1. Warm-Up 𝒙=𝟑 𝒐𝒓 𝒙=𝟔 𝒙≈𝟏.𝟑 𝒐𝒓 𝒙≈−𝟓.𝟑
Solve each quadratic using any method. (Factoring or completing the square.) Round to the nearest tenth if necessary. Explain why you chose the method you did for each problem.
1.) −1= 𝑥 2 −9𝑥 ) 𝑥 2 +4𝑥−7=0
0= 𝑥 2 −9𝑥+18
0= 𝑥 2 −6𝑥−3𝑥+18
0= 𝑥 2 −6𝑥 −3𝑥+18
0=𝑥 𝑥−6 −3 𝑥−6
0= 𝑥−3 𝑥−6
𝑥−3=0 𝑜𝑟 𝑥−6=0
𝒙=𝟑 𝒐𝒓 𝒙=𝟔
𝑥 2 +4𝑥−7=0
𝑥 2 +4𝑥=7
𝑥 2 +4𝑥+4=7+4
𝑥+2 2 =11
𝑥+2=± 11
𝑥=−2± 11
𝑥=− 𝑜𝑟 𝑥=−2− 11
𝒙≈𝟏.𝟑 𝒐𝒓 𝒙≈−𝟓.𝟑

2. The Quadratic Formula and the Discriminant
Mr. Riddle

3. Review
We’ve learned three different ways to find the zeros/roots of a function.
Solving by Graphing
Solving by Factoring
Solving by Completing the Square
and now…

4. The Quadratic Formula 𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
The Quadratic formula is another way to help us find the solutions or the roots/zeroes of a quadratic function.
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎

5. Where does Quadratic Formula come from?
Quadratic formula comes from completing the square when a quadratic is in standard form and set equal to 0.
Solve by completing the square.
𝑎 𝑥 2 +𝑏𝑥+𝑐=0
𝑎 𝑥 2 +𝑏𝑥=−𝑐 (Subtract c)
𝑎 𝑥 2 + 𝑏 𝑎 𝑥 =−𝑐 (Factor out a)
𝑎 𝑥 2 + 𝑏 𝑎 𝑥+ 𝑏 2 4 𝑎 2 =−𝑐+ 𝑏 2 4 𝑎 2 (𝑎) (Complete the square)
𝑎 𝑥+ 𝑏 2𝑎 2 =−𝑐+ 𝑏 2 4𝑎 (Factor the Square)
𝑥+ 𝑏 2𝑎 2 =− 𝑐 𝑎 + 𝑏 2 4 𝑎 (Divide by a)
𝑥+ 𝑏 2𝑎 2 = 𝑏 2 −4𝑎𝑐 4 𝑎 (Get a common denominator and Combine)
𝑥+ 𝑏 2𝑎 = ± 𝑏 2 −4𝑎𝑐 2𝑎 (Square root both sides)
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 (Subtract 𝑏 2𝑎 from both sides)

6. Example 1: Using the Quadratic Formula
Solve by using the quadratic formula.
a. 𝑥 2 +4𝑥−32=0
𝑎=1;𝑏=4;𝑐=−32
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
𝑥= − 4 ± −4 1 −
𝑥= −4±
𝑥= −4±
𝑥= −4±12 2
𝑥= − and −4−12 2
𝑥=4 and 𝑥=−8
b. 5 𝑥 2 +6𝑥=−1
5 𝑥 2 +6𝑥+1=0
𝑎=5;𝑏=6;𝑐=1
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
𝑥= − 6 ± −
𝑥= −6± 36−20 10
𝑥= −6±
𝑥= −6±4 10
𝑥= − and −6−4 10
𝑥=−0.2 and 𝑥=−1

7. You Try! a. 10 𝑥 2 −5𝑥=25 10 𝑥 2 −5𝑥−25=0 𝑎=10 𝑏=−5 𝑐=−25
Solve by using the quadratic formula. Round your answer to the nearest tenth.
a 𝑥 2 −5𝑥=25
10 𝑥 2 −5𝑥−25=0
𝑎=10 𝑏=−5 𝑐=−25
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
𝑥= (5)± −5 2 −4 10 −
𝑥= 5±
𝑥= 5±
𝑥≈1.9 𝑎𝑛𝑑 −1.4
b. 4 𝑥 2 −24𝑥+35=0
𝑎=4 𝑏=−24 𝑐=35
𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎
𝑥= (24)± −24 2 −4(4)
𝑥= 24± 576−560 8
𝑥= 24±
𝑥= 24±4 8
𝑥=3.5 𝑎𝑛𝑑 2.5

8. Example 2: Using Multiple Methods
Solve the quadratic 𝑥 2 −8𝑥=−16 by:
Factoring
B. Completing the Square
C. The Quadratic Formula

9. You Try Solve the quadratic 𝑥 2 −2𝑥−15=0 by: Factoring
B. Completing the Square
C. The Quadratic Formula

10. The Discriminant
The Discriminant of a quadratic tell us how many solutions there are to a quadratic.
𝐷𝑖𝑠𝑐𝑟𝑖𝑚𝑖𝑛𝑎𝑛𝑡 → 𝑏 2 −4𝑎𝑐
If 𝑏 2 −4𝑎𝑐 is positive, there are 2 real solutions or roots.
If 𝑏 2 −4𝑎𝑐 is zero, there is 1 solution or root.
If 𝑏 2 −4𝑎𝑐 is negative, there are no real solutions or roots and 2 imaginary roots.

11. Example 3: Using the Discriminant
State the value of the discriminant of 4 𝑥 2 +5𝑥=−3. Then determine the number of real solutions of the equation.
4 𝑥 2 +5𝑥=−3
4 𝑥 2 +5𝑥+3=0
𝑎=4;𝑏=5;𝑐=3
𝑏 2 −4𝑎𝑐
5 2 −4 4 3
25−48
−23
The quadratic has zero real solutions since the discriminant is negative and two imaginary roots.

12. You Try! 𝒂. 𝟐 𝒙 𝟐 +𝟏𝟏𝒙+𝟏𝟓=𝟎 𝒃. 𝟗 𝒙 𝟐 −𝟑𝟎𝒙+𝟐𝟓=𝟎 𝑎=2;𝑏=11;𝑐=15 𝑏 2 −4𝑎𝑐
Find the value of the discriminant and state the number of real solutions.
𝒂. 𝟐 𝒙 𝟐 +𝟏𝟏𝒙+𝟏𝟓=𝟎
𝒃. 𝟗 𝒙 𝟐 −𝟑𝟎𝒙+𝟐𝟓=𝟎
𝑎=2;𝑏=11;𝑐=15
𝑏 2 −4𝑎𝑐
11 2 −
121−120 1
The quadratic has two real solutions since the discriminant is positive.
𝑎=9;𝑏=−30;𝑐=25
𝑏 2 −4𝑎𝑐
−30 2 −
900−900
The quadratic has one real solution since the discriminant is zero.

13. Imaginary Numbers? – Part 1

14. Imaginary Numbers – Part 2

15. Imaginary Numbers – Part 3

16. Imaginary Numbers? Part 1 - Introduction
Part 2 - Why we need imaginary numbers!
Part 3 - "i" think I can!

17. Imaginary and Complex Numbers
An imaginary number is the product of a real number and the square root of -1.
Example: −25 = − 1=5 −1 or 5𝑖.
𝑖= −1
𝑖 2 =−1
A complex number is the sum of a real number and an imaginary number written in the form 𝑎+𝑏𝑖
Example: 3+4𝑖 or 3+4 −1

18. Example 4: Imaginary Roots
Solve using quadratic formula:
𝑥 2 +4𝑥+5=0
𝑎=1;𝑏=4;𝑐=5
𝑥= − 4 ± (4) 2 −4 1 (5) 2(1)
𝑥= −4± 16−20 2
𝑥= −4± −4 2
𝑥= −4± 4 −1 2
𝑥= −4±2𝑖 2
𝒙=−𝟐±𝒊

19. You Try! Solve using quadratic formula: 𝑥 2 −6𝑥+25=0 𝑎=1;𝑏=−6;𝑐=25
𝑥= − −6 ± −4(1)(25) 2(1)
𝑥= 6± 36−100 2
𝑥= 6± −64 2
𝑥= 6± −1 2
𝑥= 6±8𝑖 2
𝒙=𝟑±𝟒𝒊