1. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.8 – Slide 1

2. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.8 – Slide 2 Factoring and Applications Chapter 6

3. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.8 – Slide 3 6.8 Applications of Quadratic Equations

4. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.8 – Slide 4 Objectives 1.Solve problems about geometric figures. 2.Solve problems about consecutive integers. 3.Solve problems using the Pythagorean formula. 4.Solve problems using given quadratic models. 6.8 Applications of Quadratic Equations

5. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.8 – Slide 5 6.8 Applications of Quadratic Equations Solving an Applied Problem Step 1Read the problem, several times if necessary, until you understand what is given and what is to be found. Step 2Assign a variable to represent the unknown value, using diagrams or tables as needed. Write a statement that tells what the variable represents. Express any other unknown values in terms of the variable. Step 3Write an equation using the variable expression(s). Step 4Solve the equation. Step 5State the answer. Does it seem reasonable? Step 6Check the answer in the words of the original problem.

6. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.8 – Slide 6 Solving Problems about Geometric Figures 6.8 Applications of Quadratic Equations Example 1 Amy designs silver jewelry. Currently she is working on a triangular pendant. She prefers that the height be twice the base. She has enough silver to make the area of the triangle 4 in 2. What will the dimensions of her pendant be? Step 1 Read. Find the base and height of Amy’s pendant. Step 2 Assign a variable. Let h = height and b = base. h b

7. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.8 – Slide 7 Solving Problems about Geometric Figures 6.8 Applications of Quadratic Equations Step 3 Write an equation. Since the height is twice the base, h = 2b. We know that the formula for the area of a triangle is A = ½bh. Since the area is 4 in 2, 4 = ½bh. We have two equations: h = 2b and 4 = ½bh. Use the first equation to rewrite the second equation, substituting 2b in for h. We are left with one equation. 4 = ½b(2b) 4 = b 2 Example 1 (continued)

8. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.8 – Slide 8 Solving Problems about Geometric Figures 6.8 Applications of Quadratic Equations Example 1 (continued) Step 4 Solve. Step 5 State the answer. Since a base of –2 makes no sense, the base of the triangular pendant must be 2 in. The height of the pendant must be 4 in. (since h = 2b). 0 = b 2 – 4 0 = (b + 2)(b – 2) 4 = b 2 b + 2 = 0 or b – 2 = 0 b = –2b = 2

9. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.8 – Slide 9 Solving Problems about Geometric Figures 6.8 Applications of Quadratic Equations Example 1 (concluded) Step 6 Check. In Step 5, we assured that the height was twice the base. Now all we must check is that the area is 4 in. 4 = ½bh 4 = ½(2)(4) ? 4 = 4 ?

10. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.8 – Slide 10 Problem-Solving Hint In consecutive integer problems, if x represents the first integer, then for two consecutive integers, use x, x+1; three consecutive integers, use x, x+1,x+2; two consecutive even or odd integers, usex, x+2; three consecutive even or odd integers, use x, x+2, x+4. Solving Problems about Consecutive Integers 6.8 Applications of Quadratic Equations

11. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.8 – Slide 11 Solving Problems about Consecutive Integers 6.8 Applications of Quadratic Equations Example 2 John opened his phonebook to view the city map, which was printed across two adjacent pages. John noticed that the product of these two page numbers was 812. On what two pages was the city map printed? Step 1 Read. We are looking for two consecutive page numbers. Step 2 Assign a variable. Let x = the 1 st page and let x + 1 = the 2 nd page. x x + 1

12. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.8 – Slide 12 x – 28 = 0 or x + 29 = 0 6.8 Applications of Quadratic Equations Step 3 Write an equation. Since the product of the two page numbers is 812, 812 = x(x + 1). Example 2 (continued) Solving Problems about Consecutive Integers Step 4 Solve. 0 = x 2 + x – 812 0 = (x – 28)(x + 29) 812 = x(x + 1) x = 28x = –29 812 = x 2 + x Factors of –812 Sums of Factors –4, 203199 –14, 5844 –28, 291

13. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.8 – Slide 13 6.8 Applications of Quadratic Equations Example 2 (concluded) Solving Problems about Consecutive Integers Step 5 State the answer. Since page number –29 makes no sense, the first page number must be 28. Since the pages are consecutive, the second page number must be 29. Step 6 Check.812 = x(x + 1) 812 = (28)(28 + 1) ? 812 = 812 ? 812 = (28)(29)?

14. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.8 – Slide 14 6.8 Applications of Quadratic Equations Pythagorean Formula If a right triangle (a triangle with a 90° angle) has longest side of length c and two other sides of length a and b, then a 2 + b 2 = c 2. The longest side, the hypotenuse, is opposite the right angle. The two shorter sides are the legs of the triangle. Solving Problems Using the Pythagorean Formula

15. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.8 – Slide 15 6.8 Applications of Quadratic Equations Example 3 Carrie and Diego left Oahu at the same time. Diego sailed directly north, and Carrie sailed directly west. At the exact same time, they both dropped anchor to fish. At that moment, Carrie was 20 miles further from Hawaii than Diego was, and the distance between them was 100 miles. How far from Oahu were Carrie and Diego? Step 1 Read. We are looking for Carrie’s distance from Oahu and Diego’s distance from Oahu. Solving Problems Using the Pythagorean Formula

16. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.8 – Slide 16 6.8 Applications of Quadratic Equations Example 3 (continued) Step 2 Assign a variable. Let x = Diego’s distance and x + 20 = Carrie’s distance. Solving Problems Using the Pythagorean Formula x x + 20 100 Oahu Step 3 Write an equation. Since Carrie and Diego’s positions form a right triangle, we can use the Pythagorean formula. x 2 + (x + 20) 2 = 100 2

17. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.8 – Slide 17 6.8 Applications of Quadratic Equations Example 3 (continued) Step 4 Solve. x 2 + (x + 20) 2 = 100 2 Solving Problems Using the Pythagorean Formula x 2 + x 2 + 40x + 400 = 10,000 2x 2 + 40x – 9600 = 0 2(x 2 + 20x – 4800) = 0 x 2 + 20x – 4800 = 0 (x + 80)(x – 60) = 0 x + 80 = 0 or x – 60 = 0 x = –80x = 60

18. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.8 – Slide 18 6.8 Applications of Quadratic Equations Example 3 (concluded) Solving Problems Using the Pythagorean Formula Step 5 State the answer. Since a distance of –80 mi makes no sense, Diego must be 60 mi from Oahu. Then, Carrie is 80 mi from Oahu. Step 6 Check. 60 2 + (60 + 20) 2 = 100 2 3600 + 6400 = 10,000 ? 10,000 = 10,000 ? x 2 + (x + 20) 2 = 100 2

19. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.8 – Slide 19 6.8 Applications of Quadratic Equations Example 4 A pebble falls off the roof of a 560 ft building according to the quadratic equation h = –16t 2 + 560, where h = the height of the Pebble and t = the time in seconds. After how many seconds will the pebble be 160 ft above the ground? Step 1 Read. Find the time, t when h = 160. Step 2 Assign a variable. The variables were assigned by the model. Solving Problems Using Given Quadratic Models 560 160 t = ? 560 160 t = ? 160 t = ? 560 160 t = ?

20. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.8 – Slide 20 6.8 Applications of Quadratic Equations Step 3 Write an equation. Let h = 160 in the given equation. 160 = –16t 2 + 560 Example 4 (continued) Step 4 Solve. 160 = –16t 2 + 560 Solving Problems Using Given Quadratic Models 0 = (t + 5)(t – 5) 0 = –16t 2 + 400 0 = –16(t 2 – 25) 0 = t 2 – 25 0 = t + 5 or 0 = t – 5 t = –5 or t = 5

21. Copyright © 2010 Pearson Education, Inc. All rights reserved. 6.8 – Slide 21 6.8 Applications of Quadratic Equations Example 4 (concluded) Solving Problems Using Given Quadratic Models Step 5 State the answer. Since a time of –5 seconds makes no sense, the rock must be 160 ft above the ground 5 seconds after it leaves the roof. Step 6 Check.160 = –16t 2 + 560 ? 160 = 160 ? 160 = –16(5) 2 + 560 160 = – 400 + 560