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Lesson 4 Menu

Five-Minute Check (over Lesson 9-3) Main Ideas and Vocabulary Targeted TEKS Key Concept: The Quadratic Formula Example 1: Solve Quadratic Equations Concept Summary: Solving Quadratic Equations Example 2: Use the Quadratic Formula to Solve a Problem Key Concept: Using the Discriminant

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Lesson 4 Menu

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  1. Five-Minute Check (over Lesson 9-3) Main Ideas and Vocabulary Targeted TEKS Key Concept: The Quadratic Formula Example 1: Solve Quadratic Equations Concept Summary: Solving Quadratic Equations Example 2: Use the Quadratic Formula to Solve a Problem Key Concept: Using the Discriminant Example 3: Use the Discriminant Lesson 4 Menu

  2. Solve quadratic equations by using the Quadratic Formula. • Use the discriminant to determine the number of solutions for a quadratic equation. • Quadratic Formula • discriminant Lesson 4 MI/Vocab

  3. Key Concept 9-4a

  4. Solve Quadratic Equations A. Solve x2 – 2x – 35 = 0. Round to the nearest tenth if necessary. Method 1Factoring x2 – 2x – 35 = 0 Original equation (x –7)(x + 5) = 0 Factor x2 –2x – 35. x –7 = 0 or x + 5 = 0 Zero Product Property x = 7 x = –5 Solve for x. Lesson 4 Ex1

  5. Solve Quadratic Equations Method 2 Quadratic Formula Quadratic Formula a = 1, b = –2, and c = –35 Multiply. Lesson 4 Ex1

  6. Separate the solutions. or Solve Quadratic Equations Add. Simplify. = 7 = –5 Answer: The solution set is {–5, 7}. Lesson 4 Ex1

  7. A. Solve x2 + x – 30 = 0. Round to the nearest tenth if necessary. • A • B • C • D A. {6, –5} B. {–6, 5} C. {6, 5} D.Ø Lesson 4 CYP1

  8. Solve Quadratic Equations B. Solve 15x2 – 8x = 4. Round to the nearest tenth if necessary. Step 1Rewrite the equation in standard form. 15x2 – 8x = 4 Original equation 15x2 – 8x – 4 = 4 – 4 Subtract 4 from each side. 15x2 – 8x – 4 = 0 Simplify. Lesson 4 Ex1

  9. Solve Quadratic Equations Step 2Apply the Quadratic Formula. Quadratic Formula a = 15, b = –8, and c = –4 Multiply. Lesson 4 Ex1

  10. or Solve Quadratic Equations Add. Separate the solutions. Simplify. Lesson 4 Ex1

  11. Solve Quadratic Equations Check the solutions by using the CALC menu on a graphing calculator to determine the zeros of the related quadratic function. Answer: To the nearest tenth, the set is {–0.3, 0.8}. Lesson 4 Ex1

  12. B. Solve 20x2 – 4x = 8. Round to the nearest tenth if necessary. • A • B • C • D A. {0.5, 0.7} B. {–0.5, –0.7} C. {–0.5, 0.7} D.Ø Lesson 4 CYP1

  13. Concept Summary 9-4b

  14. Use the Quadratic Formula to Solve a Problem SPACE TRAVELTwo possible future destinations of astronauts are the planet Mars and a moon of the planet Jupiter, Europa. The gravitational acceleration on Mars is about 3.7 meters per second squared. On Europa, it is only 1.3 meters per second squared. Using the information and equation from Example 2 in the textbook, find how much longer baseballs thrown on Mars and on Europa will stay above the ground than similarly thrown baseballs on Earth. In order to find when the ball hits the ground, you must find when H = 0. Write two equations to represent the situation on Mars and on Europa. Lesson 4 Ex2

  15. Use the Quadratic Formula to Solve a Problem Baseball Thrown on Mars Baseball Thrown on Europa These equations cannot be factored, and completing the square would involve a lot of computation. Lesson 4 Ex2

  16. Use the Quadratic Formula to Solve a Problem To find accurate solutions, use the Quadratic Formula. Lesson 4 Ex2

  17. Use the Quadratic Formula to Solve a Problem Since a negative number of seconds is not reasonable, use the positive solutions. Answer: A ball thrown on Mars will stay aloft 5.6 – 2.2 or about 3.4 seconds longer than the ball thrown on Earth. The ball thrown on Europa will stay aloft 15.6 – 2.2 or about 13.4 seconds longer than the ball thrown on Earth. Lesson 4 Ex2

  18. SPACE TRAVEL The gravitational acceleration on Venus is about8.9 meters per second squared, and on Callisto, oneof Jupiter’s moons, it is 1.2 meters per secondsquared. Suppose a baseball is thrown on Callistowith an upward velocity of 10 meters per second fromtwo meters above the ground. Findhow much longerthe ball will stay in air than a similarly-thrownball on Venus.Use the equation where H isthe height of anobject t seconds after it is thrown upward, v is the initial velocity, g is the gravitational pull, and h is the initial height. • A • B • C • D A. about 10 seconds B. about 25.2 seconds C. about 12.5 seconds D. about 14.5 seconds Lesson 4 CYP2

  19. Key Concept 9-4c

  20. Use the Discriminant A.State the value of the discriminant for 3x2 + 10x = 12. Then determine the number of real roots of the equation. Step 1 Rewrite the equation in standard form. 3x2 + 10x = 12 Original equation 3x2 + 10x– 12 = 12 – 12 Subtract 12 from each side. 3x2 + 10x – 12 = 0 Simplify. Step 2 Find the discriminant. b2 – 4ac = (10)2 – 4(3)(–12) a = 3, b = 10, and c = –12 Lesson 4 Ex3

  21. Use the Discriminant = 244 Simplify. Answer: The discriminant is 244. Since the discriminant is positive, the equation has two real roots. Lesson 4 Ex3

  22. A. State the value of the discriminant for the equation x2 + 2x + 2 = 0. Then determine the number of real roots for the equation. • A • B • C • D A. –4; no real roots B. 4; 2 real roots C. 0; 1 real root D. cannot be determined Lesson 4 CYP3

  23. Use the Discriminant B.State the value of the discriminant for 4x2 – 2x + 14 = 0. Then determine the number of real roots of the equation. b2 – 4ac = (–2)2 – 4(4)(14) a = 4, b = –2, and c = 14 = –220 Simplify. Answer: The discriminant is –220. Since the discriminant is negative, the equation has no real roots. Lesson 4 Ex3

  24. B. State the value of the discriminant for the equation –5x2 + 10x = –1. Then determine the number of real roots for the equation. • A • B • C • D A. –120; no real roots B. 120; 2 real roots C. 0; 1 real root D. cannot be determined Lesson 4 CYP3

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