Task 1. How many solutions does each equation have? what are the solutions? 1. a2 = 9 2. x² = 0 3. 2? – 1 = 3 %3D 4. 2x² 50 %3D 5. (x + 1)(x + 1) = 0 6. #(x – 6) = 0 7. (x – 1)(x – 1) = 4 8. 12x2 + 24x = 0 9. x2 - 8x + 15 = 0 10. x2 - 8x + 15 = 65

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.2: Graphs Of Equations
Problem 72E
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Lesson 4.5 Solving Polynomial Equations of Second
Degrees. Connecting Zeros, Factors, and X -
Intercepts.
Sunday, April 11, 2021 7:52 PM
Learning Targets:
• I know how to find the zeros of a polynomial function of second degree
using graph or algebra.
I know how the zeros of a polynomial function relate to factored form of
a polynomial, and to x-intercepts.
Task 1. How many solutions does each equation have? what are the solutions?
1. 22
2. x²
3. x2 – 1 = 3
9
%D
%3D
4. 2x2
5. (x + 1)(x + 1) = 0
6. 2 (х — 6) — 0
7. (х — 1)(ӕ — 1) — 4
50
8. 12x2 + 24x = 0
9. x2 - 8x + 15 = 0
10. x2 - 8x + 15 = 65
Transcribed Image Text:Lesson 4.5 Solving Polynomial Equations of Second Degrees. Connecting Zeros, Factors, and X - Intercepts. Sunday, April 11, 2021 7:52 PM Learning Targets: • I know how to find the zeros of a polynomial function of second degree using graph or algebra. I know how the zeros of a polynomial function relate to factored form of a polynomial, and to x-intercepts. Task 1. How many solutions does each equation have? what are the solutions? 1. 22 2. x² 3. x2 – 1 = 3 9 %D %3D 4. 2x2 5. (x + 1)(x + 1) = 0 6. 2 (х — 6) — 0 7. (х — 1)(ӕ — 1) — 4 50 8. 12x2 + 24x = 0 9. x2 - 8x + 15 = 0 10. x2 - 8x + 15 = 65
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