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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 Example 3: Use the Discriminant

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

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Key Concept 9-4a

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

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

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

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A.A B.B C.C D.D Lesson 4 CYP1 A.{6, –5} B.{–6, 5} C.{6, 5} D.Ø A. Solve x 2 + x – 30 = 0. Round to the nearest tenth if necessary.

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Lesson 4 Ex1 Solve Quadratic Equations B. Solve 15x 2 – 8x = 4. Round to the nearest tenth if necessary. Step 1 Rewrite the equation in standard form. 15x 2 – 8x = 4Original equation 15x 2 – 8x – 4 = 4 – 4 Subtract 4 from each side. 15x 2 – 8x – 4 = 0 Simplify.

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

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

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Lesson 4 Ex1 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}.

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A.A B.B C.C D.D Lesson 4 CYP1 A.{0.5, 0.7} B.{–0.5, –0.7} C.{–0.5, 0.7} D.Ø B. Solve 20x 2 – 4x = 8. Round to the nearest tenth if necessary.

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Concept Summary 9-4b

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Lesson 4 Ex2 SPACE TRAVEL Two 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. Use the Quadratic Formula to Solve a Problem

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

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

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Lesson 4 Ex2 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. Use the Quadratic Formula to Solve a Problem

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Lesson 4 CYP2 1.A 2.B 3.C 4.D A.about 10 seconds B.about 25.2 seconds C.about 12.5 seconds D.about 14.5 seconds SPACE TRAVEL The gravitational acceleration on Venus is about 8.9 meters per second squared, and on Callisto, one of Jupiter’s moons, it is 1.2 meters per second squared. Suppose a baseball is thrown on Callisto with an upward velocity of 10 meters per second from two meters above the ground. Find how much longer the ball will stay in air than a similarly-thrown ball on Venus. Use the equation where H is the height of an object t seconds after it is thrown upward, v is the initial velocity, g is the gravitational pull, and h is the initial height.

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Key Concept 9-4c

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

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

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1.A 2.B 3.C 4.D Lesson 4 CYP3 A.–4; no real roots B.4; 2 real roots C.0; 1 real root D.cannot be determined A. State the value of the discriminant for the equation x 2 + 2x + 2 = 0. Then determine the number of real roots for the equation.

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

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1.A 2.B 3.C 4.D Lesson 4 CYP3 A.–120; no real roots B.120; 2 real roots C.0; 1 real root D.cannot be determined B. State the value of the discriminant for the equation –5x 2 + 10x = –1. Then determine the number of real roots for the equation.

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Splash Screen. Lesson Menu Five-Minute Check (over Lesson 9–1) CCSS Then/Now New Vocabulary Key Concept: Solutions of Quadratic Equations Example 1: Two.

Splash Screen. Lesson Menu Five-Minute Check (over Lesson 9–1) CCSS Then/Now New Vocabulary Key Concept: Solutions of Quadratic Equations Example 1: Two.

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