1. Algebra

2. EXPANDING - Does 6 × (3 + 5) = 6 × 3 + 6 × 5 ? YES 6 × 8 = 18 + 30
48 = 48
- The removal of the brackets is known as the distributive law and can also be applied to algebraic exressions
- When expanding, simply multiply each term inside the bracket by the term directly in front
e.g. Expand
a) 6(x + y)
= 6 × x
+ 6 × y
b) -4(x – y)
= -4 × x
- -4 × y
= 6x
+ 6y
= -4x
+ 4y
c) -4(x – 6)
= -4 × x
- -4 × 6
d) 7(3x – 2)
= 7 × 3x
- 7 × 2
= -4x
+ 24
= 21x
- 14
e) x(2x + 3y)
= x × 2x
+ x × 3y
f) -3x(2x – 5)
= -3x × 2x
- -3x × 5
= 2x2
+ 3xy
= -6x2
+ 15x
Don’t forget to watch for sign changes!

3. - If there is more than one set of brackets, expand them all then collect any like terms.
e.g. Expand and simplify
a) 2(4x + y) + 8(3x – 2y)
= 2 × 4x
+ 2 × y
+ 8 × 3x
- 8 × 2y
= 8x
+ 2y
+ 24x
- 16y
= 32x
- 14y
b) -3(2a – 3b) – 4(5a + b)
= -3 × 2a
- -3 × 3b
- 4 × 5a
+ -4 × 1b
= -6a
+ 9b
- 20a
- 4b
= -26a
+ 5b

4. EXPANDING TWO BRACKETS
- To expand two brackets, we must multiply each term in one bracket by each in the second
Remember integer laws when multiplying
e.g. Expand and simplify
a) (x + 5)(x + 2)
= x2
+ 2x
+ 5x
+ 10
b) (x - 3)(x + 4)
= x2
+ 4x
- 3x
- 12
= x2 + 7x + 10
= x2 + 1x – 12
To simplify, combine like terms
c) (x - 1)(x - 3)
= x2
- 3x
- 1x
+ 3
c) (2x + 1)(3x - 4)
= 6x2
- 8x
+ 3x
- 4
= x2 – 4x + 3
= 6x2 – 5x – 4

5. PERFECT SQUARES - When both brackets are exactly the same
To simplify, combine like terms
e.g. Expand and simplify
Watch sign change when multiplying
a) (x + 8)2
= (x + 8)(x + 8)
b) (x - 4)2
= (x – 4)(x – 4)
= x2
+ 8x
+ 8x
+ 64
= x2
- 4x
- 4x
+ 16
= x2 + 16x + 64
= x2 - 8x + 16
Write out brackets twice BEFORE expanding
c) (3x - 2)2
= (3x – 2)(3x – 2)
= 9x2
- 6x
- 6x
+ 4
= 9x2 – 12 x + 4

6. DIFFERENCE OF TWO SQUARES
- When both brackets are the same except for signs (i.e. – and +)
e.g. Expand and simplify
a) (x – 3)(x + 3)
= x2
+ 3x
- 3x
- 9
b) (x – 6)(x + 6)
= x2
+ 6x
- 6x
- 36
= x2 – 9
= x2 – 36
Like terms cancel each other out
c) (2x – 5)(2x + 5)
= 4x2
+ 10x
- 10x
- 25
= 4x2 – 25

7. MORE THAN TWO BRACKETS - Expand two brackets at a time
e.g. Expand and simplify
a) (x + 2)(x - 3)(x + 5)
= (x2
- 3x
+ 2x
- 6)
(x + 5)
= (x2 – x – 6)
(x + 5)
= x3
+ 5x2
- x2
- 5x
- 6x
- 30
= x3 + 4x2 – 11x - 30
b) (x - 2)3
= (x – 2)(x – 2)(x – 2)
= (x2
- 2x
- 2x
+ 4)
(x - 2)
= (x2 – 4x + 4)
(x – 2)
= x3
- 2x2
- 4x2
+ 8x
+ 4x
- 8
= x3 - 6x2 + 12x - 8

8. FACTORISING - Factorising is the reverse of expanding - To factorise:
1) Look for a common factor to put outside the brackets
2) Inside brackets place numbers/letters needed to make up original terms
You should always check your answer by expanding it
e.g. Factorise
a) 2x + 2y
= 2( ) x
+ y
b) 2a + 4b – 6c
= 2( ) a
+ 2b
- 3c
- Always look for the highest common factor
e.g. Factorise
a) 6x - 15
= 3( ) 2x
- 5
b) 30a + 20
= 10( ) 3a
+ 2
- Sometimes a ‘1’ will need to be left in the brackets
e.g. Factorise
a) 6x + 3
= 3( ) 2x
+ 1
b) 20b - 10
= 10( ) 2b
- 1

9. a) 5a2 – 7a5 = a2( ) 5 - 7a3 b) 4b2 + 6b3 = 2b2( ) 2 + 3b
- Letters can also be common factors
e.g. Factorise
a) cd - ce
= c( ) d
- e
b) xyz + 2xy – 3yz
= y( ) xz
+ 2x
- 3z
c) 4ad – 8a
= 4 ( ) a d
- 2
- Powers greater than 1 can also be common factors
e.g. Factorise
a) 5a2 – 7a5
= a2( ) 5
- 7a3
b) 4b2 + 6b3
= 2b2( ) 2
+ 3b

10. Factorising by grouping
- When two groups within an expression have their own common factor
e.g. Factorise
As both 2c and 3d are being multiplied by (a + b) we place them in separate brackets
2ac + 2bc + 3ad + 3bd
= 2c( ) a
+ b
+ 3d( ) a
+ b
= (2c + 3d)(a + b)

11. FACTORISING QUADRATICS
The general equation for a quadratic is ax2 + bx + c
When a = 1
- You need to find two numbers that multiply to give c and add to give b
e.g. Factorise
To check answer, expand and see if you end up with the original question!
a) x2 + 11x + 24
= (x + 3)(x + 8)
1, 24
List pairs of numbers that multiply to give 24 (c)
Check which pair adds to give 11 (b)
Place numbers into brackets with x
2, 12
3, 8
4, 6
b) x2 + 7x + 6
= (x + 1)(x + 6)
1, 6
List pairs of numbers that multiply to give 6 (c)
Check which pair adds to give 7 (b)
Place numbers into brackets with x
2, 3

12. - Expressions can also contain negatives e.g. Factorise
a) x2 + x – 12
= (x - 3)(x + 4)
b) x2 – 6x – 16
= (x + 2)(x - 8) -
1, 12
1, 16 -
As the end number (c) is , one of the pair must negative.
As the end number (c) is , one of the pair must negative. -
2, 6
2, 8 - -
3, 4
4, 4 -
Check which pair now adds to give b
Make the biggest number of the pair the same sign as b
Check which pair now adds to give b
Make the biggest number of the pair the same sign as b
c) x2 – 9x + 20
= (x - 4)(x - 5) -
1, 20 -
As the end number (c) is , but b is – 9, both numbers must be negative -
2, 10 - -
4, 5 -
Check which pair now adds to give b

13. SPECIAL CASES 1. Perfect Squares e.g. Factorise a) x2 + 10x + 25
b) 4x2 - 16x + 16
= 4(x2 – 4x + 4)
1, 25
= (x + 5)2
= 4(x – 2)(x – 2) -
1, 4 -
5, 5
= 4(x – 2)2 -
2, 2 -
2. Difference of Two Squares
e.g. Factorise
a) x2
+ 0x
- 25
= (x - 5)(x + 5)
b) 9x2 – 100
= (3x )(3x )
-5, 5
c) 2x2 – 72
= 2(x2 – 36)
Add in a zero x term and factorise
= 2(x )(x )
OR: factorise by using A2 – B2 = (A – B)(A + B)

14. TWO STAGE FACTORISING When a ≠ 1 1. Common factor
- Always try to look for a common factor first.
e.g. Factorise
a) 2x2 + 12x + 16
= 2( )
x2 + 6x + 8
b) 3x2 – 6x – 9
= 3( )
x2 – 2x – 3
1, 8
= 2(x + 2)(x + 4)
1, 3 -
= 3(x + 1)(x – 3)
2, 4
c) 3x2 + 24x
= 3x( )
x + 8
d) 4x2 – 36
= 4( )
x2 – 9
= 4(x )(x )

15. 2. No common factor (HARD)
- Use the following technique
e.g. Factorise
a) 3x2 – 10x – 8
= 3x2 + 2x – 12x – 8
= x( )
3x + 2
-4( )
3x + 2
= (x – 4)(3x + 2)
1, 24 -
Multiply constants of first and last terms
Find two nos. that multiply to -24 but add to -10
Replace 10x with the two new terms
Factorise 2 terms at a time.
2, 12 -
3, 8 -
4, 6 -
3 × - 8 = -24
Write in two brackets
b) 2x2 + 7x + 3
= 2x2 + x + 6x + 3
= x( )
2x + 1
+ 3( )
2x + 1
= (x + 3)(2x + 1)
2 × 3 = 6
1, 6
2, 3

16. RATIONAL EXPRESSIONS
- Look to factorise the numerator and/or denominator and then remove the common factor by dividing through.
Cancel out common factors e.g. x + 3 = 1
x + 3
e.g. Simplify
a) x2 + 7x + 12
x + 3
= (x + 3)(x + 4)
x + 3
= (x + 4)
As the numerator cancels out, you must leave a ‘1’ on top to signal this
b) x
49 – x2
= x
(7 – x)(7 + x) =
(7 – x)
- Beware of disguised sign changes i.e. (x – 7) cancels with (7 – x) leaving -1
e.g. Simplify x2 – 5x + 6
8 – 2x – x2
= (x – 3)(x – 2)
(x + 4)(2 – x)
= -1
(x – 3)
(x + 4)
Better written as: -(x – 3)
(x + 4)

17. POWER RULES a) p10 × p2 = p(10 + 2) b) a3 × a2 × a = a(3 + 2 + 1) 1
1. Multiplication
- Does x2 × x3 = x × x × x × x × x ?
YES
- Therefore x2 × x3 = x5
- How do you get = 5 ? +
- When multiplying index (power) expressions with the same letter, ADD the powers.
No number = 1 i.e. p = 1p1
e.g. Simplify
a) p10 × p2
= p(10 + 2)
b) a3 × a2 × a
= a( ) 1
= p12
= a6
- Remember to multiply any numbers in front of the variables first
e.g. Simplify
a) 2x3 × 3x4
= 2 × 3
x(3 + 4)
b) 2a2 × 3a × 5a4 1
= 6 x7
= 2 × 3 × 5
a( )
= 30 a7

18. 2. Division
- Does 6 = 1 ? 6
YES
- Therefore x = 1 x
- Does x5 = x × x × x × x × x ?
x x × x × x
YES
= x × x × 1 × 1 × 1
- Therefore x5 = x2 x3
- How do you get = 2 ? -
- When dividing index (power) expressions with the same letter, SUBTRACT the powers.
e.g. Simplify
a) p5 ÷ p
= p(5 - 1)
b) x7 x4
= x(7 - 4) 1
= p4
= x3
- Remember to divide any numbers in front of the variables first
e.g. Simplify
a) 12x5 ÷ 6x4
= 12 ÷ 6
x(5 - 4)
b) a7
15a2
÷ 5
= 1 5
a(7 - 2)
= 2 x
= 1 5 a5
or a5 5
If the power remaining is 1, it can be left out of the answer

19. a) (c4)6 = c(4 × 6) b) (a3)3 = a(3 × 3) = c24 = a9 a) (3d2)3 = 33
3. Powers of powers
- Does (x2)3 = x2 × x2 × x2 ?
YES
- Does x2 × x2 × x2 = x6 ?
YES
- Therefore (x2)3 = x6
- How do you get = 6 ? ×
- When taking a power of an index expression to a power, MULTIPLY the powers
e.g. Simplify
a) (c4)6
= c(4 × 6)
b) (a3)3
= a(3 × 3)
= c24
= a9
- If there is a number in front, it must be raised to the power, not multiplied
e.g. Simplify
a) (3d2)3
= 33
× d(2 × 3)
b) (2a3)4
= 24
× a(3 × 4)
= 27 d6
= 16
a12
- If there is more than one term in the brackets, raise all to the power
e.g. Simplify
b) (4b2c5)2
= 42
b(2 × 2)
c(5 × 2)
a) (x3y z4)3
= x(3 × 3)
y(1 × 3)
z(4 × 3) 1
= x9 y3
z12
= 16 b4
c10

20. NEGATIVE POWERS In General: x-n = 1 xn
e.g. Write with a positive index
a) x-4 = 1 x4
b) 3x-1 = 3 x
e.g. Evaluate
a) (3)-3 = 1 33
b) (0.5)-2 = 4
Therefore = 2
2 2 1
= 1 27
= 4
e.g. Write with a positive index and then evaluate 5
5 2 2
= 25 4

21. NEGATIVE POWER LAWS FRACTIONAL INDICES In General:
- Same laws learned previously apply to negative powers
e.g. Simplify the following and write with a positive index
a) 3x-2 × 4x-3 =
3 × 4 x
b) (xy2)-4 =
x1 × -4
y2 × -4
= 12x-5
= x-4y-8
= 12 x5 =
x4y8
FRACTIONAL INDICES
- Express roots of numbers
e.g. The power of a ½ is the same as square root.
In General:
Index form
Surd form

22. COMBINING POWERS & ROOTS
e.g. Write x1/4 in surd form
e.g. Evaluate 641/3
e.g. Write in index form
= x1/5
e.g. Simplify (64x6)1/2
= √64
x6 x 1/2
= 8x3
COMBINING POWERS & ROOTS
In General:
e.g. Evaluate 642/3
e.g. Write in index form

23. FRACTIONAL & NEGATIVE POWERS
e.g. Write x-2/3 in surd form
e.g. Evaluate

24. ALGEBRAIC FRACTIONS a) 3ab 9bd = 1a 3d b) 12x2y3z 27xz2 = 4xy3 9z = a
1. Simplifying
- Always take out any common factors.
a) 3ab
9bd
= 1a 3d
b) 12x2y3z
27xz2
= 4xy3 9z
= a 3d
2. Multiplying Fractions
- Multiply top and bottom terms separately then simplify.
e.g. Simplify:
a) 4a × 3e
5e 2a
= 4a × 3e
b) 5x2y × 6y2
4x x3y
= 5x2y × 6y2
5e × 2a
4x × 5 x3y
= 30x2y3
20x4y
= 12ae
10ae
= 3y2
2x2
= 6 5

25. a) 2a ÷ a2 5 3 = 2a 5 × 3 a2 b) 6r ÷ 3r2 5 = 6r 5 × 1 3r2 = 2a × 3
2. Dividing Fractions
- Multiply the first fraction by the reciprocal of the second, then simplify
Note: b is the reciprocal of 2 b
e.g. Simplify:
a) 2a ÷ a2
= 2a 5 × 3 a2
b) 6r ÷ 3r2 5
= 6r 5 × 1
3r2
= 2a × 3
= 6r × 1
5 × a2
5 × 3r2
= 6a
5a2
= 6r
15r2
= 6 5a
= 2 5r

26. a) 3x + 3x 10 10 = 3x + 3x 10 b) 6a - b 5 5 = 6a - b 5 = 6x 10 ÷ 2
4. Adding/Subtracting Fractions
a) With the same denominator:
- Add/subtract the numerators and leave the denominator unchanged. Simplify if possible.
e.g. Simplify:
a) 3x + 3x
= 3x + 3x 10
b) 6a - b
5 5
= 6a - b 5
= x 10
÷ 2
= x 5
b) With different denominators:
- Multiply denominators to find a common term.
- Cross multiply to find equivalent numerators.
- Add/subtract fractions then simplify.
e.g. Simplify:
b) 3 – 2
a b
b×3 =
a×b
- a×2
a) a + 2a =
2×3
3×a
+ 2×2a
= 3a + 4a 6
= 3b – 2a ab
= 7a 6

27. Always look for common factors to simplify answer!
Harder examples:
a) 3x
4x x
6x(3x + 2) =
4x×6x
+ 4x×2
b) –
x – x + 6 =
(x – 1)(x + 6)
5(x + 6)
- 2(x – 1)
= 18x2 + 12x + 8x
24x2
= 5x x + 2
x2 + 5x - 6
= 18x2 + 20x
24x2
= 3x
x2 + 5x - 6
= 2x(9x + 10)
2x×12x
= 9x + 10
12x
Always look for common factors to simplify answer!

28. LOGARITHMS 1. Formal definition: If bp = q then logb(q) = p
Where b = base, p = logarithm and q = number
For simple questions use guess and check.
e.g. a) Write 25 = 32 in log form
log2(32) = 5
b) Write log5(125) = 3 in index form
53 = 125
As less than 1, power is negative.
e.g. Calculate
a) log4(1024)
4x = 125
b) log10(0.01)
10x = 0.01
x = 5
x = -2
e.g. Solve
a) log6(x) = 3
63 = x
b) logx(243) = 5
x5 = 243
x = 216
x = 5√243
x = 3

29. 2. Properties of Logarithms:
- log (ab) = log(a) + log (b)
- log (a/b) = log(a) – log (b)
- log (an) = nlog(a)
e.g. Simplify
a) log(4) + log(6)
= log(4 × 6)
b) log(27) – log(9)
= log(27 ÷ 9)
= log(24)
= log(3)
c) Write 4log(2) as the log of a single number
= log(24)
= log(16)
Remember your BEDMAS!
e.g. Simplify
a) 3log(2) + log(4) – log(8)
log(27)
log(3)
= log(33)
log(3)
= log(23) + log(4) – log(8)
= 3log(3)
log(3)
= log(8) + log(4) – log(8)
= log(32) – log(8)
= 3
= log(4)

30. 3. Index Equations: Use the properties of logs to move the unknown
- When x is the ‘power’ and it isn’t a simple whole number answer
- Solved by taking the logs of both sides
Simply use log button to find answer
e.g. Solve
a) 3x = 21
b) 2x = 30
log3x = log21
log2x = log30
xlog3 = log21
xlog2 = log30
x = log21
log3
x = log30
log2
x = 2.77 (2 d.p.)
x = 4.91 (2 d.p.)
c) 134x - 5 = 6
d) Evaluate log4(128)
log134x - 5 = log6
4x = 128
(4x – 5)log13 = log6
log4x = log128
4x – 5 = log6
log13
xlog4 = log128
x = log128
log4
4x – 5 = 0.699
4x = 5.699
x = 3.5
x = 1.42 (2 d.p.)
Make sure to use full, not rounded answers throughout question

31. SOLVING LINEAR EQUATIONS
- Remember that addition/subtraction undo each other as do multiplication/division
e.g. Solve
a) 3x – 1 = 3 – x
b) 3(x – 1) = 5x + 17 +x +x
3x – 3 = 5x + 17
4x – 1 = 3
-5x
-5x +1 +1
-2x – 3 = 17
4x = 4 +3 +3 ÷4 ÷4
-2x = 20
x = 1
÷-2
÷-2
x = -10
- Not every linear equation has a unique solution
1) Some have an infinite number of solutions (Solution set is R, the set of real numbers)
e.g. Solve 6x + 5 = 2(3x + 5) – 5
6x + 5 = 6x + 10 – 5
As the answer is considered a ‘true statement’, x can be any value possible
6x + 5 = 6x + 5
-6x
-6x
0 + 5 = 5 -5 -5
0 = 0

32. -2x -2x +3 +3 a) x = 9 4 2 b) 3x - 1 = x + 3 5 2 2x = 36 2(3x - 1)
2) Some have no solutions (Solution set is an empty set)
e.g. Solve 2x – 3 = 2x + 7
As the answer is NOT considered a ‘true statement’, there are NO possible values for x
-2x
-2x
0 – 3 = 7 +3 +3
0 = 10
Equations in Fraction Form
- If equations are fraction only, cross multiply, then solve
e.g. Solve
a) x = 9
4 2
b) 3x - 1 = x + 3 2x
= 36
2(3x - 1)
= 5(x + 3) ÷2 ÷2
x = 18 6x
- 2
= 5x
+ 15
-5x
-5x
x - 2 = 15 +2 +2
x = 17

33. e.g. Solve 4x - 2x = 10 5 3 ×15 ×15 ×15 60x 5 - 30x 3 = 150 12x - 10x
- For two or more fractions, find a common denominator, multiply it by each term, then solve
e.g. Solve 4x x = 10
×15
×15
×15
5 × 3 = 15
60x 5
- 30x 3
= 150
Simplify terms by dividing numerator by denominator
12x
- 10x
= 150
2x = 150 ÷2 ÷2
x = 75
e.g. Solve 5x (x + 1) = 2x
×24
×24
×24
120x 6
- (24x + 24) 4
= 48x
20x
- 6x – 6
= 48x
14x – 6 = 48x
-48x
-48x
-34x – 6 = 0
+ 6
+ 6
-34x = 6
÷ -34
÷ -34
x = -6 34

34. SOLVING INEQUATIONS a) 3x + 8 > 24 b) -2x - 5 ≤ 13 -8 -8 +5 +5
- Inequations contain one of four inequality signs: < > ≤ ≥
- To solve follow the same rules as when solving equations
- Except: Reverse the direction of the sign when dividing by a negative
e.g. Solve
a) 3x + 8 > 24
b) -2x - 5 ≤ 13 -8 -8 +5 +5
3x > 16
-2x ≤ 18
As answer not a whole number, leave as a fraction ÷3 ÷3
÷-2
÷-2
Sign reverses as dividing by a negative
x > 16 3
x ≥ -9
c) 4x – 3(x + 6) ≥ 5x – 22
4x – 3x – 18 ≥ 5x – 22
x – 18 ≥ 5x – 22
+18
+18
x ≥ 5x – 4
-5x
-5x
Sign reverses as dividing by a negative
-4x ≥ – 4
÷-4
÷-4
x ≤ 1

35. WRITING AND SOLVING x + 5 + x + x + 5 + x = 58 4x + 10 = 58 -10 -10
e.g. Write an equation for the following information and solve
a) A rectangular pool has a length 5m longer than its width. The perimeter of the pool is 58m. Find its width
x + 5
x x + x x = 58
4x + 10 = 58
Draw a diagram
-10
-10 x x
4x = 48
Let x = width ÷4 ÷4
x + 5
x = 12
Therefore width is 12 m
b) I think of a number and multiply it by 7. The result is the same as if I multiply this number by 4 and add 15. What is this number?
Let n = a number 7 n
= n 4
+ 15
-4n
-4n
3n = 15 ÷3 ÷3
n = 5
Therefore the number is 5

36. REARRANGING FORMULA +2 +2 y + 2 = 6x ÷6 ÷6 y + 2 = x 6 ÷I ÷I R = V I
- Involves rearranging the formula in order to isolate the new ‘subject’
- Same rules as for solving are used
Remember: When rearranging or changing the subject you are NOT finding a numerical answer
1. Linear
a) Make x the subject of y = 6x - 2 +2 +2
y + 2 = 6x ÷6 ÷6
All terms on the left must be divided by 6
y + 2 = x 6
b) Make R the subject of IR = V
Treat letters the same as numbers! ÷I ÷I
R = V I

37. 3y(2x – 1) = 5 y(x + 3) = 5x – 2 6xy – 3y = 5 xy + 3y = 5x – 2
2. Rational Expressions
- Steps: Cross multiply
All x terms to the left, others to the right
Factorise, then divide
e.g. Make x the subject of:
3y(2x – 1) = 5
y(x + 3) = 5x – 2
6xy – 3y = 5
xy + 3y = 5x – 2
6xy = 5 + 3y
xy – 5x = – 2 – 3y
x(y – 5) = – 2 – 3y
3. Square Roots
- Squaring undoes square roots
e.g. Make x the subject of:
y2 = 3x – 1
y2 + 1 = 3x
(3y + 4)2 = 2x + 5
9y2 + 24y + 16 = 2x + 5
9y2 + 24y + 11 = 2x
Make sure the square root is on it’s own

38. 4. Squares
a) Perfect Squares
- If there’s a single perfect square term in x take the square root
e.g. Make x the subject of:
Don’t forget the +/- sign!
b) Non Perfect Squares (harder!)
- If there are single terms of x we must ‘complete the square’
e.g. Write x2 + 10x – 3 in perfect square form:
- Steps: -Take half the co-efficient of x
10 ÷ 2 = 5
-Square it
52 = 25
-Add square to the expression and factorise perfect square
= (x2 + 10x + 25) – 3
= (x + 5)2 – 3
= (x + 5)2 – 3 – 25
-Subtract square to keep original expression
= (x + 5)2 – 28

39. e.g. Make x the subject of y = x2 – 8x + 5
First turn into a perfect square
-8 ÷ 2 = -4
(-4)2 = 16
y = (x2 – 8x + 16) + 5
Then rearrange equation as before
y = (x – 4)2 + 5
y = (x – 4)2 + 5 – 16
y = (x – 4)2 – 11
y + 11 = (x – 4)2

40. SOLVING QUADRATICS To solve use the following steps:
1. Move all of the terms to one side, leaving zero on the other
2. Factorise the equation
3. Set each factor to zero and solve.
e.g. Solve
a) (x + 7)(x – 2) = 0
b) x2 – 5x – 6 = 0
1, 6 -
2, 3 -
x + 7 = 0
x – 2 = 0
(x + 1)(x – 6) = 0 -7 -7 +2 +2
x + 1 = 0
x – 6 = 0
x = - 7
x = 2 -1 -1 +6 +6
x = -1
x = 6
c) 3x2 – 21x + 36 = 0 -
1, 12 -
d) x2 = 4x + 5 -
2, 6 -
-4x -5
-4x -5
1, 5 -
3(x2 – 7x + 12) = 0 -
3, 4 -
x2 – 4x – 5 = 0
3(x – 3)(x – 4) = 0
(x + 1)(x – 5) = 0
x – 3 = 0
x – 4 = 0 +3 +3 +4 +4
x + 1 = 0
x – 5 = 0
x = 3
x = 4 -1 -1 +5 +5
x = -1
x = 5

41. e) 2x2 = 5x
f) x2 = 49
x2 – 49 = 0
2x2 – 5x = 0
(x )(x ) = 0
x( ) = 0
2x – 5
x - 7 = 0
x + 7 = 0
x = 0
2x – 5 = 0 +7 +7
- 7
- 7
+ 5
+ 5
x = 7
x = -7
2x = 5
÷ 2
÷ 2
x = 5/2

42. QUADRATIC EQUATIONS
- Involves writing an equation from the information then solving
e.g. The product of two consecutive numbers is 20. What are they?
If x = a number, then the next consecutive number is
x + 1
x + 5
x(x + 1) = 20 -
1, 20
x2 + x = 20 -
2, 10
-20
-20
A = 150 m2 x -
4, 5
x2 + x – 20 = 0
(x – 4)(x + 5) = 0
x – 4 = 0
x + 5 = 0 +4 +4 -5 -5
x(x + 5) = 150
x = 4
x = -5
x2 + 5x = 150
The numbers are 4, 5 and -5, -4
-150
-150
x2 + 5x – 150 = 0
(x – 10)(x + 15) = 0
e.g. A paddock of area 150 m2 has a length 5 m longer than its width. Find the dimensions of this paddock.
x – 10 = 0
x + 15 = 0
+10
+10
-15
-15
x = 10
x = -15
The dimensions are
10 m by 15 m

43. SOLVING BY COMPLETING THE SQUARE
- When certain equations cannot be factorised
- Steps to follow for equations of the form ax2 + bx + c = 0
- Move constant to the other side of the equation
- Add (b ÷ 2)2 to both sides
- Write left side as a perfect square
- Take square root and rearrange
e.g. Solve
x2 – 8x – 17 = 0
Don’t forget the +/- sign!
x2 – 8x = 17
x2 – 8x + (-4)2 = 17 + (-4)2
It is OK to leave answer without getting two decimal results
x2 – 8x + 16 =
x2 – 8x + 16 = 33
(x – 4)2 = 33
It is important when completing the square that a = 1
x – 4 = ±√33
x = 4 ± √33

44. QUADRATIC FORMULA
- An important concept to know, especially when equations do not factorise easily (i.e. factors are not whole numbers)
- For the general equation ax2 + bx + c = 0 use:
Solutions are also know as ROOTS
Substitute into formula
Make equation equal to zero
e.g. Use the quadratic formula to solve
a) x2 + 5x + 1 = 0
b) 4x2 – 7x = -2
4x2 – 7x + 2 = 0
a = 1
b = 5
c = 1
a = 4
b = -7
c = 2
x = or
x = or 0.360
It is OK to leave answer without getting two decimal results

45. NATURE OF ROOTS - Quadratic expressions have parabolas for graphs
- The solutions (roots) of an equation are where the parabola cuts the x axis.
- There are 3 possibilities
1. Parabola cuts x axis at 2 points
2. Parabola touches x axis at a single point
Equation has 2 real roots
(solutions) e.g. x = 1 or 4
Equation has 1 repeated root
(solution) e.g. x = 3

46. 3. Parabola does not cross x axis at all
Equation has no real roots
(solutions)

47. DISCRIMINANT OF ROOTS
- Tells us the number of real roots an equation has.
- For the general equation ax2 + bx + c = 0 the discriminant is:
Note: If b2 – 4ac > 0 and a perfect square (i.e. 9, 16, 25…) there are 2 rational roots. This means both roots can be written as a fraction and the quadratic equation factorises.
1. If b2 – 4ac > 0 there are two real roots
2. If b2 – 4ac = 0 there is one repeated root
3. If b2 – 4ac < 0 there are no real roots
e.g. Calculate the discriminant of the following and hence state the number of roots
a) x2 + 5x + 1 = 0
b) 3x2 – 2x + 8 = 0
c) 4x2 + 5x + 1 = 0
a = 1
b = 5
c = 1
a = 3
b = -2
c = 8
a = 4
b = 5
c = 1
= 21
= -92
= 9
As b2 – 4ac > 0 there are
two real roots
As b2 – 4ac < 0 there are no real roots
As b2 – 4ac > 0 there are
two real roots
As b2 – 4ac = 9 there are
also two rational roots

48. PROPERTIES OF DISCRIMINANT
e.g. The equation 3x2 – 4x + q = 0 has two real roots. Find the range of possible values of q.
As there are two real roots, b2 – 4ac > 0
a = 3
b = -4
c = q
Sign reverses as dividing by a negative

49. SIMULTANEOUS EQUATIONS
- Are pairs of equations with two unknowns
Revision of Linear-Linear Equations using two methods:
ELIMINATION METHOD
- Line up equations and either add or subtract so one variable disappears
e.g. Solve
To remove the ‘y’ variable we add as the signs are opposite.
To remove the ‘y’ variable we subtract as the signs are the same.
2x + y = 20
x – y = 4
b) 2x + y = 7
x + y = 4
+ ( )
- ( )
3x = 24
x = 3 ÷3 ÷3
Now we substitute x-value into either equation to find ‘y’
2 × 3 + y = 7
Now we substitute x-value into either equation to find ‘y’
x = 8
6 + y = 7
2 × 8 + y = 20 -6 -6
16 + y = 20
y = 1
-16
-16
y = 4
Check values in either equation
Check values in either equation
8 – 4 = 4
3 + 1 = 4

50. - You may need to multiply an equation by a number to be able to eliminate a variable
e.g. Solve
2x – y = 0
x + 2y = 5
× 2
× 1
4x – 2y = 0
x + 2y = 5
b) 4x – 2y = 28
3x + 3y = 12
× 3
× 4
12x – 6y = 84
12x + 12y = 48
+ ( )
- ( )
Multiply the 1st equation by ‘2’ then add to eliminate the y
5x = 5
Multiply the 1st equation by ‘3’ and the 2nd by ‘4’ then subtract
to eliminate the x
-18y = 36 ÷5 ÷5
÷-18
÷-18
x = 1
y = -2
2 × 1 – y = 0
4x – 2 × -2 = 28
2 – y = 0
4x + 4 = 28
Now we substitute x-value into either equation to find ‘y’
Now we substitute y-value into either equation to find ‘x’ -2 -2 -4 -4
– y = -2
4x = 24
÷-1
÷-1 ÷4 ÷4
y = 2
x = 6
Check values in either equation
Check values in either equation
1 + 2 × 2 = 5
3 × × -2 = 12
Note that it was possible to eliminate the ‘x’ variable by multiplying second equation by 2 and then subtracting
Note that it was possible to eliminate the ‘y’ variable by multiplying the 1st equation by 3 and the 2nd by 2 and then adding

51. SUBSTITUTION METHOD
Substitute what the subject equals in for that variable in the other equation
- Make x or y the subject of one of the equations
- Substitute this equation into the second
e.g. Solve
a) y = 3x + 1
9x – 2 y = 4
b) x – y = 2
y = 2x + 3
As we are subtracting more than one term, place in brackets and put a one out in front.
9x – 2(3x + 1) = 4
x – (2x + 3) = 2 1
9x – 6x – 2 = 4
x – 2x – 3 = 2
3x – 2 = 4
-x – 3 = 2 +2 +2 +3 +3
3x = 6
-x = 5 ÷3 ÷3
÷-1
÷-1
x = 2
x = -5
Now we substitute x-value into first equation to find ‘y’
y = 3×2 + 1
Now we substitute x-value into second equation to find ‘y’
y = 2×-5 + 3
y = 7
y = -7
To check values you can substitute both values into second equation
To check values you can substitute both values into second equation

52. LINEAR-NON LINEAR EQUATIONS
- Used to find where a line crosses or touches a curve
As it contains an x2 use rules for solving a quadratic
- Easiest to use the substitution method
e.g. Where does the line y = 2x – 2 cross the parabola y = x2 – x – 6
2x – 2 = x2 – x – 6
Now we substitute x-values into first equation to find ‘y’
0 = x2 – 3x – 4
0 = (x – 4)(x + 1)
y = 2×4 – 2
y = 2×-1 – 2
x = 4 or -1
y = 6
y = -4
The line crosses at (4, 6) and (-1, -4)
e.g. Find the points of intersection of the line y = x – 3 and the circle x2 + y2 = 5
Now we substitute x-values into first equation to find ‘y’
x2 + (x – 3)2 = 5
x2 + x2 – 6x + 9 = 5
y = 1×2 – 3
y = 1×1 – 3
2x2 – 6x + 4 = 0
y = -1
y = -2
2(x2 – 3x + 2) = 0
2(x – 2)(x – 1) = 0
The points of intersection are (2, -1) and (1, -2)
x = 2 or 1

53. LINES AND CURVES There are three possibilities:
- There is no intersection between the line and curve (no real roots)
- The line is a tangent to the curve (one repeated root)
- The line cuts at two points (two real roots)
e.g. Show that there are no points of intersection between the graphs
(hyperbola) and y = -x
Form an equation and use the discriminant!
a = 1
b = 0
c = 1
= -4
As b2 – 4ac < 0 there are no real roots or points of intersection