Mathematics 9709 · Pure Mathematics 2

The modulus function

$|x|$ just means how far x is from 0. Drop the minus sign, keep the number.
This one idea runs the whole topic. Every modulus equation splits into two cases. Some answers are fakes you throw out. The graph is a V that never goes below the axis.

Read every modulus as a distance. Anything with $|\,\cdots\,|$ asks “how far is this, and which side is it on?” That gives two answers.

What |x| really means: distance from zero

The modulus (or absolute value) of a number is the number with no minus sign: $|3|=3$ and $|-3|=3$ .
It is never negative. A positive number stays the same; a negative one flips to positive.

$|x|=\begin{cases} x & \text{if } x\ge 0\\[2pt] -x & \text{if } x<0\end{cases}$

y = |x| is the line y = x with the part below the axis folded up: a V with its point at 0.

One fact saves marks: a modulus is never negative.
So $|x|=-2$ has no answer. If your working ends with a modulus equal to a negative number, you made a mistake.

See |…| = a number, split into ±

$|x|=k$ (with $k\ge 0$ ) means x is $k$ away from 0. That can be either side, so $x=k$ or $x=-k$ .
A bracket works the same way: $|ax+b|=k$ splits into $ax+b=k$ or $ax+b=-k$ .

$|\text{stuff}|=k$ (a non-negative number) $\Rightarrow$ set the inside to $+k$ and to $-k$ , then solve each one. Two equations, two answers.

Worked example

Solve $|2x-1|=3$ (Worked example 1.1, textbook)

The inside is 3 away from 0, so it is $+3$ or $-3$ .

First case: $2x-1=3\ \Rightarrow\ 2x=4\ \Rightarrow\ x=2$ .
Second case: $2x-1=-3\ \Rightarrow\ 2x=-2\ \Rightarrow\ x=-1$ .

Answer

x=2\quad\text{or}\quad x=-1

See |…| = |…|, square both sides

Two moduli are equal when their sizes match: $|a|=|b|\iff a^2=b^2$ .
Squaring removes both modulus signs at once and leaves a normal equation.

Worked example

Solve $|3x+4|=|x+5|$

Square both sides to remove both modulus signs.

$(3x+4)^2=(x+5)^2\ \Rightarrow\ 9x^2+24x+16=x^2+10x+25$ .
Collect everything on one side: $8x^2+14x-9=0$ .
Factorise: $(2x-1)(4x+9)=0$ .

Answer

x=\tfrac12\quad\text{or}\quad x=-\tfrac94

Why squaring is safe here (and when it isn't)

Squaring is safe when both sides are moduli. Both are zero or positive, so squaring keeps the right answers and adds no fake ones. The problem is when one side is a plain expression like $cx+d$ that could be negative. Then squaring can add a fake root. That is why the next section makes you check. Rule: square only when both sides are zero or positive for sure.

See |…| = something that can go negative: check every root

When the other side is an expression like $cx+d$ , still split into $ax+b=cx+d$ or $ax+b=-(cx+d)$ . But now an “answer” can be a fake.
A modulus is never negative. So throw out any root that makes the other side negative.

Worked example

Solve $|x-4|=2x+1$ (Worked example 1.1b, textbook)

First case: $x-4=2x+1\ \Rightarrow\ -5=x$ , so $x=-5$ .
Second case: $x-4=-(2x+1)\ \Rightarrow\ x-4=-2x-1\ \Rightarrow\ 3x=3$ , so $x=1$ .
Check both in the original. At $x=-5$ the right side is $2(-5)+1=-9$ — negative, impossible for a modulus, so reject it. At $x=1$ : $|1-4|=3$ and $2(1)+1=3$ ✓.

Answer

x=1

Common mistake

Splitting or squaring can add fake roots. Whenever the side opposite the modulus can be negative, put every answer back in and keep only the ones that work. Skipping this check is the most common way to lose the last mark here.

Now you try

Solve $|2x-3|=|5x+1|$ (a two-moduli equation, like the textbook review). (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

Both sides are moduli, so square them: $(2x-3)^2=(5x+1)^2$ , i.e. $4x^2-12x+9=25x^2+10x+1$ . Collect: $21x^2+22x-8=0\Rightarrow(3x+4)(7x-2)=0$ .

Answer

x=-\tfrac43\quad\text{or}\quad x=\tfrac27

Now one that needs a check: solve $|x+3|+|x+5|=10$ . (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

Work it region by region (or get one modulus on its own and check). The two answers that pass the check:

Answer

x=1\quad\text{or}\quad x=-9

NextGraphs of y = |f(x)|

The modulus function

$|x|$ just means how far x is from 0. Drop the minus sign, keep the number.
This one idea runs the whole topic. Every modulus equation splits into two cases. Some answers are fakes you throw out. The graph is a V that never goes below the axis.

Read every modulus as a distance. Anything with $|\,\cdots\,|$ asks “how far is this, and which side is it on?” That gives two answers.

What |x| really means: distance from zero

The modulus (or absolute value) of a number is the number with no minus sign: $|3|=3$ and $|-3|=3$ .
It is never negative. A positive number stays the same; a negative one flips to positive.

$|x|=\begin{cases} x & \text{if } x\ge 0\\[2pt] -x & \text{if } x<0\end{cases}$

y = |x| is the line y = x with the part below the axis folded up: a V with its point at 0.

One fact saves marks: a modulus is never negative.
So $|x|=-2$ has no answer. If your working ends with a modulus equal to a negative number, you made a mistake.

See |…| = a number, split into ±

$|x|=k$ (with $k\ge 0$ ) means x is $k$ away from 0. That can be either side, so $x=k$ or $x=-k$ .
A bracket works the same way: $|ax+b|=k$ splits into $ax+b=k$ or $ax+b=-k$ .

$|\text{stuff}|=k$ (a non-negative number) $\Rightarrow$ set the inside to $+k$ and to $-k$ , then solve each one. Two equations, two answers.

Worked example

Solve $|2x-1|=3$ (Worked example 1.1, textbook)

The inside is 3 away from 0, so it is $+3$ or $-3$ .

First case: $2x-1=3\ \Rightarrow\ 2x=4\ \Rightarrow\ x=2$ .
Second case: $2x-1=-3\ \Rightarrow\ 2x=-2\ \Rightarrow\ x=-1$ .

Answer

x=2\quad\text{or}\quad x=-1

See |…| = |…|, square both sides

Two moduli are equal when their sizes match: $|a|=|b|\iff a^2=b^2$ .
Squaring removes both modulus signs at once and leaves a normal equation.

Worked example

Solve $|3x+4|=|x+5|$

Square both sides to remove both modulus signs.

$(3x+4)^2=(x+5)^2\ \Rightarrow\ 9x^2+24x+16=x^2+10x+25$ .
Collect everything on one side: $8x^2+14x-9=0$ .
Factorise: $(2x-1)(4x+9)=0$ .

Answer

x=\tfrac12\quad\text{or}\quad x=-\tfrac94

Why squaring is safe here (and when it isn't)

See |…| = something that can go negative: check every root

When the other side is an expression like $cx+d$ , still split into $ax+b=cx+d$ or $ax+b=-(cx+d)$ . But now an “answer” can be a fake.
A modulus is never negative. So throw out any root that makes the other side negative.

Worked example

Solve $|x-4|=2x+1$ (Worked example 1.1b, textbook)

First case: $x-4=2x+1\ \Rightarrow\ -5=x$ , so $x=-5$ .
Second case: $x-4=-(2x+1)\ \Rightarrow\ x-4=-2x-1\ \Rightarrow\ 3x=3$ , so $x=1$ .
Check both in the original. At $x=-5$ the right side is $2(-5)+1=-9$ — negative, impossible for a modulus, so reject it. At $x=1$ : $|1-4|=3$ and $2(1)+1=3$ ✓.

Answer

x=1

Common mistake

Now you try

Solve $|2x-3|=|5x+1|$ (a two-moduli equation, like the textbook review). (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

Both sides are moduli, so square them: $(2x-3)^2=(5x+1)^2$ , i.e. $4x^2-12x+9=25x^2+10x+1$ . Collect: $21x^2+22x-8=0\Rightarrow(3x+4)(7x-2)=0$ .

Answer

x=-\tfrac43\quad\text{or}\quad x=\tfrac27

Now one that needs a check: solve $|x+3|+|x+5|=10$ . (9709-style)

Your turn— tap to reveal the worked answer (9709-style)

Work it region by region (or get one modulus on its own and check). The two answers that pass the check:

Answer

x=1\quad\text{or}\quad x=-9