Contact
- Teacher Chat
- teacher@amonline.ie
Our Websites
Junior Maths – Junior Cycle Maths
L.C. Maths – Leaving Cert Maths
A.M. Online – Leaving Cert Applied Maths
© A.M. Online. All Rights Reserved.
A.M. Online membership includes world-class, interactive lessons, both for students that have an Applied Maths teacher and those that are learning the subject from home.
Schoolbooks take a “fill in the gaps” approach, whereby a teacher is expected to expand on the core concepts of the book.
As we are not limited by what can fit into a schoolbook, our lessons instead describe every concept within Applied Maths with as much detail as possible.
Throughout each lesson, we also highlight three important aspects:
Vectors have both a magnitude and direction.
Magnitudes are always positive. Coefficients, on the other hand, can be either positive or negative.
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.
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)
\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)
\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)
\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}
\[\,\]
\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.
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.
Wouldn’t it again be awesome if your book instead provided those full details?
As you might have guessed, A.M. Online has got that covered!
We provide detailed, written solutions to EVERY exercise question, and we have just as many questions to practice with as you’d find in a schoolbook!
Can you imagine having a detailed written solution for every single exercise question in your schoolbook?! Wow!
In fact, we go even further! We provide a total of four different methods for students to check their understanding of every question:
These “stepping stones” greatly reduce the likelihood of a student “giving up” if they don’t understand the solution to a question.
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.
Before jumping directly into answering full exercise questions, we first test each student’s basic understanding of the lesson content with quick-fire questions in the form of Knowledge Checks.
The questions within our 50+ Knowledge Checks questions 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.
We make no assumptions about what level of Leaving Cert Maths a student is taking, what a student has 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 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
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!
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.
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.
This method is summarised as follows:
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!
This method is instead summarised as follows:
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!
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.
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}
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}
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!
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.
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!
Solve the following equation:
\begin{align}2x^2-2x-12=0\end{align}
using the quadratic formula.
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!
Interactive Lessons ✓
Practice Questions ✓
Animations ✓
Walkthrough Solutions ✓
Knowledge Checks ✓
and more!
Get access to all of the above until June 30th 2023 with one payment of €79.
Similar websites typically charge hundreds of euro per year for an online course that contains 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, even though our online course is more feature-rich than any other online course that you can find, our weekly grinds and our Teacher Chat feature are also included with a membership!
For the same price as a typical grind or two, we therefore provide everything needed to ace the exam for an entire school year, all on one website, without the need to even purchase a schoolbook.
And remember, we provide lots of sample content for free so that students can see exactly what we offer – no registration required! Check the “free resources” at the bottom of this page!
| Alternative | A.M. Online |
|---|---|
|
Only answers at the back of the schoolbook. |
Full, detailed solutions to every question in addition to detailed walkthroughs. |
|
Visualise the concepts using dull, static images. |
Visualise the concepts using vibrant animations. |
|
Receive delayed feedback after your homework has been corrected. |
Receive immediate feedback via Knowledge Checks. |
|
Ensure that you already understand the required maths. |
No assumptions made about your understanding of maths. |
Junior Maths – Junior Cycle Maths
L.C. Maths – Leaving Cert Maths
A.M. Online – Leaving Cert Applied Maths
© A.M. Online. All Rights Reserved.