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Many quadratic equations can not be solved by factoring. Other techniques are required to solve them.
8.1 – Solving Quadratic Equations
x2 = 20 5x2 + 55 = 0
Examples:
( x + 2)2 = 18 ( 3x – 1)2 = –4
x2 + 8x = 1 2x2 – 2x + 7 = 0
22 5 0x x 44 2 xx
If b is a real number and if a2 = b, then a = ±√¯‾.
20
8.1 – Solving Quadratic EquationsSquare Root Property
b
x2 = 20
x = ±√‾‾
x = ±√‾‾‾‾4·5
x = ± 2√‾ 5 –11
5x2 + 55 = 0
x = ±√‾‾‾
5x2 = –55
x2 = –11
x = ± i√‾‾‾11
If b is a real number and if a2 = b, then a = ±√¯‾.
18
8.1 – Solving Quadratic EquationsSquare Root Property
b
( x + 2)2 = 18
x + 2 = ±√‾‾
x + 2 = ±√‾‾‾‾9·2
x +2 = ± 3√‾ 2
x = –2 ± 3√‾ 2
–4
( 3x – 1)2 = –4
3x – 1 = ±√‾‾
3x – 1 = ± 2i
3x = 1 ± 2i
3
21 ix
ix3
2
3
1
Review:
8.1 – Solving Quadratic EquationsCompleting the Square
( x + 3)2
x2 + 2(3x) + 9
x2 + 6x
2
6 23
x2 + 6x + 9
3 9
x2 + 6x + 9
( x + 3) ( x + 3)
( x + 3)2
x2 – 14x
2
14 277 49
x2 – 14x + 49
( x – 7) ( x – 7)
( x – 7)2
8.1 – Solving Quadratic EquationsCompleting the Square
x2 + 9x
2
9
2
2
94
81
x2 – 5x
4
8192 xx
2
9
2
9xx
2
2
9
x
2
5
2
2
54
25
4
2552 xx
2
5
2
5xx
2
2
5
x
8.1 – Solving Quadratic EquationsCompleting the Square
x2 + 8x = 1
2
824 16
1611682 xx
174 2 x
174 2 x
174 x
174 x
4
x2 + 8x = 1
8.1 – Solving Quadratic EquationsCompleting the Square
5x2 – 10x + 2 = 0
2
2 21 1
5
5
5
31 x
5
5
5
21 2 x
5
31 2 x
5
31 x
5
31x
1
5x2 – 10x = –2
5
2
5
10
5
5 2
xx
5
222 xx
15
2122 xx
5
31 2 x
5
151x
5
155x
or
8.1 – Solving Quadratic EquationsCompleting the Square
2x2 – 2x + 7 = 0
2
1
2
2
1
4
1
2
13
2
1 ix 4
1
4
14
2
12
x
4
13
2
12
x
4
13
2
1 x
2
13
2
1 x
2
1
2x2 – 2x = –7
2
7
2
2
2
2 2
xx
2
72 xx
4
1
2
7
4
12 xx
4
13
2
12
x
2
131 ix
or
The quadratic formula is used to solve any quadratic equation.
2 4
2x
cb b a
a
The quadratic formula is:
Standard form of a quadratic equation is:2 0x xba c
8.2 – Solving Quadratic EquationsThe Quadratic Formula
2 4
2x
cb b a
a
8.2 – Solving Quadratic EquationsThe Quadratic Formula
02 cbxax
cbxax 2
a
cx
a
bx
a
a 2
a
cx
a
bx
2
a
b
a
b
22
1 2
22
42 a
b
a
b
a
c
a
b
a
bx
a
bx
2
2
2
22
44
a
a
a
c
a
b
a
bx
a
bx
4
4
44 2
2
2
22
2 4
2x
cb b a
a
8.2 – Solving Quadratic EquationsThe Quadratic Formula
22
2
2
22
4
4
44 a
ac
a
b
a
bx
a
bx
2
2
2
22
4
4
4 a
acb
a
bx
a
bx
2
2
2
22
4
4
4 a
acb
a
bx
a
bx
2
22
4
4
2 a
acb
a
bx
2
2
4
4
2 a
acb
a
bx
a
acb
a
bx
2
4
2
2
a
acb
a
bx
2
4
2
2
a
acbbx
2
42
The quadratic formula is used to solve any quadratic equation.
2 4
2x
cb b a
a
The quadratic formula is:
Standard form of a quadratic equation is: 2 0x xba c
2 4 8 0x x
a 1 c b4 8
23 5 6 0x x
a 3 c b 5
22 0x x
a 2 c b1 0
2 10x a 1 c b0 106
2 10 0x
8.2 – Solving Quadratic EquationsThe Quadratic Formula
2 4
2x
cb b a
a
2 0x xba c
2 3 2 0x x
2x 1x
1x 2x 0
1 0x 2 0x
8.2 – Solving Quadratic EquationsThe Quadratic Formula
2 4
2x
cb b a
a
2 0x xba c
2 3 2 0x x a 1 c b 3 2
23 3 1 24
12x
3 9 8
2x
3 1
2x
3 1
2x
3 1
2x
3 1
2x
4
2x
2x
2
2x
1x 3 1
2x
8.2 – Solving Quadratic EquationsThe Quadratic Formula
2 4
2x
cb b a
a
2 0x xba c
22 5 0x x
a 2 c b 1 5
24
22
1 521x
1 1 40
4x
1 41
4x
8.2 – Solving Quadratic EquationsThe Quadratic Formula
2 4
2x
cb b a
a
8.2 – Solving Quadratic EquationsThe Quadratic Formula
44 2 xx
044 2 xx
42
44411 2 x
8
6411 x
8
631 x
8
631 ix
8
391
ix
8
731 ix
ix
8
73
8
1
2 4
2x
cb b a
a
8.2 – Solving Quadratic EquationsThe Quadratic Formula and the Discriminate
The discriminate is the radicand portion of the quadratic formula (b2 – 4ac).
It is used to discriminate among the possible number and type of solutions a quadratic equation will have.
b2 – 4ac Name and Type of SolutionPositive
Zero
Negative
Two real solutions
One real solutions
Two complex, non-real solutions
2 4
2x
cb b a
a
8.2 – Solving Quadratic EquationsThe Quadratic Formula and the Discriminate
2143 2
89
b2 – 4ac Name and Type of SolutionPositive
Zero
Negative
Two real solutions
One real solutions
Two complex, non-real solutions
2 3 2 0x x a 1 c b 3 2
1
Positive
Two real solutions
2x 1x
2 4
2x
cb b a
a
8.2 – Solving Quadratic EquationsThe Quadratic Formula and the Discriminate
4441 2
641
b2 – 4ac Name and Type of SolutionPositive
Zero
Negative
Two real solutions
One real solutions
Two complex, non-real solutions
a c b
63
Negative
Two complex, non-real solutions
044 2 xx
4 1 4
ix8
73
8
1
2 4
2x
cb b a
a
8.2 – Solving Quadratic Equations
The Quadratic Formula
Given the diagram below, approximate to the nearest foot how many feet of walking distance a person saves by cutting across the lawn instead of walking on the sidewalk.
20 feet
x + 2
x
2 4
2x
cb b a
a
8.2 – Solving Quadratic Equations
The Quadratic Formula
Given the diagram below, approximate to the nearest foot how many feet of walking distance a person saves by cutting across the lawn instead of walking on the sidewalk.
20 feet
x + 2
x
The Pythagorean Theorem
a2 + b2 = c2
(x + 2)2 + x2 = 202
x2 + 4x + 4 + x2 = 400
2x2 + 4x + 4 = 400
2x2 + 4x – 369 = 0
2(x2 + 2x – 198) = 0
2 4
2x
cb b a
a
8.2 – Solving Quadratic Equations
The Quadratic Formula
Given the diagram below, approximate to the nearest foot how many feet of walking distance a person saves by cutting across the lawn instead of walking on the sidewalk.
20 feet
x + 2
x
The Pythagorean Theorem
a2 + b2 = c2
2(x2 + 2x – 198) = 0
12
1981422 2 x
2
79242 x
2
7962 x
2 4
2x
cb b a
a
8.2 – Solving Quadratic Equations
The Quadratic Formula
Given the diagram below, approximate to the nearest foot how many feet of walking distance a person saves by cutting across the lawn instead of walking on the sidewalk.
20 feet
x + 2
x
The Pythagorean Theorem
a2 + b2 = c2
2
7962x
2
2.282
2
2.282 x
2
2.282 x
2
2.26x
1.13x
2
2.30x
1.15xft
2 4
2x
cb b a
a
8.2 – Solving Quadratic Equations
The Quadratic Formula
Given the diagram below, approximate to the nearest foot how many feet of walking distance a person saves by cutting across the lawn instead of walking on the sidewalk.
20 feet
x + 2
x
The Pythagorean Theorem
a2 + b2 = c2
1.13x
ft2.28
ft
21.131.13
28 – 20 = 8 ft