Modulus as Distance and Equations | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Modulus is distance from zero

The two bars in $|u|$ mean the modulus, or absolute value, of $u$ : its distance from zero. Both 5 and −5 are five units away. Zero is zero units away.

\begin{gathered}|5|=5,\quad|-5|=5\\|0|=0\end{gathered}

The result is non-negative: positive or zero. Work out the whole expression inside the bars before taking its distance.

|2-7|=|-5|=5\ne |2|-|7|

This gives a rule with two cases. Read the brace as “use the line whose condition is true”:

|u|=\begin{cases}u&u\ge0,\\-u&u<0.\end{cases}

Here $-u$ makes a negative input positive: if $u=-5$ , then $-u=-(-5)=5$ . It does not mean the modulus is negative.

Your turnPractice

Find $|3-8|$ and $-|3-8|$ . Why do they have different signs?

Show worked answer

Answer

5\quad\text{and}\quad-5

The second minus sign is outside the bars, so it acts after the modulus.

A positive distance gives two positions

$|x-a|$ is the distance between $x$ and $a$ . For example, $|x-3|=2$ puts $x$ two units to either side of 3.

|x − 3| = 2 gives x = 1 or x = 5.

Worked example

Solve |2x − 3| = 7.

The inside can be 7 or −7. Solve both linear equations.

\begin{aligned}2x-3=7&\Rightarrow x=5,\\2x-3=-7&\Rightarrow x=-2.\end{aligned}

Check: the inside becomes 7 and −7; both have modulus 7.

If the right side is zero, the inside must be zero: one position, not two different roots. A negative right side is impossible.

\begin{aligned}|x-3|=0&\Rightarrow x=3,\\|x-3|=-2&\Rightarrow\text{no solution}.\end{aligned}

Change $k$ to zero and then below zero. Predict the number of intersections before moving it.

Centre a3Comparison value k2

Two solutions: x = 1 or x = 5

At k = 0 the two intersections coincide; below zero there are none.

Your turnPractice

Solve $|3-2x|=5$ . What changes if the right side is 0 or −5?

Show worked answer

3-2x=5\ \text{or}\ 3-2x=-5

Answer

x=-1\quad\text{or}\quad x=4

With 0, $x=3/2$ only. With −5, there is no real solution.

Modulus is distance from zero

The two bars in $|u|$ mean the modulus, or absolute value, of $u$ : its distance from zero. Both 5 and −5 are five units away. Zero is zero units away.

\begin{gathered}|5|=5,\quad|-5|=5\\|0|=0\end{gathered}

The result is non-negative: positive or zero. Work out the whole expression inside the bars before taking its distance.

|2-7|=|-5|=5\ne |2|-|7|

This gives a rule with two cases. Read the brace as “use the line whose condition is true”:

|u|=\begin{cases}u&u\ge0,\\-u&u<0.\end{cases}

Here $-u$ makes a negative input positive: if $u=-5$ , then $-u=-(-5)=5$ . It does not mean the modulus is negative.

Your turnPractice

Find $|3-8|$ and $-|3-8|$ . Why do they have different signs?

Show worked answer

Answer

5\quad\text{and}\quad-5

The second minus sign is outside the bars, so it acts after the modulus.

A positive distance gives two positions

$|x-a|$ is the distance between $x$ and $a$ . For example, $|x-3|=2$ puts $x$ two units to either side of 3.

|x − 3| = 2 gives x = 1 or x = 5.

Worked example

Solve |2x − 3| = 7.

The inside can be 7 or −7. Solve both linear equations.

\begin{aligned}2x-3=7&\Rightarrow x=5,\\2x-3=-7&\Rightarrow x=-2.\end{aligned}

Check: the inside becomes 7 and −7; both have modulus 7.

If the right side is zero, the inside must be zero: one position, not two different roots. A negative right side is impossible.

\begin{aligned}|x-3|=0&\Rightarrow x=3,\\|x-3|=-2&\Rightarrow\text{no solution}.\end{aligned}

Change $k$ to zero and then below zero. Predict the number of intersections before moving it.

Centre a3Comparison value k2

Two solutions: x = 1 or x = 5

At k = 0 the two intersections coincide; below zero there are none.

Your turnPractice

Solve $|3-2x|=5$ . What changes if the right side is 0 or −5?

Show worked answer

3-2x=5\ \text{or}\ 3-2x=-5

Answer

x=-1\quad\text{or}\quad x=4

With 0, $x=3/2$ only. With −5, there is no real solution.