## Topic outline

- Cambridge International AS & A Level Mathematics 9709Videos

- Pure Mathematics 1

Video transcript1.1 Quadratics.

Welcome to this short ‘insights video’ where learners can better understand how to solve quadratic equations using the ‘completing the square’ technique. This technique can solve ‘perfect square’ quadratic equations like this...

...where the coefficient of the squared term is ‘one’. However, equations do not always appear in this form, for example in the following equation the squared term coefficient is two...

but, the equation can be rearranged like this…

...So, leaving the rest of the equation alone, learners can use the ‘completing the square’ technique to solve the ‘x squared plus two x’ part within the brackets first. An important mistake learners often make is to correctly solve this, but then, forget to replace this back into the original quadratic equation to score full marks.

Simple geometry can help students visualise how the completing the square technique works...

...If our ‘general formula’ is to solve x squared plus b times x. Learners can think of this equation as two rectangles and we need to find out the area!

The first rectangle is x wide by x height. It’s area represents the x2 part of the equation, and is a square. The second rectangle is b long and x wide. It’s area represents the b times x part of the equation. Now the learners could rearrange these shapes like this…

And if we were to ‘complete the square’ it would look like this…

So, the total area is easy to work out. It’s x plus b over two all squared. But we don't want the blue square - it’s not part of our equation. So we have to subtract b over two, times b over two. Which is b squared over four. So, the answer to the general solution becomes...

Then learners can replace this general solution back in their original equation within the brackets. And because it only contains one x function now the original quadratic equation is easy to rearrange.

I hope this short insights video has been useful to you to help explain to your learners the types of equations ‘completing the square’ solves and a very visual way to explain how to use the ‘completing the square’ method.

Thank you.**Completing the square****Solving quadratic inequalities****Solving quadratic functions in***x***Sketching quadratic functions**

Video transcriptWelcome to this short ‘insights video’ into helping learners better understand the connections between graphs and algebra. Many learners are very capable of plotting coordinates and drawing a straight line graph, but are sometimes lost seeing the connection between a line, and an algebraic equation. Here using the two coordinates ‘x1 and y1’ and ‘x2 and y2’ a straight line is drawn.

But the idea of the gradient - the steepness of the incline - the inc-’line’, is not always appreciated. This can easily thought of as the amount they go ‘up’ (the y direction) over how far they have to go along (the x direction).

In this example we go ‘up’ from y equals one to five, an increase of four units. And we go ‘along’ from x equals zero to two, an increase of two units.

So the gradient, how steep the slope is, is four over two, which is two.

From this, learners will be able to visualise the general gradient formula, the change in y coordinates over the change in x coordinates.

Often written as y2 minus y1 over x2 minus x1.

Then when they have the found the gradient they can use the following formula to find the general equation of a line. Y equals m times x plus c…

...where m is the gradient and c is the value of y where the line crosses the y axis.

In this case the gradient they have calculated is two and the y intercept point is one. So the equation becomes...

Y equals two x plus one.

A common misunderstanding is that learners use the general formula and immediately assume the gradient is the coefficient of x and the intersection is the number on its own. But they need to check that the equation starts with ‘one y’ on the left hand side and if not, organise the equation so that it is ‘one y’ before making the substitutions.

I hope you have found this insights video useful to help your learners visualise better the connections between graphs and algebra.

Thank you.- Pure Mathematics 2

Video transcriptWelcome to this short ‘insights video’ showing some of the misunderstandings learners have when solving questions with trigonometric quadratic, or cubic, equations.

Solving ‘trigonometric’ quadratic equations uses the same techniques as ordinary quadratic equations, for example, factorising.

However, there is an added layer of complication because learners also need to apply trigonometric identities or transformations to simplify these equations…

… and when learners apply these identities, they often forget to be as rigorous about the more basic quadratic rules.

They know that the square root of x squared can be either a plus or minus solution. But they often forget to apply the same logic, to say ‘cos squared x’ and only use the positive answer and don’t include the negative solution, so, will have an incomplete answer.

A second misunderstanding around solving trigonometric equations between a range of angles is not to factorise, but cancel terms.

In this expression - students might cancel the cotangent on both sides of the equation…

...But that would lead to valid answers not being given...

...Each term leads to a solution!

And finally, when learners see questions involving two trigonometric terms, to be solved within the specified range. They assume there will always be two valid solutions per term. This is not a rule!

Rather, solutions are cyclical and there can be more than two solutions per term that fit within the range. So learners need to check a good range of answers to see if others fall within the target range.

I hope this short insights video has been useful to you to see some of the challenges learners face when solving these kind of trigonometric equations.

Thank you.- Pure Mathematics 3

Video transcriptWelcome to this short insights video to help your learners better understand vector equations.

One of the challenges for learners working with vector equations is grasping the difference between regular equations, using Cartesian coordinates, and vector equations to describe lines.

Learners are used to expressing lines in terms of cartesian x and y points.

However, vectors are described differently. They have two components..

…a starting point and a direction component.

They understand what this cartesian equation for a line will look like…

...Where the m term is the gradient or incline of the line and c acts like an offset to the y crossing point.

But a vector equation for a line is described like this…

...This line starts at cartesian point x one, y one then travels ‘towards’ cartesian point x two, y two. And the t function describes the fact that the line will continue in that direction forever.

Which is quite different as they have a cartesian point to start from, but the second component, direction is expressed using a point the line goes through but continues for all values of t.

For example...

This line starts at cartesian point x equals two, y equals minus one in terms of x and y coordinates, but then travels ‘towards’ cartesian point x equals one, y equals five.

Once this transformation from cartesian space to vector space is understood transforming back to Cartesian space becomes much easier. We now know that for any point on the line…

x equals 2 plus one times t

And

Y equals minus 1 plus five times t.

Then, for any value of t in vector space we can know what the values for x and y will be in Cartesian space.

Then learners can take this concept forward, looking at more than one equation, to see where vector lines cross by finding out where the vector equations are equal to each other.

So, I hope this short insights video has been useful to you to help your learners better understand vector equations.

Thank you.**Matching different notations****Vector paths****Creating lines in 3D****Vectors and lines**- Mechanics

Video transcriptWelcome to this short ‘insights video’ where we are going to look some of the challenges learners face when analysing the forces that act on objects in different situations.

A Force is not just as how strong it is, but also, what direction it acts in. So, forces are best described as vectors, with magnitude and direction.

Using vector theory, forces can be resolved into a horizontal and a vertical components. Then learners can use these components to see what will happen to objects when these forces are applied.

The complication of many exam questions is to involve objects on slopes. In this situation the ‘horizontal’ and ‘vertical’ components need to be ‘twisted’ to align perpendicularity with the included surface to calculate the forces on an object.

Learners need to compare ‘like with like’ and calculating resultant vectors in the horizontal and vertical directions. Allowing them to resolve forces from different directions into one resulting vector force.

In more complex problems, learners are asked to analyse more then one object, with more than one force acting on them.

For example, in this situation learners need to look at each on the objects in turn and work out the forces acting on each object in turn.

They need to think about the horse on its own, the cart on its own and finally the harness connecting the horse to the cart" everything would be 'resolved’. Then using vector theory resolve the three sets of horizontal forces, the horse pulling (F), the cart having a drag force (f), and the tension in the harness force between the horse and the cart (P).

It is clear that the horse needs to provide enough force to pass through the tension in the harness to overcome the frictional drag of the cart, which of course it usually can, and the whole system moves forward.

I hope this short insights into resolving questions around force and equilibrium has been useful to you and your learners.

Thank you.- Probability & Statistics 1

Video transcriptWelcome to this short ‘insights video’ where we are going to look at arrangements, permutations and combinations and some of the challenges learners face in solving these kind of problems.

A common misconception when sorting, or arranging objects, is to think:

‘Aha - there are six objects, so I could start sorting by choosing any one of the six. Then, there are five left, so I could choose any one of the five and so on…

...giving 6 choices plus 5 choices plus 4 choices plus 3 choices plus 2 choices plus 1 choice…

...So that makes 21 ways…

...Which is incorrect. This error of ‘adding’ instead of ‘multiplying’ means they have not really grasped the mathematical process of making multiple selections.

Using simple examples of ‘selections’ quickly shows learners how to build up a general mathematical rule to the problem of arrangements, and then applying this rule is so much quicker, than listing all the possible outcomes particularly for more complex problems

Then learners are not always clear about the difference between a question asking then to make a selection, and making a selection in a particular order.

For example, a question about competitors in a schools sports competition. Guessing who will win the first three places is hard, but guessing the winners and the order they will win in is harder still The chance or ‘probability’ of guessing the winners in the order right too is less than just guessing the winners.

Examples like this let learners see that choosing, or ‘selecting’, from a series of options, is a very different answer from choosing, or ‘selecting’, from a series of options in a particular order!

They need to decide: are they being asked ‘how many ways they can select particular objects (using combinations) or how many ways they can arrange particular objects (and use permutations).

Finally, they must answer using the correct notation and correct formula when solving problems like this.

Learners often use nCr when they mean nPr, from not understanding the topic completely.

I hope this short insights video on permutations and combinations has been useful to you and your learners.

Thank you.- Probability and Statistics 2

Video transcriptWelcome to this short ‘insights video’ where we are going to look at some of the misunderstanding that learners have around the Poisson distribution. And whether, or not, they can use the Poisson distribution as a method to predict outcomes in a situation, or particular events happening over a period of time.

There are two sets of criteria learners need to understand and apply to see if events can be predicted using this method.

The first is looks at the situation itself and asks questions about the kind of events that are occurring.

And the second is a mathematical test.

If both sets of conditions are true then the situation, or events, can be predicted using this method.

First, looking at the kind of events…

Events occur at a constant rate’. What are constant rate events?

Learners sometimes struggle to understand what this means and real world examples are a good way of helping them.

Looking at the number of traffic accidents on a road over a period of time, or the number of defects in a material roll, per metre, are both examples of constant rate events.

Whereas predicting the number of customers going into a coffee shop might seem to go in at constant rate, but when you think about it, the results would be very different at say coffee time, or lunchtime compared to the middle of the afternoon, so this would not be suitable events.

Then, are events ‘singular’ ?

A good example is predicting the number of light bulbs that will fail in a factory, each bulb failure can only happen once. These are therefore singular events.

How about, are the events ‘independent’ of each other? For this, maybe think about an insurance company and the claims the office receives over a period of time. The claims from different customers at different times are going to be independent of each one another.

Finally learners need to mathematically test the situation by looking at the mean and variance of the data they are given. For the Poisson distribution to be valid the mean and the variance of the data set have to be the same.

Learners don’t always realise the importance of this. If the mean is not the same as the variance in the data, then the formula will not give accurate outcomes.

I hope this short insights video on helping learners decide if the Poisson is the right method for modelling outcomes for a given situation over a period of time.

Thank you.