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Core Mathematics 1 Notes
Algebra and Functions
Indices
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𝑎 𝑚 × 𝑎 𝑛 = 𝑎 𝑚+𝑛
𝑎 𝑚 ÷ 𝑎 𝑛 = 𝑎 𝑚−𝑛
(𝑎 𝑚 ) 𝑛 = 𝑎 𝑚𝑛
1
𝑎−𝑚 = 𝑎 𝑚

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

𝑎𝑚= √𝑎𝑛
𝑎0 = 1

𝑛

𝑚

Surds


√(𝑎𝑏) = √ 𝑎 × √𝑏

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√𝑏 =

𝑎

√𝑎
√𝑏

Rationalising the denominator of a fraction when it is a surd
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1
1
√𝑎
√𝑎
→
×
=
𝑎
√𝑎
√𝑎
√𝑎
1
1
𝑎− √𝑏
𝑎− √𝑏
→
×
= 2
𝑎+ √𝑏
𝑎+ √𝑏
𝑎− √𝑏
(𝑎 )− 𝑏
1
1
𝑎+ √𝑏
𝑎+ √𝑏
→ 𝑎− × 𝑎+ = (𝑎2 )− 𝑏
𝑎− √𝑏
√𝑏
√𝑏
1
1
(√ 𝑎− √𝑏)
(√ 𝑎− √𝑏)
→
×
=
(√ 𝑎+ √𝑏)
(√ 𝑎+ √𝑏)
(√ 𝑎− √𝑏)
𝑎+𝑏

Quadratic Functions
Factorising
2𝓍 2 + 8𝓍 + 8 = 0
Which 2 integers create the product of +16 and sum of +8?
2𝓍 2 + 4𝓍 + 4𝓍 + 8 = 0
2𝓍(𝑥 + 2) + 4(𝑥 + 2) = 0
(2𝑥 + 4)(𝓍 + 2) = 0
𝑥 = −2

Completing the square

2𝓍 2 + 8𝓍 + 8 = 0
2(𝓍 2 + 4) + 8 = 0
2((𝓍 + 2)2 − 4) + 8 = 0

2(𝓍 + 2)2 − 8 + 8 = 0
2(𝓍 + 2)2 = 0
(𝓍 + 2)2 =

0
2

𝑥 + 2 = √0
𝑥 = −2 ± √0
𝑥 = −2

Solutions
If x is equals to 0, y is equals to

𝑦 = 2𝓍 2 + 8𝓍 + 8
𝑦=8
If y is equals to 0, x is equals to
(2𝑥 + 4)(𝑥 + 2) = 0
2𝓍 + 4 = 0 𝑜𝑟 𝓍 + 2 = 0
2𝑥 = −4 𝑜𝑟 𝓍 = −2
𝑥 = −2 𝑜𝑟 𝓍 = −2
𝑥 = −2

Quadratic graph
The y-intercept will always be y-solutions of the equation

𝑦=8
The roots will always be the x- solutions of the equation
𝑥 = −2 𝑜𝑟 𝓍 = −2
The minimum or maximum will always be the negative of the ‘completing the square’ equation:
2(𝑥 + 2)2 ± 0 = 0
𝑥 = +2 𝑎𝑛𝑑 𝑤𝑖𝑙𝑙 𝑏𝑒𝑐𝑜𝑚𝑒 − 2
𝑦 = ± 0 𝑎𝑛𝑑 𝑤𝑖𝑙𝑙 𝑏𝑒𝑐𝑜𝑚𝑒 ± 0

Quadratic Formula

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−𝑏±√𝑏2 −4𝑎𝑐
2𝑎

Discriminants

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𝒃 𝟐 − 𝟒𝒂𝒄
o
o
o
o
o
o

If
If
If
If
If
If

𝑏2
𝑏2
𝑏2
𝑏2
𝑏2
𝑏2

> 4𝑎𝑐
= 4𝑎𝑐
< 4𝑎𝑐
> 4𝑎𝑐
= 4𝑎𝑐
< 4𝑎𝑐

𝑎𝑛𝑑 𝑎 > 0 then it is a ∪ shaped quadratic graph with 2 different roots
𝑎𝑛𝑑 𝑎 > 0 then it is a ∪ shaped quadratic graph with equal roots
𝑎𝑛𝑑 𝑎 > 0 then it is a ∪ shaped quadratic graph with no real roots
𝑎𝑛𝑑 𝑎 < 0 then it is a ∩ shaped quadratic graph with 2 different roots
𝑎𝑛𝑑 𝑎 < 0 then it is a ∩ shaped quadratic graph with equal roots
𝑎𝑛𝑑 𝑎 < 0 then it is a ∩ shaped quadratic graph with no real roots

Equation and Inequalities
Solving simultaneous linear equations by substitution
2𝑥 − 𝑦 = 6 [1]
4𝑥 + 3𝑦 = 22 [2]
Rearrange [1] to make 𝒙 the subject
2𝑥 = 6 + 𝑦 [1′ ]
Substitute [1’] into [2]
2(2𝑥) + 3𝑦 = 22
12 + 2𝑦 + 3𝑦 = 22
12 + 5𝑦 = 22
5𝑦 = 10
𝑦=2

Substitute 𝑦 into [1]
2𝑥 − 2 = 6
2𝑥 = 8
𝑥=4

Solving simultaneous linear equations by elimination
2𝑥 − 𝑦 = 6 [1]
4𝑥 + 3𝑦 = 22 [2]
Multiply [1] to make 𝑡ℎ𝑒 𝑥 𝑜𝑟 𝑦 𝑒𝑞𝑢𝑎𝑙
4𝑥 − 2𝑦 = 12 [1′]
Subtract [2] from [1’]
4𝑥 + 3𝑦 = 22 [2]
4𝑥 − 2𝑦 = 12 [1′]
5𝑦 = 10
𝑦=2
Substitute 𝑦 into [1]
2𝑥 − 2 = 6
2𝑥 = 8
𝑥=4

Linear inequalities
5𝑥 + 3 < 7
5𝑥 < 4
4
𝑥<
5
Two Inequalities
3𝑥 − 4 < 𝑥 + 8 𝑎𝑛𝑑 5𝑥 > 𝑥 − 8
3𝑥 − 4 < 𝑥 + 8

5𝑥 > 𝑥 − 8

3𝑥 < 𝑥 + 12

4𝑥 > −8

2𝑥 < 12

4𝑥 > −8

𝑥<6

𝑥 > −2

𝑜𝑣𝑒𝑟𝑙𝑎𝑝𝑝𝑖𝑛𝑔 𝑣𝑎𝑙𝑢𝑒𝑠 𝑠𝑜 − 2 < 𝑥 < 6
...

If 2 line are perpendicular to each other, the product of their gradients is -1
...


Proof
𝑆 =

+ (𝑎 + 𝑑)

𝑎

+ ⋯ + (𝑎 + (𝑛 − 2)𝑑)

𝑆 = (𝑎 + (𝑛 − 1)𝑑 + (𝑎 + (𝑛 − 2)𝑑) + ⋯ + (𝑎 + 𝑑)

+ (𝑎 + (𝑛 − 1)𝑑)
+ 𝑎

2𝑆 = (2𝑎 + (𝑛 − 1)𝑑 + (2𝑎 + (𝑛 − 2)𝑑) + ⋯ + (2𝑎 + (𝑛 − 1)𝑑) + (2 𝑎 + (𝑛 − 1)𝑑)
Each term is the same and there are 𝑛 terms so you multiply by 𝑛
2𝑆 = 𝑛 × (2𝑎 + (𝑛 − 1)𝑑
𝑛
𝑆 = (2𝑎 + (𝑛 − 1)𝑑
2

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∑10 2𝑛 → sum of 2n from n=1 to n=10
𝑛=1

Differentiation
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The gradient of a curve 𝑦 = 𝑓(𝑥) at any specific point is equal to the gradient of the tangent to
the curve at that point

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