A modulus measures distance from zero, so \(|X| < 3\) means \(X\) is within 3 of zero and \(|X| > 3\) means it is further away. That one idea, plus squaring and a quick sketch, solves every modulus inequality in this course.
For \(c > 0\): \(|X| < c\) means \(-c < X < c\) (one interval), and \(|X| > c\) means \(X < -c\) or \(X > c\) (two pieces). Divide by \(k\) first. When both sides are moduli, square both sides to get a quadratic inequality. When one side is not a modulus, find the critical values from the equation, check them, and use a sketch to choose the regions.
Fig. 1: y = |2x − 3| (orange) and y = |x + 1| (blue) cross at x = 2/3 and x = 4. Between those values the orange V lies on or below the blue one, so |2x − 3| ≤ |x + 1| exactly when 2/3 ≤ x ≤ 4 (marked in teal on the x-axis)
Section 01
Less than means between.
\(|X|\) is the distance of \(X\) from zero. So \(|X| < 3\) says \(X\) is less than 3 away from zero: it lies between \(-3\) and \(3\). And \(|X| > 3\) says \(X\) is more than 3 away: it is below \(-3\) or above \(3\). For \(k|ax + b| > c\) or \(k|ax + b| \le c\), divide by \(k\) first, then use the rule with \(X = ax + b\).
Inequality (\(c > 0\))
Means
Shape of the answer
\(|X| < c\)
\(-c < X < c\)
one interval
\(|X| \le c\)
\(-c \le X \le c\)
one interval, ends included
\(|X| > c\)
\(X < -c\) or \(X > c\)
two separate pieces
\(|X| \ge c\)
\(X \le -c\) or \(X \ge c\)
two pieces, ends included
The edge case \(c = 0\): \(|X| > 0\) is true for every value except \(X = 0\), so \(|x - 2| > 0\) means \(x \ne 2\).
Worked example A 'less than' inequality
Solve \(3|2x - 1| < 9\).
Divide by 3: \(|2x - 1| < 3\).
'Less than' means between: \(-3 < 2x - 1 < 3\).
Add 1 throughout: \(-2 < 2x < 4\).
Divide throughout by 2.
Answer: \(-1 < x < 2\)
Worked example A 'greater than' inequality
Solve \(2|3x + 1| > 10\).
Divide by 2: \(|3x + 1| > 5\).
'Greater than' means outside: \(3x + 1 < -5\) or \(3x + 1 > 5\).
\(-7 \le 5 - 2x \le 7\), so \(-12 \le -2x \le 2\). Dividing by \(-2\) reverses both signs: \(6 \ge x \ge -1\), that is \(-1 \le x \le 6\).
Examiner trap
A 'greater than' answer is two pieces joined by 'or'. Never write \(-5 > 3x + 1 > 5\) or \(-2 > x > \frac{4}{3}\): no number is both below \(-2\) and above \(\frac{4}{3}\). And when you multiply or divide through by a negative number, reverse every inequality sign.
Section 02
Modulus against modulus: square both sides.
For \(k|ax + b| \le |cx + d|\) with \(k > 0\), both sides are never negative, so squaring keeps the inequality true and removes the modulus signs: \(|X|^{2} = X^{2}\). Expand, collect everything on one side and you have a quadratic inequality. Factorise it to find the critical values, then decide between 'between the roots' and 'outside the roots' as in topic 2.
Square both sides, including \(k\): \(k|X| \le |Y|\) becomes \(k^{2}X^{2} \le Y^{2}\).
Collect to \(px^{2} + qx + r \le 0\) (or \(> 0\)) with \(p > 0\).
Factorise to get the critical values.
Choose: \(\le 0\) or \(< 0\) is between the roots; \(\ge 0\) or \(> 0\) is outside them.
Worked example The inequality in Fig. 1
Solve \(|2x - 3| \le |x + 1|\).
Square both sides: \((2x - 3)^{2} \le (x + 1)^{2}\).
Critical values \(x = \dfrac{2}{3}\) and \(x = 4\). A positive quadratic is \(\le 0\) between its roots, which is where the orange V is below the blue one in Fig. 1.
Answer: \(\frac{2}{3} \le x \le 4\)
Worked example A constant in front
Solve \(2|x - 1| > |x + 4|\).
Square both sides, squaring the 2 as well: \(4(x - 1)^{2} > (x + 4)^{2}\).
Critical values \(-\dfrac{2}{3}\) and \(6\); '\(> 0\)' means outside the roots.
Check \(x = 0\): \(2 > 4\) is false, so the middle is correctly excluded.
Answer: \(x < -\frac{2}{3}\) or \(x > 6\)
Your turn
Solve \(|x - 2| < |x + 4|\).
Square: \(x^{2} - 4x + 4 < x^{2} + 8x + 16\). The \(x^{2}\) terms cancel: \(-12x < 12\), so \(x > -1\). (Every point closer to 2 than to \(-4\).)
Examiner trap
Square the constant too: \(2|x - 1|\) squared is \(4(x - 1)^{2}\), not \((x - 1)^{2}\) or \(2(x - 1)^{2}\). Only square when BOTH sides are moduli (or are known to be non-negative); the next section shows what goes wrong otherwise.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
Solve \(|x - 2| < 5\).
Quick check+10 XP
Solve \(|2x + 1| > 7\).
Quick check+10 XP
Squaring both sides of \(|x + 1| < |x - 3|\) leads to which answer?
Quick check+10 XP
Solve \(3|x - 4| \le 6\).
Section 03
Critical values, then a sketch.
When one side is not a modulus, as in \(|ax + b| \le cx + d\), squaring is not safe because \(cx + d\) can be negative. The safest exam method works for every type: (1) replace the inequality sign by \(=\) and solve, taking \(+\) and \(-\) cases; (2) check each critical value in the original equation and reject any that fail; (3) sketch both graphs and read off where one is above or below the other. For \(|ax^{2} + bx + c| \le d\), the answer can be two separate intervals.
\(|X| \le d\) with \(X\) a quadratic: solve \(-d \le X \le d\) as two quadratic inequalities and keep the values that satisfy BOTH.
\(|X| > d\): solve \(X > d\) and \(X < -d\) separately and keep values from EITHER.
Write answers in the standard forms, such as \(-3 < x < 4\) or \(x < 1\) or \(x > 6\).
Worked example A V-shape against a line
Solve \(|2x - 1| < x + 4\).
Critical values from \(|2x - 1| = x + 4\). Case \(+\): \(2x - 1 = x + 4\) gives \(x = 5\). Case \(-\): \(-(2x - 1) = x + 4\) gives \(-3x = 3\), \(x = -1\).
Check: \(x = 5\): \(|9| = 9\) and \(5 + 4 = 9\). \(x = -1\): \(|-3| = 3\) and \(-1 + 4 = 3\). Both are real crossings.
Sketch: the V \(y = |2x - 1|\) has its vertex at \((\frac{1}{2}, 0)\); the line \(y = x + 4\) crosses the V at \(x = -1\) and \(x = 5\) and lies above it in between.
Test \(x = 0\): \(1 < 4\), true, so the middle region is the answer.
Answer: \(-1 < x < 5\)
Worked example A modulus of a quadratic
Solve \(|x^{2} - 5x + 5| < 1\).
'Less than' means between: \(-1 < x^{2} - 5x + 5 < 1\). Split into two inequalities.
Both must hold: inside \(1 < x < 4\) but not in \(2 \le x \le 3\). On a sketch of \(y = |x^{2} - 5x + 5|\), these are the two dips of the curve below the line \(y = 1\).
Answer: \(1 < x < 2\) or \(3 < x < 4\)
Your turn
Solve \(|x^{2} - 2x| > 3\).
Either \(x^{2} - 2x > 3\): \((x - 3)(x + 1) > 0\), so \(x < -1\) or \(x > 3\). Or \(x^{2} - 2x < -3\): \(x^{2} - 2x + 3 < 0\), but its discriminant is \(4 - 12 < 0\) and it is always positive, so this gives nothing. Answer: \(x < -1\) or \(x > 3\).
Examiner trap
Squaring \(|x - 3| > 2x\) gives \(3x^{2} + 6x - 9 < 0\), that is \(-3 < x < 1\), which is WRONG: \(x = -5\) also works, since \(8 > -10\). Squaring fails because \(2x\) is negative there. With critical values, \(x = -3\) fails the check (\(|-6| = 6\) but \(2x = -6\)), leaving only \(x = 1\), and the sketch gives \(x < 1\). Always check critical values when one side is not a modulus.
Stage 2 · Exam
Exam-style questions
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Solve the inequality \(2|3x - 1| > |x + 3|\).
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Solve \(|x - 3| > 2x\).
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Solve \(|x^{2} - 8| < 1\), giving exact values.
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Find the set of values of \(x\) for which \(|2x + 1| \le x + 5\).
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