1. 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

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

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

4. Lesson 2 Ex1 Two Roots Make a table of values to find other points to sketch the graph. To solve x 2 – 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.

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

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

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

8. Lesson 2 Ex2 A Double Root Graph the related function f(x) = x 2 – 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.

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

10. Lesson 2 CYP2 1.A 2.B 3.C 4.D A.{1} B.{–1} C.{–1, 1} D.Ø Solve x 2 + 2x = –1 by graphing.

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

12. 1.A 2.B 3.C 4.D Lesson 2 CYP3 A.{1, 5} B.{–1, 5} C.{5} D.Ø Solve x 2 + 4x + 5 = 0 by graphing.

13. Lesson 2 Ex4 Solve x 2 – 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) = x 2 – 4x + 2. Rational Roots Notice that the value of the function changes from negative to positive between the x values of 0 and 1 and between 3 and 4.

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

15. A.A B.B C.C D.D Lesson 2 CYP4 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 Solve x 2 – 2x – 5.

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

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

18. A.A B.B C.C D.D Lesson 2 CYP5 A.between 7 and 8 seconds B.between 6 and 7 seconds C.between 8 and 9 seconds D.between 0 and 1 second GOLF Martin hits a golf ball with an upward velocity of 120 feet per second. The function y = –16t 2 + 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?