1. Solving Quadratic Equations – More Examples
product is (3)(6)
Solve for x:
3x2 + 11x + 6 = 0
=18
sum is 11
1) Factor
3x2 + 9x + 2x + 6 = 0
3x(x + 3)
+ 2(x + 3) = 0
(x + 3)(3x + 2) = 0
2) Solve

2. Solve for x: 2x2 – x – 3 = 3 1) Equate to 0 2x2 – x – 3 – 3 = 0
2) Factor
2x2 – x – 6 = 0
product (2)(–6)
2x2 – 4x + 3x – 6 = 0
= –12
2x(x – 2)
+ 3(x – 2) = 0
sum: –1
– 4, 3
(x – 2)(2x + 3) = 0
3) Solve

3. Determine the values of x that satisfy the equation.
a) y = x2 – 2x – 14, when y = 10
Equate right side to 10.
x2 – 2x – 14 = 10
Equate left side to 0.
x2 – 2x – 14 – 10 = 0
x2 – 2x – 24 = 0
Factor.
(x – 6)(x + 4) = 0
6, – 4

4. The area of a rectangle is given by the equation
A = 2w2 + 18w, where w is the width in centimetres and A is the area in square centimetres.
a) Determine the area of the rectangle, when its width is 5 cm.
A = 2w2 + 18w
A = 2(5)2 + 18(5)
A = 2(25) + 90
A = 
A = 40
The area of the rectangle is 40 cm2

5. The width is either 7 cm or 2 cm.
The area of a rectangle is given by the equation
A = 2w2 + 18w, where w is the width in centimetres and A is the area in square centimetres.
b) Determine the width of the rectangle, when the area
is 28 cm2.
A = 2w2 + 18w
28 = 2w2 + 18w
2w2  18w + 28 = 0
2(w2  9w + 14) = 0
2(w  7)(w  2) = 0
w = 7, 2
The width is either 7 cm or 2 cm.

6. Problem : (Try this first on your own then
see the solution on the next slide)
Aidan’s new house has a rectangular garden
that measures 20 m by 30 m. He wants to
construct a walkway around the garden of
uniform width so that the total area of the
garden and the walkway will be 1200 m2.
How wide should the walkway be?

7. Equation: Solution Area of garden: 20 m × 30 m x = 600 m2
Area of walkway & garden: 1200 m2
2x + 30
Let x = width of the walkway.
30 m
Equation:
(2x + 20)(2x + 30) = 1200 m2
20 m
4x2 + 60x + 40x = 1200
4x x – = 0
2x + 20
4x x – 600 = 0
4(x2 + 25x – 150) = 0
The width is 5 m.
4(x + 30)(x – 5) = 0
x = – 30 or x = 5