1. 3.3: The Quadratic Formula
Mention how important the quadratic formula is because we can’t always factor every quadratic equation. Show why 3x^2+2x-7 doesn’t work.

2. Review
A quadratic equation is an equation that can be written in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐=0, where a, b, and c are real numbers and 𝑎≠0.

3. 𝒙= −(𝒃)± (𝒃) 𝟐 −𝟒(𝒂)(𝒄) 𝟐(𝒂)
Important Properties
Quadratic Formula: The solutions of 𝑎 𝑥 2 +𝑏𝑥+𝑐=0 are given by:
𝒙= −(𝒃)± (𝒃) 𝟐 −𝟒(𝒂)(𝒄) 𝟐(𝒂)
The Quadratic Formula can be used to solve any quadratic equation!!!
This formula finds the x-intercepts/zeros/roots of a quadratic function

5. Example #1 2 𝑥 2 −𝑥=4 2 𝑥 2 −𝑥−4=0 𝒂=𝟐, 𝒃=−𝟏, 𝒄=−𝟒
𝑥= −(𝑏)± (𝑏) 2 −4(𝑎)(𝑐) 2(𝑎)
Example #1
Remember, we need it to look like 𝑎 𝑥 2 +𝑏𝑥+𝑐=0 in order to use the Quadratic Formula. Otherwise, IT DOESN’T WORK!!!
What are the solutions? Use the Quadratic Formula.
2 𝑥 2 −𝑥=4
2 𝑥 2 −𝑥−4=0
𝒂=𝟐, 𝒃=−𝟏, 𝒄=−𝟒
𝑥= −(−𝟏)± (−𝟏) 2 −4(𝟐)(−𝟒) 2(𝟐)
𝒙= 1± = 1±

6. On Your Own 𝒂=𝟔, 𝒃=−𝟓, 𝒄=−𝟒 6 𝑥 2 −5𝑥−4=0
𝑥= −(𝑏)± (𝑏) 2 −4(𝑎)(𝑐) 2(𝑎)
On Your Own
Remember, we need it to look like 𝑎 𝑥 2 +𝑏𝑥+𝑐=0 in order to use the Quadratic Formula. Otherwise, IT DOESN’T WORK!!!
What are the solutions? Use the Quadratic Formula.
6 𝑥 2 −5𝑥−4=0
𝒂=𝟔, 𝒃=−𝟓, 𝒄=−𝟒
𝑥= −(−𝟓)± (−𝟓) 2 −4(𝟔)(−𝟒) 2(6)
𝒙= 5± = 5± = 5±11 12
= and 5−11 12
= and − 6 12
= 𝟒 𝟑 and −𝟏 𝟐

7. Example #2: Check your answer by factoring!!! 𝑥 2 +6𝑥+9=0
𝑥= −(𝑏)± (𝑏) 2 −4(𝑎)(𝑐) 2(𝑎)
Example #2:
Remember, we need it to look like 𝑎 𝑥 2 +𝑏𝑥+𝑐=0 in order to use the Quadratic Formula and notice that it’s already done for us!
What are the solutions? Use the Quadratic Formula.
𝑥 2 +6𝑥+9=0
𝒂=𝟏, 𝒃=𝟔, 𝒄=𝟗
𝑥= −(𝟔)± (𝟔) 2 −4(𝟏)(𝟗) 2(𝟏)
𝑥 2 +6𝑥+9=0
𝑥+3 𝑥+3 =0
𝑥+3=0 means 𝑥=−3
𝒙= −6± 36−36 2 = −6± =−𝟑
Check your answer by factoring!!!

8. Example #3 3 𝑥 2 −4𝑥=−10 𝒂=3, 𝒃=−4, 𝒄=𝟏𝟎 3 𝑥 2 −4𝑥+10=0
𝑥= −(𝑏)± (𝑏) 2 −4(𝑎)(𝑐) 2(𝑎)
Example #3
Remember, we need it to look like 𝑎 𝑥 2 +𝑏𝑥+𝑐=0 in order to use the Quadratic Formula. Otherwise, IT DOESN’T WORK!!!
What are the solutions? Use the Quadratic Formula.
3 𝑥 2 −4𝑥=−10
3 𝑥 2 −4𝑥+10=0
𝒂=3, 𝒃=−4, 𝒄=𝟏𝟎
𝑥= −(−4)± (−4) 2 −4(𝟑)(𝟏𝟎) 2(𝟑)
𝒙= 4± 16− = 4± − = 4±𝑖 = 4±2𝑖 = 𝟐±𝒊 𝟐𝟔 𝟑

9. On Your Own 7 𝑥 2 +2𝑥+4=0 𝒂=7, 𝒃=𝟐, 𝒄=𝟒 𝑥= −(𝟐)± (𝟐) 2 −4(𝟕)(𝟒) 2(𝟕)
𝑥= −(𝑏)± (𝑏) 2 −4(𝑎)(𝑐) 2(𝑎)
On Your Own
Remember, we need it to look like 𝑎 𝑥 2 +𝑏𝑥+𝑐=0 in order to use the Quadratic Formula. Otherwise, IT DOESN’T WORK!!!
What are the solutions? Use the Quadratic Formula.
7 𝑥 2 +2𝑥+3=−1
7 𝑥 2 +2𝑥+4=0
𝒂=7, 𝒃=𝟐, 𝒄=𝟒
𝑥= −(𝟐)± (𝟐) 2 −4(𝟕)(𝟒) 2(𝟕)
𝒙= −2± 4− = −2± − = −2±𝑖 = −2±6𝑖 = −𝟏±𝟑𝒊 𝟑 𝟕

10. Definition
The discriminant of a quadratic equation in the form 𝑎 𝑥 2 +𝑏𝑥+𝑐=0 is the value of the expression 𝑏 2 −4𝑎𝑐. So it’s the value inside the square root 
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
The discriminant
If 𝑏 2 −4𝑎𝑐<𝟎, then there are no real solutions to the quadratic equation
If 𝑏 2 −4𝑎𝑐=𝟎, then the quadratic equation has only one real zero.
If 𝑏 2 −4𝑎𝑐>𝟎, then the quadratic equation has two real solutions.

12. Examples of 0, 1, and 2 Solutions:
Two real solutions
𝑥 2 =4 gives 𝑥=−2, 2
One real solution
𝑥 2 =0 gives 𝑥=0
No real solutions
𝑥 2 =−4

13. Example #4: Discriminant = 𝑏 2 −4𝑎𝑐= −𝟑 2 −4 −𝟐 𝟓 =9+40=49
What is the number of real solutions of −2 𝑥 2 −3𝑥+5=0?
Hint: This means we have to know the value of the discriminant!
-2 𝑥 2 −3𝑥+5=0
𝒂=−𝟐, 𝒃=−𝟑, 𝒄=𝟓
Discriminant = 𝑏 2 −4𝑎𝑐= −𝟑 2 −4 −𝟐 𝟓 =9+40=49
Since this number is positive (i.e., > 0), we have two real solutions!

14. Example #5: Discriminant = 𝑏 2 −4𝑎𝑐= −𝟑 2 −4 𝟐 𝟓 =9−40=−31
What is the number of real solutions of 2 𝑥 2 −3𝑥+5=0?
Hint: This means we have to know the value of the discriminant!
2 𝑥 2 −3𝑥+5=0
𝒂=𝟐, 𝒃=−𝟑, 𝒄=𝟓
Discriminant = 𝑏 2 −4𝑎𝑐= −𝟑 2 −4 𝟐 𝟓 =9−40=−31
Since this number is negative (i.e., < 0), we have no real solutions!

15. Note We have now seen three ways to solve 𝑎 𝑥 2 +𝑏𝑥+𝑐=0:
By factoring
By completing the square
Using the Quadratic Formula
Order you should try first:
Factoring
Quadratic Formula
Completing the Square