astresnotes-maths

[astresnotes-maths]

Lesson 1.1 — Factorising Quadratics — 20220906

My first maths class! And the first class of pure maths. We mainly just reviewed things this class, as we’re doing in all the maths classes. Most of the lessons so far involve the teacher reading to us some explanations and showing us some examples, and then all of us working to answer some questions on our own, which we then check as a class.

Regardless of the work so far being things I already know, I spent a few hours typing out a tutorial on factorising quadratics nonetheless. Probably a huge waste of my time, but I hope at least one person will come across this post in their time of need and find it helpful.

Because of how difficult it is to represent maths equations accurately by typing, and because tumblr’s layout does not support things like subscripts and indices, I will not include every question we did in all the lessons. Don’t worry though, I will include examples for every explanation, as well as some additional questions when we did more of them in class, and any homework we’re given.

Topic 1: Quadratics

Lesson 1.1 — Solving by Factorising

The standard quadratic equation has the structure — ax^2+bx+c=0

In quadratics, and only in quadratics, this is called standard form

So always convert the question into this form before trying to solve

Factorising quadratics — means turning them into the form (rx+p)(tx+q)=0

This helps you solve it by then taking each parentheses in turn and setting it equal to zero (because seeing as they multiply together to form zero, either one of them must equal zero in order for the equation to be true)

The basic principles of factorising quadratics:

pq must equal c, because these are the only solely numerical digits (no variables) in the brackets, so they multiply to give the solely numerical digit in the standard form (this being c)

q + p equal b, because you multiply each by x, making them like terms, so you then combine them through addition/subtraction to form a singular coefficient of x — which is b in the standard from

rt must equal a, because they multiply to form the coefficient of x^2

If b and c are positive, then the signs within the parentheses will both be positive too

If c is negative, then one of the signs will be positive and one will be negative:

If b is positive when c is negative, then the number (out of p and q) which multiplies (with the respective coefficient of x) to give the larger value (i.e out of rq and tp) will be positive, while the other one will be negative

If b is negative when c is also negative, then the number (out of p and q) which multiplies (with the respective coefficient of x) to give the smaller value (i.e rq or tp) will be positive, while the other one is negative

I probably worded that very confusingly, but don’t worry I will include examples and explain the steps as I do them so you can see what I mean in action

Factorising when a = 1:

Start by laying your foundation — (x    )(x    )

List the factors of c 

Find which two factors of c add/subtract to form b — these will be p and q

Place them in the parentheses — (x   p)(x   q)

Figure out which signs go where — in the case of a = 1, if c is negative, then you don’t need to worry about “multiplying with the respective coefficient of x”, because the coefficient of x is 1, so simply the value of p and q dictate the signs

Always check by expanding the brackets before you move on to solve

Then solve by setting each set of brackets in turn to zero

Example: x^2 + 3x - 10 = 0

Lay the foundation — (x   )(x   )

List the factors of c — 5&2, 1&10

Find which factors of c add/subtract to give b — 5 - 2 = 3, so p and q must be 5 and 2

Substitute into parentheses — (x   5)(x   2)

Figure out the signs — in this case, c is negative, so one of the signs will be positive while the other will be negative

b is positive, so the larger digit out of p and q will be the positive one — in this case it’s the 5

So x^2 + 3x - 10 = (x + 5)(x - 2)

Check — (x + 5)(x - 2) = x^2 + 5x - 2x - 10 = x^2 + 3x -10  ✓

Now solve by setting each bracket to zero:

x + 5 = 0,  x = -5

x - 2 = 0,   x = 2

Factorising when a ≠ 1:

Start by laying your foundation, this time with a space before the x’s — ( x    )( x    )

See if you can divide every term by a factor of a to simplify the equation

Now find the factors of your new value of a

List the factors of c 

Find which two factors of c add/subtract to form b when multiplied with a set of factors of a — these will be p, q, r and t, which form b when rq and tp are added/subtracted

Place all numbers in the parentheses — (rx   p)(tx   q)

Figure out which signs go where

If c is negative, then you need to see which value, out of p and q form the larger or smaller value (depending on whether b is pos. or neg.) when multiplied with the respective coefficient of x (r and t)

Check by expanding the brackets before you move on to solve

Then solve by setting each set of brackets in turn to zero

Example: 9x^2 - 39x - 30 = 0

Lay the foundation — ( x    )( x    )

See if you can divide every term by a factor of a to simplify the equation — all terms can be divided by 3 to give 3x^2 - 13x - 10

Now find the factors of your remaining a — 3&1

List the factors of c — 1&10, 5&2

Find which two factors of c add/subtract to form b when multiplied with a set of factors of a — 5&2 multiply with 3&1 to give 15&2, which subtract to give 13

Place all numbers in the parentheses — (3x  2)(x  5)

Figure out which signs go where — c is negative, so it must be one positive and one negative

b is also negative, so the digit (of q or p) which multiplies with the respective coefficient of x to give the larger value (rq or tp) must be negative — the values formed are: 5 x 3 = 15, and 2 x 1 = 2; the larger value is 15, which is given by 5 x 3, so 5 must be the negative value

Therefore 3x^2 - 13x - 10 = (3x + 2)(x - 5)

Check — (3x + 2)(x - 5) = 3x^2 - 15x + 2x -10 = 3x^2 - 13x - 10  ✓

Solve by setting each bracket to zero

3x + 2 = 0, 3x = -2, x = -⅔

x - 5 = 0, x = 5

**Tip: At first glance, when trying to find the correct factors of c to use, you may find multiple solutions

For instance, in the example above, you may have thought of using 1&10 as opposed to 5&2, because they multiply with 3&1 to give 3&10, which add to give the required 13

However, you have to consider the signs; because both c and b were negative, the larger of the multiplied values (rq and tp) had to be negative, while the other one positive

10&3 would both need to be positive, or both need to be negative, to give an absolute value of 13

But in order for the correct result to be obtained when the signs were different, 5&2 had to be used (as they gave rq = -15 and tp = +2)

I’m sorry if this makes things more complicated-sounding than they actually are. I hope the examples make it clearer. Feel free to send me an ask on my main blog if you don’t understand what I meant.

No homework this class, seeing as it’s the first one this year (and we have another one on Friday).

Anyways please remember to take care of yourselves and drink water!

[astresnotes-maths]

Lesson 2.1 — Velocity and Displacement — 20220907

This was my first lesson in mechanics. As with pure maths, the first few lessons are just reviews of what we already learned in previous years, so don’t jump at me if this is really easy stuff. I wrote out guide notes anyways, because I’m dedicated like that (at least for now — chances are this enthusiasm will die down as homework, exams, and extracurricular activities pile up on me).

Fun story: I was late to this lesson! I forgot to set my alarm the night before (I knew I was forgetting something, but I had no idea what and I was tired so the little nagging feeling was completely overlooked) and, thanks to timezone differences, my class was at 6:30AM. I woke up at 6:40. I’m surprised I even managed to wake up on my own that early, but that doesn’t matter. Point is, my first impression with the mechanics teacher (who is the slightly stricter of the two maths teachers, and head of the maths department) was less than optimal. (It’s ok though, I made sure to participate a lot and show I was invested in the class to hopefully rank up some points.)

Alright, sorry for the big introduction. Without further ado, here are the notes from this lesson.

**“∆” is the Greek symbol ‘Delta’. In maths and physics, it means “a change in”. For example, “∆velocity” just means “a change in velocity”.

Topic 2: Velocity & Acceleration

Lesson 2.1 — Displacement & Velocity

Vector — a value that has both magnitude (size) and direction

There are two measurements which can be used to indicate how far something has traveled:

Distance

Displacement

Distance — a scalar quantity

used to measure the total length of a path travelled

A scalar quantity only has a magnitude, and no direction

Displacement — a vector quantity

it gives the location of an object relative to a fixed point or origin

An origin is a starting point — sometimes it is given to you, sometimes you have to establish it on your own.

It has both magnitude and direction

Displacement in one direction will be positive, and negative in the opposite direction

There are two ways to measure how fast an object is moving:

Speed

Velocity

Speed — a scalar quantity

The speed of an object is equal to the distance it has travelled, decided by the time it took to travel that distance

Or more simply, speed=distance/time

s=d/t

This is only valid for objects moving at a constant speed

For objects moving at changing speeds, consider average speed

Average speed=total distance covered/total time taken

SI Unit: m/s (can also be written as ms-1)

Other units include km/h and mph, but m/s is the standard

Velocity — a vector quantity

It measures how quickly the displacement of an object changes

Velocity=∆displacement/time taken

“s” is usually used as the symbol for displacement (confusing, right?)

s=vt or ∆x=vt

Because velocity is a vector, it will be positive in one direction and negative in the opposite

You cannot have a negative speed, but you can have a negative velocity

For this reason, the magnitude of a vector is indicated by vertical lines (|x|).

So speed = |velocity|

What this basically means is that, seeing as speed can’t be negative but velocity can, the speed of an object is just the magnitude of its velocity (the number without the direction)

Average velocity=net ∆ displacement/total time taken

Net ∆ displacement just means your end position compared to your origin

For example, if I start at point A and travel 10m to the right, then turn 180º and move 20m to the left, my net displacement will be +10 - 20 = -10m from point A

Remember, one direction is positive, and the opposite is negative. In this case, I designated moving to the right as positive, and to the left as negative

SI Unit: m/s (can also be written as ms-1)

Examples:

1. A car travels 9km in 15 minutes at constant speed. Find its speed in m/s

First, we need to convert 9km into m and 15 minutes into seconds because our final answer is in m/s

9 x 1000 = 9000

15 x 60 = 900

Then, divide the distance by the time to obtain speed:

9000 / 900 = 10

So the car’s speed is 10 m/s

2. A cyclist travels at 5 m/s for 30s, then turns back, travelling at 3 m/s for 10s. Find her displacement in the original direction of motion from her starting position

What this is asking us to do is find her displacement compared to her origin at the end of her cycle

In order to do this, we need to find the her net ∆ displacement (the distances she travelled both times, and their directions)

s1 = 5 x 30 = 150

s2 = -3 x 10 = -30

Remember, -3 because this is her displacement in the opposite direction

s = 150 - 30 = 120

So her final displacement is 120m from her origin

3. A cyclist spends some of his journey going downhill at 15 m/s and the rest of the time going uphill at 5 m/s. In 1 minute he travels 540m. Find how long he spent going downhill

Let t be the amount of time spent going downhill

60 - t is the amount of time going uphill

Total distance = 15t + 5(60-t) = 540

15t + 300 - 5t = 540

10t + 300 = 540

10t = 240

t = 24

So the cyclist spent 24 seconds going downhill

4. A speed skater averages 11 m/s over the first 5s of a race. Find the average speed required over the next 10s to average 12 m/s overall

Average speed = total distance travelled / total time taken

So the total distance the skater skated, divided by 15s (because the total time is the first 5s plus the remaining 10s) needs to give 12 m/s

Let x represent the average speed of the skater for the last 10s of the race

[(11 x 5) + 10x] / 15 = 12

(55 + 10x) / 15 = 12

55 + 10x = 180

10x = 125

x = 12.5

So the skater needs to skate at a speed of 12.5 m/s in order to average 12 m/s in the race overall

In calculations like these, you will always be making assumptions about your data

For example, if the question asks you how long it takes a cheetah running at 25 m/s to catch a gazelle who’s 150m away, you assume in your calculations that the cheetah travels at a constant 25 m/s, in a straight line towards the gazelle, who is stationary

You need to make sure your assumptions are reasonable, and always be aware of them when concluding your results

Homework: Answer the following questions. Show all your working clearly.

Due date: Friday 16/9/22

We actually got further than this in class and started lesson 2.2 (I already have the notes form as far as we got written down) but to avoid confusion between lessons I’ve decided to save them for the next set of notes from mechanics class.

Edit from the future: Just added the homework Mrs. released, and I’ve realised that it might require some of the notes which I’m saving for next class. That’s not a problem now, because no one is using this blog yet, but I’m just giving a heads-up to whoever reads this in the future, that you might have to take a look at the first half of lesson 2.2’s notes in order to understand some things :)

Remember to drink water and take care of yourselves!

[astresnotes-maths]

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Hi! Welcome to A-Level Mathematics!

Here I will be cataloguing my AS/A-Level Maths classes in order to make them accessible for anyone interested in learning the subject.

This blog is part of an umbrella account documenting all my other classes too (I take psychology, art, and maths). For this reason, asks are only open on my main blog. See the links in the menu on my page to access my other blogs. I will include a link to my main below.

Important: My maths course is split into two classes: pure maths (which focuses on your typical algebra and solving equations) and mechanics (which is more akin to a physics class, with more practical formulas and statistics). As you will find out when you read the tag guide linked below, my tags are a mess (sorry). It will become infinitely more difficult if I distinguish between the two courses like I do between my subjects. This will make it both harder for me to categorise, and harder for anyone to find the resources they need, so overall it’d be stupidly counterproductive. For this reason, I am going to simply put them in separate topics, like I would if the material was being taught in the same class. This is a heads up as to why you might see two ongoing topics for this subject, as opposed to just one like in my other courses. It might also mean that the topics are wildly different between them, and why a particular theme may seem to alternate to every other topic. It’s because I have two classes under the same subject.

One more thing to note is that, other than when labelling lessons with the universal YYYYMMDD, I use the British dating system, dd/mm/yy. So whenever you see a date written in that format, the first digit is the day and the second is the month :)

Some useful links:

Main Blog (Astre’s Notes)

Tag Guide — important!

Ask box

Notes list under the cut

Mathematics

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Topic 1: Quadratics 1/?

Lesson 1.1 — Solving by Factorising Quadratics — 20220906 Lesson 1.2 — Practicing factorising quadratics — 20220909 Lesson 1.3 — Completing the square — 20220913 Lesson 1.4 — Harder quadratic equations and solving through substitution — 20220920

Topic 2: Velocity & Acceleration 1/?

Lesson 2.1 — Displacement & Velocity — 20220907 Lesson 2.2 — Acceleration and displacement-time graphs — 20220914

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