comes from reflecting the line \( y=x \) across the \(x\)-axis for negative \(x\). Translating gives \( y = |x-a| \) (shift right by \(a\)), and vertical shifts give \( y = |x| + k \) (move up by \(k\)).

Piecewise, \( y=|x| \) is \[ y=\begin{cases} x, & x\ge 0,\\ -x, & x<0. \end{cases} \]

Using \( |x| = \sqrt{x^2} \)

The identity \( |x| = \sqrt{x^2} \) is sometimes the quickest route in algebraic work (for example, when completing the square or differentiating in later courses). However, for GCSE problem solving, the case-split method is normally the clearest: remove the bars by considering \(x\ge 0\) and \(x<0\).

Triangle inequality — why it matters

The inequality \( |a+b| \le |a| + |b| \) says: the direct distance from \(0\) to \(a+b\) is at most the journey that goes via \(a\) and then \(b\). On the number line (or in vectors), “straight there” is never longer than “there via somewhere else”.

A useful corollary is \( \bigl||a|-|b|\bigr| \le |a-b| \), often used to compare magnitudes.

Worked examples (concept checks)

1. Evaluate: \( |-18| + |5-13| \).
   \( |-18|=18 \), \( |5-13|=|-8|=8 \). Total \( =26 \).
2. Solve: \( |x|=0 \).
   Only \( x=0 \).
3. Solve: \( |x-3|=1 \).
   \( x-3=1 \) or \( x-3=-1 \Rightarrow x=4 \) or \( x=2 \).
4. Solve: \( |2x+5|=9 \).
   \( 2x+5=9 \) or \( 2x+5=-9 \Rightarrow x=2 \) or \( x=-7 \).
5. Describe the region: \( |x-2| \le 3 \).
   \( -3 \le x-2 \le 3 \Rightarrow -1 \le x \le 5 \).
6. Describe the region: \( |x+4| > 2 \).
   \( x+4 > 2 \) or \( x+4 < -2 \Rightarrow x>-2 \) or \( x<-6 \).

Absolute value in contexts

• Temperature change: A drop from \(18^\circ\text{C}\) to \(5^\circ\text{C}\) is a change of \( |5-18|=13^\circ\text{C} \).
• Deviation from a target: If a machine should produce length \(L=10\) cm, then \( |x-10| \) measures the error of a piece of length \(x\).
• Distance between points on a line: The distance between \(x=a\) and \(x=b\) is \( |b-a| \).

Common pitfalls (and fixes)

• Forgetting that absolute values are non-negative. If a question says \( |x| = -3 \), write “no solution” immediately.
• Dropping a solution in case splits. When solving \( |x-a|=b \), always write both cases and solve each one cleanly.
• Mishandling inequalities. Remember: \( |x-a|\le b \Rightarrow a-b\le x\le a+b \); \( |x-a|\ge b \Rightarrow x\le a-b \) or \( x\ge a+b \).
• Not simplifying inside the bars first. Calculate inside before taking absolute value: \( |3-10|=|-7|=7 \) — don’t try to do both in one jump.

Exam checklist

• Underline the absolute value expression(s) and decide whether you need two cases or the distance interpretation.
• For equations: check the right-hand side is \(\ge 0\) before solving.
• For inequalities: translate to an interval (\(\le\)) or two rays (\(\ge\)). Sketch a quick number line.
• Simplify inside bars first, then apply the definition.
• Verify your answers quickly by substitution.

Why the definition makes sense

Distance on a line is the length of the shortest path between two points. If we’re at \(x\) and want to reach \(0\), the length is simply the magnitude of \(x\), regardless of direction. That is why \( |x| = x \) for \( x\ge 0 \) and \( |x| = -x \) for \( x<0 \), and why \( |x| \) is always non-negative. The graph’s V-shape is the geometric picture of “distance from \(0\)”.

Mini practice (try mentally)

1. Compute \( |-25| \), \( |8-19| \), and \( |3x| \) when \( x=-5 \).
2. Solve \( |x+2|=7 \).
3. Find all \(x\) with \( |x-3| \le 4 \).
4. Describe the solution set of \( |x| \ge \tfrac{5}{2} \).

Summary

Absolute value measures distance: \( |x| \) is how far \(x\) is from \(0\). Use the piecewise definition to evaluate and to solve equations by splitting into cases, or use the distance interpretation for quick inequalities. Remember the core patterns:

• \( |x|=a \Rightarrow x=\pm a \) (when \( a\ge 0 \); otherwise no solution).
• \( |x-a|=b \Rightarrow x=a\pm b \).
• \( |x-a|\le b \Rightarrow a-b \le x \le a+b \) (an interval).
• \( |x-a|\ge b \Rightarrow x\le a-b \) or \( x\ge a+b \) (two rays).
• Properties: non-negativity, symmetry, product/quotient rules, and triangle inequality.

With these tools, absolute value questions become straightforward: simplify inside the bars, choose the right pattern, solve cleanly, and check.