1. 4-8 The Quadratic Formula
and the Discriminant
p.278-281

2. We have a number of different way
of finding the roots if a quadratic
equations
#1. Making a table
#2. Factoring
#3. Completing the Square
Now a new way that comes from completing
the square.
The Quadratic Formula

3. The Quadratic Formula
Solve for x by completing the square.
ax 2 + bx + c = 0
ax 2 + bx +  = −c + 
b
−c
2
x + x +=
+
a
a
2
b
−c  b 
 b 
x + x+  =
+ 
a
a  2a 
 2a 
2
2

4. The Quadratic Formula
Solve for x by completing the square.
ax 2 + bx + c = 0
ax 2 + bx +  = −c + 
b
−c
x2 + x +  =
+
a
a
2
b
−c  b 
 b 
x2 + x +   =
+ 
a
a  2a 
 2a 
b
b2
b 2 − 4ac
x2 + x + 2 =
a
4a
4a 2
2
2
b  b 2 − 4ac

x+  =
2a 
4a 2


5. The Quadratic Formula
Solve for x by completing the square.
b
b2
b 2 − 4ac
x + x+ 2 =
a
4a
4a 2
2
2
b  b 2 − 4ac

x+  =
2a 
4a 2

b
b 2 − 4ac
x+
=±
2a
4a 2
−b
b 2 − 4ac
x=
±
2a
2a
− b ± b 2 − 4ac
x=
2a

6. The Quadratic Formula
Solve for x by completing the square.
− b ± b − 4ac
x=
2a
2

7. How does it work
Equation:
3x 2 + 5 x + 1 = 0
a=3
b=5
c =1
− b ± b 2 − 4ac
x=
2a

8. How does it work
Equation:
3x + 5 x + 1 = 0
a=3
b=5
c =1
2
− b ± b 2 − 4ac
x=
2a
x=
− ( 5) ±
( 5) 2 − 4( 3)(1)
2( 3)
− 5 ± 25 − 12
x=
6
x=
− 5 ± 13 − 5
13
=
±
6
6
6

9. The Discriminant
The number in the square root of the
quadratic formula.
b − 4ac
2
Given x − 5 x + 6 = 0
2
( − 5)
− 4(1)( 6 )
25 − 24 = 1
2

10. The Discriminant
b − 4ac
2
The Discriminant can be negative, positive or zero
If the Discriminant is positive then there are:
2 real answers.
If the square root is not a prefect square
( for example 25 ),
then there will be 2 irrational roots
( for example 2 ± 5 ).

11. The Discriminant
b 2 − 4ac
The Discriminant can be negative, positive or zero
If the Discriminant is positive,
there are 2 real answers.
If the Discriminant is zero,
there is 1 real answer.
If the Discriminant is negative,
there are 2 complex answers.
complex answer have i.

12. Let’s put all of that
b
information in a chart.
Value of Discriminant
Type and
Number of Roots
D > 0,
D is a perfect square
2 real,
rational roots
(ex: x= 2 and x= -4)
D > 0,
D NOT a perfect square
2 real,
Irrational roots
(x = √13 x= -√13)
D=0
1 real, rational root
(double root)
(ex: x = 5)
D<0
2 complex roots
(complex conjugates)
(x = 2 ± 3i )
2
− 4ac
Sample Graph
of Related Function

13. Describe the roots
Tell me the Discriminant and the type of roots
x2 + 6x + 9 = 0

14. Describe the roots
Tell me the Discriminant and the type of roots
x2 + 6x + 9 = 0
0, One rational root

15. Describe the roots
Tell me the Discriminant and the type of roots
x2 + 6x + 9 = 0
0, One rational root
x + 3x + 5 = 0
2
-11, Two complex roots
x + 8x − 4 = 0
2
80, Two irrational roots

16. Describe the roots
Tell me the Discriminant and the type of roots
x2 + 6x + 9 = 0
0, One rational root
x + 3x + 5 = 0
2

17. Describe the roots
Tell me the Discriminant and the type of roots
x2 + 6x + 9 = 0
0, One rational root
x + 3x + 5 = 0
2
-11, Two complex roots

18. Describe the roots
Tell me the Discriminant and the type of roots
x2 + 6x + 9 = 0
0, One rational root
x + 3x + 5 = 0
2
-11, Two complex roots
x + 8x − 4 = 0
2

19. Solve using the Quadratic formula
x − 8 x = 33
2

20. Solve using the Quadratic formula
x − 8 x = 33
2
x − 8 x − 33 = 0
2
x=
− ( − 8) ±
( − 8) − 4(1)( − 33)
2(1)
2

21. Solve using the Quadratic formula
x 2 − 8 x = 33
x 2 − 8 x − 33 = 0
x=
− ( − 8) ±
( − 8) 2 − 4(1)( − 33)
2(1)
8 ± 196 8 ± 14
x=
=
2
2
8 + 14 22
x=
=
= 11
2
2
8 − 14 − 6
x=
=
= −3
2
2

22. Solve using the Quadratic formula
x − 34 x + 289 = 0
2

23. Solve using the Quadratic formula
x − 34 x + 289 = 0
2
x=
− ( − 34 ) ±
( − 34)
2(1)
2
− 4(1)( 289 )

24. Solve using the Quadratic formula
x − 34 x + 289 = 0
2
x=
− ( − 34) ±
( − 34) 2 − 4(1)( 289)
2(1)
34 ± 1156 − 1156
x=
2
34 ± 0 34
x=
=
= 17
2
2

25. Solve using the Quadratic formula
x − 6x + 2 = 0
2

26. Solve using the Quadratic formula
x − 6x + 2 = 0
2
x=
− ( − 6) ±
( − 6)
2(1)
2
− 4(1)( 2 )
6 ± 36 − 8 6 ± 28
x=
=
2
2
6 2 7
x= ±
= 3± 7
2
2

27. Solve using the Quadratic formula
x 2 + 13 = 6 x
x 2 − 6 x + 13 = 0
x=
− ( − 6) ±
( − 6) 2 − 4(1)(13)
2(1)

28. Solve using the Quadratic formula
x 2 + 13 = 6 x
x 2 − 6 x + 13 = 0
x=
− ( − 6) ±
( − 6) 2 − 4(1)(13)
2(1)
6 ± 36 − 52 6 ± − 16
x=
=
2
2

29. Solve using the Quadratic formula
x 2 + 13 = 6 x
x 2 − 6 x + 13 = 0
x=
− ( − 6) ±
( − 6) 2 − 4(1)(13)
2(1)
6 ± 36 − 52 6 ± − 16
x=
=
2
2
6 ± 4i 6 4
x=
= ± i
2
2 2
x = 3 ± 2i