1. GCSE Revision 101
Maths
Quadratics
© Daniel Holloway

2. The Basics
Factorisation of quadratics involves putting quadratic expressions back into brackets (if possible). To do this, the expression can be in the form: x2 + ax + b Where a and b are integers

3. Simple Factorisation
We can tell how to start off the brackets by looking at the signs of the expression:
x2 + ax + b = (x + ?)(x + ?) because everything is positive
x2 - ax + b = (x - ?)(x - ?) because negative x negative = positive

4. Simple Factorisation
When the second sign is a plus, both brackets are the same as the first sign x2 + ax + b = (x + ?)(x + ?) x2 - ax + b = (x - ?)(x - ?)
When the second sign is a minus, both brackets are different signs x2 + ax - b = (x + ?)(x - ?) x2 - ax - b = (x + ?)(x - ?)

5. Simple Factorisation
When a quadratic expression is in the form x2 + ax + b, we must make sure that the numbers in the brackets satisfy these rules:
the b must be the product of the two numbers
the a must be the sum of the numbers when the two brackets are the same sign
the a must be the difference between the numbers when the two brackets are different signs

6. Simple Factorisation
Take this example: x2 + 5x + 6 Both signs are positive, so we can begin with (x + ?)(x + ?) The numbers need to multiply to make 6 and add together to make 5 They must be 2 and 3 Since both signs are positive, the order does not matter: (x + 2)(x + 3)

7. Simple Factorisation
Take this example: y2 – y – 90 Both signs are negative, so we can begin with (x + ?)(x - ?) The numbers need to multiply to make 90 and add together to make 1 They must be 9 and 10 Because we need to do to get -1: (y + 9)(y - 10)

8. Test Yourself
Factorise the following… a) x2 + 5x + 4 b) x2 + 7x + 10 c) y2 + 14y + 24 d) z2 + 9z + 18 e) a2 - 6a f) x2 - x - 12 g) k2 + 4k + 3 h) s2 - 18c + 32
Factorise x2 – 24x + 144

9. Difference of Two Squares
As a rule, (a + b)(a – b) multiplies out to get a2 - b2 This type of quadratic expression where there are only two parts to it, both of which are perfect squares, separated by a minus sign, is called the difference of two squares. Examples include: a2 - 9 a a2 – 100 To factorise them, we simply root them and put them in the format shown above. e.g. a2 - 9 factorises to (a + 3)(a - 3)

10. Solving Quadratics
We can solve quadratic equations in the form x2 + ax + b by first factorising it, and finding the two answers using the double brackets. e.g. Solve x2 + 6x + 5 = 0 This factorises into (x + 5)(x + 1) If x + 5 = 0, then x = -5 If x + 1 = 0, then x = -1

11. Factorising ax2 + bx + c
We use the same method of factorisation to factorise quadratic expressions in the form ax2 + bx + c but we have to take into account the a which will be in a bracket The following slide explains how it’s done

12. Factorising ax2 + bx + c
Factorise 3x2 + 8x + 4. The same bracket rules apply, so both brackets are positive. As 3 only has the factors 3 and 1: (3x + ?)(x + ?) Next we notice the factors of 4 are 1x4 and 2x2. The only pair here which combine with 3x1 to make 8 is 2x2: (3 x 2) + (1 x 2) = 8 So the factorised expression is: (3x + 2)(x + 2)

13. Solving ax2 + bx + c
There are THREE ways we can solve quadratic equations in the form ax2 + bx + c as shown on the next few slides.

14. Via Factorisation Solve 12x2 - 28x = -15.
Rearrange the equation to equal zero 12x2 - 28x + 15 = 0
Factorise the expression (2x - 3)(6x - 5)
Find the two answers as before x = 1.5 and x = 25/6

15. Via Quadratic Formula -b ± √b2 - 4ac x = 2a
Many quadratic equations cannot be factorised because they are too complicated or do not have convenient integer factors between them. This is when the quadratic formula is used:
x =
-b ± √b2 - 4ac 2a

16. Via Quadratic Formula -b ± √b2 - 4ac x = 2a Solve 5x2 - 11x – 4 = 0.
Remember the quadratic formula:
Substitute the values from the equation as a, b and c
Solve the equation using the values, shown on the following slide
x =
-b ± √b2 - 4ac 2a

17. Via Quadratic Formula x = -b ± √b2 - 4ac 2a x = 11 ± √121 – 4(5)(-4)
Solve 5x2 - 11x – 4 = 0. We know that a = 5, b = -11 and c = -4
x =
-b ± √b2 - 4ac 2a
x =
11 ± √121 – 4(5)(-4) 10
x = 2.52 =
11 ± √ 201 10
x = 0.32

18. Via Completing the Square
Remember that (x + a)2 = x2 + 2ax + a2 This means that x2 + 2ax = (x + a)2 - a2 This is the principle behind completing the square, a rule used to solve quadratic equations There are two steps to completing the square, as shown on the following slide

19. Via Completing the Square
Step 1: Rewrite x2 + 4x – 7 in the form (x + a)2 – b Note that x2 + 4x = (x + 2)2 – 4 Therefore x2 + 4x – 7 = (x + 2)2 – 4 – 7 = (x + 2)2 – 11

20. Via Completing the Square
Step 2: Hence solve the equation x2 + 4x – 7 = 0 We have: x2 + 4x –7 = (x + 2)2 – 11 When x2 + 4x – 7 = 0, we know: (x + 2)2 – 11 = 0 (x + 2)2 = 11 Now square root both sides: x + 2 = ±√11 x = -2 ±√11 So, x = 1.32 and x = 5.32 (to 2 d.p.)