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Section 9.5 Day 1 Solving Quadratic Equations by using the Quadratic Formula Algebra 1.

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Presentation on theme: "Section 9.5 Day 1 Solving Quadratic Equations by using the Quadratic Formula Algebra 1."— Presentation transcript:

1 Section 9.5 Day 1 Solving Quadratic Equations by using the Quadratic Formula
Algebra 1

2 Learning Targets Solve quadratic equations by using the Quadratic Formula Define discriminant Use the discriminant to determine the number of solutions of a quadratic equation Summarize the differences between solving a quadratic equation by graphing, factoring, using square roots, completing the square, and the quadratic formula.

3 𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 Quadratic Formula
For 𝑎 𝑥 2 +𝑏𝑥+𝑐=0, the solutions are 𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎

4 Solving using Quadratic Equations - Example 1
Solve 𝑥 2 −12𝑥=−20 using the Quadratic Formula 1. 𝑥 2 −12𝑥+20=0: 𝑎=1, 𝑏=−12, 𝑐=20 2. Plug in: 𝑥= − −12 ± −12 2 − 3. Simplify: 𝑥= 12± 144−80 2 = 12± = 12±8 2 𝑥= =10 and 𝑥= 12−8 2 =2 𝑥=10 𝑎𝑛𝑑 𝑥=2

5 Solving using Quadratic Equations - Example 2
Solve 10 𝑥 2 −5𝑥=25 by using the Quadratic Formula 1. 10 𝑥 2 −5𝑥−25=0, 𝑎=10, 𝑏=−5, 𝑐=−25 2. 𝑥= − −5 ± −5 2 −4 10 − = 5± = 5± 𝑥= 5± = 1±

6 Solving using Quadratic Equations - Example 3
Solve 6 𝑥 2 −2𝑥+1=0 1. 𝑎=6, 𝑏=−2, 𝑐=1 2. Plug in: 𝑥= − −2 ± −2 2 − 3. Simplify: 𝑥= 2± 4−24 4 = 2± −20 4 4. No Solutions

7 Solving using Quadratic Equations - Example 4
Solve 9 𝑥 2 +12𝑥=−4 1. 9 𝑥 2 +12𝑥+4=0;𝑎=9, 𝑏=12, 𝑐=4 2. Plug in: 𝑥= − 12 ± − 3. Simplify: 𝑥= −12± 144− = −12± =− 12 18 4. 𝑥=− =− 2 3

8 Discriminant The discriminant of the quadratic formula is the expression under the radical. 𝑏 2 −4𝑎𝑐 It is used to determine the number of solutions a quadratic equation will have

9 Discriminant It comes from 𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2 𝑎
If 𝑏 2 −4𝑎𝑐>0, there are 2 solutions If 𝑏 2 −4𝑎𝑐=0, there is 1 solution If 𝑏 2 −4𝑎𝑐<0, there are no real solutions

10 Determine the number of solutions Example 1
How many solutions does 𝑥 2 +2𝑥+5=0 produce? 𝑎=1, 𝑏=2, 𝑐=5 𝑏 2 −4𝑎𝑐= − =4−20=−16<0 No real solutions

11 Determine the number of solutions Example 2
How many solutions does 𝑥 2 +10𝑥+25=0 produce? 𝑎=1, 𝑏=10, 𝑐=25 𝑏 2 −4𝑎𝑐= − =100−100=0 One real solution

12 Determine the number of solutions Example 3
How many solutions does 2 𝑥 2 −7𝑥+2=0 produce? 𝑎=2, 𝑏=−7, 𝑐=2 𝑏 2 −4𝑎𝑐= −7 2 − =49−16=33>0 Two Real Solutions

13 Summary Factoring Graphing Using Square Roots Completing the Square
GCF, “X” Method, Difference of Squares Zero Product Property Graphing Standard, Vertex, Intercept Roots, Zeros, X-Intercepts Using Square Roots Taking the square root of each side Completing the Square 𝑏 2𝑎 2 , taking the square root of each side Quadratic Formula 𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎


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