Absolute value equations and inequalities
Absolute value measures distance, so equations and inequalities involving it describe points a fixed distance from a centre. Each one splits into two ordinary cases.
Distance meaning
Since \(|x|\) is the distance from x to 0, \(|x| = 5\) means x is 5 units from zero, so \(x = 5\) or \(x = -5\). More generally, \(|x - c|\) is the distance from x to c, so \(|x - 2| = 3\) asks for points 3 units from 2: 5 and -1.
Solving absolute value equations
First isolate the absolute value. Then, if \(|A| = k\) with \(k > 0\), solve \(A = k\) and \(A = -k\). For \(|2x - 1| + 4 = 9\), isolate to get \(|2x - 1| = 5\), then solve \(2x - 1 = 5\), giving \(x = 3\), and \(2x - 1 = -5\), giving \(x = -2\).
Special cases
If the isolated absolute value equals 0, there is one solution: \(|x + 4| = 0\) only when x = -4. If it equals a negative number, there is no solution, since distances are never negative: \(|3x - 2| = -1\) has no solution. Always check this before splitting into cases.
Less-than inequalities
\(|A| < k\) means A is within k of zero, so \(-k < A < k\), a single interval. For \(|x - 3| < 2\), write \(-2 < x - 3 < 2\), giving \(1 < x < 5\): all numbers within 2 units of 3.
Greater-than inequalities
\(|A| > k\) means A is more than k from zero, so \(A < -k\) or \(A > k\), two separate pieces. For \(|2x + 1| \ge 7\), solve \(2x + 1 \le -7\), giving \(x \le -4\), or \(2x + 1 \ge 7\), giving \(x \ge 3\). The solution is \((-\infty, -4] \cup [3, \infty)\).
A memory aid
'Less than' absolute value inequalities give an 'and' (a sandwich, one interval around the centre); 'greater than' gives an 'or' (two outer rays). Checking a test point, such as the centre, quickly confirms which region is correct.
Tolerances
Absolute value inequalities describe tolerances in manufacturing and science. A bolt meant to be 12.0 mm wide with a tolerance of 0.2 mm must satisfy \(|w - 12.0| \le 0.2\), so \(11.8 \le w \le 12.2\). Opinion polls quoting a 3-point margin of error describe a similar band around the estimate.
Worked example
A thermostat set to 21°C keeps the room within 1.5°C of the setting. The temperature T satisfies \(|T - 21| \le 1.5\), so \(19.5 \le T \le 22.5\). If the room reaches 23°C, then \(|23 - 21| = 2 > 1.5\), so the heating or cooling should switch on.
Key takeaways
Absolute value is distance, so |x - c| = k describes the points k units from c. Isolate the absolute value, then split |A| = k into A = k and A = -k; a result of 0 gives one solution and a negative value gives none. |A| < k becomes -k < A < k, a single interval; |A| > k becomes A < -k or A > k, two rays. Absolute value inequalities model tolerances and margins of error.
Flashcard
Distance from x to c
Check your understanding
Solve \(|x - 3| < 2\).