Modulus Inequalities

A2 · P3 · 14 min

A modulus inequality asks for every $x$ that makes something like $|2x - 3| \le 7$ or $|x - 4| < 2|3x + 1|$ true. The answer is an interval, or a pair of intervals, never a single number. These appear on almost every P3 paper as a 4 or 5 mark opening question, and they reward one habit above all: find the critical values from the equation first, then use a sketch to decide which side of them you want.

Read the modulus function and modulus equations first if $|a| = |b| \iff a^2 = b^2$ is not yet automatic.

The distance picture

Since $|x - a|$ is the distance from $x$ to $a$ :

$|x - a| < b$ means " $x$ is less than $b$ away from $a$ ". That is the interval from $a - b$ to $a + b$ .
$|x - a| > b$ means " $x$ is more than $b$ away from $a$ ". That is everything outside that interval.

y = abs(x - 3) y = 2 (1, 0) -- (5, 0)

The V $y = |x - 3|$ is below the line $y = 2$ exactly when $1 < x < 5$ , the interval of numbers within $2$ of $3$ .

Key result

For $b > 0$ :

|x - a| < b \iff a - b < x < a + b

|x - a| > b \iff x < a - b \ \text{ or } \ x > a + b

The same holds with $\le$ and $\ge$ . If $b \le 0$ , then $|x - a| < b$ has no solutions and $|x - a| > b$ is true for every $x$ (except $x = a$ when $b = 0$ ).

For $|ax + b| < c$ , write it as the double inequality $-c < ax + b < c$ and solve all three parts at once. For $|ax + b| > c$ , write the two separate inequalities $ax + b > c$ or $ax + b < -c$ .

Within a distance

Solve $|2x - 3| \le 7$ .

Solution

-7 \le 2x - 3 \le 7

Add $3$ throughout: $-4 \le 2x \le 10$ . Divide by $2$ : $-2 \le x \le 5$ .

Check with the distance picture: $|2x - 3| = 2\left|x - \tfrac{3}{2}\right|$ , so the inequality says $x$ is within $3.5$ of $1.5$ , which is $-2$ to $5$ .

Outside a distance

Solve $|3x + 1| > 5$ .

Solution

Either $3x + 1 > 5$ , giving $x > \tfrac{4}{3}$ , or $3x + 1 < -5$ , giving $x < -2$ .

x < -2 \quad\text{or}\quad x > \tfrac{4}{3}

Watch out

Never write the answer to a "greater than" modulus inequality as $-2 > x > \tfrac{4}{3}$ . That chain claims $-2 > \tfrac{4}{3}$ , which is false, and it is marked wrong. Two separate pieces need the word "or" (or set notation $\{x : x < -2\} \cup \{x : x > \tfrac{4}{3}\}$ ).

Three methods for harder inequalities

When both sides contain $x$ , use one of these.

1. Square both sides. If both sides are known to be non-negative (two moduli, or a modulus and a positive multiple of a modulus), then $|A| < |B| \iff A^2 < B^2$ . You get a quadratic inequality, solved as in quadratics: find its roots, sketch the parabola, read off the region.

2. Critical values and a sketch. Solve the corresponding equation to find the $x$ -values where the two sides are equal. Sketch both graphs. Read off where one is above the other.

3. Test a value in each region. Once the critical values split the number line, pick one number in each region and test the original inequality. This is a reliable check on methods 1 and 2.

Solving a modulus inequality

Decide whether squaring is safe (both sides non-negative). If so, square and solve the quadratic inequality.
Otherwise, find critical values by solving the two cases of the equation, rejecting any that do not satisfy the original equation.
Sketch both sides on one diagram, or test a value in each region, to choose the correct intervals.
Write the answer with correct inequality signs: strict ( $<$ ) stays strict, non-strict ( $\le$ ) includes the critical values.

Two moduli: square

Modulus at least modulus

Solve $|x - 3| \ge |2x + 1|$ .

Solution

Both sides are non-negative, so squaring is valid:

(x - 3)^2 \ge (2x + 1)^2

x^2 - 6x + 9 \ge 4x^2 + 4x + 1

0 \ge 3x^2 + 10x - 8 = (3x - 2)(x + 4)

The parabola $y = (3x - 2)(x + 4)$ opens upwards with roots $-4$ and $\tfrac{2}{3}$ , so it is $\le 0$ between them:

-4 \le x \le \tfrac{2}{3}

Test $x = 0$ : $|-3| = 3 \ge |1| = 1$ . True, and $0$ is inside the interval, as expected.

y = abs(x - 3) y = abs(2x + 1) (-4, 0) -- (2/3, 0)

The narrower V, $y = |2x + 1|$ , lies below the wider one only between the two crossing points.

A multiple of a modulus, then hence

(a) Solve the inequality $|x - 4| < 2|3x + 1|$ .

(b) Hence solve the inequality $|3^y - 4| < 2|3^{y + 1} + 1|$ , giving your answer correct to 3 significant figures.

Solution

(a) Both sides are non-negative, so square:

(x - 4)^2 < 4(3x + 1)^2

x^2 - 8x + 16 < 36x^2 + 24x + 4

0 < 35x^2 + 32x - 12

Solve $35x^2 + 32x - 12 = 0$ : the discriminant is $32^2 + 4(35)(12) = 1024 + 1680 = 2704 = 52^2$ , so

x = \frac{-32 \pm 52}{70}, \qquad x = \tfrac{2}{7} \ \text{ or } \ x = -\tfrac{6}{5}.

The parabola opens upwards, so it is positive outside the roots:

x < -\tfrac{6}{5} \quad\text{or}\quad x > \tfrac{2}{7}

(b) Since $3^{y + 1} = 3 \times 3^y$ , this is part (a) with $x = 3^y$ . So we need $3^y < -\tfrac{6}{5}$ or $3^y > \tfrac{2}{7}$ .

$3^y > 0$ for all $y$ , so $3^y < -\tfrac{6}{5}$ is impossible. From $3^y > \tfrac{2}{7}$ :

y \ln 3 > \ln\tfrac{2}{7} \quad\Rightarrow\quad y > \frac{\ln(2/7)}{\ln 3} = -1.14 \ \text{(3 s.f.)}

Dividing by $\ln 3 > 0$ keeps the direction of the inequality.

Tip

When the coefficients of $x$ inside the two moduli have the same size, the $x^2$ terms cancel on squaring and you get a linear inequality. For example $|x - 2| > |x + 1|$ squares to $-4x + 4 > 2x + 1$ , so $x < \tfrac{1}{2}$ . Geometrically, $x$ is closer to $-1$ than to $2$ : everything left of the midpoint $\tfrac{1}{2}$ .

A modulus and a line: sketch

When one side is not a modulus, squaring is unsafe. Find where the line meets the V, then read the sketch.

Line against a V (syllabus example)

Solve the inequality $2x + 5 < |x + 1|$ .

Solution

Sketch $y = |x + 1|$ , a V with vertex $(-1, 0)$ , and the line $y = 2x + 5$ , gradient $2$ , $y$ -intercept $5$ .

Find where they meet.

Right arm, $y = x + 1$ for $x \ge -1$ : $2x + 5 = x + 1 \Rightarrow x = -4$ , but $-4$ is not on the right arm, so no meeting there.

Left arm, $y = -(x + 1)$ for $x < -1$ : $2x + 5 = -x - 1 \Rightarrow 3x = -6 \Rightarrow x = -2$ , which is on the left arm. Meeting point $(-2, 1)$ .

The line is steeper than the left arm and goes from below the V to above it as $x$ increases through $-2$ . So the line is below the V for $x < -2$ .

Check $x = -3$ : $2(-3) + 5 = -1$ and $|-2| = 2$ , so $-1 < 2$ . True.

x < -2

y = abs(x + 1) y = 2x + 5 (-2, 1)

Rejecting a critical value

Solve $|x - 2| < 2x - 1$ .

Solution

Critical values from the two cases of $|x - 2| = 2x - 1$ :

$x - 2 = 2x - 1 \Rightarrow x = -1$ . But then $2x - 1 = -3 < 0$ , so this is not a solution of the equation. Reject.

$-(x - 2) = 2x - 1 \Rightarrow 3 = 3x \Rightarrow x = 1$ . Check: $|-1| = 1 = 2(1) - 1$ . Valid.

So the graphs meet only at $x = 1$ . Test $x = 2$ : $|0| = 0 < 3$ . True. Test $x = 0$ : $2 < -1$ . False.

x > 1

Watch out

Squaring $|x - 2| < 2x - 1$ gives $(x - 2)^2 < (2x - 1)^2$ , i.e. $3x^2 - 3 > 0$ , so $x < -1$ or $x > 1$ . The piece $x < -1$ is wrong: there the right-hand side is negative and the original inequality fails. This is exactly the error examiners see most often. Squaring is only valid when both sides are non-negative.

When the line is parallel to an arm

A parallel arm

Solve $|x - 1| < 3 - x$ .

Solution

The line $y = 3 - x$ has gradient $-1$ , the same as the left arm $y = 1 - x$ . Parallel lines never meet, and $3 - x$ is always $2$ more than $1 - x$ , so on the left of the vertex the line is always above the V.

On the right arm, $x - 1 = 3 - x \Rightarrow x = 2$ . To the right of $x = 2$ the V rises and the line falls, so the V is above.

x < 2

Check $x = 0$ : $1 < 3$ , true. Check $x = 3$ : $2 < 0$ , false.

Several moduli (stretch)

With two moduli added together, split the number line at each critical value, exactly as for equations.

Sum of moduli

Solve $|x - 1| + |x + 2| < 5$ .

Solution

Critical values $x = 1$ and $x = -2$ .

For $x \ge 1$ : $(x - 1) + (x + 2) = 2x + 1 < 5 \Rightarrow x < 2$ . Combined with $x \ge 1$ : $1 \le x < 2$ .

For $-2 \le x < 1$ : $(1 - x) + (x + 2) = 3 < 5$ , always true. So all of $-2 \le x < 1$ .

For $x < -2$ : $(1 - x) - (x + 2) = -2x - 1 < 5 \Rightarrow x > -3$ . Combined: $-3 < x < -2$ .

Joining the pieces: $-3 < x < 2$ .

y = abs(x - 1) + abs(x + 2) y = 5

Common mistakes

Watch out

Using "and" logic for a "greater than" inequality. $|x| > 3$ is $x > 3$ or $x < -3$ . No single number is in both.

Watch out

Getting the region of the quadratic wrong. After squaring you have, say, $(3x - 2)(x + 4) \le 0$ . For an upward parabola, " $\le 0$ " is between the roots and " $\ge 0$ " is outside them. Sketch it every time.

Watch out

Dividing by a negative without flipping. $-2x < 6$ gives $x > -3$ . This bites when you rearrange a squared inequality and move everything to the side with a negative $x^2$ coefficient; it is safer to move terms so that the $x^2$ coefficient stays positive.

Watch out

Including or excluding endpoints wrongly. A strict inequality ( $<$ , $>$ ) gives strict answers; a non-strict one ( $\le$ , $\ge$ ) gives non-strict answers.

Exam tip

Mark schemes for these questions typically award: a mark for squaring (or forming two linear cases), a mark for a correct three-term quadratic, a mark for the critical values, and a final mark for the correct intervals with correct signs.
If you use the sketch method, a clear sketch of both graphs is itself credited. State the critical values you found from the sketch.
In "hence" questions, discard an impossible branch such as $3^y < -\tfrac{6}{5}$ with the reason " $3^y > 0$ ".
Express answers as simply as possible: $x < -\tfrac{6}{5}$ or $x > \tfrac{2}{7}$ , with exact fractions unless a decimal is requested.

Summary

$|x - a| < b \iff a - b < x < a + b$ ; $|x - a| > b \iff x < a - b$ or $x > a + b$ .
Solve $|ax + b| < c$ as $-c < ax + b < c$ , and $|ax + b| > c$ as two separate inequalities.
Two moduli (non-negative sides): square, solve the quadratic inequality, sketch the parabola to choose between or outside.
Modulus against a line: find critical values from the two cases, reject false ones, then sketch or test values.
If the inside coefficients match in size, squaring gives a linear inequality.
Answers with two pieces use "or", never a chained inequality.

Practice

Question

Solve $|x + 4| < 3$ .
Solve $|5 - 2x| \ge 1$ .
Solve $|x - 2| > |x + 1|$ .
Solve $|2x + 1| < |x - 3|$ .
Solve $3|x| > |x - 4|$ .
Solve $|x + 2| \le 2x + 1$ .
Solve $|3x - 1| > x + 3$ .
(a) Solve $|x - 3| < 2|x|$ . (b) Hence solve $|5^y - 3| < 2 \times 5^y$ .
Solve $|2x - 1| \ge |x + 1| + 1$ .

Answers

$-3 < x + 4 < 3$ , so $-7 < x < -1$ .
$5 - 2x \ge 1 \Rightarrow x \le 2$ , or $5 - 2x \le -1 \Rightarrow x \ge 3$ . Answer: $x \le 2$ or $x \ge 3$ .
Square: $x^2 - 4x + 4 > x^2 + 2x + 1 \Rightarrow 3 > 6x \Rightarrow x < \tfrac{1}{2}$ .
Square: $4x^2 + 4x + 1 < x^2 - 6x + 9 \Rightarrow 3x^2 + 10x - 8 < 0 \Rightarrow (3x - 2)(x + 4) < 0$ , so $-4 < x < \tfrac{2}{3}$ .
Square: $9x^2 > x^2 - 8x + 16 \Rightarrow 8x^2 + 8x - 16 > 0 \Rightarrow x^2 + x - 2 > 0 \Rightarrow (x + 2)(x - 1) > 0$ , so $x < -2$ or $x > 1$ .
Equation cases: $x + 2 = 2x + 1 \Rightarrow x = 1$ (check $3 = 3$ , valid); $-(x + 2) = 2x + 1 \Rightarrow x = -1$ , but then $2x + 1 = -1 < 0$ , reject. Test $x = 2$ : $4 \le 5$ , true; test $x = 0$ : $2 \le 1$ , false. Answer $x \ge 1$ .
Cases: $3x - 1 = x + 3 \Rightarrow x = 2$ (check $5 = 5$ ); $-(3x - 1) = x + 3 \Rightarrow -4x = 2 \Rightarrow x = -\tfrac{1}{2}$ (check $\left|-\tfrac{5}{2}\right| = \tfrac{5}{2} = -\tfrac{1}{2} + 3$ , valid). The V has arms of gradient $\pm 3$ , steeper than the line, so the V is above the line outside the critical values: $x < -\tfrac{1}{2}$ or $x > 2$ . (Squaring happens to give the same answer for this one, but you cannot know that in advance, so the safe method is cases plus a sketch.)
(a) Square: $x^2 - 6x + 9 < 4x^2 \Rightarrow 3x^2 + 6x - 9 > 0 \Rightarrow x^2 + 2x - 3 > 0 \Rightarrow (x + 3)(x - 1) > 0$ , so $x < -3$ or $x > 1$ . (b) With $x = 5^y$ (and $|5^y| = 5^y$ ): $5^y < -3$ is impossible, so $5^y > 1$ , giving $y > 0$ .
Critical values $x = \tfrac{1}{2}$ and $x = -1$ . For $x \ge \tfrac{1}{2}$ : $2x - 1 \ge x + 2 \Rightarrow x \ge 3$ . For $-1 \le x < \tfrac{1}{2}$ : $1 - 2x \ge x + 2 \Rightarrow x \le -\tfrac{1}{3}$ , giving $-1 \le x \le -\tfrac{1}{3}$ . For $x < -1$ : $1 - 2x \ge -x - 1 + 1 = -x \Rightarrow 1 \ge x$ , always true here, giving $x < -1$ . Combined: $x \le -\tfrac{1}{3}$ or $x \ge 3$ .

The distance picture

Three methods for harder inequalities

Two moduli: square

A modulus and a line: sketch

When the line is parallel to an arm

Several moduli (stretch)

Common mistakes

Summary

Practice

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