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EECM 3714

Lecture 3: Unit 3
Non-linear equations
Renshaw, Ch
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4
• Quadratic functions and equations
• Simultaneous quadratic equations
• Quadratic functions in economics
• Demand and supply
• Cost, revenue and profit

• Examples

QUADRARATIC EXPRESSIONS
• Consider an expression: (𝑎 + 𝑏)(𝑐 + 𝑑)
• Expanding the expression by removing the brackets?
(𝑎 + 𝑏)(𝑐 + 𝑑) ≡ 𝑎(𝑐 + 𝑑) + 𝑏(𝑐 + 𝑑)
• Use the expansion rule and multiple out the brackets
(𝑎 + 𝑏)(𝑐 + 𝑑) ≡ 𝑎𝑐 + 𝑎𝑑 + 𝑏𝑐 + 𝑏𝑑

When the expression contains an unknown, x
(𝑥 + 𝑎)(𝑥 + 𝑏) ≡ 𝑥(𝑥 + 𝑏) + 𝑎(𝑥 + 𝑏)
...

When a = b: (𝑥 + 𝑎)(𝑥 + 𝑎) ≡ 𝑥 2 + 2𝑎𝑥 + 𝑎2

QUADRATIC EXPRESSIONS
• If we expand the expression (𝑥 + 𝑎)(𝑥 + 𝑏) to get 𝑥 2 + (𝑎 + 𝑏)𝑥 + 𝑎𝑏 this is called
a quadratic expression
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• So given 𝑥 2 + 9𝑥 + 20, we look for 2 numbers a and b such that a + b = 9 and ab =
20

• After trial and error, we fix on a = 5, b = 4
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• From above, we know 𝑥 2 + 9𝑥 + 20 = (𝑥 + 5)(𝑥 + 4)
• Because the above is an identity, solving (𝑥 + 5)(𝑥 + 4) = 0 is
equivalent to solving 𝑥 2 + 9𝑥 + 20 = 0
• Now, (𝑥 + 5)(𝑥 + 4) = 0 is easy to solve: x = −5 or x = −4
• So solutions to

𝑥 2 + 9𝑥 + 20 = 0 are x = −5 or x = −4 (check)

• However, solving a quadratic equation is easy with a formula
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• Based on a given quadratic equation, we can have 3 situations:
• If 𝑏 2 > 4𝑎𝑐, there ate 2 distinct roots
• 𝑏 2 = 4𝑎𝑐, there is only 1 root, described as a repeated root
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FORMULA FOR SOLVING QUADRATIC EQUATIONS
Example, if 𝑏 2 > 4𝑎𝑐: 𝑥 2 + 9𝑥 + 20 = 0 Here, a = 1, b = 9, c = 20

𝑥=

−9± 92 −4×1×20
2

=

−9± 1
2

= −5 or − 4
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root sign, and negative numbers have no
(real) roots
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• The general form of a quadratic function is: 𝑦 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
• where 𝑎, 𝑏 and 𝑐 parameters
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16, page 140 and example 4
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6-4
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133 and example 4
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140

SIMULTANEOUS QUADRATIC EQUATIONS
• Given 2 simultaneous quadratic equations, we can solve these equations using the same
techniques that we used to solve simultaneous linear equations
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19, p
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• Supply and demand:

• 𝑃 = −𝑎𝑞𝑑2 − 𝑏𝑞𝑑 + 𝑐 and 𝑃 = 𝑑𝑞𝑠2 − 𝑒𝑞𝑠 + 𝑓
• Equilibrium: 𝑞𝑑 = 𝑞𝑠 = 𝑞 and since P = P → −𝑎𝑞 2 − 𝑏𝑞 + 𝑐 = 𝑑𝑞 2 − 𝑒𝑞 + 𝑓

• Use the quadratic formula to solve for 𝑞
• Note: if one of the answers is negative, it can’t be a solution for 𝑞 (𝑞 ≥ 0)

• Examples 4
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146-8

EXAMPLE
• The demand and supply of sweets are given by
𝑃 = −2𝑞𝑑2 − 2𝑞𝑑 + 35 and 𝑃 = 0
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5
• Equilibrium: 𝑞𝑑 = 𝑞𝑠 = 𝑞 and since P = P →
• −2𝑞 2 − 2𝑞 + 35 = 0
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5
• Collect like terms to have −2
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5=0
• This can easily be solved using the formula to get 𝑞=3 or 𝑞=-4
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• What can you say about these quantities? No economic meaning for a negative quantity,
• Hence, we only have one solution 𝑞=3 and at this quantity 𝑝 = −2(9) − 2(3) + 35= 11
• Therefore: 𝑞=3 and 𝑝=11
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• Note TC − Total Cost, TFC − Total Fixed Cost, TVC − Total Variable Cost , 𝑇𝑅 − Total Revenue , q −

Quantity produced or sold
• 𝑇𝐶 = 𝑇𝐹𝐶 + 𝑇𝑉𝐶; where 𝑇𝑉𝐶 = 𝑎𝑞 2 + 𝑏𝑞, 𝑇𝐹𝐶 = 𝑐
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148-50

EXAMPLE
• The demand and supply of cigarettes are given by: 𝑞𝑑 = 20 − 5𝑝 and 𝑞 𝑠 = 𝑝2 − 3
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Find the equilibrium
price and quantity of cigarettes
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5𝑝 + 10
• ⇒ 𝑝2 + 1
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5; 𝑐 = −10
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5± 1
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2(1)

∴ 𝑝 = 2
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• Since 𝑝 ≥ 0: 𝑝 = 2
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5 = 7
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5 and 𝑞 = 7
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