A.M. ONLINE

A school licence can be purchased both by schools either with or without an Applied Maths teacher.

For schools with such a teacher, a licence entitles any Applied Maths teacher of that school to a free teacher account. This account contains a number of unique features that greatly reduces the day-to-day teacher workload.

One such feature is the ability to show and hide different content to students at the press of a button!

For example, as described below, every exercise question that we provide includes full, detailed solutions which, by default, are initially not visible on student accounts associated with a school licence.

However, with the flick of a switch, the teacher can instead make these full solutions visible and accessible to all of their students!

This could then allow students to self-assess their work, thereby removing the need for teachers to spend time correcting schoolwork or homework.

### Key Point

Vectors have both a magnitude and a direction.

### Cause of Confusion

Magnitudes are always positive. Coefficients, on the other hand, can be either positive or negative.

### Exam Tip

Even if you are not asked to draw a diagram, it is always a good idea to draw one in your actual exam whenever possible so that you can more easily visualise the question.

## INTERACTIVE LESSONS

Our online course is composed of professional, interactive lessons that makes the subject of Applied Maths an enjoyable experience both for teachers to teach and for students to learn.

As we are not limited by what can fit into a schoolbook, our lessons describe every concept within Applied Maths with as much detail as possible.

Throughout each lesson, we also highlight three important aspects:

• Key Points – the most important parts of the lesson.
• Causes of Confusion – common mistakes that students make.
• Exam Tips – tips to give students that extra edge on the exam!

## TEACHER CHAT

Having that extra bit of guidance is always a plus for students!

As such, we provide all of our students with the ability to contact an Applied Maths teacher, Mr. Kenny, throughout the day with their questions using what we refer to as Teacher Chat (which you may have already spotted in the bottom right corner).

This can be thought of as a student “raising their hand” any time they need help, even though they’re not in a classroom!

This is of particular benefit to students at schools that do not have an Applied Maths teacher and who therefore need as much guidance as possible with the subject.

Of course, this is also of great benefit to students with Applied Maths teachers as they no longer have to wait until the next school day to have their questions answered.

And don’t worry, we don’t do a student’s homework for them! We only nudge them in the right direction in the same manner that is done by any teacher, parent or tutor.

## Practice Question

A car uniformly accelerates from rest to a speed of $$20\mbox{ m/s}$$ in $$8$$ seconds.

Immediately afterwards, it uniformly decelerates to rest, covering a distance of $$150\mbox{ m}$$.

(a) What was the acceleration rate in the first stage?

(b) What was the deceleration rate in the second stage?

(c) At what times was the car’s speed equal to $$10\mbox{ m/s}$$?

(a) $$2.5\mbox{ m/s}^2$$

(b) $$\displaystyle\frac{4}{3}\mbox{ m/s}^2$$

(c) $$4\mbox{ s}$$ and $$15.5\mbox{ s}$$

(a)

##### First Stage

\begin{align} u = 0 && v=20 && t = 8 && a=\mathord{?} \end{align}

\begin{align}\downarrow\end{align}

$v=u+at$

\begin{align}\downarrow\end{align}

\begin{align}20=0+a(8)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}a&=\frac{20}{8}\\&=2.5 \mbox{ m/s}^2\end{align}

(b)

##### Second Stage

\begin{align} u = 20 && v=0 && s = 150 &&a=\mathord{?} \end{align}

\begin{align}\downarrow\end{align}

$v^2=u^2+2as$

\begin{align}\downarrow\end{align}

\begin{align}0^2=20^2+2a(150)\end{align}

\begin{align}\downarrow\end{align}

\begin{align}a&=-\frac{20^2}{(2)(150)}\\&=-\frac{4}{3} \mbox{ m/s}^2\end{align}

The deceleration rate is therefore $$\dfrac{4}{3}\mbox{ m/s}^2$$.

(c)

##### First Stage

\begin{align} u = 0 && v=10 && a = 2.5 && t=\mathord{?} \end{align}

\begin{align}\downarrow\end{align}

$v=u+at$

\begin{align}\downarrow\end{align}

\begin{align}10=0+(2.5)t\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t&=\frac{10}{2.5}\\&=4 \mbox{ s}\end{align}

$\,$

##### Second Stage

\begin{align} u = 20 && v=10 && a = -\frac{4}{3} && t=\mathord{?} \end{align}

\begin{align}\downarrow\end{align}

$v=u+at$

\begin{align}\downarrow\end{align}

\begin{align}10=20+\left(-\frac{4}{3}\right)t\end{align}

\begin{align}\downarrow\end{align}

\begin{align}t&=\frac{10-20}{-\frac{4}{3}}\\&=7.5 \mbox{ s}\end{align}

The time since the beginning of the trip is therefore $$7.5+8=15.5$$ seconds.

$\,$

Hence, the two times at which the car’s speed is $$10\mbox{ m/s}$$ are $$4$$ seconds and $$15.5$$ seconds into the journey.

Although we are not asked to draw a time-velocity graph in this question, it is still a good idea to draw a rough graph for our own benefit.

In the first stage, the car accelerates from rest to a speed of $$20\mbox{ m/s}$$ in a period of $$8$$ seconds.

In the second stage, the car decelerates to rest, covering a distance of $$150\mbox{ m}$$.

(a) To find the acceleration rate of the first stage, we can use one of our SUVAT equations.

Step 1: State the quantities that are given in the question and the quantity that we are looking for. Be careful with signs!

We are given the following

\begin{align} u = 0 && v=20 && t = 8 \end{align}

and we are looking for $$a$$.

$\,$

Step 2: Choose the most suitable equation either related to the time-velocity graph (if possible) or from the SUVAT equations based on Step 1.

Here, we can use either the slope formula or a SUVAT equation. We shall randomly choose the SUVAT equation approach.

As we are not told $$s$$ nor do we want to find $$s$$, we should choose the SUVAT equation which does not have $$s$$ in it, namely

$v=u+at$

$\,$

Step 3: Insert the quantities into the equation.

Subbing in our known values, we obtain

\begin{align}20=0+a(8)\end{align}

Rearranging for $$a$$, this gives

\begin{align}a&=\frac{20}{8}\\&=2.5 \mbox{ m/s}^2\end{align}

(b) To find the deceleration rate of the second stage, we can again use one of our SUVAT equations.

Step 1: State the quantities that are given in the question and the quantity that we are looking for. Be careful with signs!

We are given the following

\begin{align} u = 20 && v=0 && s = 150 \end{align}

and we are looking for $$a$$.

$\,$

Step 2: Choose the most suitable equation either related to the time-velocity graph (if possible) or from the SUVAT equations based on Step 1.

As we are interested in both $$s$$ and $$a$$, we cannot use the graph and must instead use a SUVAT equation.

As we are not told $$t$$ nor do we want to find $$t$$, we should choose the SUVAT equation which does not have $$t$$ in it, namely

$v^2=u^2+2as$

$\,$

Step 3: Insert the quantities into the equation.

Subbing in our known values, we obtain

\begin{align}0^2=20^2+2a(150)\end{align}

Rearranging for $$a$$, this gives

\begin{align}a&=-\frac{20^2}{(2)(150)}\\&=-\frac{4}{3} \mbox{ m/s}^2\end{align}

Therefore, the deceleration rate of the second stage is $$\displaystyle\frac{4}{3}\mbox{ m/s}^2$$.

(c) The maximum speed of the car is $$20\mbox{ m/s}$$. Therefore, there will be two separate times that the car’s speed will be $$10\mbox{ m/s}$$ – as the car accelerates towards the maximum speed and as it decelerates away from the maximum speed.

To find this time in the first stage, we can again use a SUVAT equation.

Step 1: State the quantities that are given in the question and the quantity that we are looking for. Be careful with signs!

We are given the following

\begin{align} u = 0 && v=10 && a = 2.5 \end{align}

and we are looking for $$t$$.

$\,$

Step 2: Choose the most suitable equation either related to the time-velocity graph (if possible) or from the SUVAT equations based on Step 1.

Here, we can use either the slope formula or a SUVAT equation. We shall randomly choose the SUVAT equation approach.

As we are not told $$s$$ nor do we want to find $$s$$, we should choose the SUVAT equation which does not have $$s$$ in it, namely

$v=u+at$

$\,$

Step 3: Insert the quantities into the equation.

Subbing in our known values, we obtain

\begin{align}10=0+(2.5)t\end{align}

Rearranging for $$t$$, this gives

\begin{align}t&=\frac{10}{2.5}\\&=4 \mbox{ s}\end{align}

$\,$

To find this time in the second stage, we can again use a SUVAT equation.

Step 1: State the quantities that are given in the question and the quantity that we are looking for. Be careful with signs!

We are given the following

\begin{align} v = 10 && u=20 && a = -\frac{4}{3} \end{align}

and we are looking for $$t$$.

$\,$

Step 2: Choose the most suitable equation either related to the time-velocity graph (if possible) or from the SUVAT equations based on Step 1.

Here, we can use either the slope formula or a SUVAT equation. We shall randomly choose the SUVAT equation approach.

As we are not told $$s$$ nor do we want to find $$s$$, we should choose the SUVAT equation which does not have $$s$$ in it, namely

$v=u+at$

$\,$

Step 3: Insert the quantities into the equation.

Subbing in our known values, we obtain

\begin{align}10=20+\left(-\frac{4}{3}\right)t\end{align}

Rearranging for $$t$$, this gives

\begin{align}t&=\frac{10-20}{-\frac{4}{3}}\\&=7.5 \mbox{ s}\end{align}

Note that this is the time only from the start of the second stage, and therefore the time since the beginning of the trip is $$7.5+8=15.5$$ seconds.

$\,$

Hence, the two times at which the car’s speed is $$10\mbox{ m/s}$$ are $$4$$ seconds and $$15.5$$ seconds into the journey.

## FULL SOLUTIONS

Schoolbooks typically only provide the final answers to practice questions in the back pages rather than the full details of how those answers are obtained.

Teacher handbooks typically only contain very condensed solutions to those same questions that are often very difficult to follow.

On A.M. Online, we instead provide detailed, written solutions to EVERY exercise question, and we have just as many questions for students to practice with as found in a schoolbook!

In fact, we go even further! We provide a total of four different methods for students to check their understanding of every question:

• First, check to see if they got the right Answer. If they didn’t…
• Read the Solution, i.e. what they should write down when answering the question. If they don’t understand the solution…
• Read the Walkthrough, i.e. what a teacher would say if they were explaining the solution to them. If they don’t understand the walkthrough…
• Ask Mr. Kenny for further guidance on Teacher Chat!

These “stepping stones” greatly reduce the likelihood of a student “giving up” if they don’t understand the solution to a question.

## ANIMATIONS

Applied Maths is an incredibly dynamic subject with a significant portion of the curriculum dealing with the motion of objects. Animations, therefore, make for a much more helpful and beneficial learning experience compared to the static images of a schoolbook.

We therefore provide rich, custom-made animations, together with thousands of illustrations, throughout the lessons in order to provide students with what we believe are the best learning resources available for Applied Maths.

## KNOWLEDGE CHECKS

Before jumping directly into answering full exercise questions, we first test each student’s basic understanding of lesson content with quick-fire questions in the form of Knowledge Checks.

The questions within our 50+ Knowledge Checks give each student immediate feedback as to whether a lesson “clicked” or not, thereby indicating whether they are ready to answer the corresponding exercise set or whether they need to do some quick revision first.

Also, as Knowledge Checks are marked automatically, they provide another way of reducing the time that a teacher spends correcting and thereby another way of reducing a teacher’s workload!

/20

$\,$Knowledge Check 8

$\,$A quick check on your understanding of:

Displacement

Velocity

Acceleration

SUVAT equations

$\,$

The rate of change of a vector is itself a vector.

1 / 20

The rate of change of velocity with respect to time is the definition of

The rate of change of a vector is itself a vector.

2 / 20

The rate of change of displacement with respect to time is the definition of

What quantity has that unit?

3 / 20

"A car brakes from $$\mathbf{15\mbox{ m/s}}$$ east to $$0\mbox{ m/s}$$ east at a rate of $$5\mbox{ m/s}^2$$ within $$3$$ seconds". The quantity in bold refers to:

What quantity has this unit?

4 / 20

"A car brakes from $$15\mbox{ m/s}$$ east to $$\mathbf{0}\mbox{ m/s}$$ east at a rate of $$5\mbox{ m/s}^2$$ within $$3$$ seconds". The quantity in bold refers to:

Displacement is a vector quantity.

5 / 20

A pedestrian walks $$25$$ metres west from $$o$$ followed by $$10$$ metres east. Their total displacement from $$o$$ is

Distance is a scalar quantity.

6 / 20

A pedestrian walks $$25$$ metres west from $$o$$ followed by $$10$$ metres east. Their total distance travelled is

Speed is a scalar quantity.

7 / 20

A cyclist travels $$60$$ metres east in $$2$$ seconds at a constant acceleration. The cyclist's speed is

Velocity is a vector quantity.

8 / 20

A cyclist travels $$60$$ metres east in $$2$$ seconds at a constant acceleration. The cyclist's velocity is

Speed is equivalent to the magnitude of velocity.

9 / 20

Speed and velocity have the same units.

Acceleration is the rate of change of velocity.

10 / 20

A cyclist increases her velocity from $$10\mbox{ m/s}$$ east to $$20\mbox{ m/s}$$ east before reducing it again to $$10\mbox{ m/s}$$ east. As her starting and final velocities are the same, she did not accelerate during this motion.

A vector has both a magnitude and a direction.

11 / 20

When a car uses a roundabout, it must accelerate.

Acceleration is the rate of change of velocity.

12 / 20

If an object is moving, it must be accelerating.

If an object is not moving, its velocity is zero.

13 / 20

In order for a moving car to come to a stop, it must decelerate.

Displacement is a vector quantity.

14 / 20

In the below image, the car is parked and is not moving. What is its displacement from $$o$$ after 10 seconds?

$\,$

$\,$

If there is no acceleration, then there is no change in velocity.

15 / 20

In the SUVAT vector equations, if $$\vec{a}=0$$, then $$\vec{u}=\vec{v}$$.

What SUVAT variable is not included in this triangle?

16 / 20

"Dad's Silly Triangle" can be used:

What quantities do we know and what quantity are we looking for?

17 / 20

A cyclist increases his velocity from $$15\mbox{ m/s}$$ east to $$25\mbox{ m/s}$$ east over a period of $$5$$ seconds. Which of the following SUVAT equations should we use to find the cyclist's acceleration in that time?

What quantities do we know and what quantity are we looking for?

18 / 20

A cyclist increases his velocity from $$15\mbox{ m/s}$$ east to $$25\mbox{ m/s}$$ east over a period of $$5$$ seconds. Which of the following SUVAT equations should we use to find the cyclist's displacement in that time?

What happens when we multiply a scalar by a vector?

19 / 20

$$5t^3\vec{a}$$ is a vector

What assumption was made when deriving the SUVAT equations?

20 / 20

If the acceleration of an object is gradually changing over time, we cannot use the SUVAT equations.

0%

### Key Point

A linear equation is any equation of the form

\begin{align}ax+b=0\end{align}

where $$a$$ and $$b$$ are constants.

All of the following are therefore considered linear equations.

\begin{align}2x+3=0 && x-7=0\end{align}

\begin{align}\frac{x}{3}+2=0 && 1.7x-\pi=0\end{align}

In fact, even the following equation

\begin{align}5x=x+3\end{align}

can be considered a linear equation since, as we shall see shortly, it can be arranged as

\begin{align}4x-3=0\end{align}

In general, as long as the highest power of the unknown variable is $$1$$, the equation is considered linear.

To solve such equations, we perform the following steps

• Move all of the terms containing $$x$$ to one side of the equation and all of the terms not containing $$x$$ to the other side. When switching any term from one side to the other, we must flip the sign of that term!
• Add/subtract all of the terms on each side.
• Divide both sides by any coefficient in front of the final term containing $$x$$.

## Guided Example

Solve the following equation:

\begin{align}6x-7 = 2x+5\end{align}

To begin, we place all of the terms containing $$x$$, i.e. $$6x$$ and $$2x$$, on one side, e.g. the left side, and all of the terms not containing $$x$$, i.e. $$-7$$ and $$+5$$, on the other side.

To do that, we need to move the $$2x$$ term from the right side to the left side, causing its sign to switch to $$-2x$$. We also need to move the $$-7$$ term from the left side to the right side, causing its sign to instead switch to $$+7$$.

\begin{align}6x-7 = 2x+5\end{align}

\begin{align}\downarrow\end{align}

\begin{align}6x-2x = 5+7\end{align}

We now add/subtract the term on both sides

\begin{align}6x-2x = 5+7\end{align}

\begin{align}\downarrow\end{align}

\begin{align}4x = 12\end{align}

Finally, as our goal is to find $$x$$, i.e. to write $$x=\mbox{‘something’}$$, we need to remove the coefficient in front of the $$x$$ by dividing both sides by $$4$$.

\begin{align}4x = 12\end{align}

\begin{align}\downarrow\end{align}

\begin{align}\frac{4x}{4}=\frac{12}{4}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x=3\end{align}

Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.

In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!

### Exam Tip

If you’re ever unsure as to whether you solved an equation correctly, you can quickly check by reinserting the value that you obtained for the unknown variable back into the original equation.

For example, if we reinsert $$x=3$$ back into our original equation above, we obtain

\begin{align}6x-7 = 2x+5\end{align}

\begin{align}\downarrow\end{align}

\begin{align}6(3)-7 = 2(3)+5\end{align}

\begin{align}\downarrow\end{align}

\begin{align}11 = 11\end{align}

As this is of course true, our solution must indeed be correct!

Consider the following statement:

“The sum of two numbers equals five. What are those numbers?”

In this case, as we have two distinct unknown numbers, we shall represent them using two variables $$x$$ and $$y$$.

Answering this question is then equivalent to solving the following equation:

\begin{align}x+y=5\end{align}

So, what is the solution to this equation?

Well, we can easily see that $$x=2$$ and $$y=3$$ satisfies this equation.

\begin{align}2+3=5\end{align}

However, $$x=1$$ and $$y=4$$ also satisfies this equation

\begin{align}1+4=5\end{align}

as does $$x=1.7$$ and $$y=3.3$$

\begin{align}1.7+3.3=5\end{align}

and so on. This equation in fact has an infinite number of solutions!

Let us therefore instead consider the following different statement.

“The sum of two numbers equals five. The difference between those two numbers is three. What are those numbers?”

This information is instead represented by the following two equations:

\begin{align}x+y&=5\\ x-y&=3\end{align}

To answer this question, we need to solve these two equations at the same time, i.e. simultaneously.

This time, rather than obtaining an infinite number of solutions, we will instead only obtain one solution. This fact is indeed true more generally if we have a larger number of unknown variables.

### Key Point

If we wish to find $$n$$ unknown variables, we need to solve $$n$$ simultaneous equations.

For now, we shall consider only solving two simultaneous equations for two unknowns.

So, how do we solve two simultaneous equations? More, specifically how do we solve such equations if they are both linear equations?

\begin{align}x+y&=5\\ x-y&=3\end{align}

Well, there are in fact two methods that we can use.

##### 1) Substitution

This method is summarised as follows:

• Rewrite one of the equations as either $$x=\mbox{‘stuff’}$$ or $$y=\mbox{‘stuff’}$$.
• Substitute the $$\mbox{‘stuff’}$$ into the other equation.

## Guided Example

Solve the following simultaneous equations using substitution:

\begin{align}2x+3y&=7\\ 4x-5y&=3\end{align}

To begin, we shall rewrite one of the equations as either $$x=\mbox{‘stuff’}$$ or $$y=\mbox{‘stuff’}$$. As it seems equally as easy to do that for either equation, we shall randomly choose to do it for the first equation.

For that equation, it also seems equally as easy to rearrange it as either $$x=\mbox{‘stuff’}$$ or $$y=\mbox{‘stuff’}$$. We shall randomly choose to rearrange it as $$y=\mbox{‘stuff’}$$.

\begin{align}2x+3y&=7\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3y=7-2x\end{align}

\begin{align}\downarrow\end{align}

\begin{align}y=\frac{7-2x}{3}\end{align}

Using this, we can now replace $$y$$ in our second equation.

\begin{align}4x-5y=3\end{align}

\begin{align}\downarrow\end{align}

\begin{align}4x-5\left(\frac{7-2x}{3}\right)=3\end{align}

We now have an equation just for $$x$$ which we can solve as usual!

Multiplying the above equation by $$3$$, we obtain

\begin{align}12x-5(7-2x)=9\end{align}

\begin{align}\downarrow\end{align}

\begin{align}12x-35+10x=9\end{align}

\begin{align}\downarrow\end{align}

\begin{align}22x=44\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x=2\end{align}

Now that we know what $$x$$ is, we can find $$y$$ by inserting this value into any of our equations that contain both $$x$$ and $$y$$, e.g.

\begin{align}y=\frac{7-2x}{3}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}y&=\frac{7-2(2)}{3}\\&=\frac{3}{3}\\&=1\end{align}

Therefore, the solution to this set of simultaneous equations is $$x=2$$ and $$y=1$$.

Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.

In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!

##### 2) Subtraction

This method is instead summarised as follows:

• rewrite one of the equations such that the coefficient of either the $$x$$ term or the $$y$$ term is the same in both equations
• subtract one equation from the other

## Guided Example

Solve the following simultaneous equations using subtraction:

\begin{align}2x+3y&=7\\ 4x-5y&=3\end{align}

Currently, both equations have different coefficients in front of their $$x$$ and $$y$$ terms ($$2$$ and $$4$$ for the $$x$$ terms and $$3$$ and $$-5$$ for the $$y$$ terms).

For either pair of coefficients to be the same, we need to multiply one (or both) equations by a constant. We shall choose to make the $$x$$ coefficients the same as this only requires multiplying the first equation by $$2$$

\begin{align}4x+6y&=14\\ 4x-5y&=3\end{align}

Now, the coefficient of the $$x$$ term is the same in both equations. Therefore, if we now subtract one equation from the other, the $$x$$ terms will disappear, leaving us with an equation just for $$y$$!

\begin{align}4x+6y=14&\\\underline{-\,\,\,(4x-5y=3)}&\\11y=11&\end{align}

since $$4x-4x=0$$, $$6y-(-5y)=11y$$ and $$14-3=11$$.

This equation can now be easily solved for $$y$$ to give

\begin{align}y&=\frac{11}{11}\\&=1\end{align}

Again, we can now insert this value for $$y$$ into any equation involving $$x$$ and $$y$$ to find $$x$$, e.g.

\begin{align}2x+3y&=7\end{align}

\begin{align}\downarrow\end{align}

\begin{align}2x+3(1)&=7\end{align}

\begin{align}\downarrow\end{align}

\begin{align}2x=4\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x=2\end{align}

As expected, the solution is the same as that obtained using the first method, i.e. $$x=2$$ and $$y=1$$.

Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.

In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!

### Key Point

A quadratic equation is any equation of the form

\begin{align}ax^2+bx+c=0\end{align}

where $$a$$, $$b$$ and $$c$$ are constants.

All of the following are therefore considered quadratic equations.

\begin{align}x^2+5x+3=0 && 2x^2-6=0\end{align}

\begin{align}8x^2+17x=0 && \frac{3x^2}{5}-12x+1=0\end{align}

The following equation

\begin{align}12x=-5+4x^2\end{align}

is also a quadratic equation as it can be arranged as

\begin{align}4x^2-12x-5=0\end{align}

In general, as long as the highest power of the unknown variable is $$2$$, i.e. the degree of the polynomial is $$2$$, the equation is considered quadratic.

So, how do we solve equations of this type?

Well, first note that if $$a=0$$, the equation is now longer quadratic but instead linear

\begin{align}bx+c=0\end{align}

and we already know how to solve such equations.

If instead $$b=0$$, i.e. if the equation is of the form

\begin{align}ax^2+c=0\end{align} then the equation can easily be solved by rearranging.

## Guided Example

Solve the following equation:

\begin{align}3x^2-48=0\end{align}

This equation can be rearranged as follows

\begin{align}3x^2-48=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}3x^2=48\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x^2=16\end{align}

Note that there are in fact two numbers that satisfy this equation: $$-4$$ and $$4$$ (since both $$(-4)^2=16$$ and $$4^2=16$$).

Therefore, this quadratic equation has two solutions: $$x=-4$$ and $$x=4$$. (This can also be written in one line as $$x=\pm 4$$.)

In general, quadratic equations will have (at most) two solutions. The solution(s) to quadratic equations (and indeed equations of an even higher degree) are often referred to as the roots of the equation.

Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.

In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!

In all other cases, we cannot simply rearrange the equation to find the solution. For example, consider the most general case of all three coefficients being non-zero, e.g.

\begin{align}x^2+3x+2=0\end{align}

There is no way to rearrange this equation to obtain $$x=\mbox{‘stuff’}$$ (or even $$x^2=\mbox{‘stuff’}$$).

Of course, we can rearrange it as

\begin{align}x^2=-3x-2\end{align}

but that is of no help as we still have an $$x$$ term on the right hand side!

So, how do we proceed?

Well, we can use factorisation! We can factorise the following quadratic expression

\begin{align}x^2+3x+2\end{align}

as

\begin{align}(x+1)(x+2)\end{align}

Therefore, we can rewrite this quadratic equation as

\begin{align}(x+1)(x+2)=0\end{align}

How does that help us solve this equation?

Well, we now have ‘something’ multiplied ‘something else’ equalling zero. We know with regular numbers that this can only be the case if (at least) one of the two numbers is zero!

\begin{align}0\times 6=0 && 4 \times 0=0\end{align}

There is no way of multiplying two numbers together to get zero unless (at least) one of those numbers is zero!

This is also true with algebra – (at least) one of these factors must also be zero, i.e.

\begin{align}x+1=0 && \mbox{or}&& x+3=0\end{align}

This therefore gives us our two solutions (roots)

\begin{align}x=-1 && \mbox{or} && x=-3\end{align}

## Guided Example

Solve the following equation:

\begin{align}2x^2-2x-12=0\end{align}

The quadratic expression on the left hand side can be factorised, giving

\begin{align}(2x+4)(x-3)=0\end{align}

and therefore

\begin{align}2x+4=0 && \mbox{or}&& x-3=0\end{align}

Hence, the two roots of this equation are $$x=-2$$ and $$x=3$$.

Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.

In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!

We can also use factorisation if $$c=0$$, i.e. if the equation is of the form

\begin{align}ax^2+bx=0\end{align}

## Guided Example

Solve the following equation:

\begin{align}3x^2-12x=0\end{align}

The quadratic expression on the left hand side can be factorised, giving

\begin{align}(2x+4)(x-3)=0\end{align}

and therefore

\begin{align}2x+4=0 && \mbox{or}&& x-3=0\end{align}

Therefore, the two roots of this equation are $$x=-2$$ and $$x=3$$.

Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.

In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!

Unfortunately, it is not possible to factorise all quadratic expressions!

For example, consider the following expression

\begin{align}2x^2-5x+2=0\end{align}

The quadratic expression on the left cannot be factorised using usual methods.

However, let us check if $$x=0.5$$ is a possible root.

\begin{align}2(0.5)^2-5(0.5)+2=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}0.5-2.5+2=0\end{align}

\begin{align}\downarrow\end{align}

\begin{align}0=0\end{align}

Therefore, $$x=0.5$$ is indeed one of the roots!

Of course, it makes sense that this root cannot be found using usual factoring methods as such methods don’t consider the possibility that factor pairs may be decimals, fractions etc. And for good reason – we would then have an infinite number of factor pairs to check!

So, how do we determine that $$x=0.5$$ is a root without guessing? And what is the other root of this equation?

Well, it turns out that there is in fact a formula that we can use!

### Key Point

The roots of a quadratic equation of the form

\begin{align}ax^2+bx+c=0\end{align}

are given by

\begin{align}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\end{align}

where $$a$$, $$b$$ and $$c$$ are constants.

Note the $$\pm$$ sign in this equation: for one of the roots, we use a $$+$$ sign, and for the other, we use a $$-$$ sign.

## Guided Example

Solve the following equation:

\begin{align}2x^2-5x+2=0\end{align}

This is a quadratic equation of the form

\begin{align}ax^2+bc+c=0\end{align}

with $$a=2$$, $$b=-5$$ and $$c=2$$.

Inserting these values into our quadratic formula, we obtain

\begin{align}x&=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\&=\frac{-(-5)\pm\sqrt{(-5)^2-4(2)(2)}}{2(2)}\\&=\frac{5\pm \sqrt{25-16}}{4}\\&=\frac{5\pm 3}{4}\end{align}

Therefore, our two roots are

\begin{align}x=\frac{5-3}{4} && \mbox{and} && x=\frac{5+3}{4}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x=0.5 && \mbox{and} && x=2\end{align}

Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.

In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!

This quadratic formula can in fact always be used to find the roots of a quadratic equation, even when factorisation is possible!

## Guided Example

Solve the following equation:

\begin{align}2x^2-2x-12=0\end{align}

Note that this is the same quadratic equation that was solved in the second Guided Example above using factorisation.

The coefficients in this case are $$a=2$$, $$b=-2$$ and $$c=-12$$.

Inserting these values into our quadratic formula, we obtain

\begin{align}x&=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\&=\frac{-(-2)\pm\sqrt{(-2)^2-4(2)(-12)}}{2(2)}\\&=\frac{2\pm \sqrt{100}}{4}\\&=\frac{2\pm 10}{4}\end{align}

Therefore, our two roots are

\begin{align}x=\frac{2-10}{4} && \mbox{and} && x=\frac{2+10}{4}\end{align}

\begin{align}\downarrow\end{align}

\begin{align}x=-2 && \mbox{and} && x=3\end{align}

which are the same roots that we obtained by using factorisation!

Video Walkthroughs are currently in the process of being gradually uploaded, with that process expected to be finished by the end of October.

In the meantime, if you don't understand something in the Walkthrough above, feel free to reach out to Mr. Kenny on Teacher Chat!

## NO PRIOR KNOWLEDGE

We make no assumptions about what level of Leaving Cert Maths students are taking, what they have learned so far in their maths lessons etc.

Instead, any time we believe that there may be some students who need to know additional content outside of Applied Maths, we provide that material in the form of a prerequisite at the beginning of a lesson.

This therefore removes any possibility of any student not having the skills needed to understand any lesson on this website.

A.M. Online is highly community driven by the feedback that we receive from both students and teachers alike. As such, we frequently update this website based on that feedback.

For example, although we provide what we believe are more than enough exercise questions for each lesson, a teacher may have a need for additional questions.

After a quick talk on Teacher Chat, we get to work writing those questions, uploading them over the next day or two!

No need to wait years for a schoolbook to publish a new edition with those extra questions, presuming that they listen to your feedback at all!

## School Licence

$\,$

One payment of

### €40

per student for access until June 30th 2023.

(€30 for DEIS schools)

$\,$

#### ONLINE COURSE:

Interactive Lessons

Practice Questions

Animations

Walkthrough Solutions

Knowledge Checks

and more!

## CHEAP AND CHEERFUL

Similar websites typically charge hundreds of euro per year for a membership while also including significantly less features. These are businesses whose primary goal is to make a profit.

A.M. Online, on the other hand, was created by a teacher whose sole ambition is to give all students access to the best Applied Maths resources, all in one place.

As such, we have priced our membership at what we believe is the minimum needed to keep the website up and running.

Every school licence also includes free teacher accounts for any Applied Maths teacher at that school. Maths teachers can also obtain free accounts if they wish to use this website to upskill.

For the same price as a schoolbook, we therefore provide all of the resources that both teachers and students need for Applied Maths for an entire school year, all on one website.

And remember, we provide lots of sample content for free so that both students and teachers can see exactly what we offer – no registration required! Check the “free resources” at the bottom of this page!

A.M. Online Schoolbook

Explanations

Illustrations

Exercises

Solutions

all

examples only

Walkthroughs

Grinds included

60

None

Show/Hide Content

Teacher Chat

Animations

Knowledge Checks

Sample Papers

Updated...

daily

> yearly

Price

€30/€40

≈ €40

#### “A.M. Online brings the concept of a schoolbook into the 21st century.”

Is a school licence recommended for schools with or without an Applied Maths teacher?

Both!

For many schools, hiring an Applied Maths teacher is not a financially viable option due to low student interest.

As such, A.M. Online is a great alternative for students in such schools whose only other options are 1) to learn at home using a schoolbook with little help or guidance and/or 2) to spend a significant amount of money on additional support, which is often not possible.

For schools with an Applied Maths teacher, A.M. Online offers everything that can be found in a schoolbook, and much more! This website therefore makes for a much better classroom learning experience than the static pages of a schoolbook.

Even if our website is not used specifically in the classroom, it’s still incredibly beneficial as a secondary resource for Applied Maths teachers and their students by, for example, giving students another teacher they can ask questions to, even when they are at home!

School term has already started. Can I still obtain a school licence?

Yes! There are no restrictions on when you can obtain a school licence.

Of course, the earlier, the better!

I would like a full tour of the website first. Is that possible?

Good idea! You should certainly have a look around the website first to ensure that it is as incredible as we imply!

Once we have launched, you can request a temporary account for that very purpose by emailing us at teacher@amonline.ie using your school email address.

How should a school licence be funded?

That is entirely dependent on the school!

Some schools fund it themselves, some schools fund it through parents, and others a bit of both.

The choice is up to you!