[mkcoldwolf] 1st Standard Test

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mkcoldwolf
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[mkcoldwolf] 1st Standard Test

Post by mkcoldwolf » Sun Oct 21, 2018 7:55 am

Formula Notes:
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(a+b)^2 = a^2 + 2ab + b^2
(a-b)^2 = a^2 - 2ab + b^2

a^3 + b^3 = (a+b)(a^2 - ab + b^2)
a^3 - b^3 = (a-b)(a^2 + ab + b^2)
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When α,β is the root of ax^2 + bx + c = 0
Sum of root = α+β = -(b/a)
Product of root = αβ = c/a

Special Case:
(α-β)^2 = (α+β)^2 - 4αβ
α^2 + β^2 = (α+β)^2 - 2αβ

To from a quadratic equation in x when roots are α,β
x^2 - (Sum of root)x + (Product of root)
= x^2 - (α+β)x + αβ
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Quadratic Equation:
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ax^2 ± bx ± c is a Quadratic Equation

To solve Quadratic Equation when it is Irrational number
x = (-(b) ± √(b^2 - 4ac))/2a

Since the type of root would be determined by the surd,
So we have created △ to represent the formula inside the surd
△ = b^2 - 4ac

To determine the type of root
△ >= 0 = Has real roots
△ > 0 = Has two unequal(distinct) real roots
△ = 0 = Has one equal roots
△ < 0 = No real root
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Complex Number:
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"i" means a (-1) inside a square root, so
√(-1) = 1i
√(-9) = 3i
√(-1024) = 32i

also "i" can be convert to a real number some times
i = i
i^2 = -1
i^3 = -i
i^4 = 1

Therefor, If the i^x is larger than 4
we can just get the remainder of 2016/4 instand

i^9
= i^1
= i

i^22
= i^2
= -1

i^2016
= i^0
= 1
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Function:
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Very easy la, no need me to teach but give some example

1. f(x) = 5x + 6

a) f(2) = 5(2) + 6 = 16

b) f(3) + f(-2)
= 5(3) + 6 + (5(-2) + 6)
= 21 + (-4)
= 17

c) f(u) = 126
5u + 6 = 126
5u = 120
u = 14
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Exercise:
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Answer:
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Extra:
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