There is a master rule of mathematics if you like. We haven’t directly mentioned it yet but you may have already picked it up. Simply: there must always be balance in the force.
What does balance in the force mean outside of a Star Wars movie? Well, in mathematics there is an important symbol that separates one side of an equation from another (usually the equals sign…). You can change anything in either side of an equation provided you do the same to both sides to maintain the force (of equalness) between both sides. For example:
$$ 5 \color{red}{+5} = 5 \color{red}{+5} $$
Is a very basic example. Whilst, in that situation it’s pretty pointless in algebra it is a master tool that allows us to easily find whatever value we are looking for or rearrange the symbols in order to find the formula we want for example. Another fairly powerful tool is called collecting like terms which is where we can simplify the same letter together when we’re adding or subtracting letters for example:
$$ a + 2a - b + 3b - b = 3a + 3b - 2b = 3a + b $$
Now it’s time to do some proper work. Remember what we learnt about brackets? Well we need to expand and simplify the brackets in equations like this:
$$ 3x (2x + 5) = 2x \color{red}{\times 3x} + 5 \color{red}{\times 3x} = 6x^2 + 15x $$
Remember that x multiplied by x is the same as x^2. Also we now need to recall what we learnt about common factors. We can have common factors that have the same letters in them as well. Using common factors, we can factorise an equation. Factorising is when we put an equation into brackets whilst Expanding is when we take an equation out of brackets. Let’s try factorising:
$$ 8xy + 12y^2 = \color{red}{4y}(2x + 3y)$$
The highest common factor of 8 and 12 is 4. However we have letters, and the common letter is y in both terms. So our common factor is 4y. We can always check if we got a question like this right by factorising the equation out again and seeing if we get what we started with.
We also need to be able to expand double brackets with letters in it:
$$(x + 8)(x - 5) = x \color{red}{\times x} + 8 \color{red}{\times x} - 5 \color{red}{\times 8} - 5 \color{red}{\times x} = x^2 + 8x - 40 - 5x = x^2 + 3x - 40$$
We have discovered a very interesting type of algebraic equation known as a quadratic equation. Basically, they follow the general form:
$$ax^2 + bx + c = 0 $$
where a, b and c are coefficients or a certain number value. A less fancy way to think of quadratics is that they have an x squared term. Quadratics are very interesting numbers especially when you graph them in later tutorials.
We now need to know how to factorise quadratics (put them back into brackets). To do this we need to put the whole equation into 2 brackets. We are able to do this we must get the quadratic into the above form where the other side must equal 0.
We must find a pair of numbers which add up to b and multiply together to make ac (a multiplied by c). Let’s try doing this with a simple quadratic:
$$x^2 + 10x + 30 = 6 $$ $$x^2 + 10x + 30 \color{red}{- 6} = 6 \color{red}{- 6} $$ $$(1 \times x^2) + 10x + 24 = 0$$
In this case we are looking for a pair of numbers that add up to make 10 and multiply up to make 24 (we can tell the value for a is 1). We can look for the potential numbers by looking at the factors for 24:
$$ 24 \times 1, 12 \times 2, 6 \times 4 $$
we have found the pair of numbers we satisfy what we’re looking for; 6 and 4. We can now plug this into separate brackets and since our value for a is 1 we don’t need to worry about anything else:
$$ (x + \color{red}{6})(x + \color{red}{4}) = 0 $$
This is one way to find the value for x in quadratics. Because we are looking for the 2 numbers for x that make both sides of the equation equal. We can do this by making either bracket equal to 0 because anything multiplied by 0 is still 0. So the value for x in this case must be -6 and -4. We can always check by substituting the values we’ve found into the initial equation:
$$ \color{red}{-6}^2 + 10(\color{red}{-6}) + 24 = 0 $$
We can find this is indeed the case, and also the case for negative 4 as well. But… what do we do if a does not equal 1? Well we can use the same method with a slight tweak.
Let’s look at this example:
$$ 6x^2 - 5x - 6 = 0 $$
Firstly, we find the same pair of numbers; they must add up to make -5 and multiply together to make -36. Let’s look at the factors of positive 36 and since we know that the end result is negative we know that one of the pair of numbers must be negative, the other must be positive:
$$ 36 \times 1, 18 \times 2, 12 \times 3, 9 \times 4 $$
Seems like negative 9 and 4 would add together to make negative 5. Now, we must do things differently. We must replace bx with our 2 pairs of numbers and try and factor the equation into 2 identical brackets. This allows us to factorise the equation like we would for single brackets:
$$ 6x^2 - 5x - 6 = 6x^2 \color{red}{-9x} \color{red}{+4x} - 6 = 0 $$ $$ 3x (2x - 3) + 2 (2x - 3) = 0 $$
We now make everything outside of the brackets one bracket and the 2 identical brackets form another single bracket:
$$ (3x + 2)(2x - 3) = 0 $$
The only problem with this method is it may not work if you put the pair of numbers into the equation in the wrong order, so you must try both orders for it to work properly.
Before we say farewell to learning about nasty brackets, we need to learn how to expand (not factorise!) triple brackets. The simplest method to do this is to expand 2 of the brackets to give us a large bracket and a smaller bracket and then expand again to leave us with the entire equation. It means we don’t really have to learn anything new:
$$(x + 2)(x + 3)(x - 1) = (x \color{red}{\times x} + 2 \color{red}{\times x} + 3 \color{red}{\times 2} + 3 \color{red}{\times x})(x - 1) = (x^2 + 5x + 6)(x - 1) $$
We must multiply everything in one bracket by everything in the other bracket.
$$ = (x^2 \color{red}{\times x} + 5x \color{red}{\times x} + 6 \color{red}{\times x}) + (x^2 \color{red}{\times -1} + 5x \color{red}{\times -1} + 6 \color{red}{\times -1}) $$
$$ = x^3 + 5x^2 + 6x - x^2 - 5x - 6 = x^3 + 4x^2 + x - 6 $$
Remember the powerful tool(s) we started off with learning? We now need to use all of this and the other techniques we’ve covered so far in learning mathematics to solve fractions with algebra (or more commonly known as algebraic fractions). Let’s simplify:
$$ \dfrac{3x + 1}{x + 2} - \dfrac{x - 2}{x - 1} $$
We start by finding a common denominator; looks like that will be x + 2 multiplied by x - 1. This allows us to express the 2 fractions as 1 fraction:
$$ \dfrac{(3x + 1)\color{red}{(x - 1)}}{(x + 2)\color{red}{(x - 1)}} - \dfrac{(x - 2)\color{red}{(x + 2)}}{(x - 1)\color{red}{(x + 2)}} = \dfrac{(3x + 1)(x - 1) - (x - 2)(x + 2)}{(x - 1)(x + 2)} $$
Now we need to expand the brackets and collect the like terms:
$$ = \dfrac{(3x \color{red}{\times x} + 1 \color{red}{\times x} + 1 \color{red}{\times -1} + 3x \color{red}{\times -1}) - (\color{red}{x^2} - \color{red}{2^2})}{x \color{red}{\times x} - 1 \color{red}{\times x} + x \color{red}{\times 2} + - 1 \color{red}{\times 2}} = \dfrac{3x^2 + x - 1 - 3x - x^2 + 4}{x^2 - x + 2x - 2} $$ $$ = \dfrac{2x^2 - 2x + 3}{x^2 - x - 2} $$
And now we’ve simplified it into 1 fraction as much as possible.
Another cool trick we can do with quadratics is to complete the square with them. Completing the square is very important when graphing quadratics but we’ll look more on that in other lesson. For now, completing the square is an alternative method for solving quadratics. The general form for “completing the square” is:
$$ ax^2 + bx + c \Rightarrow a(x + d)^2 + e $$
It will come in handy to learn this formula for solving them:
$$ d = \dfrac{b}{2a} $$ $$ e = (c - \dfrac{-b^2}{4a}) $$
This is probably the longest equation you have to learn in all of your GCSEs. An equation might ask for you to put a normal quadratic equation into the correct form. This is super easy, all you need to do is substitute the correct values into the equation.
Finally, we need to learn about using algebra to make proofs. Since symbols and letters can represent any number (variables) we can use them to make a general rule about something in maths. This is what “professional” mathematicians are dedicated to doing; finding new rules in maths through proofs. Firstly we need to know how to show certain things.
Firstly, an even number will always be 2n whilst an odd number will always be 2n + 1 or 2n - 1. Consecutive numbers are numbers next to each other on the number line so you need to add or subtract 1. For example these would be consecutive odd numbers:
$$ 2n + 1, 2n + 3, 2n + 5 $$
These would be consecutive numbers:
$$ n + 1, n + 2, n + 3 $$
remember that n is the same number.
An example question might be “prove the sum of three consecutive even integers is divisible by 6”. The hardest part is to write this as algebra and then simplify:
$$ 2n, 2n + 2, 2n + 4 $$
would be our 3 consecutive even integers.
$$ 2n + 2n + 2 + 2n + 4 = 6n + 6 = 6(n + 1) $$
since we were able to factorise the entire number by 6 it means that the number must be divisible by 6.
More coming soon