Lesson 4 Menu : Lesson 4 Menu Five-Minute Check (over Lesson 9-3)
Main Ideas and Vocabulary
California Standards
Key Concept: The Quadratic Formula
Example 1: Solve Quadratic Equations
Concept Summary: Solving Quadratic Equations
Example 2: Real-World Example: Use the Quadratic Formula to Solve a Problem
Key Concept: Using the Discriminant
Example 3: Use the Discriminant
Lesson 4 MI/Vocab : 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.
Lesson 4 CA : Lesson 4 CA Standard 19.0 Students know the quadratic formula and are familiar with its proof by completing the square. (Key)
Standard 20.0 Students use the quadratic formula to find the roots of a second-degree polynomial and to solve quadratic equations. (Key)
Standard 22.0 Students use the quadratic formula or factoring techniques or both to determine whether the graph of a quadratic function will intersect the x-axis in zero, one, or two points
Key Concept 9-4a : Key Concept 9-4a
Lesson 4 Ex1 : Lesson 4 Ex1 Solve Quadratic Equations A. Solve x2 – 2x – 35 = 0. Round to the nearest tenth if necessary. Method 1 Factoring 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 : Lesson 4 Ex1 Solve Quadratic Equations Method 2 Quadratic Formula Quadratic Formula a = 1, b = –2, and c = –35 Multiply.
Lesson 4 Ex1 : Lesson 4 Ex1 Solve Quadratic Equations Add. Simplify. Answer: The solution set is {–5, 7}. = 7 = –5
Lesson 4 Ex1 : Lesson 4 Ex1 Solve Quadratic Equations B. Solve 15x2 – 8x = 4. Round to the nearest tenth if necessary. Step 1 Rewrite 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 : Lesson 4 Ex1 Solve Quadratic Equations Step 2 Apply the Quadratic Formula. Quadratic Formula a = 15, b = –8, and c = –4 Multiply.
Lesson 4 Ex1 : Lesson 4 Ex1 Solve Quadratic Equations Add. Separate the solutions. Simplify.
Lesson 4 Ex1 : 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}.
Slide 12 : A
B
C
D A. {6, –5}
B. {–6, 5}
C. {6, 5}
D. Ø A. Solve x2 + x – 30 = 0. Round to the nearest tenth if necessary.
Slide 13 : A
B
C
D A. {0.5, 0.7}
B. {–0.5, –0.7}
C. {–0.5, 0.7}
D. Ø B. Solve 20x2 – 4x = 8. Round to the nearest tenth if necessary.
Concept Summary 9-4b : Concept Summary 9-4b
Lesson 4 Ex2 : 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
Lesson 4 Ex2 : 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
Lesson 4 Ex2 : Lesson 4 Ex2 To find accurate solutions, use the Quadratic Formula. Use the Quadratic Formula to Solve a Problem
Lesson 4 Ex2 : 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
Slide 19 : A
B
C
D A. about 10 seconds
B. about 25.2 seconds
C. about 12.5 seconds
D. about 14.5 seconds
Key Concept 9-4c : Key Concept 9-4c
Lesson 4 Ex3 : Lesson 4 Ex3 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.
Lesson 4 Ex3 : Lesson 4 Ex3 Use the Discriminant = 244 Simplify. Answer: The discriminant is 244. Since the discriminant is positive, the equation has two real roots. Step 2 Find the discriminant. b2 – 4ac = (10)2 – 4(3)(–12) a = 3, b = 10, and c = –12
Lesson 4 Ex3 : Lesson 4 Ex3 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.
Slide 24 : A
B
C
D 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 x2 + 2x + 2 = 0. Then determine the number of real roots for the equation.
Slide 25 : A
B
C
D 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 –5x2 + 10x = –1. Then determine the number of real roots for the equation.
End of Lesson 4 : End of Lesson 4