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

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


Factoring and Applications

Chapter 6


6.8

Applications of Quadratic Equations


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



Solving an Applied ProblemStep 1 Read the problem, several times if necessary,

until you understand what is given and what is to be found.

Step 2 Assign 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 3 Write an equation using the variable expression(s).Step 4 Solve the equation.Step 5 State the answer. Does it seem reasonable?Step 6 Check the answer in the words of the original

problem.


Solving Problems about Geometric Figures


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 in2. 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




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 in2, 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 = b2

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 = b2 – 4

0 = (b + 2)(b – 2)

4 = b2

b + 2 = 0 or b – 2 = 0

b = –2 b = 2




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

?


Problem-Solving HintIn consecutive integer problems, if x represents the first integer, then fortwo consecutive integers, use x, x+1;three consecutive integers, use x, x+1,x+2;two consecutive even or odd integers, use x, x+2;three consecutive even or odd integers, use x, x+2, x+4.

Solving Problems about Consecutive Integers





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 1st page and let x + 1 = the 2nd page.

x x + 1


x – 28 = 0 or x + 29 = 0


Step 3 Write an equation. Since the product of the two page numbers is 812, 812 = x(x + 1).



Step 4 Solve.

0 = x2 + x – 812

0 = (x – 28)(x + 29)

812 = x(x + 1)

x = 28 x = –29

812 = x2 + xFactors of –812

Sums of Factors

–4, 203 199

–14, 58 44

–28, 29 1





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)?



Pythagorean FormulaIf 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

a2 + b2 = c2.

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



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.





Step 2 Assign a variable. Let x = Diego’s distance and x + 20 = Carrie’s distance.


xx + 20

100

Oahu

Step 3 Write an equation. Since Carrie and Diego’s positions form a right triangle, we can use the Pythagorean formula. x2 + (x + 20)2 = 1002




Step 4 Solve.

x2 + (x + 20)2 = 1002


x2 + x2 + 40x + 400 = 10,000

2x2 + 40x – 9600 = 0

2(x2 + 20x – 4800) = 0

x2 + 20x – 4800 = 0

(x + 80)(x – 60) = 0

x + 80 = 0 or x – 60 = 0x = –80 x = 60





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.

602 + (60 + 20)2 = 1002

3600 + 6400 = 10,000?

10,000 = 10,000

?

x2 + (x + 20)2 = 1002



Example 4A pebble falls off the roof of a 560 ft building according to the quadratic equation h = –16t2 + 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

160t = ?

560

160t = ?

160t = ?

560

160t = ?



Step 3 Write an equation. Let h = 160 in the given equation. 160 = –16t2 + 560


Step 4 Solve. 160 = –16t2 + 560


0 = (t + 5)(t – 5)

0 = –16t2 + 400

0 = –16(t2 – 25)

0 = t2 – 25

0 = t + 5 or 0 = t – 5

t = –5 or t = 5





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 = –16t2 + 560

?

160 = 160

?

160 = –16(5)2 + 560

160 = – 400 + 560

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