2. Equations & Inequalities
Factorizing Quadratic Equations
The Quadratic Formula
“k”- Substitution
Simultaneous Equations
Completing the Square
Word Problems
Quadratic Inequalities
The Number System
Nature of Roots 1
Factorizing Quadratic Equations
Standard form: ax² + bx + c = 0
Solve for x:
1) x² = 25
=
x = + 5
or x² - 25 = 0
(x – 5)(x +5) = 0
x = 5 or x = -5
Difference of Squares
Two answers when you take the square root 2
Solve for x:
2) x² - 5x + 4 = 0 Trinomial
(x-4)(x-1) = 0
x = 4 or x = -1
3) 2x(x - 2) = 16
2x² - 4x = 16
2x² - 4x – 16 = 0
2(x² - 2x – 8) = 0
2(x – 4)(x+2) = 0
x = 4 or x = -2
First take out a
common factor
before factorizing
the trinomial
3
Solve for x:
4) x² - 2x = 1
x² - 2x – 1 = 0
This trinomial can’t be
factorized …
Now what ???
Standard form of a
Quadratic Equation
4
THE Quadratic Formula
The quadratic equation must always be in
standard form i.e. 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0
To solve for 𝑥, substitute 𝑎, 𝑏 𝑎𝑛𝑑 𝑐 into THE
Quadratic Formula …
𝑥 =−𝑏 ± 𝑏2 − 4𝑎𝑐
2𝑎
5
4) x² - 2x = 1
x² - 2x – 1 = 0
So, if we look at the previous example …
ax² => a = 1
bx => b = -2
c => -1
Quadratic Formula Calculator
𝒙 =−𝒃 ± 𝒃𝟐 − 𝟒𝒂𝒄
𝟐𝒂
6
Solve for x:
5)
LCD: x² + 2x =>
complicated!
Let x² + 2x = k
k² - 6 = k
k² - k – 6 = 0
(k - 3)(k + 2) = 0
k = 3 or k = -2
Times through by
LCD = k
Have we solved
for x yet?
Rather let repeated
expression = k
“k”- Substitution
7
So, we said … Let x² + 2x = k
AND we’ve just solved for k … k = 3 or k = -2
x² + 2x = k
x² + 2x = 3
x² + 2x - 3 = 0
(x + 3)(x - 1) = 0
x = -3 or x = 1
OR …
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x² + 2x = k
x² + 2x = -2
x² + 2x + 2 = 0
Can’t be
factorized …
The Quadratic
Formula!
a = 1
b = 2
c = 2
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Simultaneous Equations
1. Solve for x and y: 2y = 4x
y² - 3y – x = 2
2y = 4x
y = 2x
1
Subst. into
y² - 3y – x = 2
(2x)² - 3(2x) – x = 2
4x² - 6x – x – 2 = 0
4x² - 7x - 2 = 0
(4x + 1)(x – 2) = 0
2
1
3
3 2
x = - ¼ or x = 2
Are we done
yet? 10
Solve for x and y: 2y = 4x
y² - 3y – x = 2
1
Subst. x = - ¼ or x = 2 into simplest equation
2y = 4x
2y = 4(- ¼ )
2y = -1
y = - ½
2
or 2y = 4x
2y = 4(2)
2y = 8
y = 4
Don’t forget to solve for y!
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Simultaneous Equations
2. Solve for x and y: -2 + y – 2x = 0
xy = 12
-2 + y – 2x = 0
y = 2x + 2
1
Subst. into
xy = 12
x(2x+2) = 12
2x² + 2x – 12 = 0
x² + x - 6 = 0
(x + 3)(x – 2) = 0
2 1
3
3 2
x = -3 or x = 2
Are we done
yet? 12
Solve for x and y: -2 + y – 2x = 0
xy = 12
1
Subst. x = -3 or x = 2 into simplest equation
xy = 12
(-3)y = 12
y = -4
2
or xy = 12
(2)y = 12
y = 6
Don’t forget to solve for y!
Solving a hyperbola
and a straight line
Solving a circle
and a straight line 13
Solve for x by Completing the Square
1. x2 - 8 x – 7 = 0
x2 – 8x = 7
x2 - 8x + ( −8
2 )2 = 7 + (
−8
2 )2
x2 - 8x + (-4)² = 7 + 16
(x – 4) 2 = 23
x – 4 = ± 23
x = 4 ± 23
x = 8,80 or x = -0,80
Graphical Representation of Completing the Square
Add half the
coefficient of
x & square it
3 terms form the
perfect square
Take the
square root
Two answers!
Completing the Square Method 14
2. 3x2 - 12x - 9 = 0
x2 – 4x - 3 = 0
x2 – 4x = 3
x2 - 4x + ( −4
2 )2 = 3 + (
−4
2 )2
x2 - 4x + (-2)² = 7 + 4
(x – 2) 2 = 11
x – 2 = ± 11
x = 2 ± 11
x = 5,32 or x = -1,32
Add half the
coefficient of
x & square it
3 terms form the
perfect square
Take the
square root
Two answers!
NB! Coefficient of x2 must
be 1 before completing
the square.
Practice Completing the Square 15
Exercise
Solve for x by completing the square:
1. x2 + 6x + 1 = 0
2. x2 – 5x = - 3
3. 2x2 + 8x – 4 = 0
4. 3x2 = 9x - 2
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Completing the Square & The
Quadratic Formula
The Quadratic Formula Proof
Solving Quadratic Equations
Different Methods of Solving Quadratic Equations
Quadratic Equation Problems
17
Quadratic Word Problems
1. A farmer has a chicken enclosure which
has an area of 12m2. If he increases the
length by 5m and the breadth by 1m, then the
area of the enclosure is 3 times larger than
the original enclosure.
a) If the length is x metres,
write down an expression
for the breadth, in terms of x.
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a) If the length is x metres, write down an
expression for the breadth, in terms of x.
𝐴 = 𝑙 × 𝑏
12 = 𝑥 × 𝑏
𝑏 = 12
𝑥
b) Write down an equation which models this
scenario.
𝑙 + 5 𝑏 + 1 = 12 × 3
𝑥 + 512
𝑥+ 1 = 36
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c) Determine the length of the enclosure, if it
is given that x < 10.
𝑥 + 512
𝑥+ 1 = 36
12 + 𝑥 + 60
𝑥+ 5 = 36
𝑥 + 60
𝑥− 19 = 0
𝑥2 − 19𝑥 + 60 = 0
𝑥 − 4 𝑥 − 15 = 0
𝑥 = 4 𝑜𝑟 𝑥 = 15
But! 𝑥 ≠ 15 𝑎𝑠 𝑥 < 10, 𝑠𝑜 𝑥 = 4 ∴ 𝑙𝑒𝑛𝑔𝑡ℎ = 4𝑚.
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2. A farmer stays 120 km from the biggest
city. If he travels by car, he travels 20 km/h
faster than by train and he saves 18 minutes.
Determine how fast the farmer travels by car.
Time (train) – T (car) = 18 minutes = 3
10 ℎ𝑜𝑢𝑟
120
𝑥 −
120
𝑥 + 20=
3
10
1200 𝑥 + 20 − 1200𝑥 = 3𝑥 𝑥 + 20
1200𝑥 + 24000 − 1200𝑥 = 3𝑥2 + 60𝑥
𝑥2 + 20𝑥 − 8000 = 0
𝑥 + 100 𝑥 − 80 = 0
𝑥 = −100 𝑜𝑟 𝑥 = 80
𝑇𝑖𝑚𝑒 =𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑆𝑝𝑒𝑒𝑑
∴ 𝐹𝑎𝑟𝑚𝑒𝑟 𝑡𝑟𝑎𝑣𝑒𝑙𝑠 𝑎𝑡 80𝑘𝑚/ℎ
Speed,
Distance &
Time Word
Problem
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