I’m going to write a series of posts about quadratic equations over the next few weeks which will hopefully clear up the confusion about them and lead to you, the reader, becoming an expert on the topic of quadratic equations. This is my plan anyhow…
When you first start studying algebra the equations you will be looking at will no doubt be all linear equations. I’ll write some examples below:


 x + 5 = 8
 3a 10 = 26
 6w – 3 = 7 – 4w

Feel free to try and solve these equations as a little warm up. They get gradually more difficult, but are a lot easier, and less weird, than quadratic equations. One thing they all have in common is that the variable, which is the unknown value that has been represented by a letter in bold, are all to the power of 1. That is that nothing has been squared, cubed, square rooted or anything like that. In quadratic equations the variable could also be squared, but not cubed or any other power. I have put some examples below:
IMPLICIT DIFFERENTIATION AND ITS USE IN DERIVATIVES
What makes quadratic equations special?
The strange thing about quadratic equations is that they can have more than one solution! What?? I know right… How could that possibly be?? Let’s imagine a very simple equation:
x^{2} = 9
How would you solve this equation? Well, we want to find what x is, and we currently know what x^{2} is. Like all equations, we can split this equation into two sides: Left of equal sign and right of the equal sign. I shall refer to these sides as the left and right sides.
The golden rule for all equations is:
“Whatever you do to one side, you must do to the other side of the equation.”
So to change x^{2} to x I just need to square root it.
However, following the above rule, I would have to also square root 9. That’s easy, it’s 3 right? 3^{2} = 9 and so the square root of 9 is 3. Ta da!
Really? Why not? Well… 3^{2} does in fact equal 9, but (3)^{2} also equals 9! So there are two solutions to x^{2} = 9:
x = 3 or x = 3
I know right?? Mind blown!
Both solutions satisfy the equation, so there really are two solutions.
INTEGRATION IN MATHS: THE PADDINGTON BEAR APPROACH
The trick to solving quadratic equations
Let’s expand the equation below:
(x3)(x5) = 0
I’m going to presume you know how to expand double brackets here, but if not you might want to brush up. Below is the equation expanded and simplified:
x^{2} – 8x +15 = 0
You may wonder why I’m having zero on the right hand side of the equation. All I shall say for now is that it is very useful to do so, but read on and you shall see why.
Because all I have done is expand the equation, they are in fact the same equation, but arranged differently. Once we expand out the brackets we see the equation is a quadratic because it has a maximum power of 2.
Now in the first rearrangement we have two algebraic expressions being multiplied by each other and equaling zero. The first expression: x3, and the second expression: x5. At the risk of making this more complicated, let’s say:
a = x – 3 and b = x – 5
So…
ab = 0
The only way this could be true is if a equals zero, b = zero or they both equal zero.
a = 0
or
b = 0
Which means that:
x – 3 = 0
or
x – 5 = 0
In other words…
x = 3
or
x = 5
Two solutions! Mind blown… No wait, I’ve done that bit…
Ok, let’s see if this actually works, or have I just broken a load of Maths rules?? We’re going to put the values for x back into the left side of the equation (x^{2} – 8x +15) to see if it does in fact equal zero:
x = 3: 3^{2} – (8×3) + 15
= 9 – 24 +15 = 0
x = 5: 5^{2} – (8×5) + 15
= 25 – 40 +15 = 0
No way!!
Hang on though… That’s all well and good, but we started with: (x3)(x5) = 0, which made it a lot easier. What if we started with: x^{2} – 8x +15 = 0?
IMPROVING NUMBER CONFIDENCE – A FRACTION AT A TIME
Factorising quadratics
Without going into too much of an explanation, I shall teach you the method of how to factorise, which is the opposite of expanding.
Factorising: x^{2} – 8x +15 → (x3)(x5)
Expanding: (x3)(x5) → x^{2} – 8x +15
We always want our quadratic equation to be in the form of:
ax^{2} + bx + c = 0
In our equation: x^{2} – 8x +15, a = 1, b = 8 and c = 15.
We are looking for two numbers, let’s call them m and n, where:
m + n = b
and
mn = c
In our case:
m + n = 8
and
mn = 15
With a little bit of thinking we see the numbers are: 3 and 5. It doesn’t matter which one is n and which one m. We then put that into the form of
(x+n)(x+m)
In our case: (x3)(x5)
So, there you have it. That’s how to factorise and you already know why to factorise.
Important note… This only works when a = 1. We shall look at other situations in a future post.
One more thing before I sign off… Here is an AWESOME resource which helps you solve any quadratic equation. Check it out. I shall see you next post!
A bit about the author, Paul H:
Paul is a qualified and experienced Physics, Maths, and Science teacher, now working as a fulltime tutor, providing online tuition using a variety of hitech resources to provide engaging and interesting lessons. He covers Physics, Chemistry, Biology, and Science from Prep and Key Stage 3 through to GCSE and IGCSE. He also teaches Physics, Maths, and Chemistry to ALevel across all the major Exam Boards.
You can enquire about tutoring with Paul here