A LevelFurther MathsEdexcelFurther Pure 1Revision NotesInequalitiesInequalitiesSolving Inequalities involving Modulus Functions

Solving Inequalities involving Modulus Functions (Edexcel A Level Further Maths: Further Pure 1): Revision Note

Exam code: 9FM0

Written by: Mark Curtis

Updated on 8 January 2026

Solving inequalities involving modulus functions

How do I solve modulus inequalities where the graphs are easy to sketch?

The easiest way to solve a modulus inequality
- e.g. $| x^{2} - 1 | < x$
  - is to sketch both sides of the inequality on the same axes
Recall that to sketch $y = | f (x) |$
- first sketch $y = f (x)$
  - then reflect anything below the $x$ -axis
Then use simultaneous equations to find the critical points
- The $x$ -coordinates of intersection
Use the sketch
- to help select the correct equation to use out of $| f (x) | = \{\begin{cases} f (x) \\ - f (x) \end{cases}$
- to identify
  - the ranges of $x$ that satisfy the inequality
  - any solutions that must be excluded
  - any asymptotes

Examiner Tips and Tricks

Exam questions may refer to this method as "using algebra", even though it requires a sketch!

Worked Example

Use algebra to determine the values of $x$ for which

(a) $| x^{2} - 25 | < 2 x + 1$

(b) $| x^{2} - 25 | < | 2 x + 1 |$

Answer:

(a)

Sketch the graph of $y = | x^{2} - 25 |$

It is the graph of $y = (x + 5) (x - 5)$
with the part below the $x$ -axis reflected

A graph of y=|x^2-25| which is the negative quadratic y=25-x^2 between -5 and 5 and the positive quadratic y=x^2-25 either side of this region.

Add the graph of $y = 2 x + 1$ on to the same axes

A graph of y=|x^2-25| which is the negative quadratic y=25-x^2 between -5 and 5 and the positive quadratic y=x^2-25 either side of this region. The graph of y=2x+1 is shown, with a positive gradient and y-intercept of 1, intersecting the other graph twice in the positive quadrant.

Now use algebra to find the $x$ -coordinates of the points of intersection

Use the graph to identify where the modulus function is the positive equation and where it is the negative equation

$| x^{2} - 25 | = \{\begin{cases} x^{2} - 25 for x \leq - 5 or x \geq 5 \\ - (x^{2} - 25) for - 5 < x < 5 \end{cases}$

Solve the simultaneous equations $y = | x^{2} - 25 |$ and $y = 2 x + 1$ in the two different cases above

Case 1 is when $x \leq - 5 or x \geq 5$

i.e. $| x^{2} - 25 | = x^{2} - 25$

$\begin{array}{rcl} x^{2} - 25 & = & 2 x + 1 \\ x^{2} - 2 x - 26 & = & 0 \end{array}$

Solve the quadratic equation, e.g. by completing the square

$\begin{array}{rcl} {(x - 1)}^{2} - 1 - 26 & = & 0 \\ {(x - 1)}^{2} & = & 27 \\ x - 1 & = & \pm \sqrt{27} \\ x - 1 & = & \pm 3 \sqrt{3} \\ x & = & 1 \pm 3 \sqrt{3} \end{array}$

Check the solutions satisfy the requirement of case 1, i.e. $x \leq - 5 or x \geq 5$

$1 + 3 \sqrt{3} = 6.19615 . . . \geq 5$

but $1 - 3 \sqrt{3} = - 4.19615 . . .$ (not $x \leq - 5 or x \geq 5$ )

Case 2 is when $- 5 < x < 5$

i.e. $| x^{2} - 25 | = - (x^{2} - 25) = 25 - x^{2}$

$\begin{array}{rcl} 25 - x^{2} & = & 2 x + 1 \\ 0 & = & x^{2} + 2 x - 24 \end{array}$

Solve the quadratic equation, e.g. by factorisation

$\begin{array}{rcl} 0 & = & (x + 6) (x - 4) \\ x & = & - 6 or x = 4 \end{array}$

Check the solutions satisfy the requirement of case 2, i.e. $- 5 < x < 5$

$- 5 < 4 < 5$

but $- 6$ is not in $- 5 < x < 5$

The solutions $x = 1 + 3 \sqrt{3}$ and $x = 4$ are included

The solutions $x = 1 - 3 \sqrt{3}$ and $x = - 6$ are not included

This can be seen by extending the quadratics beyond their ranges

From the two solutions that are included, look at the sketch of the graph and identify the region where $y = 2 x + 1$ is greater than or equal to (i.e. vertically above) $y = | x^{2} - 25 |$

The original inequality in the question was strict, so give the final answer as a strict inequality

$4 < x < 1 + 3 \sqrt{3}$

(b)

This time, the graph of $y = 2 x + 1$ has changed to $y = | 2 x + 1 |$

A graph of y=|x^2-25| which is the negative quadratic y=25-x^2 between -5 and 5 and the positive quadratic y=x^2-25 either side of this region. The V-shaped graph of y=|2x+1| is shown. The two points of intersection on the straight line with the positive gradient are x=4 and x=1+3*sqrt(3) are shown. Two points of intersection on the straight line with the negative gradient are shown.

As well as the critical points $4$ and $1 + 3 \sqrt{3}$ in part (a), two more critical points appear due to the line $y = - (2 x + 1)$

Solve simultaneously $y = - (2 x + 1)$ and $y = x^{2} - 25$

$\begin{array}{rcl} - (2 x + 1) & = & x^{2} - 25 \\ - 2 x - 1 & = & x^{2} - 25 \\ 0 & = & x^{2} + 2 x - 24 \end{array}$

Solve this quadratic by factorisation (or use the working from part (a))

$(x + 6) (x - 4) = 0 x = - 6 or x = 4$

Choose the correct critical point for this region, $x \leq - 5$ (use the sketch to help)

$x = - 6$

Next, solve simultaneously $y = - (2 x + 1)$ and $y = - (x^{2} - 25)$

$\begin{array}{rcl} - (2 x + 1) & = & - (x^{2} - 25) \\ - 2 x - 1 & = & - x^{2} + 25 \\ x^{2} - 2 x - 26 & = & 0 \end{array}$

Solve this quadratic equation, e.g. by completing the square (or use the working from part (a))

$\begin{array}{rcl} {(x - 1)}^{2} - 1 - 26 & = & 0 \\ {(x - 1)}^{2} & = & 27 \\ x - 1 & = & \pm \sqrt{27} \\ x - 1 & = & \pm 3 \sqrt{3} \\ x & = & 1 \pm 3 \sqrt{3} \end{array}$

Choose the correct critical point for this region $- 5 < x < 5$ (use the sketch to help)

$x = 1 - 3 \sqrt{3}$

Look at the sketch of the graph and identify the regions where $y = | 2 x + 1 |$ is greater than or equal to (i.e. vertically above) $y = | x^{2} - 25 |$

$- 6 < x < 1 - 3 \sqrt{3}$ or $4 < x < 1 + 3 \sqrt{3}$

How do I solve modulus inequalities where the graphs are not easy to sketch?

If the graph of a modulus inequality is hard to sketch
- the graph will be given to you in the exam
However, you still need to
- use simultaneous equations to find the critical points
  - The $x$ -coordinates of intersection
Use the sketch
- to help select the correct equation to use out of $| f (x) | = \{\begin{cases} f (x) \\ - f (x) \end{cases}$
- to identify
  - the ranges of $x$ that satisfy the inequality
  - any solutions that must be excluded
  - any asymptotes

Examiner Tips and Tricks

Exam questions may refer to this method as "using algebra", even though a sketch is given!

Worked Example

The graph of $y = \frac{x - 1}{2 - | x - 1 |}$ is shown below.

The graph of y=(x-1)/(2-|x-1|) shown in three parts. A left-hand backwards L branch above the x-axis up to the vertical asymptote x=-1. A middle cubic-looking part between the asymptotes of x=-1 and x=3. A final right-hand upside-down L branch between the asymptote x=3 below the x-axis.

Use algebra to determine the values of $x$ for which

$5 - 2 x \geq \frac{x - 1}{2 - | x - 1 |}$

Answer:

Sketch the line $y = 5 - 2 x$ to see where the points of intersection are

Don't worry if it's not exactly in the right position

Now use algebra to find the $x$ -coordinates of the points of intersection

First, identify where the modulus function is the positive equation and the negative equation

$| x - 1 | = \{\begin{cases} x - 1 for x \geq 1 \\ - (x - 1) for x < 1 \end{cases}$

Solve the simultaneous equations $y = 5 - 2 x$ and $y = \frac{x - 1}{2 - | x - 1 |}$ in the two different cases above

Case 1 is when $x \geq 1$ (i.e. $| x - 1 | = x - 1$ )

$\begin{array}{rcl} 5 - 2 x & = & \frac{x - 1}{2 - (x - 1)} \\ 5 - 2 x & = & \frac{x - 1}{2 - x + 1} \\ 5 - 2 x & = & \frac{x - 1}{3 - x} \end{array}$

Multiply both sides by $3 - x$ and simplify

$\begin{array}{rcl} (5 - 2 x) (3 - x) & = & x - 1 \\ 15 - 5 x - 6 x + 2 x^{2} & = & x - 1 \\ 2 x^{2} - 12 x + 16 & = & 0 \\ x^{2} - 6 x + 8 & = & 0 \end{array}$

Solve the quadratic, e.g. by factorisation

$\begin{array}{rcl} (x - 2) (x - 4) & = & 0 \\ x & = & 2 or x = 4 \end{array}$

Check the solutions satisfy the requirement of case 1, i.e. $x \geq 1$

$2 > 1$ and $4 > 1$

Case 2 is when $x < 1$ (i.e. $| x - 1 | = - (x - 1)$ )

$\begin{array}{rcl} 5 - 2 x & = & \frac{x - 1}{2 - (- (x - 1))} \\ 5 - 2 x & = & \frac{x - 1}{2 + (x - 1)} \\ 5 - 2 x & = & \frac{x - 1}{x + 1} \end{array}$

Multiply both sides by $x + 1$ and simplify

$\begin{array}{rcl} (5 - 2 x) (x + 1) & = & x - 1 \\ 5 x + 5 - 2 x^{2} - 2 x & = & x - 1 \\ 0 & = & 2 x^{2} - 2 x - 6 \\ 0 & = & x^{2} - x - 3 \end{array}$

Solve the quadratic, e.g. by using the quadratic formula

$\begin{array}{rcl} x & = & \frac{1 \pm \sqrt{13}}{2} \end{array}$

Check the solutions satisfy the requirement of case 2, i.e. $x < 1$

$\frac{1 - \sqrt{13}}{2} = - 1.30277 . . . < 1$ but $\frac{1 + \sqrt{13}}{2} = 2.30277 . . . > 1$

only $\frac{1 - \sqrt{13}}{2}$ is a solution

Write out all the critical points ( $x$ -coordinates of intersection) in order

$x = \frac{1 - \sqrt{13}}{2}, 2 and 4$

Now look at the sketch of the graph and identify the regions where $y = 5 - 2 x$ is greater than or equal to (i.e. vertically above) $y = \frac{x - 1}{2 - | x - 1 |}$

The two vertical asymptotes are shown

They can also be found by setting the denominator equal to zero

$\begin{array}{rcl} 2 - | x - 1 | & = & 0 \\ 2 & = & | x - 1 | \\ 2 & = & \pm (x - 1) \\ \pm 2 + 1 & = & x \\ x & = & - 1 or x = 3 \end{array}$

If the end points of the ranges of $x$ -values that satisfy the inequality include an asymptote, this value must be excluded

$x \neq - 1$ and $x \neq 3$

This gives the solution with the correct inequality signs

$x \leq \frac{1 - \sqrt{13}}{2}$ or $- 1 < x \leq 2$ or $3 < x \leq 4$

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