Solving Quadratic Equations

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