Algebra

Written on 27 June 2020.

Negative Numbers

Sections:

0, Chapter 1, Introduction,
Introduction

You are no doubt familiar with negative numbers. In this unit we will review the rules for manipulating such numbers.

The unit aims to give participants an understanding of
- the concept of negative numbers
- the addition and subtraction of negative numbers
- multiplication and division of negative numbers.
There are three sections in this Unit, namely:
1. Negative Numbers
2. Addition and Subtraction
3. Multiplication and Division
0, Chapter 2, Negative Numbers,
Negative Numbers

We start with a quick review of key facts:

Remember

Positive numbers are numbers greater than zero.
Negative numbers are numbers less than zero.
Zero is neither positive nor negative.

A number line illustrates positive and negative numbers. Negative numbers are always written with a '−' sign in front of them. They are counted down from zero to the left on a number line.

Note: You can extend this number line as far as you like/need in either direction.
1, Chapter 2, Task 1, Examples of Negative Numbers,
Examples of Negative Numbers

Here are some examples of negative numbers:

Temperature:
Thermometers measure both negative and positive values.
For example, the maximum temperature reached in Hull last year was
\(32\)°C, and the minimum \(- 9\)°C.

Altitude:
Places below sea level (such as the Dead Sea in Israel, and Death
Valley in California) have negative values for altitude; places above
sea level have positive values for altitude.

Sports:
For example, in golf ‘\(3\) below par’ is recorded as \(-3\).

Money:
If the amount of money in a bank account goes ‘into the red’, then the
bank account holder owes the bank money (that is, they are in debt);
this amount is shown by a negative value, for example: \(-\)£\(234.25\).
1, Chapter 2.1, Inequalities Signs,
Inequalities Signs

We also use inequality signs with negative numbers.

The sign \(\lt\) means “is less than” as, for example \(4 \lt 7\).
The sign \(\gt\) means “is greater than” as in \(8 \gt 4\).

You can see, using the number line that:

\(\begin{align*} -2 &\lt 5\\ -7 &\lt -3\\ 5 &\gt -4\\ -2 &\gt -8 \end{align*}\)

Based on the above, is this statement true or false?

\(3 \gt -4\)
1, Chapter 2.1, Task 1, Fitness Check 1,
Exercises

Exercise 1

Put inequality signs, \(\lt\) or \(\gt\), between each pair of numbers to give a true statement:

a. \(-3\)       \(4\)

b. \(-6\)       \(-7\)

c. \(2\)       \(-4\)

Exercise 2

Write the set of numbers in order with the smallest first:

\(3\), \(-2\), \(8\), \(0\), \(-1\), \(1\), \(-3\)

Exercise 3

Write down all integers that could go in box below to make the statement true:

\(-2 \lt\) ............... \(\lt 4\)
0, Chapter 3, Adding and Subtraction,
Adding and subtracting positive and negative numbers

Remember

To add and subtract numbers always begin counting from zero.
To add a positive number, move to the right on a number line.
To add a negative number, move to the left on a number line.

Negative Numbers: Adding

Let's calculate \(-3 + 8\)

To add -3 and 8, draw the starting number on a number line and count to the right when adding numbers. We start at \(-3\) and move \(8\) to the right so, \(-3 + 8 = 5\)

Negative Numbers: Substracting

What is \(5 - 7\)?

To subtract \(7\) from \(5\), start at \(5\) on a number line, then count \(7\) to the left. This will give you the answer \(-2\)

Here is another example, let's calculate \(4 - \left(-2\right)\).

We start at \(4\) and move \(2\) to the right so \(4 - \left(-2\right) = 6\)

1, Chapter 3, Task 1, Exercises 2,

Exercises

Exercise 1

Calculate

a. \(-5 + 8\)

b. \(-6 + 6\)

c. \(5 - 7\)

d. \(5 - (-6)\)

Exercise 2

Write done two possible calculations shown below:

Exercise 3

Copy and complete this addition table; the first row has been done.

\(+\)	\(-4\)	\(-2\)	\(0\)	\(2\)	\(4\)
\(-3\)	\(-7\)	\(-5\)	\(-3\)	\(-1\)	\(1\)
\(-1\)
\(1\)
\(3\)

Exercise 4

Overnight the temperature dropped from \(5\)°C to \(−14\)°C.
How many degrees did it fall?

0, Chapter 4, Multiplication and Division,
Negative Numbers: Multiplication & Division

The rule for multiplying and dividing is very similar to the rule for adding and subtracting.
- When the signs are different the answer is negative.
- When the signs are the same the answer is positive.
Remember

A positive number \(\times\) a negative number gives a negative;
a negative number \(\times\) a negative number gives a positive ;

The table on the left shows what happens to the sign of the answer when positive and negative numbers are multiplied; the same table can be used for division of positive and negative numbers.

For example, \(5 \times (-7) = -35\) and \((-8) \times (-7) = 56\).

Similarly, \(18 \div (-3) = -6\) and \((-21) \div (-3) = 7\).

1, Chapter 4, Task 1, Exercises 3,

Exercises

Exercise 1

Copy and complete this multiplication grid:.

\(\times\)	\(1\)	\(2\)	\(3\)	\(4\)
\(-1\)
\(-2\)
\(-3\)
\(-4\)

Exercise 2

Calculate these multiplications and divisions:

a. \((-6) \times 3\)

b. \((-3) \times (-8)\)

c. \((-6) \div 3\)

d. \(8 \div (-2)\)

Exercise 3

Calculate \((-6 + 10) \div (-2)\)

0, Chapter 5, Summary,
Summary

Number line

A number line is shown below.

Note that we can use inequalities signs, \(\lt\) and \(\gt\) to write that

\(2 \gt −1\) (2 is greater than -1)

\(-3 \lt 3\) (-3 is less than 3)

\(-5 \lt 2\) (-5 is less than -2)

\(-1 \gt 4\) (-1 is greater than -4)

Addition and subtraction

A number line can be used to help in calculation. For example,

\(2-3=-1\)

\(-3+6=3\)

\(-2+-3=-5\)

Multiplication and division

Note that

\(\left(−1\right) \times \left(−1\right) = 1\)
\(\left(-1\right) \times 1 = -1\)
\(1\times \left(−1\right) = -1\)
\(1\times 1=1\)

and similarly for division; for example,

\(\left(−4\right) \times 2= -8,\;\;\left(−4\right) \times \left(−2\right)= 8\)
\(4\times \left(−2\right) = -8,\;\;4\times 2=8\)
\(\left(−4\right) \div 2=-2,\;\;\left(−4\right) \div \left(−2\right)= 2\)
\(4\div \left(−2\right) = -2,\;\;4\div2=2\)

Interactive Exercises:

Negative Numbers Interactive Exercises, https://www.cimt.org.uk/sif/number/n6/interactive.htm
Negative Numbers, https://www.cimt.org.uk/sif/number/n6/interactive/s1.html
Addition and Substraction, https://www.cimt.org.uk/sif/number/n6/interactive/s2.html
Multiplication and Division, https://www.cimt.org.uk/sif/number/n6/interactive/s3.html

YouTube URL: https://youtu.be/osu0a5vRokA

External Links:

Negative Numbers Quiz, https://www.bbc.co.uk/bitesize/articles/zjbk8xs, images/Images/bitesize-logo.png, For more on negative numbers and a fun quiz try this link to BBC bitesize
Multiplying Positive and Negative Numbers, https://youtu.be/QmarTb7wXro, images/Images/math-club.png, Think your family is weird? This family's eccentricities get multiplied, like, tenfold.

File Attachments: media/course-resources/Negative-Numbers_Essential-Information.pdfmedia/course-resources/Negative-Numbers_Learning-Objectives.pdfmedia/course-resources/Negative-Numbers_PowerPoint-Presentation.pptxmedia/course-resources/Negative-Numbers_Text.pdfmedia/course-resources/Negative-Numbers_Answers.pdf

Written on 30 June 2020.

Sine and Cosine Rules

Sections:

0, Chapter 1, Introduction,
Introduction

This unit extends the theme of finding angles and lengths in a right angle triangle to finding them in any triangle. After completing this unit you should
- understand how to use the sine and cosine rules in non right angled triangles
- be able to find the area of any triangles, given two sides and the included angle
- be able to use Heron's formula to calculate the area of any triangle, given the lengths of its three sides.
You have five sections to work through.
1. Sine Rule
2. Cosine Rule
3. Sine and Cosine
4. Application: Area of Any Triangle
5. Heron's Formula
6. Angles larger than \(90^{\circ}\)
0, Chapter 2, Sine Rule,
Sine Rule

In the triangle \(ABC\) shown below, the side opposite angle \(A\) has length \(a\), the side opposite angle \(B\) has length \(b\) and the side opposite angle \(C\) has length \(c.\)
Remember
The sine rule states \(\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}\)

The sine rule can be used in any triangle to calculate:
- a side when two angles and an opposite side are known \(\left(AAS\right)\)
- an angle when two sides and an opposite angle are known \(\left(SSA\right)\)
The cosine rule states:
\( a^{2} = b^{2} + c^{2} - 2bc \; cos \; A \)
\( b^{2} = a^{2} + c^{2} - 2ac \; cos \;B \)
\( c^{2} = a^{2} + b^{2} - 2ab \; cos \; C \)
Note that when, for example \( A = 90^{\circ} \) , the formula becomes, as expected, Pythagoras’ Theorem: \( a^{2} = b^{2} + c^{2} \)

The cosine rule can be used in any triangle to calculate:
- a side when two sides and the angle in between them are known \( (SAS) \)
- an angle when three sides are known \( (SSS) \)
1, Chapter 2, Task 1, Worked Example 1,
Worked Example

Use the slider to explore worked examples.
- Worked Example
  
  Find the unknown angles and side length of the triangle shown.
- Worked Example
  
  Find two solutions for the unknown angles and side of the triangle shown.
- Worked Example
  
  Find the unknown side and angles of the triangle shown.
0, Chapter 3, Cosine Rule,
Cosine Rule

In the triangle \(ABC\) shown below, the side opposite angle \(A\) has length \(a\), the side opposite angle \(B\) has length \(b\) and the side opposite angle \(C\) has length \(c\).

Remember

The cosine rule states:
\(a^2=b^2+c^2-2bc\cos A\)
\(b^2=a^2+c^2-2ac\cos B\)
\(c^2=a^2+b^2-2ab\cos C\)

The cosine rule can be used in any triangle to calculate:
- a side when two sides and the angle in between them are known \(\left(SAS\right)\)
- an angle when three sides are known \(\left(SSS\right)\)
1, Chapter 3, Task 1, Worked Example 2,
Worked Example

Use the slider to explore worked examples.
- Worked Example
  
  Find the unknown side and angles of the triangle shown.
- Worked Example
  
  Find the shaded angle in the triangle shown.
1, Chapter 3, Task 2, Worked Example 3,
Worked Example
- Worked Example
  
  Find the unknown side and angles of the triangle shown.

0, Chapter 4, Sin and Cos,

Sin and Cos

sin\(\theta\) and cos\(\theta\)

The graphs of \(\sin \theta\) and \(\cos \theta\) for any angle are demonstrated iin the diagrams below which show the sine and cosine curves.

The graphs are examples of periodic functions. The graph repeats every \(360^\circ\) (This is called the period , ie the period is \(360^\circ\)). In each period the maximum and minimum value of the functions are \(1\) and \(-1\) respectively.

Below are the values of sine and cosine for angles which often appear in questions.

\( \theta\)	\(\sin \theta\)	\(\cos \theta\)
\(0^\circ\)	\(0\)	\(1\)
\(30^\circ\)	\(\frac{1}{2} \)	\(\frac{\sqrt{3} }{2} \)
\(45^\circ\)	\(\frac{1}{\sqrt{2} }\)	\(\frac{1}{\sqrt{2} }\)
\(60^\circ\)	\(\frac{\sqrt{3} }{2} \)	\(\frac{1}{2} \)
\(90^\circ\)	\(1\)	\(0\)

1, Chapter 4.1, Bearing,
Bearing

Remember

A bearing is an angle in degrees measured clockwise from North.

Hot Tip

Some questions using sine and cosine are about objects (such as cars, ships, or joggers) travelling in certain directions for periods. In these questions you use sine or cosine to determine the eventual position of the object. They usually contain the concept of bearing described above.

Here is an example of such a question.
1, Chapter 4, Task 1, Worked Example 4,
Worked Example

Use the slider to explore worked examples.
- Worked Example
  
  An oil tanker leaves Town X, and travels on a bearing of \( 050^{\circ} \) to Town Z, \( 50 \; km \) away.
  The tanker then travels to Town Y, \( 70 \; km \) away, on a bearing of \( 120^{\circ} \).
  a) Find the distance of Y from X, giving your answer to 3 sig. figs.
  b) b. Calculate the bearing of \(X\) from \(Y\), giving your answer to the nearest degree
- Worked Example
  
  Find the shaded angle in the triangle shown.
1, Chapter 4, Task 2, Exercise 1,
Exercise

Here are some questions to check your progress

Exercise 1
1. Find the unknown angle marked \(\theta\).
2. Find the unknown side marked \(a\)
Exercise 2

Find the unknown angles and sides.

Exercise 3

To calculate the height of a church tower, a surveyor measures the angle
of elevation of the top of the tower from two points 50 m apart.
1. Calculate the distance BC.
2. Hence calculate the height CD.
0, Chapter 5, Application: Area of Any Triangle,
Application: Area of Any Triangle

An important application of trigonometry is that of finding the area of a triangle with side lengths a and b and included angle \(\theta\)

Remember

Area of a triangle when the lengths of two sides and the included angle are known,

The area (A) is given by \(A=\frac{1}{2} ab\sin \theta\)

We can prove this result by constructing the perpendicular line from point \(B\) to the line \(AC\).

Area \( = \frac{1}{2} \; \times \; base \times \; prependicular height \)
\( = \frac{1}{2} \times b \times p \)
where: \( p= a \; sin \; \theta \)
So Area \( = \frac{1}{2} \times b \times a \; sin \; \theta \)
\( = \frac{1}{2} \; a \; b \; sin \; \theta \)
1, Chapter 5, Task 1, Worked Example 5,
Worked Example

Use the slider to explore worked examples.
- Worked Example
  
  The diagram shows a circle of radius \(64\;cm\). The length of the chord \(AB\) is \(100\;cm\).
  
  Find the angle \(\theta\) , to \(2 d.p\).
  
  Find the area of triangle \(OAB\).
- Worked Example
  
  Given that \(\sin \theta=\frac{\sqrt{3} }{2}\), \(0^\circ\leq \theta\leq 90^\circ\)
  Find the area of triangle \(CDE\) where
  
  \(CD = 30\;units\) and \(DE = 20\;units\).
1, Chapter 5, Task 2, Exercise 2,
Exercise

Here are some questions to check your progress

Exercise 1

Find the area of the shaded region

Exercise 2

Find the area of the shaded region

Exercise 3

In the diagram ST = 5 cm , TW = 9 cm and \(STW = 52^\circ\)

Calculate
1. the length of SW
2. the area of \(\triangle\) STW
1. 𝑆𝑊 = 7.11 cm
2. Area =17.73〖 cm〗^2
0, Chapter 6, Heron's Formula,
Heron's Formula

You have already met the formula for the area of a triangle when the lengths
of two sides and the included angle are known,

Remember

Formula for the area of a triangle when the lengths of all three sides are known.

\(A=\frac{1}{2} ab\sin C\)

We will use this result to find the formula for the area of a triangle when the lengths
of all three sides are known.

The formula is credited to Heron (or Hero) of Alexandria, and a proof can be found in his book,
Metrica, written in about AD 60.

The formula, known as Heron's formula, is given by

\(Area= \sqrt{s\left(s- a\right)\left(s- b\right)\left(s- c\right) }\)

\(S=\frac{a+ b+ c}{2}\)
1, Chapter 6, Task 1, Worked Example 6,
Worked Example
- Worked Example
  
  For the triangle shown, find
  
  the area of the triangle,
  
  angle \(\theta\).
1, Chapter 6, Task 2, Exercise 3,
Exercise

Here are some questions to check your progress

Exercise 1

Calculate the area of the triangle

Exercise 2
1. the area of the triangle,
2. the angle shown by \(\theta\)
1. Area = 27.81〖 mm〗^2
2. 𝛳 = 44°
0, Chapter 7, Angles Larger than \(90^{\circ}\),
Angles Larger than \(90^{\circ}\)

Let us explore how we define the trig rules for angles greater than \(90^{\circ}\)

First consider the \(x-y\) plane is divided into four quadrants by the \(x\) and \(y-axes\) as shown in the picture below
The angle \(\theta\) that a line \(OP\) makes with the positive \(x-axis\) lies between \(0^{\circ}\) and \(360^{\circ}\).

Angles between \(0^{\circ}\) and \(90^{\circ}\) are in the first quadrant.

Angles between \(90^{\circ}\) and \(180^{\circ}\) are in the second quadrant.

Angles between \(180^{\circ}\) and \(270^{\circ}\) are in the third quadrant.

Angles between \(270^{\circ}\) and \(360^{\circ}\) are in the fourth quadrant.

Angles larger than \(360^{\circ}\) can be reduced to lie between \(0^{\circ}\) and \(360^{\circ}\) by subtracting multiples of \(360^{\circ}\)

Remember

The trigonometric formulae, \(\cos \theta\) and \(\sin \theta\), are defined for all angles between \(0^{\circ}\) and \(360^{\circ}\) as the coordinates of a point, \(P\), where \(OP\) is a line of length \(1\), making an angle \(\theta\) with the positive \(x-axis\).

This is illustrated in the diagram below:

Here are some of the important values of \(\sin \theta\) , \(\cos \theta\) and \(\tan \theta\) which we have meet before; you should already be familiar with these.
1, Chapter 7.1, Sin and Cos 1,
Graphs of Sin\(\theta\) and Cos\(\theta\)

The graphs of \(\sin \theta\) and \(\cos \theta\) for any angle are shown below.

The graphs are examples of periodic functions. Each basic pattern repeats itself every \(360^{\circ}\). We say that the period is \(360^{\circ}\).

For any angle, note that \(\sin \left(90^{\circ}- \theta \right) = \cos \theta\)

The graph of \( \tan \theta\) has period \(180^{\circ}\). It is an example of a discontinuous graph.

The trigonometric equations \(sin \theta =a\) , \( \cos \theta =b\) and \( \tan \theta=c\) can have many solutions.

The inverse trig keys on a calculator (that is \(sin^{-1}\), \(\cos^{-1},\tan^{-1}\) give the principal value solution.

For \(\sin \theta=a\) and \(\tan \theta=c\) , the principal value solution is in the range:

\(-90^{\circ} \leq \theta \leq 90^{\circ}\)

For \(\cos \theta = b\), the principal value solution is in the range:

\( 0 \leq \theta \leq 180^{\circ}\)
1, Chapter 7.2, Tan,
Graphs of Tan\(\theta\)

The graph of tanθ has period \(180^{\circ}\). It is an example of a discontinuous graph.
1, Chapter 7.3, Trigonometric Equations,
Trigonometric Equations

The trigonometric equations \( sin \; \theta = a \) , \( cos \; \theta = b \) and \( tan \; \theta = c \) can have many solutions.

The inverse trig keys on a calculator (that is \( sin^{-1} \), \( cos^{-1} \), \( tan^{-1} \)) give the principal value solution.

For \( sin \; \theta = a \) and \( tan \; \theta = c \) , the principal value solution is in the range:
\( -90^{\circ} \; \leq \; \theta \; \leq \; 90^{\circ} \)

For \( cos \; \theta = b \), the principal value solution is in the range:
\( 0^{\circ} \; \leq \; \theta \; \leq \; 180^{\circ} \)
1, Chapter 7.3, Task 1, Worked Example 7,
Worked Example

Use the slider to explore worked examples.
- Worked Example
  
  Find a) \(cos \; 150^{\circ}\)
  b) \(sin \; 240^{\circ}\)
- Worked Example
  
  If \( cos \; \theta = - \frac{1}{2} \), how many values of the angle \(\theta\) are possible for \( 0 \leq \theta \leq 720^{\circ} \). Find these values of \(\theta\) .
- Worked Example
  
  Use a calculator to solve the equation \(sin \theta = -0.2\). Sketch the sin graph to show this solution. Give the principal value solution
  
  scroll left to see another worked example
0, Chapter 8, Summary,
Summary

the angle of elevation \(\alpha\) is the angle made with the horizontal, as shown in the diagram
(angle \(\beta\) is called the angle of depression). Angle \(\alpha = angle\; \beta\) (alternate angles).

\(Angle\; \alpha=\;angle\;\beta\) (alternate angles).

for any triangle ABC, the sine rule states
\(\frac{sin\;A}{a} = \frac{sin\;B}{b} = \frac{sin\;C}{c}\)
and the cosine rule states
\(c^{2}=a^{2}+b^{2} - 2 ab \cos C.\)
and similarly,
\(a^{2} = b^{2} + c^{2} - 2bc \cos A\)
\(b^{2} = c^{2} + a^{2} - 2ca \cos B\)

Interactive Exercises:

Sine and Cosine Rules Interactive Exercises, https://www.cimt.org.uk/sif/trigonometry/t3/interactive.htm
Sine and Cosine Rules , https://www.cimt.org.uk/sif/trigonometry/t3/interactive/s1.html
Area of Any Triangle , https://www.cimt.org.uk/sif/trigonometry/t3/interactive/s2.html
Heron's Formula, https://www.cimt.org.uk/sif/trigonometry/t3/interactive/s3.html

YouTube URL: https://youtu.be/i3Q1hvi6I5w

File Attachments: media/course-resources/Sine-Cosine-Rules_Learing-Objectives.pdfmedia/course-resources/Sine-Cosine-Rules_PowerPoint-Presentation.pptxmedia/course-resources/Sine-Cosine-Rules_Text.pdfmedia/course-resources/Sine-Cosine-Rules_Answers.pdfmedia/course-resources/Sine-Cosine-Rules_Essential-Information.pdf

Written on 06 July 2020.

Pythagoras' Theorem

Sections:

0, Chapter 1, Introduction,
Introduction

Many people have heard of Pythagoras' Theorem. In this unit we explain the theorem and give examples of its use.

After completing this unit you should;
- understand and be able to use Pythagoras' Theorem in right angled triangles
- be able to apply Pythagoras' Theorem when finding the length of an unknown side in a right angled triangle.
You have two sections to work through.
1. Pythagoras' Theorem
2. Further Work With Pythagoras’ Theorem
0, Chapter 2, Pythagoras Theorem,
Pythagoras Theorem

Pythagoras' Theorem gives a relationship between the lengths of the sides of a right-angled triangle.

Pythagoras' Theorem states that:

In any right-angled triangle, the area of the square on the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares on the other two sides (the two sides that meet at the right angle).

\(Area A + Area B = Area C\)

Remember

In any right angle triangle where \(c\) is the length of the hypotenuse and \(b\) and \(c\) are the lengths of the triangle's other two sides

\(a^{2}+ b^{2}=c^{2}\)
1, Chapter 2.1, Finding the Hypotenuse,
Finding the Hypotenuse

Hypotenuse

The hypotenuse is the side opposite the largest angle in a right-angled triangle.
In the diagram opposite "\(a\)" is opposite the 90 degree angle (largest angle).

See if you can spot the hypotenuse in the examplesbelow. The answers are at the bottom:
Solution
1. \(Side PQ\)
2. \(Side YZ\)
3. \(Side JK\)
4. \(Side TS\)
1, Chapter 2.1, Task 1, Worked Examples 1,
Worked Examples

Use the slider to explore worked examples.
- Worked Example
  
  Find the length of the hypotenuse of the triangle shown in the diagram.
  Give your answer correct to 2 decimal places.
- Worked Example
  
  Find the length of the hypotenuse of the triangles shown in the diagrams below.
  Give your answers correct to 2 decimal places.
- Worked Example
  
  Find the length of the side marked \(x\) in this triangle.
- Worked Example
  
  Find the length of the side marked \(x\) in this triangle.
- Worked Example
  
  Find the length of the side marked \(x\) in this triangle.
1, Chapter 2.1, Task 2, Exercises 1,
Exercises

Here are some questions to check your progress

Exercise

Find the length of the side x in each of the triangles below.

Exercise

Here is a problem solving question to solve.
0, Chapter 3, Further Work With Pythagoras’ Theorem,
Further Work With Pythagoras’ Theorem

In this section we look at some slightly more complicated examples of applying Pythagoras's theorem.

Use the slider to explore worked examples.
- Worked Example
  
  Find the length of the side marked \(x\) in the diagram.
- Worked Example
  
  Find the value of x as shown on the diagram.
- Worked Example
  
  Verify Pythagoras' Theorem for the right-angled triangle opposite.
  
  Solution
1, Chapter 3, Task 1, Exercises 2,
Exercises

Here are some questions to check your progress.

Exercise

Find the value of \(x\), giving your answer to 2 decimal places.

Exercise

Use Pythagoras Theorem to check that the following triangles contain a right angle.
0, Chapter 4, Summary,
Summary

Pythagoras' Theorem

In any right angled triangle, the area of the square on the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares on the other two sides (the two sides that meet at the right angle).

Note that a right angle is indicated by a small 'box' at an angle which is \(90^{\circ}\) as shown below

In a right angled triangle as labelled above you can use Pythagoras Theorem, namely

\(c^{2}=a^{2}+b^{2}\) to find the length of \(c\), the hypotenuse,

and \(a^{2}=c^{2} - b^{2}\) (or \(b^{2} = c^{2} - a^{2}\))to find the length of \(a\) (or \(b\)), given the lengths of \(b\) and \(c\) (or \(c\) and \(a\)).

Interactive Exercises:

Pythagoras' Theorem Interactive Exercises, https://www.cimt.org.uk/sif/trigonometry/t1/interactive.htm
Pythagoras' Theorem, https://www.cimt.org.uk/sif/trigonometry/t1/interactive/s1.html
Further Work with Pythagoras' Theorem, https://www.cimt.org.uk/sif/trigonometry/t1/interactive/s2.html

YouTube URL: https://youtu.be/n2D0NgiaynA

File Attachments: media/course-resources/Pythagoras-Theorem_Learning-Objectives.pdfmedia/course-resources/Pythagoras-Theorem_PowerPoint-Presentation.pptxmedia/course-resources/Pythagoras-Theorem_Text.pdfmedia/course-resources/Pythagoras-Theorem_Answers.pdfmedia/course-resources/Pythagoras-Theorem_Essential-Information.pdf

Written on 17 July 2020.

Formulae

Sections:

0, Chapter 1, Introduction,
Introduction

Formulae appear throughout science, economics, and in fact most disciplines.

After completing this unit you should
- understand the difference between a formula and an equation
- be able to substitute numbers into formulae in practical situations
- be able to change the subject of a formula
- appreciate how formulae are used to solve practical and relevant problems.
You have five sections to work through
1. Using Formulae
2. Construct and Use Simple Formulae
3. Substitution into Formulae
4. More Complex Formulae
5. Changing the Subject
0, Chapter 2, Using Formulae,
Using Formulae

In formulae, letters are used to represent numbers.

For example, the formula

Area of Rectangle

\(A = lw\) can be used to find the area of a rectangle.

Here \(A\) is the area, \(l\) the length and \(w\) the width.

In this formula, \(lw\) means \(l \times w\). In fact formulae are usually written in this way, without multiplication signs.
1, Chapter 2, Task 1, Worked Example 1,
Worked Example

Use the slider to explore worked examples.
- Worked Example
  
  The perimeter of a rectangle can be found using the formula: \(P = 2l + 2w\)
  Here \(P\) is the perimeter, \(l\) the length and \(w\) the width.
  Use the formula above to find the perimeter if \(l= 8\) and \(w= 4.\)
- Worked Example
  
  The final speed of a car is \(v\) and this can be calculated using the formula \(v = u + at\)
  where \(u\) is the initial speed, \(a\) is the acceleration and \(t\) is the time taken.
  Find \(v\) if the acceleration is \(2\;\)ms\(^{−2}\), the initial speed is \(4\;\)ms\(^{−1}\) and the time taken is \(10\) seconds.
2, Chapter 2, Task 2, Exercise 1,
Exercises

Here are some questions to check your progress; there are more practice questions if needed.

Exercise 1

If \(Q = 3x + 7y\), find the value of \(Q\) if \(x= 4\) and \(y = 2\)

Exercise 2

If \(Q = xy + 4\), find the value of \(Q\) if \(x= 3\) and \(y = 5\)

Exercise 3

A rectangle has a length of \(a\) cm and a width of \(b\) cm.
The perimeter of a rectangle is given by the formula
\(P = 2\left(a + b\right)\)
Calculate the perimeter of a rectangle when \(a = 4.5\) cm and \(b = 4.2\) cm.
0, Chapter 3, Constructing and Using Formulae,
Constructing and Using Formulae

A formula can be constructed to match a problem. For example, a formula for the perimeter of a rectangle describes how to find the perimeter, given the length and width of the rectangle.

We can name the perimeter \(P\), the length \(l\) and the width \(w\).

The perimeter is found by adding the lengths of the sides together.

So we construct the formula as follows:

\(P = l + w + l + w\) , which simplifies to the following formula

\(P = 2l + 2w\)
1, Chapter 3, Task 1, Worked Example 2,
Worked Example

Use the slider to explore worked examples.
- Worked Example
  
  Write down a formula for the perimeter of the shape shown.
- Worked Example
  
  When laying a patio, a landscape gardener charges a basic fee of \(£30\) plus \(£12\) per hour.
  
  Find a formula for calculating the gardener's charge.
1, Chapter 3, Task 2, Exercise 2,
Exercises

Here are some questions to check your progress; there are more practice questions if needed.

Exercise 1

a. Tickets for a school concert are sold at £\(6\) for adults and £\(4\) for children.
If \(p\) adults and \(q\) children buy tickets, write a formula for \(T\), the total value of the ticket sales in £s.

b. Find the total value of the ticket sales if \(p\) = \(50\) and \(q\) = \(20\).

Exercise 2

Find a formula for the perimeter of the shape shown here, and find the perimeter for the values specified.

\(a = 4\) cm, \(b = 9\) cm
0, Chapter 4, Substitution into Formulae,
Substitution into Formulae

Remember

The process of replacing the letters in a formula is known as substitution.
1, Chapter 4, Task 1, Worked Example 3,
Worked Example

Use the slider to explore worked examples.
- Worked Example
  
  The length of a metal rod is \(l\) cm. The length changes with temperature and can be found by the formula:
  \(l = 40 + 0.02T\) where \(T\) is the temperature.
  Find the length of the rod when \(T = 50^\circ\)C
- Worked Example
  
  The profit in £ made by a salesman when he sells \(n\) books is calculated by the formula:
  \(P = 4n − 50\)
  Find the profit if he makes \(30\) sales.
- Worked Example
  
  If \(z = x^{2} − 4y^{2}\) what is the value of \(z\) when \(x = 4\) , \(y = −2\) ?
1, Chapter 4, Task 2, Exercise 3,
Exercises

Here are some questions to check your progress; there are more practice questions if needed.

Exercise 1

The formula below is used to convert temperatures in degrees Celsius to degrees Fahrenheit,
where \(F\) is the temperature in degrees Fahrenheit and \(C\) is the temperature in degrees Celsius:
\(F = 1.8 C + 32\)

Calculate \(F\) if \(C = 20\)

Exercise 2

If \(p = a^{2} - b^{2}\) calculate \(p\) if \(a = 10\) and \(b = -7\)
0, Chapter 5, More Complex Formulae,
More Complex Formulae

Some formula such as \(\frac {1}{f}=\frac{1}{u}+\frac{1}{v}\) and \(z^{2}=x^{2}+y^{2}\)
do not give you a value straight away when you substitute in the variables to find the unknowns. In this case \(f\) and \(z\)

For example:

Find the value of \(z\) when \(z^{2} = x^{2} + y^{2}, \) if \(x = 3.6\) and \(y = 4.8\)

\(z^{2} = x^{2} + y^{2}\)
\(z^{2} = 3.62 + 4.82\)
\(z^{2} = 12.96 + 23.04\)
\(z^{2} = 36\)

Now the square root can be taken of both sides to give
\(z = \sqrt{36} = 6\)

Note here that \(z= - 6\) is another solution, as \(\left(- 6\right) \times \left(- 6\right) = 36\)
So we have two solutions \(z= 6\) or \(z= - 6\)
1, Chapter 5, Task 1, Worked Example 4,
Worked Example
- Worked Example
  
  Find the value of \(f\) when \(\frac {1}{f}=\frac{1}{u}+\frac{1}{v}\) if \(u= 10\) and \(v = 8\)
1, Chapter 5, Task 2, Exercise 4,
Exercises

Here are some questions to check your progress; there are more practice questions if needed.

Exercise 1

Find the value of \(z\) when \(\frac{1}{z}= \frac{2}{x}+ \frac{3}{y}\) if \(x = 3\) and \(y = 4\)

Exercise 2

Find the value of \(z\) when \(z^{2}=\frac{x+6}{y}\) if \(x = 6\) and \(y = 3\)
0, Chapter 6, Changing the Subject,
Changing the Subject

Sometimes a formula can be rearranged into a more useful format.
For example, the formula
\(F = 1.8C + 32\)
can be used to convert temperatures in degrees Celsius to degrees Fahrenheit.
It can be rearranged to make \(C\) the subject of the formula as shown in the next worked example.

Remember

To enable, for example, temperatures in degrees Fahrenheit to be converted to degrees Celsius.
We say that the formula has been “rearranged to make \(C\) the subject of the formula”
1, Chapter 6, Task 1, Worked Example 5,
Worked Example

Use the slider to explore worked examples.
- Worked Example
  
  Rearrange the formula
  
  \(F = 1.8C + 32\)
  
  to make \(C\) the subject of the formula.
- Worked Example
  
  The distance, \(s\), travelled by a car in time \(t\) from initial speed \(u\)
  to final speed \(v\) is given by the formula:
  \(s=\frac{\left(u+v\right)t }{2}\)
  
  Make \(v\) the subject of the formula.
1, Chapter 6, Task 2, Exercise 5,
Exercises

Here are some questions to check your progress; there are more practice questions if needed.

Exercise 1

Rearrange \(y = \frac{ax-c}{b}\) to make \(x\) the subject of the formula

Exercise 2

Concentration of a solution is given by the formula
\(C=\frac{A}{V}\)
where \(C\) is the concentration, \(A\) is the amount (in micrograms), and \(V\) is the volume (in cubic centimeters).
Rearrange the formula to make \(A\) the subject.

Exercise 3

Rearrange the formula:
\(v^{2}=u^{2}+2as\)
to make \(s\) the subject of the formula
0, Chapter 7, Summary,
Summary

Formula

This describes how one quantity is related to one or more other quantities.
Here are some formula as examples;

\(y=x^{2}\), \(y= 2x+1\), \(A=ab\), \(V=xyz\), \(F=\frac{9}{5} C+32\), \(T=2\pi \sqrt{\frac{l}{g} }\)

Substituting numbers into a formula

This means that letters are replaced by numbers in a formula.

For example,

\(F=\frac{9}{5} C+32\) when \(C=25\)

gives

\(F=\frac{9}{5} \times 25 + 32\) \(=45+32\) \(=77\)

Changing the subject

This means rearranging a formula so that, for example,

\(y=f\left(x\right)\) (formula in \(x\))

is rearranged to give

\(x=g\left(y\right)\) (formula in \(y\))

For example,

\(F=\frac{9}{5} C+32\)

can be rearranged to give

\(\frac{9}{5} C = F-32\)
\(C=\frac{5}{9}\left(F-32\right)\)

Be careful when using algebraic notation. For example,

\(3\times a= 3a\)
\(\left(3a\right) ^{2}=\left(3a\right) \times\left(3a\right) = \left(3\times a\right) \times \left(3\times a\right) = \left(3\times 3\right) \times \left(a\times a\right) = 9\times a^{2} = 9a^{2}\)

When changing the subject of an equation, always write down clearly the operations which have been used.
For example, make \(a\) the subject of the equation \(4 a + b = c\).

\(4a + b =c\) (add \(-b\) to each side)
\(4a + b - b = c - b\)
\(4a = c-b\) (divide both sides by \(4\))
\(a=\frac{c-b}{4}\)

Note that

\(\left(-a\right) \times b= -a\times b = -ab\) for any number \(a\), \(b\)
\(\left(-a\right) \times \left(-b\right)= a\times b = ab\) for any number \(a\), \(b\)

Interactive Exercises:

Formulae Interactive Exercises, https://www.cimt.org.uk/sif/algebra/a4/interactive.htm
Using Formulae, https://www.cimt.org.uk/sif/algebra/a4/interactive/s1.html
Construct and Use Simple Formulae, https://www.cimt.org.uk/sif/algebra/a4/interactive/s2.html
Substitution into Formulae, https://www.cimt.org.uk/sif/algebra/a4/interactive/s3.html
More Complex Formulae, https://www.cimt.org.uk/sif/algebra/a4/interactive/s4.html
Changing the Subject, https://www.cimt.org.uk/sif/algebra/a4/interactive/s5.html

YouTube URL: https://youtu.be/qOwEWK9_AP4

File Attachments: media/course-resources/Formulae_Learning-Objectives.pdfmedia/course-resources/Formulae_Answers.pdfmedia/course-resources/Formulae_Essential-Information.pdfmedia/course-resources/Formulae_Text.pdfmedia/course-resources/Formulae_Presentation.pptx

Written on 03 August 2020.

Simultaneous Linear Equations and Factorisation

Sections:

0, Chapter 1, Introduction,
Introduction

Solving equations is key part of mathematics and critical skill in numerous contexts.
This unit explores solving equations where the variables are not squared or cubed etc (called linear equations).

After completing this unit you should be able to
- solve simultaneous equations
- fully expand brackets
- factorise expressions by taking out single terms and linear terms
- factorise quadratic expressions into two linear terms.
You have four sections to work through and there are check up audits and fitness tests for each section.
1. Simultaneous Linear Equations
2. How to handle expressions with Brackets
3. Linear Factorisation
4. Quadratic Factorisation
0, Chapter 2, Simultaneous Linear Equations,
Simultaneous Linear Equations

Introduction

Try solving: \(x + y = 24\)

What could ‘\(x\)’ be?

What could ‘\(y\)’ be?

Quickly you will realise that there are simply too many possible solutions.

So if we have more than one ‘unknown’ (variable), we must have more than one equation.

With two ‘unknowns’, we must have at least two equations

And we must solve them at the same time, i.e. simultaneously.

Try again: \(x+y=24\) and \(2x = 20\)

Can you solve it now?
1, Chapter 2.1, Solving Simultaneous Linear Equations,
Solving Simultaneous Linear Equations
Remember
There are 3 ways to solve simultaneous linear equations:
- by substitution (as in the example on previous slides);
- by drawing graphs (and finding the coordinates of where the 2 lines cross);
- by elimination.
In the worked examples in the next section we will demonstate the substitution method and elimination method. In fact, to demonstrate substitution we will return to the example we discussed in the first section.
1, Chapter 2.1, Task 1, Worked Example 1,
Worked Example

Use the slider to explore worked examples.
- Worked Example
  
  Solve:
  
  \( x + y = 24\)
  \(2x \hspace{15 pt} = 20\)
- Worked Example
  
  Try solving: \(x + 2y = 5\) and \(3x + y = 5\)
- Worked Example
  
  Solve the equations
  
  \(7x - 2y = 23 \)
  \(3x + 4y = 39 \)
  
  Label the equations:
  
  \(7x - 2y = 23 \longrightarrow 1\)
  \(3x + 4y = 39 \longrightarrow 2\)
  
  Equation 1 is multiplied by 2 so that both equations have the same coefficient of the \(y\) term
  
  Label the new equation 3 so that we now have.
  
  \(14x - 4y = 46 \longrightarrow 3\)
  \(3x + 4y = 39 \longrightarrow 2\)
  
  Now, if we add the two equations 3 and 2 together we get the following;
  
  \(17x= 85\) which gives \(x=5\)
  
  We now know \(x=5\) and we substitute this value into equation 2 to give:
  \(\begin{align*} 3x + 4y &= 39 \\ 3 × 5 + 4y &= 39\\ 15 + 4y &= 39\\ 4y &= 24\\ y &= 6 \end{align*}\)
  We check these solutions in equation 1
  
  \(7x - 2y = 23\) which gives \((7 × 5) - (2 × 6) = 23\)
  
  Solutions are therefore: \(x=5\) \(y=6\)
0, Chapter 3, How to handle expressions with brackets,
How to handle expressions with Brackets

Expressions and equations may contain brackets.

For example: \(3\left(y - 5\right) = 45\)

This is shorthand for \(3\) ’lots of’ \(\left(y - 5\right)= 45\). (We simply do not write, or show, the ‘lots of’ or × symbol.)

Removing brackets is known as expanding brackets, and involves multiplying out the bracket:

\(3 \left(y - 5\right) = 45\)
\(3y - 15 = 45\)

Now we can solve:
\(3y = 60\)
\(y = 20\)

Note: The reverse of expanding brackets is the process of factorisation.

Lets look at lots of worked examples
1, Chapter 3, Task 1, Worked Example 2,
Worked Example

Use the slider to explore further examples.
- Worked examples
  
  Expand: \(2m \left(m + 5\right)\)
  
  Expand: \(-5 \left(a - 6\right)\)
  
  Expand: \(3y \left(-2y + 8\right)\)
- Worked examples
  
  Expand: \(m \left(m + 3\right)\)
  
  Expand: \(2 \left(m + 3\right)\)
  
  Expand: \(\left(m+2\right)\;\left(m+3\right)\)
- Worked examples
  
  Expand: \(\left(2y - 3\right)\left(y + 4\right)\)
  
  Expand: \(\left(m+2\right)^{2}\)
0, Chapter 4, Linear Factorisation,
Linear Factorisation

The reverse process of expanding brackets is known as factorisation.
When factorised, an expression is written as a product of terms.

To factorise an expression, it is necessary to identify what factors each term has in common.
In other words, what ‘goes into’ all the terms?

For example: Factorise \(8m + 12\)

In this example, we are looking for the biggest factor of \(8m\) and of \(12\).
The largest common factor is \(4\).
\(8m+ 12\)
is the same as
\(\left(4 \times 2m\right) + \left(4 \times 3\right)\)
\(= 4 \times \left(2m + 3\right)\)
\(= 4 (2m + 3)\)
1, Chapter 4, Task 1, Worked Example 3,
Worked Example

Use the slider to explore worked examples.
- Worked Example
  
  Factorise: \(6a - 21\)
- Worked Example
  
  Factorise: \(6y^{2}+ 8y\)
1, Chapter 4, Task 2, Examples 1,
Further Examples

Add examples from charts 26 and 27
1, Chapter 4, Task 3, Examples 2,
Further Examples

add charts 28 to 32
0, Chapter 5, Factorising Quadratic Expressions,
Factorising Quadratic Expressions

Factorising quadratic expressions takes plenty of practise

The more you do – the easier it will become!
And remember – it is often a good idea to check the answers you obtain by expanding the brackets, for example :

\(\left(3x - 4\right)\;\left(2x - 3\right)\) = \(6x^{2} - 17v +12\)

In special circumstances, there is one very important result. It is known as the difference between two squares and can be used to factorise some expressions

You may know the following expression:

\(\left(a-b\right)\;\left(a+b\right)=a^{2}-b^{2}\)

This result is known as the ‘difference between two squares’

For example
\(\left(x-4\right)\;\left(x+4\right)=x^{2}-16=x^{2}-4^{2}\)

We can use this result, in reverse, to help us factorise some expressions, i.e.
\(x^{2}-4^{2}=\left(x-4\right)\;\left(x+4\right)\)

Lets look at an example: \(4x^{2}-25\)
The expression is a difference of two squares, namely \(\left(2x\right)^{2}\) and \(5^{2}\) can be factorised as below:

\(4x^{2}-25=\left(2x-5\right)\left(2x+5\right)\)

Lets look at another example: \(9x^{2} -81\)

First look to see if both terms are ‘squares’.
The answer is yes \(\left(3x\right)^{2}\) and \(9^{2}\)

Now check to see whether they are being subtracted (i.e. a ‘difference’ is being sought).
The answer is yes

Now you can factorise easily: \(\left( 3x - 9\right)\;\left(3x + 9\right)\)

Here is another example: \(16x^{2} - 36\)
\(\left(4x -6\right)\;\left(4x+6\right)\)

Finally one more (remember 1 is a square number!!)
\(x^{2}-49\) = \(\left(x -7\right)\;\left(x +7\right)\)
0, Chapter 6, Summary,
Summary

Simultaneous linear equations

These are of the form
\(ax + by = c\)
\(dx + ey = f\)
when \(a,\;b,\;c,\;d,\;e\) and \(f\) are constants.

Expanding brackets Expressions such as \(a\left(x + b\right)\) are expanded out.

\(2\left(x+3\right) = 2x+6\)
\(x\left(4-x\right) = 4x - x^{2}\)
\(2x\left(1+3x\right) = 2x +6x^{2}\)
\(a\left(x+b\right) =ax+ab\)
- \(\left(a+b\right) \left(a-b\right)= a^{2}-b^{2}\)
- \(\left(a+b\right)^{2} = a^{2} + 2ab + b^{2}\)
- \(\left(a+b\right)^{2} = a^{2} - 2ab + b^{2}\)
- If \(x^{2} + ax + b = \left(x+p\right) \left(x+q\right)\), then \(a = p+q\) and \(b=pq\)
Linear factorisation

Single factors are taken out of an expression. For example,
\(3x+9=3\left(x+3\right)\)
\(4x^{2}+x=x\left(4x+1\right)\)
\(5x^{2}+10x=5\left(x+2\right)\)

Quadratic factorisation

Quadratic expressions are factorised into linear factors. For example,
\(x^{2}+5x+6 =\left(x+3\right) \left(x+2\right)\)
\(x^{2}- 4 =\left(x+2\right) \left(x-2\right)\)
\(\left(-2x-1\right) \left(x+2\right)= 2x^{2}+3x-2\)

Also note that, in general, simultaneous linear equations normally have unique solutions, but not when one equation is the multiple of another. For example, when
\(x+y=1\)
\(2x + 2y = 2\)

There is no unique solution. Also when there are two inconsistent equations, for example,
\(x+y=1\)
\(x+y=3\)
there are NO solutions.

Interactive Exercises:

Simultaneous Linear Equations and Factorisation Interactive Exercises, https://www.cimt.org.uk/sif/algebra/a11/interactive.htm
Simultaneous Linear Equations , https://www.cimt.org.uk/sif/algebra/a11/interactive/s1.html
Expanding Brackets, https://www.cimt.org.uk/sif/algebra/a11/interactive/s2.html
Linear Factorisation , https://www.cimt.org.uk/sif/algebra/a11/interactive/s3.html
Quadratic Factorisation, https://www.cimt.org.uk/sif/algebra/a11/interactive/s4.html

YouTube URL: https://youtu.be/clX_q7orQGw

File Attachments: media/course-resources/Simultaneous-Linear-Equations-Factorsation_Learning-Objectives.pdfmedia/course-resources/Simultaneous-Linear-Equations-Factorsation_PowerPoint-Presentation.pptxmedia/course-resources/Simultaneous-Linear-Equations-Factorsation_Text.pdfmedia/course-resources/Simultaneous-Linear-Equations-Factorsation_Answers.pdfmedia/course-resources/Simultaneous-Linear-Equations-Factorsation_Essential-Information.pdf

Written on 05 August 2020.

Linear Equations

Sections:

0, Chapter 1, Introduction,
Introduction

This unit focuses on the foundations of algebra and is crucial for developing your mathematics.

After completing this unit you should:
- be confident in simplifying algebraic expressions, including multiplying and working with brackets
- understand how to solve equations in one unknown
- be able to translate problems in context into equations.
You have three sections to work through.
1. Simplifying Expressions
2. Simple Equations
3. Solving Linear Equations
0, Chapter 2, Simplifying Expressions,
Simplifying Expressions

An expression is a group of terms.

To simplify an expression, we collect together ‘like terms’, i.e. terms with the same variable.

In the worked examples in the next section the variables are \(a\) or \(x\).

In the worked examples also note the following points
- The + and - signs go with the term which follows.
- \(x\) and \(x^{2}\) must be treated as if they were different letters.
- You can not add an \(x\) term to an \(x^{2}\) term.
1, Chapter 2, Task 1, Worked Example 1,
Worked Example

Use the slider to explore worked examples.
- Worked Example
  
  Simplify: \(5a+2a+3+6\)
- Worked Example
  
  Simplify: \(8x+3y-11x+6y-11+5\)
- Worked Example
  
  Simplify: \(x\left(x-5\right)+6x^{2}\)
0, Chapter 3, Simple Equations,
Simple Equations

An equation has terms on both sides of an equals sign.

To solve simple equations,
- we collect together ‘like terms’ on each side of the equals sign.
- we then make sure the equation always remains balanced: so whatever operation we do on one side of the equation, we must also do on the other.
Top Tip. When moving terms from one side an equation to another some people like to remember the rhyme ' cross the line, change the sign'
1, Chapter 3, Task 1, Worked Example 2,
Worked Example

Use the slider to explore worked examples.
- Worked Example
  
  Solve: \(5a+2a=14\)
- Worked Example
  
  Solve: \(12x -19= 5\)
- Worked Example
  
  Solve: \(2x-8=-18\)
0, Chapter 4, Solving Linear Equations,
Solving Linear Equations

Remember

Linear equations are simple equations which would draw a straight (linear) line if plotted.

To solve linear equations, we sometimes need to complete a number of steps along the way such as expanding brackets, collecting like terms and mutilpying or dividing both sides by the same number. These techniques are demonstrated in the next worked examples.
1, Chapter 4, Task 1, Worked Example 3,
Worked Example

Use the slider to explore worked examples.
- Worked Example
  
  Solve: \(3x + 8 = − 4\left(x + 5\right)\)
- Worked Example
  
  Solve: \(\frac{m}{2}=3m-10\)
0, Chapter 5, Summary,
Summary

Simplifying expressions

This means collecting like terms together;
e.g. \(2x + 5 + 3x - 4 -x = 4x + 1\)

Removing brackets

Multiplying through an expression to remove the brackets;
e.g. \(4\left(5 - 2x\right) = 4 \times 5 + 4 \times \left(-2x\right) = 20 - 8x\)

Note that, in general, \(a\left(x + b\right) = ax + ab\) when \(a\) and \(b\) are any numbers.

Equation

This relates an unknown quantity to known quantities, and can be solved to find the value of the unknown quantity: this is called the solution.
e.g. \(4x + 1 = 9\) (take \(1\) from both sides)
\(4x = 9-1\)
\(4x=8\) (divide both sides by 4)
\(x=\frac{8}{4}\)
i.e. \(x=2\)

So \(x = 2\) is the solution of the equation \(4x + 1 = 9\).

Balancing equations

For any equation, the right hand side (RHS) must equal the left hand side (LHS) and any operation you do to one side must be done to the other side (see example above).

Interactive Exercises:

Interactive Exercises - Linear Equations, https://www.cimt.org.uk/sif/algebra/a6/interactive.htm
Simplifying Expressions, https://www.cimt.org.uk/sif/algebra/a6/interactive/s1.html
Simple Equations, https://www.cimt.org.uk/sif/algebra/a6/interactive/s2.html
Solving Linear Equations, https://www.cimt.org.uk/sif/algebra/a6/interactive/s3.html

YouTube URL: https://www.youtube.com/watch?v=_8shj_QZktg&list=PL8ce1n6mRlCtS0rO_C7CNwtjhyqDVkr8g&index=41

File Attachments: media/course-resources/Linear-Equations_Essential-Informations.pdfmedia/course-resources/Linear-Equations_Learning-Objectives.pdfmedia/course-resources/Linear-Equations_Presentation.pptxmedia/course-resources/Linear-Equations_Text.pdfmedia/course-resources/Linear-Equations_Answers.pdf

Written on 05 August 2020.

Number Sequences

Sections:

0, Chapter 1, Introduction,
Introduction

The work in this unit is focused on number sequences. After completing this unit you should
- understand and deduce the rule that defines simple sequences
- be able to find sequences based on geometrical patterns
- be able to use formulae to find and use number sequences.
You have four sections to work through.
1. Simple Number Patterns
2. Recognising Number Patterns
3. Extending Number Patterns
4. Formulae for Number Sequences
0, Chapter 2, Simple Number Patterns,
Simple Number Patterns

A list of numbers which form a pattern is called a sequence.

To continue a number sequence, it is often very useful to look for the difference between numbers.

Remember

When looking at a sequence of numbers the first step to understand the pattern is to look at the difference between the consectutive numbers
1, Chapter 2, Task 1, Worked Example 1,
Worked Example
- Worked Example
  
  What is the difference in sequence \(7, \;11,\;15,\;19, \;23, …\) ?
0, Chapter 3, Recognising Number Patterns,
Recognising Number Patterns

Some number sequences follow different patterns. Look at the ones below.
Can you spot the patterns?

Examples of Sequences

Fibonacci Sequence:

\(1\;,1\;,2,\;3,\;5,\;8,\;13,\;21,\;34,…\)

Square Numbers Sequence:

\(1,\;4,\;9,\;16,\;25,\;36,\;49…\)

Triangular Numbers Sequence:

\(1,\;3,\;6,\;10,\;15,\;21,\;28,\;36...\)
(Note: Two consecutive Triangular Numbers sum to a Square Number, for example: \(10 + 15 = 25\))

Cube Numbers Sequence:

\(1,\;8,\;27,\;64,\;125,\;216...\)
1, Chapter 3, Task 1, Recognising Patterns,
Recognising Patterns

Number sequences sometimes require detective work.
Often they are very similar to ones previously encountered - with just an added tweak or two.

Use the slider to explore worked examples.
- Worked Example
  
  Look at this sequence \(2,\;5,\;10,\;17,\;26,\;37,\;50…\)
  
  What comes next?
- Worked Example
  
  Look at this sequence \(2,\;6,\;12,\;20,\;30,\;42,…\)
  
  What comes next?
- Worked Example
  
  Look at this sequence \(1,\;5,\;6,\;11,\;17,\;28,…\)
  
  What comes next?
1, Chapter 3, Task 2, Worked Examples,
Worked Examples

Use the slider to explore worked examples.
- Worked Example
  
  Describe how the following sequences are related, and find the next term for each:
  
  a. \(1,\;1,\;2,\;3,\;5,\;8,\;13,\;21, …\)
  
  b. \(5,\;5,\;6,\;7,\;9,\;12,\;17,\;25, …\)
  
  c. \(2,\;2,\;4,\;6,\;10,\;16,\;26,\;42, …\)
- Worked Example
  
  Find the next 2 terms in this number sequence: \(-1,\;2,\;7,\;14,\;23,\;34,…\)
0, Chapter 4, Extending Number Patterns,
Extending Number Patterns

Finding the \(n^{th}\) term in a sequence

A formula or rule can be used to work out any term of a sequence without having to work out all the terms.

For example, can we work out the \(100^{th}\) term of \(2,\;5,\;8,\;11,\;14…\) without having to work out all the interspersing terms. Lets look at a worked example to see how we can do this.
1, Chapter 4, Task 1, Worked Example 2,
Worked Example

Use the slider to explore worked examples.
- Worked Example
  
  Find a solution to the \(n^{th}\) term of the sequences \(3,\;6,\;9,\;12,\;15,\;18…\)
- Worked Example
  
  The sequence is: \(2, 5, 8, 11, 14,…\)
  
  What is the \(100\)^th term?
0, Chapter 5, Formulae for Number Sequences,
Formulae for Number Sequences

In the last section we found the \(n^{th}\) term for linear sequences, that is, for sequences which go up or down by a constant amount.
We can also find a formulae for the \(n^{th}\) term for other sequences.

For example:

\(1,\;4,\;9,\;16,\;25…\)

We can refer to the first term as \(u_{1}\), to the second term as \(u_{2}\), the third term as \(u_{3}\), and so on

\(u_{1} = 1 = 1 \times 1\)
\(u_{2}, = 4 = 2 \times 2\)
\(u_{3} = 9 = 3 \times 3\)
\(u_{4} = 16 = 4 \times 4\)
\(u_{5} = 25 = 5 \times 5\)

This sequence can be described by the general formula:

\(u_{n} = n\times n\)
\(u_{n} = n^{2}\)
1, Chapter 5, Task 1, Worked Example 3,
Worked Example
- Worked Example
  
  Use the formula below to find the first \(6\) terms of the sequence:
  \(u_{n}, = 3n^{2} - 2\)
0, Chapter 6, Summary,
Summary

Linear sequence

This is given by the formula \(u_{n}=an+b, \left(n = 1,2,3...\right)\) where \(a\) and \(b\) are constants, and \(u_{n}\) is the \(n^{th}\) term in the sequence, \(n=1,2,3...\)
For example, \(u_{n} = 3n + 4\) gives the sequence \(7,\;10,\;13,\;16,\;19, ...\)

Fibonacci Sequence

This is defined by \(u_{n} = u_{n-1} + u_{n-2} \left( n=3,\;4,\;5, ...\right)\)
with \(u_{1}=u_{2}=1\)

This gives the sequence \(1, 1, 2, 3, 5, 8, 13, ...\)

Iterative formula

When the \(n^{th}\) term in a sequence is given in a formula in terms of earlier terms, but with sufficient information to generate the complete sequence. For example,

\(u_{n} = u_{n - 1} + 1,\;u_{1}=0\) for \(n=2,3, ...\)
\(u_{n} = u_{n - 1} + u_{n - 2}, u_{1}=1,u_{2}=1\) for \(n=3,4, ...\)
\(u_{n}=\;\)\(\frac{u_{n} - 1}{2}\), \(u_{1}=2\) for \(n=2,3, ...\)

Interactive Exercises:

Number Sequences Interactive Exercises, https://www.cimt.org.uk/sif/algebra/a7/interactive.htm
Simple Number Patterns, https://www.cimt.org.uk/sif/algebra/a7/interactive/s1.html
Recognising Number Patterns, https://www.cimt.org.uk/sif/algebra/a7/interactive/s2.html
Extending Number Patterns, https://www.cimt.org.uk/sif/algebra/a7/interactive/s3.html
Formulae for Number Sequences, https://www.cimt.org.uk/sif/algebra/a7/interactive/s4.html

YouTube URL: https://youtu.be/BII5r-jUJwA

External Links:

Sequences Video, https://www.bbc.co.uk/bitesize/guides/zy82ng8/revision/1, images/Images/bitesize-logo.png, A video by BBC Bitesize explaining linear, quadratic, exponential and Fibonacci sequences are explored

File Attachments: media/course-resources/Number-Sequences_Learning-Objectives.pdfmedia/course-resources/Number-Sequences_Presentation.pptxmedia/course-resources/Number-Sequences_Text.pdfmedia/course-resources/Number-Sequences_Answers.pdfmedia/course-resources/Number-Sequences_Essential-Informations.pdf

Written on 10 August 2020.

Solving Inequalities

Sections:

0, Chapter 1, Introduction,
Introduction

This unit introduces the topic of solving inequalities, which is an important aspect of mathematics and a natural extension to solving equations.

After completing this unit you should
- be able to illustrate inequalities on a number line
- be able to solve linear inequalities
- be able to solve quadratic inequalities
- understand how to illustrate linear inequalities in two variables.
You have five sections to work through namely;
1. Inequalities on a Number Line
2. Solution of Linear Inequalities
3. Inequalities Involving Quadratic Terms
4. Graphical Approach to Inequalities
5. Dealing with more than One Inequality
0, Chapter 2, Inequalities on a Number Line,
Inequalities on a Number Line

An Inequality involves using one or more of these signs: \(\lt\; \leq\; \gt\) and \(\geq\)

For example:

\(x \gt 1 \Rightarrow x\) is greater than \(1\)
\(x \geq 2 \Rightarrow x\) is greater or equal to \(2\)
\(x \lt 10 \Rightarrow x\) is less than \(10\)
\(x \leq 12 \Rightarrow x\) is less than or equal to \(12\)

The examples below show how to represent and inequality on a number line.

a.

\(x \gt -1\)

b.

\(x \leq 2\)

c.

\(-1 \leq x \lt 5\)

Note
We use filled dot when the point is included and no fill of the dot if not
1, Chapter 2, Task 1, Worked Examples 1,
Worked Examples

Use the slider to explore worked examples.
- Worked Example
  
  Represent this inequality \(x \geq 2\)
- Worked Example
  
  Represent this inequality \(x \lt -1\)
- Worked Example
  
  Represent this inequality \(-2 \lt x \leq 4\)
0, Chapter 3, Solution of Linear Inequalities,
Solution of Linear Inequalities

We can solve linear inequalities in a similar way to linear equations as we see in the following worked examples. The basic rule to remember is that whatever you do on one side of the inequality (be it add , subtract, multilpe or divide by a number) you must do exactly the same on the other side of the inequality.
1, Chapter 3, Task 1, Examples 1,
Examples

ADD TWO WORKED EXAMPLES ON CHARTS 8 AND 9
0, Chapter 4, Solving Quadratic Inequalities,
Solving Quadratic Inequalities

We now consider how to solve inequalities that involve \(x^{2}\) terms.
For example, the inequality \(x^{2} \lt 9\) is satisfied by any number between \(-3\) and \(3\) as shown below on the number line.

\(-3 \lt x \lt 3\)

Note that if the inequality had been \(x^{2} \gt 9\) then \(x\) would be greater than \(3\) or less than \(-3\);
that is \(x\gt 3\) or \(x \lt -3\) as shown below.

\(x \lt -3\) or \(x \gt 3\)

As with linear inequalities, we solve the quadratic inequalities in a similar way we would solve quadratic equations. Lets look at some simple worked examples.
1, Chapter 4, Task 1, Examples 2,
Examples

ADD THE WORKED EXAMPLE FORMAT

Find the solutions of the inequalities:
1. \(x^{2}+6 \gt 15\)
2. \(3x^{2}-7 \leq 41\)
Solutions
1. Subtract \(6\) from each side to give \(x^{2} \gt 9\) and we now that this has solutions:
  \(x\lt -3\) or \(x \gt 3\)
2. Add 7 to both sides to give: \(3x^{2} \leq 8\)
  Now divide both sides by \(3\) to give:
  \(x^{2} \leq 16\)
  This has solution:
  \(-4 \leq x \leq 4\)
1, Chapter 4.1, Factorising the quadratic expression,
Factorising the quadratic expression

In some examples, we need to factorise the quadratic expression.

For example, to solve the inequality \(x^{2} -3x-4 \gt 0\)

We can factorise the left hand side \(\left(LHS\right)\) to give \(\left(x-4\right)\left(x+1\right)\gt 0\)

So we can see that the \(LHS\) is zero when \(x = 4\) or \(x=-1\).

These points are the cut off points between possible regions.

We can test the regions by considering points; for example:

At \(x=-2, x^{2} -3x -4=6 \gt 0\) so \(x=-2\) is in the solution.

Similarly at \(x=2\) \(x^2 -3x -4=-6 \lt 0\) so \(x=2\) is not in solution.

At \(x=5\), \(x^{2}-3x-4=6 \gt 0\) so \(x=5\) is in the solution, shown below
0, Chapter 5, Graphical Approach,
Graphical Approach

Sometimes it is more straight forward to graph the quadratic to solve the inequality

Here is an example of the graphical approach

Solve the inequality \(x^{2}+x-2\lt 0\)

Solution

We can factorise the LHS to give \(x^{2}+x-2=\left(x+2\right)(x-1)\)

Sketching the function (shown below) shows, as expected that the \(LHS\) is zero at \(x=-2\) and \(x=1\)

So in the region between \(x=-2\) and \(x=1\), the inequality is true.

Hence the solution is \(-2 \lt x \lt 1\)
1, Chapter 5.1, Graphical Approach to Inequalities,
Graphical Approach to Inequalities

When an inequality involves 2 variables, we can represent it in a region on a graph.

For example, the inequality \(x+y \geq 4\) is shown below

.

The coordinates of any point in the shaded area satisfy \(x+y\geq 4\)

Whilst the coordinates of any point on the line satisfy \(x+y=4\)
1, Chapter 5.2, Strict Inequality,
Strict Inequality

Note that if the inequality was a strict inequality, that is; \(x+y \gt 4\) then we use a dashed line to indicate that points actually on the line do not satisfy the inequality.

This is shown on the diagram below
1, Chapter 5.2, Task 1, Worked Examples 2,
Worked Examples
- Worked Example
  
  Shade the region that satisfies the inequality: \(y \geq 4x-7\)
0, Chapter 6, Dealing With More Than One Inequality,
Dealing With More Than One Inequality

Here we extend our graphical approach in the previous section and now look at problems with more than one inequality. This is the beginning of a topic called “Linear Programming” which has applications in many different contexts from Business planning and scheduling to Economic modelling.

Let us illustrate the concept with an example of a problem.

The problem

A small factory employs people at two rates of pay. The maximum number of people who can be employed is 10. More workers are employed on the lower rate than on the higher rate.

Describe this situation using inequalities, and draw a graph to show the feasible region in which they are satisfied.
The solution

Let \(x =\) number employed at the lower rate of pay and \(y =\) number employed at the higher rate of pay

Maximum number employed is \(10\) so \(x + y \leq 10\)

See diagram below, noting that we shade out the side not wanted.

More people are employed at the lower rate so \(x \gt y\) as shown in the diagram

As \(x\) and \(y\) must be positive, \(x \geq 0\) and \(y \geq 0\)

The feasible region is the lower RED triangle as shown. Only integer solutions are possible and these are marked with BLACK dots
0, Chapter 7, Summary,
Summary

\(\gt\) means 'is greater than'; for example, \(5 \gt 3\).
\(\geq\) means 'is greater than or equal to'; for example, \(5 \geq 3\),\(5 \geq 5\)
\(\lt\) means 'is less than'; for example, \(3 \lt 5\).
\(\leq\) means 'is less than or equal to'; for example, \(3 \leq 5\), \(5 \leq 5\)
\(\neq\) means 'is not equal to'; for example, \(3 \neq 5\).

You can represent inequalities on a number line using:

Linear inequalities

are of the form
\(ax+b \leq c\) or \(ax+b \lt c,\) etc.
For example, \(2x + 3 \leq 7\)

Solving linear inequalities

means manipulating the inequality to give \(x \geq ...\) or \(x \leq ...\)
For example, \(2x + 3 \leq 7 \Rightarrow 2x \leq 4 \Rightarrow x \leq 2\)

General linear inequality

in two variables is of the form \(ax + by \geq\) and can be illustrate graphically.
For example, \(x + y \leq 1\)

Interactive Exercises:

Solving Inequalities Interactive Exercises, https://www.cimt.org.uk/sif/algebra/a12/interactive.htm
Inequalities on a Number, https://www.cimt.org.uk/sif/algebra/a12/interactive/s1.html
Line Solution of Linear Inequalities (Inequations), https://www.cimt.org.uk/sif/algebra/a12/interactive/s2.html
Graphical Approach to Inequalities, https://www.cimt.org.uk/sif/algebra/a12/interactive/s3.html

YouTube URL: https://www.youtube.com/watch?v=vcYlrdGctJQ&list=PL8ce1n6mRlCtS0rO_C7CNwtjhyqDVkr8g&index=27&ab_channel=RussellGeachRussellGeach

File Attachments: media/course-resources/Solving-Linear-Inequalities_Learning-Objectives.pdfmedia/course-resources/Solving-Linear-Inequalities_PowerPoint-Presentation.pptxmedia/course-resources/SolvingLinearInequalities_Text.pdfmedia/course-resources/Solving-Linear-Inequalities_Answers.pdfmedia/course-resources/Solving-Linear-Inequalities_Essential-Informations.pdf

Written on 10 September 2020.

Algebraic Fractions and Equations

Sections:

0, Chapter 1, Introduction,
Introduction

Manipulating and solving Algebraic Fractions and Equations is an important part of algebra work.

You have five sections to work through and there are worked examples and fitness tests for each section.
1. Algebraic Manipulation
2. Algebraic Fractions: Part 1
3. Algebraic Fractions: Part 2
4. Algebraic Fractions and Quadratic Equations
5. Algebraic Solution of Simultaneous Equations - One Linear and One Quadratic
This unit extends your algebraic skills by solving more complicated equations. After completing
this unit you should
- be able to manipulate algebraic fractions
- be able to solve quadratic equations that are derived from algebraic equations with fractions
- understand how to solve simultaneous equations where one is linear and one is quadratic.
0, Chapter 2, Algebraic Manipulation,
Algebraic Manipulation

Sometimes we want to rearrange an equation to make a different variable (letter) the subject.

For example, make \(x\) the subject of this equation:

\(y = x + 24\)

We rearrange, by subtracting \(24\) from each side:

\(y - 24 = x\)

\(x = y - 24\)

Sometimes a variable may appear twice in a formula or equation.

For example: Make \(x\) the subject of this equation:

\(5x - 4 = 3x + a\)

First bring all the terms containing \(x\) to one side of the equation. Subtract \(3x\) to give: \(2x - 4 = a\)

Now add 4: \(2x = a + 4\)

Divide by 2: \(x = \frac {a+4}{2}\)
1, Chapter 2, Task 1, Worked Examples 1,
Worked Examples

Use the slider to explore worked examples.
- Worked Example
  
  Make \(x\) the subject of this equation:
  
  \(2(x − 5) = 5a − 3x\)
- Worked Example
  
  Make \(x\) the subject of the equation:
  
  \(p=\sqrt{\frac{x+10}{x} } (p\neq \pm 1)\)
1, Chapter 2, Task 2, Fitness Check 1,
Fitness Check

Make x the subject of each of the following:

7x − a = x − 5

xa − 6 = bx − 5

5(x − a) = 9(a − x)

p(x − 1) = q(x − 1)

2(x + 1) = a − x

− 3(3x − 2 − b) = 9(x − 1)

\(\frac{2x-a}{6}-\frac{x+b}{6} \)
0, Chapter 3, Algebraic Fractions: Part 1,
Algebraic Fractions: Introduction

To add or subtract fractions, we first need a common denominator.

For example: \(\frac{3}{5} + \frac {1}{3} = \frac {9}{15} + \frac {5}{15} = \frac {14}{15}\)

The same is true, when working with algebraic fractions. For example: Express \(\frac{x}{6} + \frac{x}{4}\) as a single fraction.

\(\frac{x}{6} + \frac{x}{4} = \frac {4x}{24} + \frac{6x}{24} = \frac {10x}{24} = \frac{5x}{12}\)
1, Chapter 3, Task 1, Worked Examples 2,
Worked Examples
- Worked Example
  
  Express: \(\frac{3}{x} +\frac{2}{x+4} \) as a single fraction.
1, Chapter 3, Task 2, Fitness Check 2,
Fitness Check

Simplify each expression:

\(\frac{1}{x+4}+\frac{2}{x-3} \)

\(\frac{x}{x+2}+\frac{2x}{x-4} \)

\(\frac{3x}{x}+\frac{2}{x+4} \)

\(\frac{5x}{6-x}-\frac{2}{x+4} \)
0, Chapter 4, Algebraic Fractions: Part 2,
Algebraic Fractions: Simplifying

Some algebraic fractions can be easily simplified if the numerator and the denominator are first factorised.

For example, simplify: \(\frac {x^{2} - x - 12}{2x^{2} + 8x + 6}\)

Both the top and bottom of the fraction can be factorised to give:

\(\frac{x^{2} - x - 12}{2x^{2} + 8x + 6} = \frac{x^{2} - x - 13}{2x^{2} +8x + 6 } = \frac{\left(x -4\right)\left(x + 3\right) }{\left(2x + 2\right) \left(x + 3\right) }\)

The term \(\left(x+3\right)\) appears as a factor on both the top and bottom of the fraction so it cancels to give:
\(\frac{\left(x -4\right)\left(x + 3\right) }{\left(2x + 2\right) \left(x + 3\right) } =\frac{x-4}{2x +2}\)

No further simplification is possible.
1, Chapter 4, Task 1, Worked Example 3,
Worked Example

Use the slider to explore worked examples.
- Worked Example
  
  Simplify: \(\frac{x^{2} }{x+4} \times \frac{x^{2}-16 }{x^{3} } \)
- Worked Example
  
  Simplify: \(\frac{x^{2}+5x+6 }{x-4} \div \frac{x+3}{x^{2}-2x-8 } \)
  
  HOT TIP: To divide by a fraction, we multiply by its reciprocal
- Worked Example
  
  Simplify: \(\frac{x-3}{x+2}+\frac{x-7}{x+1} \)
  
  HOT TIP: Algebraic fractions are added in the same way as ordinary fractions, by using the lowest common denominator.
0, Chapter 5, Algebraic Fractions and Quadratic Equations,
Algebraic Fractions and Quadratic Equations

Some quadratic equations may be presented in algebraic fraction form. We need to first simplify the fraction, then solve the equation.

For example: simplify and solve: \(\frac {x}{2x+1} + \frac {x}{x-1} = \frac {12}{5}\)

In this case, the lowest common denominator is \(\left(2x + 1\right) \left(x - 1\right)\)

So:

\(\frac{x}{2x+1} +\frac{x}{x -1} \longrightarrow \frac{x\left(x-1\right) + x\left(2x +1\right) }{\left(2x+1\right) \left(x -1\right) } = \frac{12}{5}\)

Expanding brackets:

\(\frac{x^{2}-x+2x^{2}+x}{2x^{2} -2x + 1x -1} \longrightarrow \frac{3x^{2}}{2x^{2} -x -1} = \frac {12}{5}\)

Simplifying by cross multiplying:

\(5 \times 3x^{2} = 12 \times \left(2x^{2} -x -1\right)\)
\(15x^{2} = 24x^{2} - 12x -12\)

Now we are ready to solve!

\(0=9x^{2} -12x -12\)

This is the same as: \(9x^{2} - 12x - 12 = 0\)

First we factorise: \(\left(3x +2\right) \left(3x-6\right) = 0\)

So: \(\left(3x+2\right) = 0\) or \(\left(3x - 6\right) = 0\)

We solve each: \(3x = -2\) or \(3x = 6\)

\(x = - \frac{2}{3}\) or \(x=2\)
0, Chapter 6, Algebraic Solution of Simultaneous Equations,
Algebraic Solution of Simultaneous Equations

Previously you will have solved simultaneous equations where both equations are linear. Now we will solve simultaneous equations where one equation is linear and one equation is quadratic.

For example, solve:

\(y=2x^{2} + x -3 \longrightarrow 1\)

\(y=3x + 1 \longrightarrow 2\)

Now the \(y\) from equation (1) has exactly the same value
as the \(y\) from equation (2), in other words:

\(y=y\)

Therefore: \(2x^{2} + x - 3 = 3x +1\)

This equation simplifies to: \(2x^{2} - 2x - 4 = 0\)

Factorising: \(\left(2x +2\right) \left(x -2\right) = 0\)

And solving: \(\left(2x +2\right) = 0\) or \(\left(x-2\right) = 0\)

\(x= -1\) or \(x =2\)

Here is another example:

The values of \(x\) are now substituted into one of the original equations to find the corresponding values of \(y\). It is usually easier to use the linear equation for this.

\(y =2x^{2} + x -3 \longrightarrow 1\)

\(y =3x +1 \longrightarrow 2\)

\(x = -1\) or \( x = 2\)

Substituting the first solution into equation (2):

\(y=3x +1\)

\(y=3x -1 +1= -2\)

\(y= 3x -1 +1= -2\)

Substituting the first solution into equation (2):

\(y = 3x +1\)

\(y =3 \times 2 + 1 =7\)

Thhe solutions are: \(x= -1, y=-2\) and \(x=2, y=7\)
1, Chapter 6, Task 1, Worked Examples 4,
Worked Examples

Use the slider to explore worked examples.
- Worked Example
  
  Solve:
  
  \(y=3x^{2}-4 \) (1)
  \(y=2x+3\) (2)
- Worked Example
  
  Solve:
  
  \(x^{2}+y^{2} =10 \) (1)
  \(y = x + 2\) (2)
  
  Note – The first equation is the equation of a circle
- Worked Example
  
  Solve:
  
  \(x + 2y = 2 \) (1)
  \(x^{2}+8y=8 \) (2)
0, Chapter 7, Summary,
Summary

Algebraic fractions

These are of the form \(\frac{ax+b}{cx+d}\)

when \(a, b, c,\) and \(d\) are constants.

For example, \(\frac{x+3}{3x-1}\), \(\frac{2y+4}{3-2y}\), \(\frac {3}{x+2}\)

Note that
- \(\frac{1}{a} + \frac{1}{b} = \frac{b+a}{ab}\)
- If \(\left(x-a\right) \left(x-b\right) = 0\) then \(x = a\) or \(x= b\)
- The formula for solving the quadratic equation \(ax^{2} + bx + c = 0\)
  is \(x = \frac{-b \pm \sqrt{b^{2}- 4ac} }{2a}\)

Interactive Exercises:

Algebraic Fractions and Equations, https://www.cimt.org.uk/sif/algebra/a1/interactive.htm
Algebraic Manipulation, https://www.cimt.org.uk/sif/algebra/a1/interactive/s1.html
Algebraic Fractions: Part 1, https://www.cimt.org.uk/sif/algebra/a1/interactive/s2.html
Algebraic Fractions: Part 2, https://www.cimt.org.uk/sif/algebra/a1/interactive/s3.html
Algebraic Fractions and Quadratic Equations, https://www.cimt.org.uk/sif/algebra/a1/interactive/s4.html
Algebraic Solution of Simultaneous Equations, https://www.cimt.org.uk/sif/algebra/a1/interactive/s5.html

YouTube URL: https://youtu.be/P4Odh4fWlvE

File Attachments: media/course-resources/Algebraic-Fractions-Equations_Essential-Info.pdfmedia/course-resources/Algebraic-Fractions-Equations_Learning-Objectives.pdfmedia/course-resources/Algebraic-Fractions-Equations_Text.pdfmedia/course-resources/Algebraic-Fractions-Equations_Answers.pdfmedia/course-resources/Algebraic-Fractions-Equations_Presentation.pptx

Written on 10 September 2020.

Vectors

Sections:

0, Chapter 1, Introduction,
Introduction

This unit introduces the important topic of vectors. Vectors are used in many applications in more
advanced studies in, for example, engineering and astrophysics; for example, fluid flow. After
completing this unit you should
- be able to understand the concept of a vector and add and subtract vectors
- be able to use vectors in 2-dimensional geometry to prove geometric results.
You have two sections to work through namely;
1. Vectors and Scalars
2. Vectors and Geometry
0, Chapter 2, Vectors and Scalars,
Vectors and Scalars

Remember

Vectors have magnitude (size) and direction. Quantities which have only size are known as scalars or scalar quantities.

Examples of vectors are: displacement, force and velocity.

Examples of scalars are: mass, length, distance, area and speed.

Notation:

A vector is written in bold, like a and b, and a scalar quantity is written normally like c

A vector can also be written using the letters if its start and finish with an arrow above, like this: \(\overrightarrow{AB} \)

\(\overrightarrow{AB} = a\)
1, Chapter 2. 1, Describng Translations,
Describng Translations

Another use of vectors is as a way to describe translations. Below we are showing using 2-dimensional vectors in the form \(\left(^{a}_{b}\right) \)

Note: The ‘image’ is the translated object

The diagram shows the translation of a triangle by a vector

\(\left(^{4}_{2}\right) \)
- \(4\;units\) to the right
- \(2\;units\) up
The vector specifies the length (how far) and the direction in which the triangle is to be moved.
1, Chapter 2. 2, Vectors and Scalars 1,
Vectors and Scalars

Clear distinctions can be made between:

Distance and Displacement

Distance is a scalar (like 5 miles).
Displacement is a vector (5 miles north-east).
You can walk a long way and the distance travelled may be great, but
if you return to the start your displacement will be zero!

Speed and Velocity

Speed is how fast something moves: it is a scalar (like 5 miles per hour).
Velocity is speed with direction: it is a vector (5 miles per hour south-west).
1, Chapter 2. 3, Using vectors,
Using vectors

We can add two vectors by joining them together:

Adding vectors is easy:
- You simply add together the top and bottom numbers
- Here is an example \(\left(^{4}_{2}\right) + \left(^{3}_{-4}\right) = \left(^{7}_{-2}\right) \)
Top Tip

think of the vector above represented in the triangle in the following way. You can walk along the line a+b by walking along the line a and the line b and either route you end in the same place
1, Chapter 2. 4, Adding or Subtracting Vectors,
Adding or Subtracting Vectors

A similar method is used for subtraction:

\(\left(^{10}_{7}\right) - \left(^{8}_{-3}\right) = \left(^{2}_{10}\right)\)

Adding or Subtracting Vectors

Remember

\(\left(^{a}_{b}\right) + \left(^{c}_{d}\right) = \left(^{a + c}_{b+d}\right)\)
\(\left(^{a}_{b}\right) - \left(^{c}_{d}\right) = \left(^{a - c}_{b-d}\right)\)
1, Chapter 2. 5, Multiplying a vector by a scalar,
Multiplying a vector by a scalar

Multiplying vectors by scalars is also easy

For example, \(5 \times \left(^{3}_{2}\right) = \left(^{15}_{10}\right)\)

Each component is simply multiplied by the scalar.

Multiplying a vector by a scalar

Remember

\(k \times \left(^{a}_{b}\right) = \left(^{ka}_{kb}\right)\)

Another example: \(-8 \times \left(^{5}_{-4}\right) = \left(^{-40}_{32}\right)\)
1, Chapter 2. 6, The length of a vector,
The length of a vector

Remember

The length of a vector is called its magnitude or modulus:
We write this as | a | .

If \(a=\left(^{4}_{3}\right)\), then using Pythagoras' Theorem:

We can calculate

\(| a | = \sqrt {4^{2} + 3^{2}}\)
\(| a | = \sqrt {25} = 5\)

Remember

In general, if \(a = \left(^{x}_{y}\right)\), then using Pythagoras' Theorem: \(| a | = \sqrt {x^{2} + y^{2}}\)
0, Chapter 3, Vectors and Geometry,
Vectors and Geometry

Vectors can be used to solve problems in geometry.

It is possible to describe any point, in 2 dimensions, using two vectors.

For example, using multiples of the vectors a and b, we can get to all the points shown on the diagram, assuming that the \(AE = EI\) and \(AB = BC = CD\)
and lines \(AI\) and \(BJ\) etc. and \(AD\) and \(EH\) etc. are parallel.

\(\overrightarrow{AC}\) = 2a

\(\overrightarrow{EH}\) = 3a

\(\overrightarrow{KJ}\) = -a

\(\overrightarrow{KG}\) = -b

\(\overrightarrow{BJ}\) = 2b

\(\overrightarrow{AG}\) = 2a + b

\(\overrightarrow{AG}\) = 3a + 2b

\(\overrightarrow{GI}\) = -2a + b

\(\overrightarrow{LA}\) = -3a - 2b

\(\overrightarrow{LA}\) = -3a + 2b

Note: to reverse the direction of a vector, reverse the sign of the vector
1, Chapter 3.1, Parallel vectors,
Parallel vectors

Remember

Parallel vectors are those with the same direction; they do not have to be the same size.

Examples of parallel vectors from thger last section are: \(\overrightarrow{AE}, \overrightarrow{CK}\) and \(\overrightarrow{HL}\)
1, Chapter 3.1, Task 1, Worked Example 1,
Worked Example

Use the slider to explore worked examples.
- Worked Example
  
  \(ABCD\) is a parallelogram such that \(\vec{DC} = 3x\) and \(\vec{DY} = 3y\).
  
  The point \(P\) is on \(DB\) such that \(DP \div PB = 1 \div 2\)
  
  Express \(\vec{CD}\)and\(\vec{DP}\) in terms of \(x\) and \(y\):
- Worked Example
  
  \(ABCD\) is a parallelogram such that \(\vec{DC} = 3x\) and \(\vec{DY}= 3y\).
  
  The point \(P\) is on \(DB\) such that \(DP \div PB = 1 \div 2\)
  
  Show that \(P\) and \(E\) are collinear.
  
  Worked Example
  
  \(ABCDEF\) is a regular hexagon, with centre \(O\). \(\vec{OA} = a\) and \(\vec{OB} = b\).
  
  The point \(P\) is on \(DB\) such that \(DP \div PB = 1 \div 2\)
  
  Express \(\vec{FC}\) in terms of \(x\) and \(y\):
- 0, Chapter 4, Summary,
  Summary
  
  Scalar this means 'a number'
  
  Vector a quantity that has magnitude and direction; for example, velocity
  
  Magnitude (modulus) is the length of the vector, denoted by | a | if
  
  \(a= \left(^{x}_{y}\right)\), then \(| a | = \sqrt {x^{2} + y^{2}}\)
  
  Zero vector \(0= \left(^{0}_{0}\right)\),
  
  Parallel vectors are vectors in the same direction but not necessarily of the same magnitude;
  for example, a and 2a
  
  Unit vector is a vector that has magnitude 1; that is \(| a | = 1\)
  
  Position vector is a vector that starts at the origin (O)
  
  Multiplying a vector by a scalar means multiplying all components of the vector by the scalar.
  For example, \(k= \left(^{x}_{y}\right) = \left(^{kx}_{ky}\right) \)
  
  The vector \(\overrightarrow{AB}\) starts at position A and finishes at position B.
  
  Vector additionis defined by \(\left(^{a}_{b}\right) + \left(^{c}_{d}\right) = \left(^{a + c}_{b + d}\right)\) and geometric interpretation – see diagram below.
  
  Multiplication of a vector by a scalar is shown in the example below
  
  For example, \(k \left(^{a}_{b}\right) = \left(^{ka}_{kb}\right)\)

Interactive Exercises:

Vectors Interactive Exercises, https://www.cimt.org.uk/sif/algebra/a13/interactive.htm
Combined Transformations, https://www.cimt.org.uk/sif/algebra/a13/interactive/s1.html
Vectors and Geometry, https://www.cimt.org.uk/sif/algebra/a13/interactive/s2.html

YouTube URL: https://www.youtube.com/watch?v=Wte0HA-yT1E&list=PL8ce1n6mRlCtS0rO_C7CNwtjhyqDVkr8g&index=17

File Attachments: media/course-resources/Vectors_Learning-Objectives.pdfmedia/course-resources/Vectors_Text.pdfmedia/course-resources/Vectors_Answers.pdfmedia/course-resources/Vectors_Essential-Informations.pdfmedia/course-resources/Vectors_PowerPoint-Presentation.pptx

Written on 11 September 2020.

Quadratic Equations

Sections:

0, Chapter 1, Introduction,
Introduction

Solving Quadratic Equations is another important part of algebra work.

This unit is focused on methods of solving quadratic equations. After completing it you should
- be able to solve quadratic equations that factorise into two linear terms
- be able to use the formula for solving quadratic equations
- understand how to complete the square for a quadratic and hence solve the equation.
You have three sections to work through and there are worked examples and fitness tests for each section.
1. Factorisation
2. Using the Formula
3. Completing the Square
1, Chapter 1.1, Quadratic Equations,
Quadratic Equations

What are quadratic equations? Here are some examples:

\(x^{2} + 6x + 5 = 0\)

\(9y^{2} + 2y + 5 = 0\)

\(2n^{2} - 2n - 5 = 0\)

\(3x^{2} - 4x = 0\)

\(4p^{2} - 49 = 0\)

Note:

\(x^{2} + 6x + 5 \) is a quadratic expression whilst

\(x^{2} + 6x + 5 = 0 \) is a quadratic equation

Remember

Equations of the form:

\(ax^{2} + bx + c = 0\)

Where \(x\) represents an unknown and \(a, b, c\) represent known numbers , where \(a\) is not equal to zero, are called quadratic equations.
1, Chapter 1.2, Solving Quadratic Equations,
Solving Quadratic Equations

Quadratic Equations can be solved in different ways, and we will look at three ways in turn, namely:
- Factorisation
- Using the Formula
- Completing the Square
0, Chapter 2, Factorisation,
Factorisation

Some quadratic equations can be factorised, that is they can be expressed as a product of two factors. This enables us to solve the quadratic equation.

For example, \(x^{2} + 5x + 6 = 0\) can be written as

\(\left(x + 2\right) \left(x + 3\right) = 0\) (You can check this by expanding the brackets yourself)

Now, for this equation to hold true, one of the factors must equal zero. i.e. either \(\left(x + 2\right) = 0\) or \(\left(x + 3\right) = 0\) (or both \(\left(x + 2\right)\) and \(\left(x + 3\right)\) equal zero).

So there are two possible solutions:

If \(\left(x + 2\right) = 0\) then \(x = -2\)

If \(\left(x + 3\right) = 0\) then \(x = -3\)

Before we solve any more quadratic equations we will recap factorising quadratic expressions.
1, Chapter 2.1, Factorising a quadratic expression,
Factorising a quadratic expression

To factorise a quadratic expression, such as \(x^{2} + 5x + 6\) we need to turn it into an expression which looks like the following

\(\left(x+?\right) \left(x+?\right)\)

(Note: if we were to expand the brackets, the \(x\) multiplied by the \(x\) will give us the \(x^{2}\).)

We now need to find two numbers which
1. when added together make \(5\) ( they could be \(\left(4,1\right)\) or \(\left(3,2\right)\)
And
1. when multiplied together make \(6\).
We can see the only pairs that work are \(2\) and \(3\)
We can now factorise \(x^{2} + 5x + 6\) by using \(2\) and \(3\)

\(x^{2} + 5x + 6 =\left(x+2\right) \left(x +3\right)\)

Note: If we now expand \(\left(x + 2\right) \left(x + 3\right)\) we will get back to where we started, i.e. \(x^{2} + 5x + 6\)
1, Chapter 2.2, Factorising expressions,
Factorising expressions

It is also possible to factorise expressions such as: \(2x^{2} + 9x - 5\).

That is expressions where we now have more than one \(x^{2}\)

We still set up a pair of brackets, but now they will look like:

\(\left(2x + ?\right) \left(x + ?\right)\)

This is because, when expanded, \(2x \times x\) will give us the \(2x^{2}\)

But how do we find the missing numbers?

We can use a ‘trial and improvement method’ called the ‘cross method’.

Note: Sometimes a few trials are necessary before we get a ‘hit’.

The next Worked Example shows how this method works
1, Chapter 2.3, Difference between two squares,
Difference between two squares

Factorising quadratic expressions takes plenty of practice. The more you do - the easier it will become!

Top Tip

It is important to check the answers you obtain by expanding the brackets and test if you get back to the original expression. If not then you will need to try different combinations of numbers, continuing until you find one that works.

In special circumstances, there is one very important result.
It is known as the ‘difference between two squares’ and can be used to factorise some expressions

Difference between two squares

You may know that \(\left(a-b\right) \left(a+b\right) = a^{2} - b^{2}\)

This result is known as the ‘difference between two squares’

For example: \(\left(x-4\right) \left(x+4\right) = x^{2} - 16 = x^{2} - 4^{2}\)

We can use this result, in reverse, to help us factorise some expressions, i.e. \(x^{2} - 4^{2} = \left(x-4\right) \left(x+4\right)\)

Now look at: \(4x^{2} - 25\)

The expression is a difference of two squares, namely \(\left(2x\right)^{2}\) abd \(5^{2}\)

and can be factorised as below:

\(4x^{2} - 25 = \left(2x -5\right) \left(2x +5\right)\)
1, Chapter 2.4, Factorisation to Solve Quadratic Equations,
Factorisation to Solve Quadratic Equations

Now we know how to factorise quadratic expressions, we have a straightforward method to solve quadratic equations.

Remember if the product of two factors is zero, at least one of those factors must itself be zero. For example;

Solve: \(2x^{2} - 3x - 5 = 0\)

We rewrite this as: \(\left(x+1\right) \left(2x-5\right) = 0\)

Now: \(\left(x+1\right) = 0\) or \(\left(2x-5\right) = 0\)

We solve each: If \(\left(x+1\right) = 0\) or \(\left(2x -5\right)= 0\)

If \(\left(x+1\right) = 0\) then \(x= -1\)

If \(\left(2x -5\right) = 0\) then \(2x = 5\) \(x=2.5\)

Therefore there are two solutions: \(x=-1\) and \(x=2.5\)
0, Chapter 3, Using the Formula,
Using the Formula

It is not possible to factorise all quadratic expressions, so we need alternative methods to solve quadratic equations. When a quadratic expression cannot be factorised
(or when it is not easy to do so), we can use the quadratic formula.

The solutions of the quadratic equation where a, b and c are numbers and a is not equal to \(0\) and the equation is given by \(ax^{2} + bx+ c = 0\) are given by:

Quadratic Formula

\(x = \frac{-b\pm \sqrt{b^{2} - 4ac} }{2a}\)

We prove this result in the next section but you can skip the proof and go straight to Worked Examples if you wish
1, Chapter 3.1, Proving the Formula 1,
Proving the Formula

We will solve \(ax^{2}\) + \(bx\) + c = 0 where \(a \left( \neq 0\right)\), \(b\) and \(c\) are numbers.

We first divide by \(a\) to give \(x^{2} + \frac{b}{a}x + \frac{c}{a} = 0\) and note that:

\(\left(x+ \frac{b}{2a}\right)^{2} = x^{2} + \frac{b}{2a}x+\frac{b}{2a} x +\frac{b^{2}}{4a^{2}}\)

Hence

\(\left(x+ \frac{b}{2a}\right)^{2} - \frac{b^{2}}{4a^{2}} = x^{2} + \frac{b}{a}x\)

Substituting back in the equation:

\(\left(x+ \frac{b}{2a}\right)^{2} - \frac{b^{2}}{4a^{2}} + \frac{c}{a} = 0\)

Rearranging gives

\(\left(x+ \frac{b}{2a}\right)^{2} = \frac{b^{2}}{4a^{2}} - \frac{c}{a} = \frac{b^{2} - 4ac}{4a^{2}}\)

Taking the square root of both sides gives

\(x+\frac{b}{2a} = \pm \frac{\sqrt{b^{2} - 4ac} }{2a} \)

Taking the \(\frac{b}{2a}\) term to the other side gives:

\(x = \frac{-b\pm \sqrt{b^{2} - 4ac} }{2a}\)
1, Chapter 3.1, Task 1, Worked Example 1,
Worked Example

Use the slider to explore worked examples.
- Worked Example
  
  Solve: \(4x^{2}-7x+3=0 \)
- Worked Example
  
  Solve: \(4x^{2}-12x+9=0 \)
- Worked Example
  
  Solve: \(x^{2}+x+5=0 \)
1, Chapter 3.2, Using the Formula 2,
Using the Formula

The three previous examples illustrate that a quadratic equation can have \(2, 1\) or \(0\) solutions.

The graphs below illustrate \(y = ax^{2} + bx + c\) in the 3 worked examples.
The solution to equation is represented on the graph by the point or points where the graph touches the \(x-axis\).The worked examples showed us that this depends on the value of \(b^{2} - 4ac\)

Remember

We call \(b^{2}−4ac\) the discriminant
0, Chapter 4, Completing the Square,
Completing the Square

Completing the square is another method for solving quadratic equations.

We can use this method for quadratic equations that do not factorise.

It is also a very useful technique for finding the minimum and maximum value of a quadratic.

A general quadratic equation \(ax^{2}+ bx+ c = 0\) is written in the form \(ax + p^{2} + q = 0\) when completing the square.

We need to find the constants \(p\) and \(q\) so that the two expressions are identical, i.e.

\(ax^{2} + bx + c =a \left(x + p\right)^{2} + q\)
1, Chapter 4, Task 1, Worked Example 2,
Worked Example

Use the slider to explore worked examples.
- Worked Example
  
  Complete the square for: \(x^{2}+10x+8 \)
- Worked Example
  
  Complete the square for: \(x^{2}-8x+2=0 \)
- Worked Example
  
  Complete the square for: \(3x^{2}+12x+7=0 \)
  
  Note: There are now more than one \(x^{2}\)
- Completing the square is also a useful technique for finding the minimum or maximum value of a quadratic.
  
  We will look at an example, to see how this is done.
  
  Worked Example
  
  (a) Complete the square for \(y=2x^{2} -8x+2\)
  
  (b) Find the minimum value of \(y\)
  
  (c) Sketch the graph of \(y=2x^{2} -8x+2\)
- Here is an example, to find the maximum turning point:
  
  Worked Example
  
  The height of a ball thrown into the air is given by
  \(h=1+20t-10t^{2}\)
  
  Find the maximum height reached by the ball.
1, Chapter 4.1, Graph,
Graph

Before sketching the graph, it is useful to know where the curve crosses the x-axis.

Remember: The curve crosses the x-axis when \(h = 0\).

That is when: \( 0 = -10 \left(t -1\right)^{2} + 11\)

Hence \(10\left(t -1\right)^{2} = 11\)

\(\left(t -1\right)^{2} = 1.1\)
\(t-1 = \pm \sqrt{1.1}\)
\(t = 1 + \pm \sqrt {1.1}\)

So the curve crosses the x-axis at \(t = 1 + \sqrt {1.1}\) and \(t = 1 - \sqrt{1.1}\) and has a maximum at \(\left(1,11\right)\)

Thos is shown on the graph below

Have you noticed anything?

When a quadratic equation is in the completed the square form, it is easy to spot the minimum or maximum “turning” point.

For example: \(y=2\left(x-3\right)^{2} - 9\)

The minimum possible value of \(y\)

\(y = -9\)

is obtained when \(\left(x-3\right) = 0\)
that is, when \(x=3\)

The turning point will be at \(\left(3, -9\right)\)
0, Chapter 5, Summary,
Summary

There are three methods of solving quadratic equations:

(i) quadratic factorisation, where we can write
\(ax^{2} + bx + c = a \left(x - p\right) \left( x - q\right)\) when \(p\) and \(q\) are rational numbers
which has solution \(x=p\) or \(x=q\)

(ii) formula \(x= \frac{-b\pm \sqrt{b^{2} - 4ac} }{2a} \)

(iii) completing the square, where we write
\(ax^{2} + bx + c = a \left(x+ \frac{b}{2a}\right)^{2} - \frac {b^{2}}{4a} +c\)

Note that if \(\left(x-a\right) \left(x-b\right) = 0\), then either \(x-a =0\) or \(x-b=0\), giving \(x=a\) or \(x=b\)

Quadratic equations have \(2, 1\) or \(0\) real roots according to the value of " \(b^{2} - 4ac\)":
(i) If \(b^{2} \gt 4ac\), there are \(2\) real distinct roots
(ii) If \(b^{2} = 4ac\), there is \(1\) (repeated) real root
(iii) If \(b^{2} \lt 4ac\), there are no real roots

Interactive Exercises:

Interactive Exercises - Quadratic Equations, https://www.cimt.org.uk/sif/algebra/a9/interactive.htm
Factorisation, https://www.cimt.org.uk/sif/algebra/a9/interactive/s1.html
Using the Formula, https://www.cimt.org.uk/sif/algebra/a9/interactive/s2.html
Completing the Square, https://www.cimt.org.uk/sif/algebra/a9/interactive/s3.html

YouTube URL: https://www.youtube.com/watch?v=I85yhiBcRm0&list=PL8ce1n6mRlCtS0rO_C7CNwtjhyqDVkr8g&index=31

File Attachments: media/course-resources/Quadratic-Equations_Learning-Objectives.pdfmedia/course-resources/Quadratic-Equations_Presentation.pptxmedia/course-resources/Quadratic-Equations_Text.pdfmedia/course-resources/Quadratic-Equations_Answers.pdfmedia/course-resources/Quadratic-Equations_Essential-Informations.pdf

Written on 18 September 2020.

Functions

Sections:

0, Chapter 1, Introduction,
Introduction

A function is an important tool to use in algebra. Functions usually have numerical inputs and outputs and are often defined by an algebraic expression. After completing this unit you should
- understand the concepts of the domain and range of a function and 1 : 1 mappings
- understand and be able to use composite functions
- have confidence in finding the inverse of a function
- be able to transform graphs of functions.
You have four sections to work through and there are worked examples and fitness tests for each section.
1. Functions, Mappings and Domain
2. Composite Functions
3. Inverse Functions
4. Transformations of Graphs of Functions
0, Chapter 2, Functions, Mappings and Domains,
Functions, Mappings and Domains

A function is a way of describing what happens to an input variable, in order to get the output result. If we think of a simple function such as add five, we can determine a set of output values.

\(Input \rightarrow FUNCTION \rightarrow Output\)
1, Chapter 2, Task 1, Worked Examples 1,
Worked Example

Use the slider to explore worked examples.
- Worked Example
  
  Given that \(f(x)=x-3\) find:
  
  a) \(f(5)\)
  
  b) \(f(-1)\)
- Worked Example
  
  Given that \(g(x)=2x^{2}-8\) find:
  
  a) \(g(3)\)
  
  b) \(g(-4)\)
- Worked Example
  
  Given that \(f(x)=x^{2}-3\) find:
  
  a) \(f(10)\)
  
  b) \(f(-3)\)
1, Chapter 2.1, Remember 1,
Remember

A function can be thought of as a ‘well behaved’ relation, in other words, given a starting point we know exactly what to do and where to go. Given an input, we get exactly one output; that is, given an \(x\) we get one and only one \(y\) and we write \(y = f \left(x \right)\)

for example , function \(y = f \left(x\right) = x^{2}, -5 \leq x \leq 5\)

In simple terms: the domain is all the values that go into the function, and the range is all the values that come out.

A mapping diagram as shown below shows the domain and the range

Remember

A function is a relation which maps values from the domain \(\left(x\right)\) to the range \(\left(y\right)\) such that there is only one (and exactly one) \(y\) that corresponds to a given \(x\).

Note: In some functions, multiple input values (\(x\) values) can lead to the same (single) output.
1, Chapter 2.2, Note 1,
Note

So to be absolutely clear!! ...

For a function:

One input can lead to one output.
This is called a \(1 : 1\) mapping.
For example:

Or...

Different input values can lead to the same output value.
This is an \(m : 1\) mapping.

But: a function can not have an input value mapping to more than one output.
It cannot be a \(1 : m\) mapping or an \(m : n \) mapping.
1, Chapter 2.3, Vertical line test,
Vertical line test

The vertical line test is a good way to decide whether a relationship is a function.

If you graphically plot a relationship you can decide whether a relationship is a function by drawing vertical lines.

If you can draw a vertical line that crosses the graph in more than one place then the relation is not a function.

For example, the graph of a circle does not show a function.

A single \(x-value\) would map to more than \(1\) \(y-value\).

This is not a function.
1, Chapter 2.3, Task 1, Worked Example 2,
Worked Examples
- Worked Example
  
  Are these functions?
  
  \(2y-3x=-1\)
  
  \(y^{2}+16=9x\)
  
  In other words, can these be solved to give a unique \(y\) value?
0, Chapter 3, Composite Functions,
Composite Functions

Composite functions are ‘functions of a function’.

In other words the output from one function is the input to another!

Suppose you are given two functions:

\(f\left(x\right) = x +1\)
and \(g \left(x\right) = 4x - 2\)

Composition means that you can plug one into the other, for example:

\(g \left( f \left( x \right) \right)\)
\(g \left( f \left( x \right) \right) = 4 \left(f\left(x\right) \right) - 2 \)

\(f \left(x\right)\) becomes the input to \( g \left(x\right)\)

And the output from \(f \left(x\right)\) becomes the input to \(g \left(x\right)\)

\(g\left(f\left(x\right) \right) = 4\left(x+1\right) -2\)
\(=4x + 4 - 2\)
\(=4x + 2\)

the output from \(f \left(x\right)\) is \(x+1\)
1, Chapter 3.1, Remember 2,
Remember

When working with composite functions you work from inside to out (or from right to left) which may, at first, feel a little counter-intuitive.

Here are another two functions:

\(f \left(x\right) = x^{2} + 3\)

and

\(g \left(x\right) = 4x + 1\)

And you want to know \(f \left( g \left( x \right) \right)\)

This time \(g \left(x\right)\) is the input to \(\left(x\right)\), and so the output from \(g \left(x\right)\) becomes the input to \(f \left(x\right)\). So you deal with \(g \left(x\right)\) first.

\(f \left(g \left( x \right) \right) ^{2} + 3\)
\( \left( 4x + 1 \right) ^{2} + 3\)
\( = 16x^{2} + 8x + 1 + 3\)
\( = 16 x^{2} + 8x + 4\)

the output from \(g \left(x\right)\) is \(4x + 1\)

Compare this with: \(g(x)\)

Remember the functions are: \(f(x)=x^{2}+3\) and \(g(x)=4x+1\)

Now \(f(x)\) is the input to \(g(x)\).

So the output from \(f(x)\) becomes the input to \(g(x)\).

\(g(f(x)) = 4(f(x)) + 1\)
\(=4(x^{2}+3)+1\)
\(=4x^{2}+12+1\)
\(=4x^{2}+13\)

the output from \(f(x)\) is \(x^{2}+3\)

Note: in this case \(g(f(x))\) does not equal \(f(g(x))\).

And the result \(f(g(x)) = g(f(x))\) is not generally true (although it can be so on occasions).

For shorthand, we often write \(f(g(x))\) as \(fg(x)\) and \(g(f(x))\) as \(gf(x)\) i.e. by removing set of brackets

But we must be careful to remember what this notation means!!
1, Chapter 3.1, Task 1, Worked Examples 3,
Worked Examples
- Worked Example
  
  Given that \(f(x)=5x+2\) and \(g(x)=3x+5\)
  
  Find:
  a) \(gf(4)\)
  b) \(fg(1)\)
0, Chapter 4, Inverse Functions,
Inverse Functions

An inverse function is a function that reverses another function. It ‘un-does’ what has been ‘done’, taking us back to where we started!

An inverse function is usually shown with a little \(-1\) after the function name, like this: \(f^{-1}(y)\) (we say "\(f\) inverse of \(y\)")

If we input \(x\) to a function \(f(x)=y\) then the inverse function \(f^{-1}(y)=x\)

Suppose we have the function: \(f(x)=2x-4\)
We can call the output: \(y\)
To ‘undo’ this, we would first have to: \(+4\) to the output (y) and then divide by \(2\)
This gives the inverse function: \(f^{-1}(y)= \frac{y+4}{2}=x\)
1, Chapter 4, Task 1, Let’s try it!,
Let’s try it!

Let’s try it!

Let us input 5 into the function: \(f(x)=2x-4\)
\(f(5)=10-4=6\)

The inverse function: \(f^{-1}(y)= \frac{y+4}{2}\)
\(= \frac{6+4}{2}=5\)

"So applying a function \(f\) and then its inverse \(f^{-1}\) gives us the original value back again.

We can also write: \(f^{-1}(y)\) as: \(f^{-1}(f(x))=x\)

By convention we can the replace the \(y\) notation with \(x\) to give the inverse function:
\(f^{-1}(x)= \frac{x+4}{2}\)

And: \(f^{-1}(f(x))=x\)

We could write this in the other order and it would still work:
\(f(f^{-1}(x))=x\)

\(f^{-1}(f(x))=x\) and \(f(f^{-1}(x))=x\)
0, Chapter 5, Transformations of Graphs of Functions,
Transformations of Graphs of Functions

Just like transformations we have may have met before (of shapes), we can move and resize the graphs of functions.

Look at the function: \(f(x)=x^{2}\)

We can plot the graph of this function (or use software to do it for us!!) and look at some simple things we can do to move the graph up and down, or left and right.

Here we have transformed the graph of \(y=f(x)\) by moving it up \(2\) places, by adding a constant of \(2\), giving us: \(y=f(x)+2\)

In general \(f(x) + a \) moves \(a\) curve up \(a\) units and \(f(x)-a\) moves it down \(a\) units, where \(a\) is a positive number.
1, Chapter 5, Task 1, Exercise 1,
Exercise

Graph the function \(f(x)=x^{2}\)

Here we have transformed the graph of \(y=f(x)\) by translating it to the left \(1\) place. We do this by by adding \(1\) to the input, \(x\), giving us:

\(y=f(x+1)\)

Try it, with an input value, \(x\), of \(1\):

Orginal: \((1,1)\)
\(f(x)=x^{2}\)
\(f(1)=1^{2}=1\)

Transformation: \((1,4)\)
\(f(x+1)=(x+1)^{2}\)
\(f(1+1)=(1+1)^{2}=4\)

Here we have transformed the graph of \(y=f(x)\) by moving it up \(2\) places, by adding a constant of \(2\), giving us: \(y=f(x)+2\)

In general \(f(x) + a \) moves \(a\) curve up \(a\) units and \(f(x)-a\) moves it down \(a\) units, where \(a\) is a positive number.
1, Chapter 5, Task 2, Exercise 2,
Exercise

Graph the function \(f(x)=x^{2}\)

Here we have transformed the graph of \(y=f(x)\) by stretching it to the \(y-direction\) by a factor of \(2\). We do this by by multiplying the output by the factor.

\(y=2f(x)\)

Try it, with an input value, \(x\), of \(2\):

Orginal: \((2,4)\)
\(f(x)=x^{2}\)
\(f(2)=2^{2}=4\)

Transformation: \((2,8)\)
\(2f(x)=2x^{2}\)
\(2f(2)=2(2)^{2}=8\)

In general the curve of \(y=kf(x) \) stretches tje graph of \(y=f(x)\) by a factor of \(k\) in the \(y\)-direction where \(k>1\)

What happens if we multiply the input or output by a negative value?:

The graph of \(y=f(x)\) is shown:

Multiplying the whole function by \(−1\), will flip it upside down:
\(y=-f(x)\) is a reflection in the \(x-axis\)

Multiplying the input, \(x\), by \(−1\), will flip it left-right:
\(y=f(-x)\) is a reflection in the \(y-axis\).
1, Chapter 5, Task 3, Worked Examples 3,
Worked Examples
- Worked Example
  
  The graph of \(y=f(x)\) is shown below.
  
  The coordinates of the maximum point of this curve are \(\left(− 2, 1\right)\).
  
  Write down the coordinates of the maximum point of the curve with equation:
  a) \(y=f(x-3)\)
  b) \(y=f(x)-2\)
  c) \(y=f(-x)\)
  d) \(y=-f(x)\)
0, Chapter 6, Summary,
Summary

The domain of a function is the value of \(x\) for which the function is defined. For example, \(f \left(x\right) = x^{2} + 1\) for \(x \geq 0\) (domain is \(x \geq 0\)) \(g \left(x\right) = \frac {1}{x-1} + \frac {1}{x-2} \) for \(1 \lt x \lt 2\) (domain is \(1 \lt x \lt 2\))

The range of a function is the set of values that the function maps onto.
For example, if \(f \left(x\right) = x^{2}\), \(0 \leq x \leq 5\),
the range of \(f\) is \(0 \leq f \left(x\right) \leq 25\).

1 : 1 mapping is a function for which every value of \(f\left(x\right)\) is unique; that is, if \(f \left(a\right) \neq f \left(b\right) \) unless \(a = b\)
For example, \(f \left(f\right) = x + 1\) is a 1 : 1 mapping, but
\(f \left(x\right) = x^{2}\)is not a 1 : 1 mapping as, for example, \(f \left(2\right) = 4 = f\left(-2\right)\)

The composite function \(f g\) or \(f g\left(x\right)\) is defined as \(f \left(g\left(x\right)\right)\)
For example, if \(f \left(x\right) = x +1\) and \(g \left(x\right) = x^{2}\), then \(f \left(x\right) = f \left(x^{2}\right) = x^{2} + 1\)
and \(gf \left(x\right) = g \left(x +1\right) = \left(x + 1\right) ^{2}\)

The inverse function of \(f\) is denoted by \(f^{-1}\) and is such that \( f f^{-1} \left(x\right) = x = f^{-1} f \left(x\right)\)
The inverse only exists if the function is a 1 : 1 mapping.

The graph of \(y = f\left(x\right)\) is mapped onto the graph of \(y= f\left(x\right) + 2\) by translating it up \(2\;units\)
In general \(f \left(x\right) + a\) moves a curve up \(a\;units\) and \(f\left(x\right) -a\) moves it down \(a\;units\), where \(a\) is a positive number.

The graph of \(y = f \left(x\right)\) is mapped onto \(f\left(x+1\right)\) by a translation of \(1\;unit\) to the left.

In general \(f \left(x+a\right)\) translates a curve \(a\) units to the left and \(f \left(x-a\right)\) translates a curve units to the right, where a is a positive number.

The curve for \(f\left(2x\right)\) is much steeper than for \(f\left(x\right)\). This is because the curve has been compressed by a factor of \(2\) in the \(x-direction\). Compare the rectangles \(ABCD\) and \(EFGH\).

In general the curve of \(y = f \left(kx\right)\) will be compressed by a factor of \(k\) in the \(x-direction\), where \(k \gt 1\).

Here the curve \(y = f \left(x\right)\) has been stretched by a factor of \(2\) in the vertical or \(y-direction\) to obtain the curve \(y = f \left(x\right)\). Compare the rectangles \(ABCD\) and \(CDFE\).

In general the curve of \(y = kf \left(x\right)\) stretches the graph of \(y = f \left(x\right)\) by a factor of \(k\) in the \(y-direction\) where \(k \gt 1\).

Note that if \(k\) is negative and \(k \lt −1\), the curve will be stretched and reflected in the \(x-axis\), while if \(−1 \lt k \lt 1\), it is compressed.

Interactive Exercises:

Functions Interactive Exercises, https://www.cimt.org.uk/sif/algebra/a5/interactive.htm
Functions, Mappings and Domains, https://www.cimt.org.uk/sif/algebra/a5/interactive/s1.html
Composite Functions, https://www.cimt.org.uk/sif/algebra/a5/interactive/s2.html
Inverse Functions, https://www.cimt.org.uk/sif/algebra/a5/interactive/s3.html
Functions Interactive Exercises, https://www.cimt.org.uk/sif/algebra/a5/interactive.htm

YouTube URL: https://youtu.be/narzR8NvBGI

File Attachments: media/course-resources/Functions_Essential-Information.pdfmedia/course-resources/Functions_Learning-Objectives.pdfmedia/course-resources/Functions_Presentation.pptxmedia/course-resources/Functions_Text.pdfmedia/course-resources/Functions_Answers.pdf

Written on 22 September 2020.

Sequences and Series

Sections:

0, Chapter 1, Introduction,
Introduction

This unit extends the algebraic themes met in earlier units, to sequences and series. After completing this unit you should understand the concept of a general geometric sequence
- be able to sum the first n terms of a geometric series and consider the convergence of this sum as n → ∞
- understand the concept of a general arithmetic sequence
- be able to sum the first n terms of an arithmetic series
- understand and use the ∑ notation of summing series.
- Sequences, Series and Sigma Notation
You have five sections to work through namely;
1. Geometrical Sequences
2. Never-Ending Sums
3. Arithmetic Series
4. Sigma Notation
5. More Series
0, Chapter 2, Arithmetic Sequence,
Arithmetic Sequence

A Sequence or Series is a set of things (often numbers) which follow a pattern.

\(5,9,13,17,21,...\)
where:
\(5\) - first item,
\(9\) - second item,
\(13\) - third item,
...... dots show that sequence continues forever

The pattern is...
add \(4\)

What is the \(10^{th}\) item?
We started with \(5\) and then ‘added \(4\)’ nine times. so the sequence is

\(5,9,13,17,21,25,29,33,37,41\)

This is an example of an Arithmetic Sequence.
0, Chapter 3, Geometric Sequences,
Geometric Sequences

In a Geometric Sequence or Series each term is found by multiplying the previous term.

\(5,15,45,135,405,...\)
where:
\(5\) - first item,
\(15\) - second item,
\(45\) - third item,
..... dots show that sequence continues forever

The pattern is...
multiply by \(3\)

What is the \(10^{th}\) item?
We started with 5 and then multiply by \(3\) nine times:

\(5,12,45,135,405,1215,3645,10935,32805,98415\)

This is an example of an Arithmetic Sequence.

1, Chapter 3, Task 1, Worked Example 1,

Worked Example

A classic legend surrounds the game of chess. The story goes that the inventor of the game was told to name her reward. The reply was something like this:

“My request is to be rewarded in grains of rice, counted out as follows:
\(1\) grain for the first square of the chessboard,
\(2\) grains for the second square of the chessboard,
\(4\) grains for the third,
\(8\) grains for the fourth,
\(16\) grains for the fifth,
and so on…

Not an unreasonable request, right?

Can you work out how many grains of rice that would actually be?

1, Chapter 3.1, Geometric Sequences,
Geometric Sequences

This series of numbers \(1,2,4,8,16,32,64\) … is an example of a geometric sequence, also sometimes called a geometric progression (GP).

In general we write a geometric sequence as:
\(a,ar,ar^{2},ar^{3}, ...,ar^{n-1},...\)
where \(a\) is the first term, and \(r\) is the factor between the terms (also known as the common ratio)

So in general we write a geometric sequence as:
\(a,ar,ar^{2},ar^{3},...,ar^{n-1},...\)*

The sequence goes forever but we can sum a chosen number of terms.
The sum of several terms of a sequence is called a series.

In the chessboard example, the sum:
\(2^{0} + 2^{1} + 2^{2} + 2^{3} + 2^{4} + 2^{5} + ... + 2^{63}\) is called a geometric series.

*We do, however, need to make sure \(r\) does not equal zero, otherwise the sequence will be:
\(a,0,0,0,...,0,...\)
1, Chapter 3.1, Task 1, Geometric Sequences: Example,
Geometric Sequences: Example

Such sequences occur in many situations.

Imagine a coronavirus with an \(R\) number ( the number of people an infected person will will on average infect) of \(2\) (not difficult to imagine!!!).

If \(R\) is \(2\), one infected person will, on average, infect \(2\) others - who will infect \(4\) others who will infect \(8\) others, and so on. Sound familiar?

This doubling of the number of new infections may happen every \(2\) to \(3\) days.

After just \(20\) ‘doublings’ the number of newly infected people will go from just one to over one million!
1, Chapter 3.2, Sequences and Series,
Sequences and Series

In the chessboard problem, the solution involved adding up the first \(64\) terms.

The sum of the first n terms of a series is often denoted by \(S_{n}\).
The formula for \(S_{n}\) is:
\(S_{n}= \frac{a(r^{n}-1)}{r-1}\)

It may be worth noting that, if \(r \lt 1\), it is easier to use:
\(S_{n}= \frac{a(1-r^{n})}{1-r}\)
We can prove these formulae when \(r \neq 0\).

Proof:
\(S_{n}=a+ar+ar^{2}+ar^{3}+...+ar^{n-1}\)
Multiply both sides by \(r\):
\(rS_{n}=a+ar+ar^{2}+ar^{3}+...+ar^{n-1} + ar^{n}\)
\(rS_{n}-S_{n}=ar^{n}-a\)
\(S_{n}(r-1)=a(r^{n}-1)\)
\(S_{n}= \frac{a(r^{n}-1)}{r-1}\)

Or:
\(S_{n}-rS_{n}=a-ar^{n}\)
\(S_{n} (1-r)=a(1-r^{n})\)
\(S_{n}= \frac{a(1-r^{n})}{1-r}\)
1, Chapter 3.2, Task 1, Worked Example 2,
Worked Example
- Worked Example
  
  After how many complete years will a starting capital of \(£10 000\) first exceed \(£30 000\) if it grows at \(4\)% per annum?
0, Chapter 4, Never-Ending Sums,
Never-Ending Sums

Sequences can be either convergent or divergent.

If the further you go along a sequence, the closer you get to a specific value, then we call this a convergent sequence.
For example: \(10,5,2.5,1.25,0.625,0.3125,0.15625,0.078125,...\) the numbers get closer and closer to zero, and zero is said to be the limit of the sequence.

If, as you go along a sequence, the terms either grow without limit, or oscillate back and forth, then we call these divergent sequences.
Any sequence that does not converge is said to be divergent.
For example: \(1,3,9,27,...\) (heads towards infinity \(\infty \))
\(2,-2,2,-2,...\) (goes up and down without settling towards some specific value)

A geometric series, \(a+ar+ar^{2}+ar^{3}+...ar^{n-1}\) converges when \(|r| \lt 1 \), that is, when \(-1 \ \lt r \lt 1\).

If \(|r| \lt 1\) then as the sequence goes further and further and the terms tend to infinity (i.e. \(n\rightarrow \infty \)), then \(r^{n}\) will get smaller and smaller and eventually tend to zero (i.e. \(r^{n} \rightarrow 0 \)).

This gives us:
\(S_{n}= \frac{a(r^{n}-1)}{r-1}\)
\(S_{n}= \frac{a(0-1)}{r-1} = \frac{-a}{r-1}\)
\(S_{n}= \frac{a}{1-r}\) as \(n \rightarrow \infty\)
\(S_{n} = \frac{a}{1-r}\)

Remember: to be able to sum to infinity, \(|r|\) must be less than \(1\)

The limit \(\frac{a}{1-r}\) is known as the ‘sum to infinity’ and is denoted by \(S+{\infty}\).
1, Chapter 4, Task 1, Worked Example 3,
Worked Example
- Worked Example
  
  Express \(0.2929292929292929...\) as an infinite geometric series and then find the fraction it represents.
0, Chapter 5, Arithmetic Series,
Arithmetic Series

Another important type of sequence is the Arithmetic sequence.

In an arithmetic sequence there is a constant difference between consecutive terms.

In other words we just add (or subtract) the same value each time.
For example: \(5,8,11,14,17,20,23,27,...\)
The sequence has a difference of \(3\) between each number.

In general, we could write an arithmetic sequence as:
\(a,a+d,a+2d,a+3d,a+4d,a+5d,...\)
where: \(a\) is the first item, and
\(d\) is the difference between terms (known as the common difference

Note: we can write the 𝑛th term as: \(a+(n-1)d\)

The sum of terms of an arithmetic sequence is called an arithmetic series or an arithmetic progression (AP).

To sum the terms of an arithmetic series we use:
\(S_{n}= \frac{n}{2}(2a+(n-1)d)\)

An alternative form for \(S_{n}\) is given by finding the average of the first \((a)\) and last terms \((l)\) and multiplying by the number of terms \((n)\).

\(S_{n}= \frac{n}{2}(a+l)\)
The last term \((l)\) can be written as: \(l=a+(n-1)d\)
so \(S_{n}\) can be written as:
\(S_{n}=\frac{n}{2}(a+(a+(n-1)d))\)
\(S_{n}=\frac{n}{2}(2a+(n-1)d)\)
1, Chapter 5, Task 1, Worked Example 4,
Worked Example
- Worked Example
  
  Sum the series: \(5+8+11+14+17+20+23+27+...\) to \(20\) terms.
0, Chapter 6, Sigma Notation,
Sigma Notation

The Greek letter \(\sum{} \) (sigma) is used to denote ‘summing up’.
It can be used as a shorthand to show what we want to sum up.

For exa,ple: \(1+2+3+4+5+6+7+8+9\)
can be denoted by:
\(\sum\limits_{n=1}^{9}{n} \)
This means we will substitute, and then sum, all values from \(1\) up to \(9\) into whatever comes after \(\sum{}\), in this case simply just \(n\) itself.
\(\sum\limits_{n=1}^{9}{n} = 1+2+3+4+5+6+7+8+9\)

Let’s look at:
\(\sum\limits_{n=1}^{5}{n^{2}}\)

This means we will substitute, and then sum, all values from \(1\) up to |(5\) into whatever comes after \(\sum\), in this case \(n^{2}\)

\(\sum\limits_{n=1}^{5}{n^{2}} = 1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2}\)
\(= 1+4+9+16+25 = 55\)

And for:

\(\sum\limits_{n=1}^{4}{2n+3}\)
This means we will substitute, and then sum, all values from \(1\) up to \(4\) into whatever comes after \(\sum{}\), in this case \(2n+3\).

\(\sum\limits_{n=1}^{4}{2n+3} = 5+7+9+11 = 32\)

[Note this an Arithmetic progression with \(a=5\) and \(l=11\) and \(n=4\) and we have seen \(S_{n}= \frac{n}{2}(a+l) =\)
\(\frac{4}{2}(5+11) = \frac{4}{2}(16) = 32\)]

We can start and end with any number, and use different letters.
For example:
\(\sum\limits_{r=3}^{7}{\frac{1}{r-1}} \)
\(\frac{1}{3-1} + \frac{1}{4-1} + \frac{1}{5-1} + \frac{1}{6-1} + \frac{1}{7-1}\)
\(=\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6}\)
1, Chapter 6, Task 1, Worked Example 5,
Worked Example
- Worked Example
  
  Shorten the following using the \(\sum{}\) notation:
  a) \(1+8+27+64+125+216\)
  b) \(4+9+14+19+24\)
  c) \(1+2+2^{2}+2^{2}+2^{4}+2^{5}+...+2^{16}\)
0, Chapter 7, More Series,
More Series

These three important results are really useful:

\(\sum\limits_{r=1}^{n}{r} = \frac{1}{2}n(n+1)\)
We have seen this one before. It is the sum of the arithmetic progression, \(1,2,3,4,...,n\)

\(\sum\limits_{r=1}^{n}{r^{2}} = \frac{1}{6}n(n+1)(2n+1)\)

\(\sum\limits_{r=1}^{n}{r^{3}} = \frac{1}{4}n^{2}(n+1)^{2}\)

Interestingly: \(\sum{} ^{n}_{r=1} r^{3}=(\sum{} ^{n}_{r=1} r)^{2}\)

Formulae do exist for higher powers, \(r^{4},r^{5},r^{6},...\) but they are not user-friendly so are seldom used.

This is a coincidence!!! (but a very useful fact!!!)
1, Chapter 7, Task 1, Worked Example 6,
Worked Example

Use the slider to explore worked examples.
- Worked Example
  
  Work out:
  \(\sum\limits_{5}^{15}{r} \)
- Worked Example
  
  Work out:
  \(\sum\limits_{11}^{24}{r^{2}} \)
- Worked Example
  
  Work out:
  \(7^{3} + 8^{3} + 9^{3} + ... + 15^{3}\)
- Worked Example
  
  Work out:
  \(\sum\limits_{0}^{16}{3+5r}\)
  
  Note: \(\sum\limits_{r=1}^{n}{1} = n\)
  as we are simply adding \(1+1+1+1+...\) \(n\) times
0, Chapter 8, Summary,
Summary

The general form of a geometric sequence of \(n\) terms is
\(a, ar, ar^{2}, ... , ar^{n-1}\)
where \(a\) is a constant and \(r\), the ratio of consecutive terms, is called the common ratio.

The sum of n terms of a geometrical series is given by
\(S_{n}= \frac {a \left(r^{n} - 1\right)}{\left(r -1\right)}\)
\(v\left(r \neq 1 \right)\)

If \(|r| \lt 1\), the sum to infinity of a geometric series is given by
\( S_{n} \rightarrow \frac {a}{\left(1-r\right)}\) as \( n \rightarrow \infty\)

The general arithmetic sequence is of the form
\(a, a + d, a + 2d, a + 3d,...\)
where \(a\) and \(d\) are constants.

The sum of \(n\) terms of an arithmetic sequence is given by
\(S_n = \frac{n}{2} \left(2a + \left( n - 1\right) d \right)\)
If \(l\) is the last term, then \(l=a + \left(n-1\right) d\) and
\(S_{n} = \frac {n}{2} \left(a + 1\right)\)

Sigma notation can be used to write a sum of series.
For example, \(2+4+8+...+2^{12} = \sum\limits_{r=1}{2^{r}} \)

Important results are
\(\sum\limits_{r=1}^{n}{r} = 1 +2+3+...+n=\frac {1}{2}n \left(n+1\right)\)
\(\sum\limits_{r}^{n}{r^{2}} = 1^{2} +2^{2}+3^{2}+...+n^{2}=\frac {1}{6}n \left(n+1\right)\left(2n+1\right) \)
\(\sum\limits_{r}^{n}{r^{3}} = 1^{3} +2^{3}+3^{3}+...+n^{3}=\frac {1}{4}n^{2} \left(n+1\right)^{2}\)

Interactive Exercises:

Sequences and Series Interactive Exercises, https://www.cimt.org.uk/sif/algebra/a10/interactive.htm
Geometrical Sequences, https://www.cimt.org.uk/sif/algebra/a10/interactive/s1.html
Never-Ending Sums , https://www.cimt.org.uk/sif/algebra/a10/interactive/s2.html
Arithmetic Sequences , https://www.cimt.org.uk/sif/algebra/a10/interactive/s3.html
Sigma Notation, https://www.cimt.org.uk/sif/algebra/a10/interactive/s4.html
More Series, https://www.cimt.org.uk/sif/algebra/a10/interactive/s5.html

YouTube URL: https://youtu.be/BII5r-jUJwA

File Attachments: media/course-resources/Sequences-Series-Sigma_Essential-Informations.pdfmedia/course-resources/Sequences-Series-Sigma_Learning-Objectives.pdfmedia/course-resources/Sequences-Series-Sigma_Text.pdfmedia/course-resources/Sequences-Series-Sigma_Answers.pdfmedia/course-resources/Sequences-Series-PowerPoint-Presentation_copy_1.pptx

Written on 02 November 2020.

Factorising Polynomials

Sections:

0, Chapter 1, Introduction,
Introduction

Factorising Polynomials is an interesting algebraic technique. It is elegant!

After studying this unit, you should
- be able to solve cubic equations
- be able to identify the number of roots of cubic and quantic polynomials
- understand and be able to use the factor theorem
- be able to find remainders when dividing a polynomial by a linear factor.
You have five sections to work through namely.
1. The Cubic Equation
2. Cubic Equation: Number of Roots
3. Factor Theorem
4. Factors of Higher Order Polynomials
5. Remainders
0, Chapter 2, The Cubic Equation,
The Cubic Equation

Previously you would have factorised quadratic equations of the form:

\(ax^{2} + bx + c = 0\)

For example \(3x^{2} - 7x +4 = 0\)

We can solve by factorising: \( \left(3x - 4\right) \left(x - 1\right) = 0\)

Therefore: \(\left(3x - 4\right) = 0\) or \(\left(x-1\right) = 0\)

\(x= \frac{4}{3}\) or \(x = 1\)

Now we want to solve cubic equations of the form:

\(ax^{3} + bx^{2} + cx + d = 0\)

For example: \(x^{3} - 20x^{2} + 100x - 125 = 0\)

We want to be able to solve in a similar way, by factorising. Let us explore ways to factorise by looking at an example.
1, Chapter 2, Task 1, Worked Example 1,
Worked Example

Use the slider to explore worked examples.
- Worked Example
  
  Factorise \( \; x^{3}-20x^{2}+100x-125=0 \)
- Worked Example
  
  Find all the solutions for: \( \; x^{3}-3x^{2}-3x+35=0 \)
0, Chapter 3, Cubic Equation: Number of Roots,
Cubic Equation: Number of Roots

In the previous examples we have see cubic equations with 3 real roots. This is not always the case.

Cubic equations can in fact have 1, 2 or 3 roots.

The root is the x-value when the graph crosses the x-axis (i.e. when the value of the cubic is zero). Below we show each case graphically

Note also that when the coefficient of the term is negative, then the graph will look like:

We can find the three roots by considering a numerical or graphical approach.

For example, to solve the equation \(x^{3} - 100x^{2} + 2000x -1500 = 0\)

it is helpful to work out a few values of the function first.

We could call this function \(f\left(x\right)\)

\(f \left(x\right) = x^{3} - 100x^{2} + 2000x - 1500\)

HOT TIP - If f(x) changes sign between two points then the graph must cross the x axis between these points - try sketching a graph to convince yourself that this is true

A table of values is shown below

The arrows show where f(x) changes from positive to negative or vice versa.

Hence the zeros of f(x) must occur
- between 0 and 10
- between 20 and 30
- between 70 and 80.
Trial and improvement (or graphs in the correct regions) give the answers 0.78, 26.40 and 72.82 (working to 2 d.p.).
0, Chapter 4, Factor Theorem,
Factor Theorem

We have considered factorising quadratics such as:
\(x^{2} - 33x + 35 = 0\)

And we have have considered factorising cubics such as :
\(x^{3}-3x^{2} - 33x + 35 = 0\)

Now we also want to consider factorising higher order polynomials such as:
\(x^{4} - x^{3} - 3x^{2} - 33x + 35 = 0\)

For this, we use the Factor Theorem.

When solving cubic equations, the ‘spotting’ of a linear root led immediately to finding a factor.

For example, \(1\) gives a zero value for the cubic:

\(x^{3} - 4x^{2} + 5x - 2\)
\(1^{3} -4 \left(1\right)^{2} + 5 \left(1\right) -2 = 0\)

So \(\left(x-1\right)\) must be a factor.

Leading us to \(\left(x-1\right) x\) (quadratic) \(\rightarrow \left(x-1\right) x \left(x^{2} - 3x + 2\right)\)

We can then solve the quadratic to find any remaining roots.
\(\rightarrow \left(x-1\right) \times \left(x-1\right) \times \left(x-2\right)\)
Two roots: \(x=1\;x=2\)

We can now apply the same technique to polynomials of any order. This is the Factor Theorem.

Remember

Factor Theorem states that a polynomial \(f\left(x\right)\) has a factor \(\left(x-k\right)\) if and only if \(f\left(k\right)=0\)
1, Chapter 4, Task 1, Example 1,
Example

For example: Find a factor of the quartic:

\(x^{4} + 3x^{3} - 2x^{2} + 5x-7\)

Using trial and improvement we can ‘spot’ a root.

Try simple values first, for example: \(0, 1, -1, 2, -2…\)
\(1+3 \left(1\right)^{3} - 2 \left(1\right)^{2} + 5\left(1\right) - 7 = 0\)

So \(\left(x-1\right)\) must be a factor.

Leading us to: \(\left(x-1\right) \times \left(cubic\right) \rightarrow \left(x-1\right) \times \left(ax^{3} + bx^{2} + cx + d\right)\)
\(\rightarrow \left(x-1\right) \times \left(x^{3} + bx^{2} + cx + 7\right)\)
\(\rightarrow x^{4} + bx^{3} + cx^{2} + 7x\)
\(-x^{3} - bx^{2} - cx - 7\)
\(x^{4} + 3x^{3} - 2x^{2} + 5x -7\)

\(\rightarrow \left(x-1\right) \times \left(x^{3} + 4x^{2} + 2x + 7\right)\)
1, Chapter 4, Task 2, Example 2,
Example

Try: \(x^{4} - 5x^{3} - 2x^{2} + 25x - 3\) divided by \( \left(x-3\right)\)

First we see how many \(x\) 'go into' \(x^{4}\)
Answer: \(x^{3}\)

\(x^{3}-2x^{2} - 8x +1\)

Then we multiply \(x^{3}\) with \(\left(x-3\right)\) to give:
\(x^{4} - 5x^{3} - 2x^{2} + 25x -3\)

\(x^{4}-3x^{3}\)

Then we substract to see what remains
\(-2x^{3} - 2x^{2}\)

And then 'bring down' the next term \(-2x^{2}\)
\(-2x^{2}+6x^{2}\)

Then we substract to see what remains, and 'bring down' the next term \(+25x\)
\(-8x^{2} + 25x\)
\(-8x^{2}+24x\)

And then: how many \(x\) 'go into' \(-8x^{2}\)?
Answer: \(-8x\)

Then we multiply \(-8x\) with \(\left(x-3\right)\) too give:
\(x-3\)
\(x-3\)

And how many \(x\) 'go into' x?
The answer: \(1\)

Then we multiply \(1\) with \(\left(x-3\right)\) to givr
\(0\)

So \(\left(x^{3} - 2x^{2} - 8x + 1\right)\) is a factor zero, and \(x^{4} - 5x^{3} - 2x^{2} + 25x -3= \left(x-3\right) \left(x^{3}\right) \left(x^{3}-2x^{2} - 8x+1\right)\)
0, Chapter 5, Factors of Higher Order Polynomials,
Factors of Higher Order Polynomials

We have seen previously that once we have spotted a linear factor, we can try and work out the ‘missing’ expression by inspection.

For example:

\(x^{3} - 4x^{2} + 5x -2 \rightarrow \left(x-1\right) \times \left(x^{2} -3x +2\right)\)

We can then solve the quadratic to find any remaining roots.

\(\rightarrow \left(x-1\right) \times \left(x-1\right) \times \left(x-2\right) \)

Two roots \(x=1\) \(x=2\)

Another method is long division of polynomials.

Once we have a linear factor, we use this to divide throughout the original polynomial to find what is ‘left’.

Let us explain long divison of polynomials by showing an example:

\(x^{4} + 5x^{3} + 7x^{2} + 6x + 8 \rightarrow \left(x+2\right) \times \left(cubic\right)\)

So we divide \(x^{4}+5x^{3}+7x^{2}+6x+8\) by \(\left(x+2\right)\)

\(\frac{x^{4}+5x^{3}+7x^{2}+6x+8}{x+2}\)

This division is carried out in a very similar fashion to regular long division:

First, we see how many \(x\) ‘go into’ \(x^{4}\). Answer: \(x^{3}\)

\(x^{4}+5x^{3} + 7x^{2} + 6x + 8\)

Then we multiply \(x^{3}\) with \(\left(x+2\right)\) to give

\(x^{4}+2x^{3}\)

Then we subtract to see what remains

\(x^{3} + 3x^{2} + x + 4\)
\(x^{4} + 5x^{3} + 7x^{2} + 6x +8\)
\(x^{4} + 2x^{3}\)

Then we substract to see what remains
\(3x^{3}+7x^{2}\)

And 'bring down' the next term \(+7x^{2}\)
\(3x^{2}+6x^{2}\)

Then we subtract to see what remains

And then ‘bring down’ the next term: +6x
\(x^{2}+6x\)

\(x^{2} + 2x\)

Then we substract to see what remains

And then ‘bring down’ the final term: +8
\(4x + 8 \)
\(4x + 8 \)
-----------------
\(0\)

So (\(x^{3} + 3x^{2} + x + 4\)) is a factor
and \(x^{4} + 5x^{3} + 7x^{2} + 6x + 8 = \left(x + 2\right) \left(x^{3} + 3x^{2} + x + 4\right)\)
1, Chapter 5, Task 1, Example 3,
Example

Try: \(x^{4} - 5x^{3} - 2x^{2} + 25x - 3\) divided by \(\left(x-3\right)\)

First see how many \(x\) 'go into' \(x^{4}\)
Answer: \(x^{3}\)

\(x^{3} -2x^{2} - 8x + 1\)

Then we multiply \(x^{3}\) with \(\left(x-3\right)\) to give:
\(x^{4} - 5x^{3} - 2x^{2} + 25x-3\)

\(x^{4}-3x^{3}\)

Then we substract to see what remains
\(-2x^{3} - 2x^{2}\)

And then 'bring down' the next term: \(-2x^{2}\)
\(-2x^{3}+6x^{2}\)

Then we substract to see what remains and ‘bring down’ the next term: +25x
\(-8x^{2} + 25x\)

\(-8x^{2} + 24x\)

And then : How many \(x\) ‘go into’ \(-8x^{2}\)
Answer: \(-8x\)

Then we multiply \(-8x\) with \(\left(x-3\right)\) and we subtract to see what remains

And: How many \(x\) ‘go into’ \(x\)
Answer: \(1\)

Then we multiply \(1\) with \(\left(x-3\right)\) to give:
\(x-3\)

And then ‘bring down’ the final term: \(-3\)
\(x-3\)

Then we subtract to see what remains
\(0\)

So \(\left(x^{3} - 2x^{2} - 8x + 1\right)\) is a factor.
and \(x^{4} - 5x^{3} - 2x^{2} + 25x - 3 = \left(x-3\right)\left(x^{3} - 2x^{2} -8x+1\right)\)
0, Chapter 6, Remainders,
Remainders

In the previous section, we found a missing factor by dividing the polynomial by a linear factor.

We could be confident, therefore, that there would be no remainder

(by knowing that the linear factor would definitely ‘go into’ the original polynomial).

But what happens if we divide a polynomial by an expression that is not a factor?

For example:

\(\left(x+3\right)\) is not a factor of \(x^{3} + 6x^{2} + 7x -4\)

In scenarios such as this, there will be a remainder.

\(x^{3}+6x^{2}+7x-4\) divided by \(\left(x+3\right)\)

First, we see how many \(x\) ‘go into’ \(x^{3}\).
Answer: \(x^{2}\)

\(x^{2}+3x-2\)
\(x^{3}+6x^{2}+7x-4\)

Then we multiply \(x^{2}\) with \(\left(x+3\right)\) to give:

\(x^{3}+3x^{2}\)

Then we subtract to see what remains

\(3x^{2}+7x\)
\(3x^{2}+9x\)
And then 'bring down' the next term: \(+7x\)

Then we subtract to see what remains
\(-2x - 4\)
and then ‘bring down’ the next term: \(-4\)
\(-2x-6\)

And then:
How many x ‘go into’ \(-2x\) ?
Answer: \(-2\)
Then we multiply \(-2\) with \(\left(x+3\right)\) to give:

Then we subtract to see what remains
\(2\)

\(x^{3}+6x^{2}+7x-4\) divided by \(\left(x+3\right)\)

We can write this as:

\(\frac{x^{3}+6x^{2}+7x-4}{\left(x+3\right)} = x^{2} + 3x -2\), remainder \(2\)

or more commonly as: = \(x^{2} + 3x - 2 + \frac{2}{x+3}\)

or: \(x^{3} + 6x^{2} + 7x -4 = \left(x^{2} + 3x - 2\right) \left(x+3\right) + 2\)
1, Chapter 6, Task 1, Example 4,
Example

Try: \(x^{4} - 5x^{3} - 2x^{2} + 25x - 8\) divided by \(\left(x-2\right)\)

First, we see how many \(x\) 'go into' \(x^{4}\).
Answer: \(x^{3}\)

\(x^{3}-3x^{2}-8x+8\)

Then we multiply \(x^{3}\) with \(\left(x-2\right)\) to give:
\(x^{4} - 5x^{3} -2x^{2} + 25x -8\)
\(x^{4}-2x^{3}\)

Then we subtract to see what remains
\(-3x^{3}-2x^{2}\)

And then ‘bring down’ the next term: \(-2x^{2}\)
\(-3x^{3} + 6x^{2}\)

And then: How many x ‘go into’ \(-8x^{2}\)
Answer: \(-8x\)

Then we multiply \(-8x\) with \(\left(x-2\right)\) to give:

\(-8x^{2}+25x\)

Then we substract to see what remainsand 'bring down' the next term \(+25x\)
\(-8x^{2}+16x\)

Then we subtract to see what remains
And then ‘bring down’ the final term: \(-8\)

And: How many \(x\) ‘go into’ \(9x\) ?
Answer: \(9\)

Then we multiply \(9\) with
\(\left(x-2\right)\) to give:
\(9x - 8\)

\(9x - 18\)

Then we substract to see what remains
\(10\)

So there is a remainder of \(10\).
1, Chapter 6, Task 2, Another example,
Another Example

\(x^{4} - 5x^{3} - 2x^{2} + 25x -8\) divided by \(\left(x-2\right)\)

We can write this as:
\(\frac{x^{4}-5x^{3} -2x^{2} + 25x -8}{\left(x-2\right) = x^{3} -3x^{2} -8x+9}\), remainder 10

or more commonly as: \( = x^{3} - 3x^{2} - 8x + 9 + \frac{10}{x-2}\)

or: \(x^{4}-5x^{3} - 2x^{2} + 25x -8=\left(x^{3} -3x^{2} - 8x + 9\right) \left(x-2\right) + 10\)
0, Chapter 7, Summary,
Summary

Polynomial function of degree n

is of the form

\(a_{n}x^{n} + a_{n-1}x^{n-1}+...+a_{2}x^{2}+a_{1}x+a_{0}\)

whre \(a_{0}, a_{1}... a_{n}\) are real numbers, and \(a_{n} \neq 0\)

Roots of a cubic

A cubic polynomial can have 1, 2 or 3 roots as shown in the diagrams below

Factor Theorem

If \(P\left(x\right)\) is a polynomial of degree n and has zero \(x-a\), that is \(P\left(a\right) = 0\), then \(\left(x-a\right)\) is a factor of \(P\left(x\right)\) and \(P\left(x\right) \times Q \left(x\right)\) factor
where \( Q\left(x\right)\) is a polynomial of degree \( \left(n-1\right)\)

Remainder Theorem

If \(P \left(x\right)\) is a polynomial of degree \(n\) and \(a\) is any number, then \(P\left(x\right) = \left(x-a\right) \times Q \left(x\right) + R\)

where \(Q \left(x\right)\) is a polynomial of degree \(\left(n-1\right)\) and \(R=P\left(a\right)\)

Interactive Exercises:

Interactive Exercises - Factorising Polynomials, https://www.cimt.org.uk/sif/algebra/a3/interactive.htm
The Cubic Equation, https://www.cimt.org.uk/sif/algebra/a3/interactive/s1.html
Cubic Equation: Number of Roots, https://www.cimt.org.uk/sif/algebra/a3/interactive/s2.html
Factor Theorem, https://www.cimt.org.uk/sif/algebra/a3/interactive/s3.html
Factors of Higher Order Polynomials, https://www.cimt.org.uk/sif/algebra/a3/interactive/s4.html
Remainders, https://www.cimt.org.uk/sif/algebra/a3/interactive/s5.html

YouTube URL: https://www.youtube.com/watch?v=fk6u8ObheLA&list=PL8ce1n6mRlCtS0rO_C7CNwtjhyqDVkr8g&index=24

File Attachments: media/course-resources/Factorising-Polynomials_PowerPoint-Presentation.pptxmedia/course-resources/Factorising-Polynomials_Learning-Objectives.pdfmedia/course-resources/Factorising-Polynomials_Text.pdfmedia/course-resources/Factorising-Polynomials_Essential-Information.pdfmedia/course-resources/Factorising-Polynomials_Answers.pdf

Written on 03 November 2020.

Binomial Theorem

Sections:

0, Chapter 1, Introduction,
Introduction

The Binomial Theorem is a significant and useful tool. It helps us to efficiently expand binomial expressions which have been raised to a power, for example \(\left(2x + 5\right)^{10}\).

A binomial is a polynomial (an expression) with just 2 terms.
In the example above, they are: \(2x\) and \(5\)

You have four sections to work through and there are check up audits and fitness tests for each section.
1. Binomial Expansion
2. Binomial Coefficients
3. Factorials
4. Binomial Theorem
In this unit we generalise algebraic expressions of the form \(\left(a + b\right)^{n}\). After completing the unit you should be able to
- understand how to calculate expressions of the form \(\left(a+b\right)^{n}\) for positive integers, \(n\)
- use binomial cofficients, including factorials, in the binomial expansion
- apply factorials to determine the number of arrangements of letters
- understand and use Binomial Theorem for \(\left(a+x\right)^{n}\)
0, Chapter 2, Binomial Expansion,
Binomial Expansion

‘Binomial’ is a word meaning ‘two terms’;

‘Polynomial’ is a word meaning ‘many terms’.

Binomials include expressions such as:

\(x+1\) \(2a+b\) \(x+y\) \(12x+5\) \(a^{2} + b^{2}\)

Binomial expansions include examples such as:

\(\left(x+1\right)^{4}\) \(\left(2a+b\right)^{10}\) \(\left(x+y\right)^{8}\) \(\left(13x+5\right)^{5}\) \( \left(a^{2} + b^{2} \right)^{2}\)

0, Chapter 3, Binomial Coefficients,

Binomial Coefficients

Now let’s think about expanding: \(\left(1+x\right)^{8}\)

Try it!

This is very tedious indeed!!!

As the power (or index, or exponent) becomes greater, so too does the effort required to expand.

Imagine expanding \(\left(1+x\right)^{20}\) or \(\left(1+x\right)^{100}\)

Fortunately, there is a more efficient method we can employ.

First - let’s look at the pattern formed when we consider these expansions:

\(\left(1+x\right)^{1}, \left(1+x\right)^{2}, \left(1+x\right)^{3}, \left(1+x\right)^{4}, \left(1+x\right)^{5},\left(1+x\right)^{6},\left(1+x\right)^{7}\)

The table below shows the binomial coefficients (the number bit) after the expansion of \(\left(1+x\right)\), for example:

\(\left(1+x\right) \rightarrow 1 + 1x\)

\(\left(1+x\right)^{2} \rightarrow 1 + 2x + 1x^{2}\)

\(\left(1+x\right)^{3} \rightarrow 1 + 3x + 3x^{2} + 1x^{2}\)


	\(1\)	\(x\)	\(x^{2}\)	\(x^{3}\)
\( (1+x)^{1} \)	1	1
\( (1+x)^{2} \)	1	2	1
\( (1+x)^{3} \)	1	3	3	1
\( (1+x)^{4} \)
\( (1+x)^{5} \)
\( (1+x)^{6} \)
\( (1+x)^{7} \)

\(\left(1+x\right)^{4} \rightarrow 1 + 4x + 6x^{2} + 4x^{3} + 1x^{4}\)

\(\left(1+x\right)^{5} \rightarrow 1 + 5x + 10x^{2} + 10x^{3} + 5x^{4} + 1x^{5}\)

\(\left(1+x\right)^{6} \rightarrow 1 + 6x + 15x^{2} +20x^{3} + 15x^{4} + 6x^{5} + 1x^{6}\)

\(\left(1+x\right)^{7} \rightarrow 1 + 7x + 21x^{2} + 35x^{3} + 35x^{4}+21x^{5} + 7x^{6} + 1x^{7}\)


	\(1\)	\(x\)	\(x^{2}\)	\(x^{3}\)	\(x^{4}\)	\(x^{5}\)	\(x^{6}\)	\(x^{7}\)
\( (1+x)^{1} \)	1	1
\( (1+x)^{2} \)	1	2	1
\( (1+x)^{3} \)	1	3	3	1
\( (1+x)^{4} \)	1	4	6	4	1
\( (1+x)^{5} \)	1	5	10	10	5	1
\( (1+x)^{6} \)	1	6	15	20	15	6	1
\( (1+x)^{7} \)	1	7	21	35	35	21	7	1

1, Chapter 3.1, Binomial Coefficients: Pascal’s Triangle,
Binomial Coefficients: Pascal’s Triangle

The binomial coefficients follow a pattern.

Did you spot it?

Each row begins and ends with a ‘1’.

Subsequent terms can be found by summing two consecutive
terms from the row above.

The coefficients actually make Pascal’s Triangle:

We can now easily expand \(\left(1+x\right)^{8}\).

We can predict the pattern of coefficients for row 8, by using row 7 from Pascal's Triangle:

We can now slot these coefficients into the pattern of the rising powers of \(x\)

\(1x^{0}+8x^{1} + 28x^{2} + 56x^{3} + 70x^{4} + 56x^{5} + 28x^{6} + 8x^{7} + 1x^{8}\)

We could add in \(x^{0}\) if we wish, but remember \(x^{0} = 1\) And remember \(x=x^{1}\)

But imagine expanding \(\left(1+x\right)^{20}\)

We can only tackle this one if we know the coefficients of \(\left(1+x\right)^{19}\)

And what about \(\left(1+x\right)^{100}\)

Do we need to write out 100 rows of Pascal’s Triangle?!!!?

We will address these questions in due course.

1, Chapter 3.2, Another Pattern,

Another Pattern

Did you notice the sums of the rows of binomial coefficients?

The sums are:

\( (1+x)^{1} \)	2	\(2^{1}\)
\( (1+x)^{2} \)	4	\(2^{2}\)
\( (1+x)^{3} \)	8	\(2^{3}\)
\( (1+x)^{4} \)	16	\(2^{4}\)
\( (1+x)^{5} \)	32	\(2^{5}\)
\( (1+x)^{6} \)	64	\(2^{6}\)
\( (1+x)^{7} \)	128	\(2^{7}\)

What would be the sum of the coefficients for \(\left(1+x\right)^{20}\)?

\(2^{20}=1048576\)

1, Chapter 3.2, Task 1, Examples 1,
Examples

Following these patterns, we can expand other binomial expressions such as:

\(\left(a+x\right)^{6}\)

The expansion row 6 of Pascal's Triangle

Row 6 1 6 15 20 15 6 1

and sum of the coefficients is: \(2^{6}=64\)

The expansion is:

\(1a^{6}x^{0}+6a^{5}x^{1}+15a^{4}x^{2}+20a^{3}x^{3} + 15a^{2}x^{4}+6a^{1}x^{5}+1a^{0}x^{6}\)

The powers of \(x\) ascend in the same pattern as previously.

The powers of \(a\) descend in the symmetrically opposite direction.

For each term the two powers sum to \(6\).
1, Chapter 3.2, Task 2, Worked Example 1,
Worked Example

Use the slider to explore worked examples.
- Worked Example
  
  Expand: \( (2+y)^{5} \)
- Worked Example
  
  Expand: \( (2a+b)^{4} \)
- Worked Example
  
  Expand: \( (2+k)^{15} \)
0, Chapter 4, Factorials,
Factorials

First we need to be familiar with factorial notation

Remember

\(n! = n\times \left(n-1\right) \times \left(n-2\right) \times \left(n-3\right) \times ... \times 3 \times 2 \times 1\)

We call this 'n factorial'

What does this mean?

Quite simply

\(9!=9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1\)

and \(15!=15\times 14\times 13\times 12\times 11\times 10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2\times 1\times \)

and \(105!=105\times 104\times 103\times 102\times 101\times 100\times 99\times 98\times ... \times 3\times 2\times 1\)

It is that straightforward!

Note: \(0!=1\)
1, Chapter 4, Task 1, Worked Example 2,
Worked Example

Use the slider to explore worked examples.
- Worked Example
  
  In the word SIMPLE all the letters are different. Including the one given, how many possible arrangements are there of these 6 letters?
- Worked Example
  
  The word EXCEL has one letter which repeats: the ‘E’ repeats twice. How many different arrangements are there of these 5 letters?
- Worked Example
  
  The word \( EXCEED \) has a letter which repeats 3 times. How many different arrangements are there of these 6 letters?
- Worked Example
  
  The word \( EXCELLENT \) has two different letters which repeat.
  
  The \(E\) repeats 3 times, the \(L\) twice.
  
  How many arrangements are there of these 9 letters?
1, Chapter 4.1, Arrangements: Notation,
Arrangements: Notation

This is going to help us find binomial coefficients more efficiently.

We will now consider cases with only 2 different letters or ‘options’ (this is the bi-nomial ‘bit’).

If we have 5 letters AABBB

We know there will be different arrangements of the letters.

\(\frac{5!^{5}C_{2}}{2!\times3}\)

Where
- \(5\) is the total number of letters
- \(2!\) is the number of times the letter 'A' was repeated
- \(3!\) is the number of times the letter 'B' was repeated
We can also write \(\left(^{5}_{2}\right)\)

And as \(^{5}C_{2} \frac{5!}{3!}\)

We can read this as '5 choose 2'.

Note also the symmetry \(\left(^{5}_{3}\right) = \frac{5!}{3!\times2!}=\frac{5!}{2!\times3!}= \left(^{5}_{2}\right)\)
1, Chapter 4.2, Notation for Arrangements,
Notation for Arrangements

Remember

In general we use the following notation to denote the number of arrangements when ‘choosing’ \(r\) objects (letters,players,options etc) from a total number of \(n\) objects

\(\left(^{n}_{r}\right)\) or \(^{n}C_{r}\)

And the formula is

\(^{n}C_{r}=\left(^{n}_{r}\right)=\frac{n!}{r! \times \left(n-r\right)!}\)

Where:

\(n!\) is the total number of letters or ‘options’

\(r!\) is the number of times the ‘option’ we are choosing is required/repeated

\(\left(n-r\right)!\) the number of times for the remaining ‘option’, i.e. the difference between \(n\) and \(r\)
1, Chapter 4.2, Task 1, Worked Example 3,
Worked Example

Use the slider to explore worked examples.
- Worked Example
  
  How many different ways are there of selecting a swimming relay-team of 4 swimmers from a squad of 7?
1, Chapter 4.2, Task 2, Calculations 1,
Calculations

We can also use the notation:

\(\left(^{7}_{4}\right)\)

Where:

\(7\) is the total number of swimmers

\(4\) is the total number of swimmers who can be in the team

\( \left(^{7}_{4}\right) = \frac{7!}{4! \times \left(7-4\right)} = \frac{7!}{4! \times 3!} \)

\(4! \times 3!\) is the number of number of swimmers who can not be in the team, i.e. the difference between \(7\) and \(4\)

\(=\frac{7\times 6\times 5\times 4\times 3\times 2\times 1\times }{\left(4 \times 3 \times 2 \times 1\right)} \left(3 \times 2 \times 1 \right)} = 35\)

Note how we can simply cancel our the \(4\)!

There are \(35\) different possible 'realy-teams'.
1, Chapter 4.3, Important Results,
Important Results

We have seen that:

\(\left(^{n!}_{r}\right) = \frac {n!}{r! x\left(n-r\right)}\)

Here are some interesting results:

\(\left(^{n!}_{0}\right) = \frac{n!}{0! x \left(n-0\right)!} = \frac{n!}{n!}=1\)

\(\left(^{n!}_{1}\right) = \frac{n!}{1! x \left(n-1\right)!} = \frac{n \times \left(n-1\right) \times \left(n-2\right) \times \left(n-3\right) ... 3 \times 2 \times 1}{\left(n -1\right) \times \left(n -2\right) \times \left(n -3\right)... 3 \times 2\times 1}=n\)

\(\left(^{n!}_{n}\right) = \frac{n!}{n! x \left(n-n\right)!} = \frac{n!}{n!}=1\)

We have also seen the symmetry of this notation, for example:

\(\left(^{8}_{3}\right) = \frac{8!}{3!\times5!} = \frac{8!}{5! \times 3!} = \left(^{8}_{5}\right)\)
1, Chapter 4.4, Summary of Results,
Summary of Results

In summary

\(\left(^{n}_{r}\right) = \frac{n!}{r!\times \left(n-r\right) !}\)

\(\left(^{n}_{0}\right) = 1\)

\(\left(^{n}_{1}\right) = n\)

\(\left(^{n}_{n}\right) = 1\)

\(\frac{n!}{r!\times \left(n-r\right) !} = \frac{n!}{\left(n-r\right)! \times r!}\)
0, Chapter 5, Binomial Theorem,
Binomial Theorem

We have previously expanded \(\left(1+x\right)^{8}\)

by using row 8 from Pascal's Triangle:

\(1\;\;8\;\;28\;\;56\;\;70\;\;56\;\;28\;\;8\;\;1\)

We slotted these coefficients into the pattern of the rising powers of \(x\)

\(1 + 8x + 28x^{2} + 56x^{3} + 70x^{4} + 56x^{5} + 28x^{6} + 8x^{7} + 1x^{8}\)

But expanding something like \(\left( 1 + x\right) ^{20}\) would require many rows of Pascal’s Triangle. And what about \(\left(1+x\right)^{100}\) ?!!!??
Fortunately, we can use our work on factorials to help.

Lets work out a few and see what happens.
Calculate: \(\left(^{8}_{0}\right) \), \(\left(^{8}_{1}\right) \), \(\left(^{8}_{2}\right) \), \(\left(^{8}_{3}\right) \), \(\left(^{8}_{4}\right) \), \(\left(^{8}_{5}\right) \), \(\left(^{8}_{6}\right) \), \(\left(^{8}_{7}\right) \), \(\left(^{8}_{8}\right) \),

The general result of expanding in this way is called the binomial theorem.

Here, some terms of the general binomial theorem are illustrated:

In summary

\(\left(a+x\right) ^{n}=\left(^{n}_{0}\right) a^{n} + \left(^{n}_{1}\right) a^{n-2}x^{2}+...+\left(^{n}_{r}\right)a^{n-r} x^{r}+...+\left(^n_{n-1}\right) a^{1}x^{n-1}+\left(^{n}_{n}\right) x^{n}\)

The general term is:

\(\left(^{n}_{r}\right)a^{n-r}x^{r}\)
1, Chapter 5, Task 1, Example 2,
Example

We can now revist Example from Section 2:

Expand: \(\left(2+k\right)^{15}\)

\(\left(^{15}_{0}\right) 2^{15} + \left(^{15}_{1}\right) 2^{14} k + \left(^{15}_{2}\right) 2^{13} k^{2} + \left(^{15}_{3}\right) 2^{12} k^{3} + \left(^{15}_{4}\right) 2^{11} k^{4} + \left(^{15}_{5}\right) 2^{10} k^{5} + \left(^{15}_{6}\right) 2^{9} k^{6} + \left(^{15}_{7}\right) 2^{8} k^{7} + \left(^{15}_{8}\right) 2^{7} k^{8} + \left(^{15}_{9}\right) 2^{6} k^{9} + \left(^{15}_{10}\right) 2^{5} k^{10} + \left(^{15}_{11}\right) 2^{4} k^{11} + \left(^{15}_{12}\right) 2^{3} k^{12} + \left(^{15}_{13}\right) 2^{2} k^{13} + \left(^{15}_{14}\right) 2^{1} k^{14} + \left(^{15}_{15}\right) k^{15} + \)

We may ask such questions as ‘What is the coefficient of \(k^{11}\)

Answer: \(\left(^{15}_{11}\right)2^{4}=\frac{15!}{11! \times 4!} \times 2^{4} = \frac {15 \times 14 \times 13 \times 12} { 4 \times 3 \times 2 \times 1} \times 16 = 21840\)
1, Chapter 5.1, Typical Use,
Typical use

One typical use of the binomial theorem is to give approximations to numbers like \(\left(1.02\right)^{7}\)

We can write \(1.02\) as \(1+0.02\)

\(\left(1+0.02\right)^{7}\)

\( =1+\left(^{7}_{1}\right) 0.02 + \left(^{7}_{2}\right) 0.02^{2} + \left(^{7}_{3}\right) 0.02^{3} + \left(^{7}_{7}\right) 0.02^{4} + \left(^{7}_{5}\right) 0.02^{5} + \left(^{7}_{6}\right) 0.02^{6} + \left(^{7}_{7}\right) 0.02^{7} + \)
\( = 1 \times 0.02 + 21 \times 0.02^{2} + 35 \times 0.02^{3} + 35 \times 0.02^{4} + 21 \times 0.02^{5} + 7 \times 0.02^{6} + 1 \times 0.02^{7} \)
\( = 1 + 0.14 + 0.0084 + 0.0000056 + ... \)
\( = 1.148686 ... \)

We may not need to calculate all the ‘bits’- depending on how many significant figures we need

\(\neq 1.148686\) (to 7 significant figures)
1, Chapter 5.1, Task 1, Worked Example 4,
Worked Example
- Worked Example
  
  Expand in full: \((3m + 2n)^{6}\)
1, Chapter 5.1, Task 2, Calculation 2,
Calculation

We can now use this notation for binomial expansions instead of Pascal’s Triangle.

We can now expand \(\left(1+x\right)^{20}\) as:

\( \left(^{20}_{0}\right) + \left(^{20}_{1}\right)x +  \left(^{20}_{2}\right)x^{2} +  \left(^{20}_{3}\right)x^{3}
+ \left(^{20}_{4}\right)x^{4}
+ \left(^{20}_{5}\right)x^{5}
+ \left(^{20}_{6}\right)x^{6}
+ \left(^{20}_{7}\right)x^{7}
+ \left(^{20}_{8}\right)x^{8}
+ \left(^{20}_{9}\right)x^{9}
+ \left(^{20}_{10}\right)x^{10}
+ \left(^{20}_{11}\right)x^{11}
+ \left(^{20}_{12}\right)x^{12}
+ \left(^{20}_{13}\right)x^{13}
+ \left(^{20}_{14}\right)x^{14}
+ \left(^{20}_{15}\right)x^{15}
+ \left(^{20}_{16}\right)x^{16}
+ \left(^{20}_{17}\right)x^{17}
+ \left(^{20}_{18}\right)x^{18}
+ \left(^{20}_{19}\right)x^{19}
+ \left(^{20}_{20}\right)x^{20} \)

And we may ask such questions as ‘What is the coefficient of \(x^{16}\)?

Answer: \(\left(^{20}_{16}\right) = \frac {20!}{16! \times 4!} = \frac {20 \times 19 \times 18 \times 17}{4 \times 3 \times 2 \times 1} = 4845\)
1, Chapter 5.1, Task 3, Example 3,
Example

Similarly we can now expand \(\left(a+x\right)^{12}\) as:

\( \left(^{12}_{0}\right)a^{12} + \left(^{12}_{1}\right)a^{11}x
+ \left(^{12}_{2}\right)a^{10}x^{2}
+ \left(^{12}_{3}\right)a^{9}x^{3}
+ \left(^{12}_{4}\right)a^{8}x^{4}
+ \left(^{12}_{5}\right)a^{7}x^{5}
+ \left(^{12}_{6}\right)a^{6}x^{6}
+ \left(^{12}_{7}\right)a^{5}x^{7}
+ \left(^{12}_{8}\right)a^{4}x^{8}
+ \left(^{12}_{9}\right)a^{3}x^{9}
+ \left(^{12}_{10}\right)a^{2}x^{10}
+ \left(^{12}_{11}\right)a^{1}x^{11}
+ \left(^{12}_{12}\right)x^{12} \)

We expand \(\left(3+x\right)^{12}\) as:

\(\left(^{12}_{0}\right)3^{12} + \left(^{12}_{1}\right)3^{11}x
+ \left(^{12}_{2}\right)3^{10}x^{2}
+ \left(^{12}_{3}\right)3^{9}x^{3}
+ \left(^{12}_{4}\right)3^{8}x^{4}
+ \left(^{12}_{5}\right)3^{7}x^{5}
+ \left(^{12}_{6}\right)3^{6}x^{6}
+ \left(^{12}_{7}\right)3^{5}x^{7}
+ \left(^{12}_{8}\right)3^{4}x^{8}
+ \left(^{12}_{9}\right)3^{3}x^{9}
+ \left(^{12}_{10}\right)3^{2}x^{10}
+ \left(^{12}_{11}\right)3^{1}x^{11}
+ \left(^{12}_{12}\right)x^{12}\)

We may ask such questions as ‘What is the coefficient of \(x^{7} ?\)

Answer: \( \left(^{12}_{7}\right)3^{5} = \frac {12!}{7! \times 5!} \times 3^{5} = \frac {12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} \times 243 = 192456\)
0, Chapter 6, Summary,
Summary

\(^{n}C_{r}\) is the number of different combinations of \(n\) objects, \(r\) of one type, \(\left(n-r\right)\) another.

\(n!=n \left(n-1\right) \left(n-2\right) ... 1\)

\(^{n}C_{r}=\frac{n!}{r!\left(n-r\right)!}=\left(^{n}_{r}\right)\)

The number of arrangements of an \(n\)-letter word, with ONE letter repeated \(p\) times is \(\left( \frac {n!}{p!} \right)\)

The Binomial Theorem is given by

\( (a+x)^{n} = a^{n} + \left(^{n}_{1}\right) a^{n-1} x + \left(^{n}_{2}\right)a^{n-2}x^{2}+...+\left(^{n}_{r}\right) a^{a-r}x^{r}+...x^{n}\)

where \(a\) and \(x\) are numbers and \(n\) is a positive integer.

For example,

\(\left(1+x\right)^{4} = 1 + 4x + 6x^{2} + 4x^{3} + x^{4}\)

So the binomial coefficients are \(1, 4, 6, 4\) and \(1\)

Note that \(1!=1\) and \(0!=1\)

Interactive Exercises:

Binomial Expansions, https://www.cimt.org.uk/sif/algebra/a2/interactive/s1.html
Binomial Coefficients, https://www.cimt.org.uk/sif/algebra/a2/interactive/s2.html
Factorials, https://www.cimt.org.uk/sif/algebra/a2/interactive/s3.html
Binomial Theorem, https://www.cimt.org.uk/sif/algebra/a2/interactive/s4.html

YouTube URL: https://youtu.be/7plaAZ1y6jw

File Attachments: media/course-resources/Binomial-Theorem_Answers.pdfmedia/course-resources/Binomial-Theorem_Essential-Information.pdfmedia/course-resources/Binomial-Theorem_Learning-Objectives.pdfmedia/course-resources/Binomial-Theorem_PowerPoint-Presentation.pptxmedia/course-resources/Binomial-Theorem_Text.pdf

\(1^{st}\) square	\(1\)	\(2^{0}\)
\(2^{nd}\) square	\(2\)	\(2^{1}\)
\(3^{rd}\) square	\(4\)	\(2^{2}\)
\(4^{th}\) square	\(8\)	\(2^{3}\)
\(5^{th}\) square	\(16\)	\(2^{4}\)
...	...	...
\(10^{th}\) square	\(512\)	\(2^{9}\)
\(20^{th}\) square	\(524 288\)	\(2^{19}\)
\(64^{th}\) square	A HUGE NUMBER!!! \(\approx 9.22 \times 10^{18}\)	\(2^{63}\)
\(n^{th}\) square		\(2^{n-1}\)

Negative Numbers

Introduction

Negative Numbers

Remember

Examples of Negative Numbers

Inequalities Signs

Exercises

Adding and subtracting positive and negative numbers

Remember

Negative Numbers: Adding

Negative Numbers: Substracting

Exercises

Negative Numbers: Multiplication & Division

Remember

Exercises

Summary

Addition and subtraction

Multiplication and division

Sine and Cosine Rules

Introduction

Sine Rule

Remember

Worked Example

Worked Example

Worked Example

Worked Example

Cosine Rule

Remember

Worked Example

Worked Example

Worked Example

Worked Example

Worked Example

Sin and Cos

sin\(\theta\) and cos\(\theta\)

Bearing

Remember

Hot Tip

Worked Example

Worked Example

Worked Example

Exercise

Application: Area of Any Triangle

Remember

Area of a triangle when the lengths of two sides and the included angle are known,

Worked Example

Worked Example

Worked Example

Exercise

Heron's Formula

Remember

Formula for the area of a triangle when the lengths of all three sides are known.

Worked Example

Worked Example

Exercise

Angles Larger than \(90^{\circ}\)

Remember

Graphs of Sin\(\theta\) and Cos\(\theta\)

Graphs of Tan\(\theta\)

Trigonometric Equations

Worked Example

Worked Example

Worked Example

Worked Example

Summary

Pythagoras' Theorem

Introduction

Pythagoras Theorem

Remember

Finding the Hypotenuse

Worked Examples

Worked Example

Worked Example

Worked Example

Worked Example

Worked Example

Exercises

Further Work With Pythagoras’ Theorem

Worked Example

Worked Example