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Lesson 2 Menu

Five-Minute Check (over Lesson 9-1) Main Ideas and Vocabulary Targeted TEKS Example 1: Two Roots Example 2: A Double Root Example 3: No Real Roots Example 4: Rational Roots Example 5: Real-World Example. Lesson 2 Menu. Solve quadratic equations by graphing.

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Lesson 2 Menu

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  1. Five-Minute Check (over Lesson 9-1) Main Ideas and Vocabulary Targeted TEKS Example 1: Two Roots Example 2: A Double Root Example 3: No Real Roots Example 4: Rational Roots Example 5: Real-World Example Lesson 2 Menu

  2. Solve quadratic equations by graphing. • Estimate solutions of quadratic equations by graphing. • quadratic equation • roots • zeros • double root Lesson 2 MI/Vocab

  3. Two Roots Solve x2 – 3x – 10 = 0 by graphing. Graph the related function f(x) = x2 – 3x – 10. Lesson 2 Ex1

  4. Two Roots Make a table of values to find other points to sketch the graph. To solve x2 – 3x – 10 = 0 you need to know where the value of f(x) is 0. This occurs at the x-intercepts. The x-intercepts of the parabola appear to be –2 and 5. Lesson 2 Ex1

  5. Two Roots Check Solve by factoring. x2 – 3x – 10 = 0 Original equation (x – 5)(x + 2) = 0 Factor. x – 5 = 0 or x + 2 = 0 Zero Product Property x = 5 x = –2 Solve for x.   Answer: The solutions of the equation are –2 and 5. Animation: Solving Quadratic Equations By Graphing Lesson 2 Ex1

  6. Solve x2 – 2x – 8 = 0 by graphing. • A • B • C • D A. {–2, 4} B. {2, –4} C. {2, 4} D. {–2, –4} Lesson 2 CYP1

  7. A Double Root Solve x2 – 6x = –9 by graphing. First, rewrite the equation so one side is equal to zero. x2 – 6x = –9 Original equation x2 – 6x+ 9 = –9 + 9 Add 9 to each side. x2 – 6x + 9 = 0 Simplify. Lesson 2 Ex2

  8. A Double Root Graph the related function f(x) = x2 – 6x + 9. Notice that the vertex of the parabola is the x-intercept. Thus, one solution is 3. What is the other solution? Try solving the equation by factoring. Lesson 2 Ex2

  9. A Double Root x2 – 6x + 9 = 0 Original equation (x – 3)(x – 3) = 0 Factor. x – 3 = 0 or x – 3 = 0 Zero Product Property x = 3 x = 3 Answer: The solution is 3. Lesson 2 Ex2

  10. Solve x2 + 2x = –1 by graphing. • A • B • C • D A. {1} B. {–1} C. {–1, 1} D.Ø Lesson 2 CYP2

  11. No Real Roots Solve x2 + 2x + 3 = 0 by graphing. Graph the related function f(x) = x2 + 2x + 3. Answer: The graph has no x-intercept. Thus, there are no real number solutions for the equation. Lesson 2 Ex3

  12. Solve x2 + 4x + 5 = 0 by graphing. • A • B • C • D A. {1, 5} B. {–1, 5} C. {5} D.Ø Lesson 2 CYP3

  13. Noticethat the value of thefunction changesfromnegative to positivebetween the x values of0 and 1 and between 3and 4. Rational Roots Solve x2 – 4x + 2 = 0 by graphing. If integral roots cannot be found, estimate the roots by stating the consecutive integers between which the roots lie. Graph the related function f(x) = x2 – 4x + 2. Lesson 2 Ex4

  14. Rational Roots The x-intercepts of the graph are between 0 and 1 and between 3 and 4. Answer: One root is between 0 and 1, and the other root is between 3 and 4. Lesson 2 Ex4

  15. Solve x2 – 2x – 5. • A • B • C • D A. one root is between 0 and 1, and the other root is between 4 and 5. B. one root is between –1 and 0, and the other root is between 3 and 4. C. one root is between –1 and –2, and the other root is between 3 and 4. D.Ø Lesson 2 CYP4

  16. MODEL ROCKETSShelly built a model rocket for her science project.The equation y = –16t2 + 250t models the flight ofthe rocket, launched from ground level at a velocity of 250 feet per second, where y is the height of the rocketin feet after t seconds. For how many seconds wasShelly’s rocket in the air? You need to find the solution of the equation 0 = –16t2+ 250t. Use a graphing calculator to graph the related function y = –16t2 + 250t. The x-intercept is between 15 and 16 seconds. Lesson 2 Ex5

  17. Answer: between 15 and 16 seconds. Lesson 2 Ex5

  18. GOLFMartin hits a golf ball with an upward velocity of 120 feet per second. The function y = –16t2 + 120t models the flight of the golf ball, hit at ground level, where y is the height of the ball in feet after t seconds. How long was the golf ball in the air? • A • B • C • D A. between 7 and 8 seconds B. between 6 and 7 seconds C. between 8 and 9 seconds D. between 0 and 1 second Lesson 2 CYP5

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