1. E) Quadratic Formula & Discriminant
Objective:
To solve quadratic equations using the quadratic formula and calculate the number of solutions using the discriminant.

2. Quadratic Formula
Discriminant
and

3. Used to determine the # of x-intercepts
Discriminant:
Used to determine the # of x-intercepts
b2 − 4ac = +
Two real
roots
b2 − 4ac = 0
One real
root
b2 − 4ac = −
No real
roots
No Solution

4. 1) y = x2 + 0x − 4 1 -4 Use the Discriminant to find the # of roots
then graph the equation to find the x-intercepts
1) y = x2
+ 0x
− 4
Discriminant
b2 – 4ac
= ( )2 – 4( )( ) 1 -4
= +
Axis of Symmetry
x = -b 2a
= −( )
2( )
= 0 1
Vertex
y = ( )2 − 4  
= -4
x = {-2, 2} 
2nd Point
y = ( )2 − 4
Choose x = 1 1
= -3

5. 2) y = x2 + 0x + 0 1 Use the Discriminant to find the # of roots
then graph the equation to find the x-intercepts
2) y = x2
+ 0x + 0
Discriminant
b2 – 4ac
= ( )2 – 4( )( ) 1
= 0
x = 0
Axis of Symmetry
x = -b 2a  
= −( )
2( )
= 0  1
Vertex
y = ( )2
= 0
2nd Point
y = ( )2
Choose x = 1
= 0

6. 3) y = x2 + 0x + 4 1 4 Use the Discriminant to find the # of roots
then graph the equation to find the x-intercepts
3) y = x2
+ 0x
+ 4
Discriminant
b2 – 4ac
No x-intercepts   
= ( )2 – 4( )( ) 1 4
= −
Axis of Symmetry
x = -b 2a
= −( )
2( )
= 0 1
Vertex
y = ( )2 + 4
= 4
2nd Point
y = ( )2 + 4
Choose x = 1 1
= 3

7. Used to find the x-intercepts
Play
Video
Quadratic Formula
Used to find the x-intercepts
Axis of Symmetry
Discriminant
x = -b 2a
Rules
1) Write problem as a quadratic equation (ax2 +bx + c = 0)
2) Determine values for a, b, and c
3) Plug values into the Quadratic Formula
4) Simplify the Discriminant (factor out “pairs” if possible)
5) Separate into 2 problems (+) and (−) and solve

8. Example 1 y = -x2 + 5x + 2 - ( ) ± √( )2 − 4( )( ) 2( ) 5 5 -1 2 = -1b c
- ( ) ± √( )2 − 4( )( )
2( ) 5 5 -1 2 = -1 =

9. Example 2 y = 2x2 + 5x − 3 - ( ) ± √( )2 − 4( )( ) 2( ) 5 5 2 -3 2 a bc
- ( ) ± √( )2 − 4( )( )
2( ) 5 5 2 -3 2

10. 3 Example 3 y = 3x2 − 4x − 1 - ( ) ± √( )2 − 4( )( ) 2( ) -4 -4 3 -1 =b c
- ( ) ± √( )2 − 4( )( )
2( ) -4 -4 3 -1 = 3 = = = = 3

11. Example 4 x2 + 3x = 7 1 - ( ) ± √( )2 − 4( )( ) 2( ) 3 3 1 -7 = 1 = ab c
- ( ) ± √( )2 − 4( )( )
2( ) 3 3 1 -7 = 1 =

12. Example 5 y = -x2 + 4x − 7 - ( ) ± √( )2 − 4( )( ) 2( ) 4 4 -1 -7 = -1b c
- ( ) ± √( )2 − 4( )( )
2( ) 4 4 -1 -7 = -1 =
No Solution

13. Example 6
Solve
a = 1
b = -6 a b c
c = 9 = =
= 2

14. 1) 2) y = 2x2 − x + 1 x = 3, 1 2 No Solution 3) y = x2 − x − 2 4)
Classwork 1) 2)
y = 2x2 − x + 1
x = 3, 1 2
No Solution 3)
y = x2 − x − 2 4)
y = x2 − 2x + 1
x = {-1, 2}
x = 1 5)
y = 3x2 − 2x + 4 6)
y = -x2 + 5x + 1
No Solution