The document discusses solving quadratic equations using various techniques like factoring, completing the square, and the quadratic formula. It provides examples of using these methods to solve equations in standard form. The quadratic formula is derived and explained. The concept of the discriminant is introduced and how it relates to the number and type of solutions. An example problem is worked through applying the Pythagorean theorem and quadratic formula to solve a real world word problem.

1. 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
2
2 5 0x x− − = 44 2
−−= xx

2. If b is a real number and if a2
= b, then a = ±√¯‾.
20
8.1 – Solving Quadratic Equations
Square Root Property
b
x2
= 20
x = ±√‾‾
x = ±√‾‾‾‾4·5
x = ± 2√‾5 –11
5x2
+ 55 = 0
x = ±√‾‾‾
5x2
= –55
x2
= –11
x = ± i√‾‾‾11

3. If b is a real number and if a2
= b, then a = ±√¯‾.
18
8.1 – Solving Quadratic Equations
Square 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 i
x
±
=
ix
3
2
3
1
±=

4. Review:
8.1 – Solving Quadratic Equations
Completing the Square
( x + 3)2
x2
+ 2(3x) + 9
x2
+ 6x
=
2
6
=2
3
x2
+ 6x + 9
3 9
x2
+ 6x + 9
( x + 3) ( x + 3)
( x + 3)2
x2
– 14x
=
−
2
14
( ) =−
2
77− 49
x2
– 14x + 49
( x – 7) ( x – 7)
( x – 7)2

5. 8.1 – Solving Quadratic Equations
Completing the Square
x2
+ 9x
2
9
=





2
2
9
4
81
x2
– 5x
4
81
92
++ xx






+





+
2
9
2
9
xx
2
2
9






+x
2
5
=





2
2
5
4
25
4
25
52
++ xx






+





+
2
5
2
5
xx
2
2
5






+x

6. 8.1 – Solving Quadratic Equations
Completing the Square
x2
+ 8x = 1
=
2
8
=2
4 16
1611682
+=++ xx
( ) 174
2
=+x
( ) 174
2
±=+x
174 ±=+x
174 ±−=x
4
x2
+ 8x = 1

7. 8.1 – Solving Quadratic Equations
Completing the Square
5x2
– 10x + 2 = 0
=
−
2
2
( ) =−
2
1 1
5
5
5
3
1 ⋅±=x( )
5
5
5
2
1
2
+−=−x
( )
5
3
1
2
±=−x
5
3
1 ±=−x
5
3
1±=x
1−
5x2
– 10x = –2
5
2
5
10
5
5 2
−=−
xx
5
2
22
−=− xx
1
5
2
122
+−=+− xx
( )
5
3
1
2
=−x
5
15
1±=x
5
155±
=x
or

8. 8.1 – Solving Quadratic Equations
Completing the Square
2x2
– 2x + 7 = 0
=
−
2
1
=





−
2
2
1
4
1
2
13
2
1 i
x ±=
4
1
4
14
2
1
2
+−=





−x
4
13
2
1
2
−±=





−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
1
2
−=





−x
2
131 i
x
±
=
or

9. The quadratic formula is used to solve any quadratic equation.
2
4
2
x
cb b a
a
− ± −
=
The quadratic formula is:
Standard form of a quadratic equation is:
2
0x xba c+ + =
8.2 – Solving Quadratic Equations
The Quadratic Formula

10. 2
4
2
x
cb b a
a
− ± −
=
8.2 – Solving Quadratic Equations
The Quadratic Formula
02
=++ cbxax
cbxax −=+2
a
c
x
a
b
x
a
a −
=+2
a
c
x
a
b
x
−
=+2
a
b
a
b
22
1
=⋅ 2
22
42 a
b
a
b
=





a
c
a
b
a
b
x
a
b
x −=++ 2
2
2
2
2
44
a
a
a
c
a
b
a
b
x
a
b
x
4
4
44 2
2
2
2
2
⋅−=++

11. 2
4
2
x
cb b a
a
− ± −
=
8.2 – Solving Quadratic Equations
The Quadratic Formula
22
2
2
2
2
4
4
44 a
ac
a
b
a
b
x
a
b
x −=++
2
2
2
2
2
4
4
4 a
acb
a
b
x
a
b
x
−
=++
2
2
2
2
2
4
4
4 a
acb
a
b
x
a
b
x
−
±=++
2
22
4
4
2 a
acb
a
b
x
−
±=





+
2
2
4
4
2 a
acb
a
b
x
−
±=+
a
acb
a
b
x
2
4
2
2
−
±=+
a
acb
a
b
x
2
4
2
2
−
±−=
a
acbb
x
2
42
−±−
=

12. The quadratic formula is used to solve any quadratic equation.
2
4
2
x
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 =b = 4 8
2
3 5 6 0x x− + =
a = 3 c =b = 5−
2
2 0x x+ =
a = 2 c =b = 1 0
2
10x = −
a = 1 c =b = 0 106
2
10 0x + =
8.2 – Solving Quadratic Equations
The Quadratic Formula

13. 2
4
2
x
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 Equations
The Quadratic Formula

14. 2
4
2
x
cb b a
a
− ± −
=2
0x xba c+ + =
2
3 2 0x x− + =
a = 1 c =b = 3− 2
( ) ( ) ( ) ( )
( )
2
3 3 1 24
12
x
− ± −−
=
−
3 9 8
2
x
± −
=
3 1
2
x
±
=
3 1
2
x
±
=
3 1
2
x
+
=
3 1
2
x
−
=
4
2
x =
2x =
2
2
x =
1x =3 1
2
x
±
=
8.2 – Solving Quadratic Equations
The Quadratic Formula

15. 2
4
2
x
cb b a
a
− ± −
=2
0x xba c+ + =
2
2 5 0x x− − =
a = 2 c =b = 1− 5−
( ) ( ) ( ) ( )
( )
2
4
22
1 521
x
−
=
− −±−−
1 1 40
4
x
± +
=
1 41
4
x
±
=
8.2 – Solving Quadratic Equations
The Quadratic Formula

16. 2
4
2
x
cb b a
a
− ± −
=
8.2 – Solving Quadratic Equations
The Quadratic Formula
44 2
−−= xx
044 2
=++ xx
( ) ( )( )
( )42
44411
2
−±−
=x
8
6411 −±−
=x
8
631 −±−
=x
8
631 i
x
±−
=
8
391 ⋅±−
=
i
x
8
731 i
x
±−
= ix
8
73
8
1
±−=

17. 2
4
2
x
cb b a
a
− ± −
=
8.2 – Solving Quadratic Equations
The 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 Solution
Positive
Zero
Negative
Two real solutions
One real solutions
Two complex, non-real
solutions

18. 2
4
2
x
cb b a
a
− ± −
=
8.2 – Solving Quadratic Equations
The Quadratic Formula and the Discriminate
( ) ( )( )2143
2
−−
89 −
b2
– 4ac Name and Type of Solution
Positive
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 =

19. 2
4
2
x
cb b a
a
− ± −
=
8.2 – Solving Quadratic Equations
The Quadratic Formula and the Discriminate
( ) ( )( )4441
2
−
641−
b2
– 4ac Name and Type of Solution
Positive
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
ix
8
73
8
1
±−=

20. 2
4
2
x
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

21. 2
4
2
x
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

22. 2
4
2
x
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

23. 2
4
2
x
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
7962
x =
±−
2
2.282
2
2.282 +−
=x
2
2.282 −−
=x
2
2.26
=x
1.13=x
2
2.30−
=x
1.15−=xft

24. 2
4
2
x
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
1.13=x
ft2.28
ft
=++ 21.131.13
28 – 20 = 8 ft