# Connexions

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# Applications

Module by: Wade Ellis, Denny Burzynski. E-mail the authors

Summary: This module is from Elementary Algebra</link> by Denny Burzynski and Wade Ellis, Jr. Methods of solving quadratic equations as well as the logic underlying each method are discussed. Factoring, extraction of roots, completing the square, and the quadratic formula are carefully developed. The zero-factor property of real numbers is reintroduced. The chapter also includes graphs of quadratic equations based on the standard parabola, y = x^2, and applied problems from the areas of manufacturing, population, physics, geometry, mathematics (numbers and volumes), and astronomy, which are solved using the five-step method. Objectives of this module: become more proficient at using the five-step method for solving applied problems.

## Overview

• The Five-Step Method
• Examples

## The Five-Step Method

We are now in a position to study some applications of quadratic equations. Quadratic equations can arise from a variety of physical (applied) and mathematical (logical) problems.

We will, again, apply the five-step method for solving word problems.

### Five-Step Method of Solving Word Problems

• Step 1:   Let x x (or some other letter) represent the unknown quantity.
• Step 2:   Translate the verbal expression to mathematical symbols and form an equation.
• Step 3:   Solve this equation.
• Step 4:   Check the solution by substituting the result into the equation found in step 2.
• Step 5:   Write a conclusion.

Remember, step 1 is very important.

ALWAYS START BY INTRODUCING A VARIABLE.

Once the quadratic equation is developed (step 2), try to solve it by factoring. If factoring doesn’t work, use the quadratic formula. A calculator may help to make some of the calculations a little less tedious.

## Sample Set A

### Example 1

A producer of personal computer mouse covers determines that the number N N of covers sold is related to the price x x of a cover by N=35x x 2 . N=35x x 2 . At what price should the producer price a mouse cover in order to sell 216 of them?

Step 1: Letx=thepriceofamousecover. Step 2: SinceNistobe216,theequationis 216=35x x 2 Step 3: 216 = 35x x 2 Rewriteinstandardform. x 2 35x+216 = 0 Tryfactoring. (x8)(x27) = 0 x8=0 or x27=0 x=8 or x=27 Checkthesepotentialsolutions. Step 4: Ifx=8, Ifx=27, 35·8 8 2 = 216 Is this correct? 35·27 27 2 = 216 Is this correct? 28064 = 216 Is this correct? 945729 = 216 Is this correct? 216 = 216 Yes, this is correct. 216 = 216 Yes, this is correct. Thesesolutionscheck. Step5: Thecomputermousecoverscanbepricedateither$8or$27inordertosell216ofthem. Step 1: Letx=thepriceofamousecover. Step 2: SinceNistobe216,theequationis 216=35x x 2 Step 3: 216 = 35x x 2 Rewriteinstandardform. x 2 35x+216 = 0 Tryfactoring. (x8)(x27) = 0 x8=0 or x27=0 x=8 or x=27 Checkthesepotentialsolutions. Step 4: Ifx=8, Ifx=27, 35·8 8 2 = 216 Is this correct? 35·27 27 2 = 216 Is this correct? 28064 = 216 Is this correct? 945729 = 216 Is this correct? 216 = 216 Yes, this is correct. 216 = 216 Yes, this is correct. Thesesolutionscheck. Step5: Thecomputermousecoverscanbepricedateither$8or$27inordertosell216ofthem.

## Practice Set A

### Exercise 1

A manufacturer of cloth personal computer dust covers notices that the number N N of covers sold is related to the price of covers by N=30x x 2 . N=30x x 2 . At what price should the manufacturer price the covers in order to sell 216 of them?

Step 1:

Step 2:

Step 3:

Step 4:

Step 5:   In order to sell 216 covers, the manufacturer should price them at either

or

.

12 or 18

### Exercise 2

It is estimated that t t years from now the population of a particular city will be

P= t 2 24t+96,000. P= t 2 24t+96,000.

How many years from now will the population be 95,865?

Step 1:

Step 2:

Step 3:

Step 4:

Step 5:

#### Solution

In 9 and 15 years, the population of the city will be 95,865.

## Sample Set B

### Example 2

The length of a rectangle is 4 inches more than twice its width. The area is 30 square inches. Find the dimensions (length and width).

Step 1:   Let x= x= the width. Then, 2x+4= 2x+4= the length.

Step 2: The area of a rectangle is defined to be the length of the  rectangle times the width of the rectangle. Thus, x( 2x+4 )=30 Step 2: The area of a rectangle is defined to be the length of the  rectangle times the width of the rectangle. Thus, x( 2x+4 )=30
Step 3: x( 2x+4 ) = 30 2 x 2 +4x = 30 2 x 2 +4x30 = 0 Divide each side by 2. x 2 +2x15 = 0 Factor. ( x+5 )( x3 ) = 0 x = 5,3 x = 5 has  no physical meaning so we disregard it. Check x=3. x = 3 2x+4=2·3+4 = 10 Step 3: x( 2x+4 ) = 30 2 x 2 +4x = 30 2 x 2 +4x30 = 0 Divide each side by 2. x 2 +2x15 = 0 Factor. ( x+5 )( x3 ) = 0 x = 5,3 x = 5 has  no physical meaning so we disregard it. Check x=3. x = 3 2x+4=2·3+4 = 10
Step 4: x(2x+4) = 30 Is this correct? 3(2·3+4) = 30 Is this correct? 3(6+4) = 30 Is this correct? 3(10) = 30 Is this correct? 30 = 30 Yes, this is correct. Step 4: x(2x+4) = 30 Is this correct? 3(2·3+4) = 30 Is this correct? 3(6+4) = 30 Is this correct? 3(10) = 30 Is this correct? 30 = 30 Yes, this is correct.
Step 5: Width=3 inches and length=10 inches. Step 5: Width=3 inches and length=10 inches.

## Practice Set B

### Exercise 3

The length of a rectangle is 3 feet more than twice its width. The area is 14 square feet. Find the dimensions.

#### Solution

width = 2 feet, length = 7 feet

### Exercise 4

The area of a triangle is 24 square meters. The base is 2 meters longer than the height. Find the base and height. The formula for the area of a triangle is A= 1 2 b·h. A= 1 2 b·h.

#### Solution

height = 6 meters, base = 8 meters

## Sample Set C

### Example 3

The product of two consecutive integers is 156. Find them.

Step 1: Let  x = the smaller integer. x+1 = the next integer. Step 2: x( x+1 ) = 156 Step 3: x( x+1 ) = 156 x 2 +x = 156 x 2 +x156 = 0 ( x12 )( x13 ) = 0 x = 12,13 Step 1: Let  x = the smaller integer. x+1 = the next integer. Step 2: x( x+1 ) = 156 Step 3: x( x+1 ) = 156 x 2 +x = 156 x 2 +x156 = 0 ( x12 )( x13 ) = 0 x = 12,13
This factorization may be hard to guess. We could also use the quadratic formula.

x 2 +x156=0 a=1, b=1, c=156 x = 1± 1 2 4( 1 )( 156 ) 2( 1 ) = 1± 1+624 2 = 1±25 2 1±25 2 = 24 2 =12and 125 2 = 26 2 =13 x = 12,13 Check 12, 13 and 13, 12. x+1 = 13,12 x 2 +x156=0 a=1, b=1, c=156 x = 1± 1 2 4( 1 )( 156 ) 2( 1 ) = 1± 1+624 2 = 1±25 2 1±25 2 = 24 2 =12and 125 2 = 26 2 =13 x = 12,13 Check 12, 13 and 13, 12. x+1 = 13,12 Step4: Ifx=12: 12(2+1) = 156 Is this correct? 12(13) = 156 Is this correct? 156 = 156 Is this correct? Ifx=13 13(13+1) = 156 Is this correct? 13(12) = 156 Is this correct? 156 = 156 Yes, this is correct. Step5: Therearetwosolutions: 12,13and13,12. Step4: Ifx=12: 12(2+1) = 156 Is this correct? 12(13) = 156 Is this correct? 156 = 156 Is this correct? Ifx=13 13(13+1) = 156 Is this correct? 13(12) = 156 Is this correct? 156 = 156 Yes, this is correct. Step5: Therearetwosolutions: 12,13and13,12.

## Practice Set C

### Exercise 5

The product of two consecutive integers is 210. Find them.

#### Solution

14 and 15, amd –14 and –15

### Exercise 6

Four is added to an integer and that sum is tripled. When this result is multiplied by the original integer, the product is −12. Find the integer.

–2

## Sample Set D

### Example 4

A box with no top and a square base is to be made by cutting out 2-inch squares from each corner and folding up the sides of a piece of a square cardboard. The volume of the box is to be 8 cubic inches. What size should the piece of cardboard be?

Step 1: Let x=the length ( and width ) of the piece of cardboard. Step 1: Let x=the length ( and width ) of the piece of cardboard.
Step 2: The volume of a rectangular box is V=( length )( width )( height ) 8=( x4 )( x4 )2 Step 2: The volume of a rectangular box is V=( length )( width )( height ) 8=( x4 )( x4 )2
Step 3: 8=( x4 )( x4 )2 8=( x 2 8x+16 )2 8=2 x 2 16x+32 2 x 2 16x+24=0 Divide each side by 2. x 2 8x+12=0 Factor. ( x6 )( x2 )=0 x=6,2 Step 3: 8=( x4 )( x4 )2 8=( x 2 8x+16 )2 8=2 x 2 16x+32 2 x 2 16x+24=0 Divide each side by 2. x 2 8x+12=0 Factor. ( x6 )( x2 )=0 x=6,2
x x cannot equal 2 (the cut would go through the piece of cardboard). Check x=6. x=6.
Step 4: (64)(64)2 = 8 Is this correct? (2)(2)2 = 8 Is this correct? 8 = 8 Yes, this is correct. Step 4: (64)(64)2 = 8 Is this correct? (2)(2)2 = 8 Is this correct? 8 = 8 Yes, this is correct.
Step 5: The piece of cardboard should be 6 inches by 6 inches. Step 5: The piece of cardboard should be 6 inches by 6 inches.

## Practice Set D

### Exercise 7

A box with no top and a square base is to be made by cutting 3-inch squares from each corner and folding up the sides of a piece of cardboard. The volume of the box is to be 48 cubic inches. What size should the piece of cardboard be?

#### Solution

10 in. by 10 in.; 2 by 2 is not physically possible.

## Sample Set E

### Example 5

A study of the air quality in a particular city by an environmental group suggests that t t years from now the level of carbon monoxide, in parts per million, in the air will be

A=0.3 t 2 +0.1t+4.2 A=0.3 t 2 +0.1t+4.2

(a) What is the level, in parts per million, of carbon monoxide in the air now?

Since the equation A=0.3 t 2 +0.1t+4.2 A=0.3 t 2 +0.1t+4.2 specifies the level t t years from now, we have t=0. t=0.

A=0.3 t 2 +0.1t+4.2 A=0.3 t 2 +0.1t+4.2
A=4.2 A=4.2

(b) How many years from now will the level of carbon monoxide be at 8 parts per million?
Step 1: t = the number of years when the level is 8. Step 1: t = the number of years when the level is 8.
Step 2: 8 = 0.3 t 2 +0.1t+4.2 Step 2: 8 = 0.3 t 2 +0.1t+4.2
Step 3: 8 = 0.3 t 2 +0.1t+4.2 0 = 0.3 t 2 +0.1t3.8 This does not readily factor, so  we'll use the quadratic formula. a = 0.3,b=0.1,c=3.8 t = 0.1± ( 0.1 ) 2 4( 0.3 )( 3.8 ) 2( 0.3 ) = 0.1± 0.01+4.56 0.6 = 0.1± 4.57 0.6 = 0.1±2.14 0.6 t = 3.4and3.73 Step 3: 8 = 0.3 t 2 +0.1t+4.2 0 = 0.3 t 2 +0.1t3.8 This does not readily factor, so  we'll use the quadratic formula. a = 0.3,b=0.1,c=3.8 t = 0.1± ( 0.1 ) 2 4( 0.3 )( 3.8 ) 2( 0.3 ) = 0.1± 0.01+4.56 0.6 = 0.1± 4.57 0.6 = 0.1±2.14 0.6 t = 3.4and3.73
t=3.73has no physical meaning. Checkt=3.4 Step 4: This value of t has been rounded to the nearest tenth. It does check ( pretty closely ). Step 5: About 3.4 years from now the carbon monoxide level will be 8. t=3.73has no physical meaning. Checkt=3.4 Step 4: This value of t has been rounded to the nearest tenth. It does check ( pretty closely ). Step 5: About 3.4 years from now the carbon monoxide level will be 8.

## Practice Set E

### Exercise 8

A study of the air quality in a particular city by an environmental group suggests that t t years from now the level of carbon monoxide, in parts per million, in the air will be

A=0.2 t 2 +0.1t+5.1 A=0.2 t 2 +0.1t+5.1

(a) What is the level, in parts per million, now?

(b) How many years from now will the level of carbon monoxide be at 8 parts per million? Round to the nearest tenth.

#### Solution

(a). 5.1 5.1 parts per million (b). 3.6 3.6 years

## Sample Set F

### Example 6

A contractor is to pour a concrete walkway around a swimming pool that is 20 feet wide and 40 feet long. The area of the walkway is to be 544 square feet. If the walkway is to be of uniform width, how wide should the contractor make it?

Step 1:  Let x x = the width of the walkway.
Step 2:  A diagram will help us to get the equation.

(Area of pool and walkway) — (area of pool) = (area of walkway)
( 20+2x )( 40+2x )20·40=544 ( 20+2x )( 40+2x )20·40=544
Step 3: ( 20+2x )( 40+2x )20·40 = 544 800+120x+4 x 2 800 = 544 120x+4 x 2 = 544 4 x 2 +120x544 = 0 Divide each term by 4. x 2 +30x136 = 0 Solve by factoring. ( x4 )( x+34 ) = 0 (This is difficult to factor so we may  wish to use the quadratic formula.)  Step 3: ( 20+2x )( 40+2x )20·40 = 544 800+120x+4 x 2 800 = 544 120x+4 x 2 = 544 4 x 2 +120x544 = 0 Divide each term by 4. x 2 +30x136 = 0 Solve by factoring. ( x4 )( x+34 ) = 0 (This is difficult to factor so we may  wish to use the quadratic formula.)
x4=0 or x+34=0 x=4 or x=34 has no physical meaning. x4=0 or x+34=0 x=4 or x=34 has no physical meaning.
Check a width of 4 feet as a solution.
Step 4: Area of pool and walkway  = ( 20+2·4 )( 40+2·4 ) = ( 28 )( 48 ) = 1344 Step 4: Area of pool and walkway  = ( 20+2·4 )( 40+2·4 ) = ( 28 )( 48 ) = 1344
Area of pool = (20)(40) = 800
Area of walkway = 1344800=544 Yes, this is correct. 1344800=544 Yes, this is correct.
This solution checks.
Step 5:  The contractor should make the walkway 4 feet wide.

## Practice Set F

### Exercise 9

A contractor is to pour a concrete walkway around a swimming pooi that is 15 feet wide and 25 feet long. The area of the walkway is to be 276 square feet. If the walkway is to be of uniform width, how wide should the contractor make it?

3 ft wide

## Exercises

Some of the following problems have actual applications and some are intended only as logic developers. A calculator may be helpful. The problems appear in groups and correspond to the noted Sample Set problem.

### Sample Set A—Type Problems

#### Exercise 10

The manufacturer of electronic fuel injectors determines that the number N N of injectors sold is related to the price x x per injector by N=22x x 2 . N=22x x 2 . At what price should the manufacturer price the injectors so that 112 of them are sold?

##### Solution

$8or$14 $8or$14

#### Exercise 11

The owner of a stained-glass shop determines that the number N N of pieces of a particular type of glass sold in a month is related to the price x x per piece by N=21x x 2 . N=21x x 2 . At what price should the shop buyer price the glass so that 162 sell?

#### Exercise 12

It is estimated that t t years from now the population of a certain city will be

P= t 2 15t+12,036 P= t 2 15t+12,036

(a) What is the population now?

(b) How many years from now will the population be 12,000?

##### Solution

(a) 12,036  (b) 3 and 12 years from now (a) 12,036  (b) 3 and 12 years from now

#### Exercise 13

It is estimated that t t years from now the population of a certain city will be

P= t 2 16t+24,060 P= t 2 16t+24,060

(a) What is the population now?

(b) How many years from now will the population be 24,000?

#### Exercise 14

If an object is thrown vertically upward, its height h h , above the ground, in feet, after t t seconds is given by h= h 0 + v 0 t16 t 2 h= h 0 + v 0 t16 t 2 , where h 0 h 0 is the initial height from which the object is thrown and v 0 v 0 is the initial velocity of the object. Using this formula and an approach like that of Sample Set A, solve this problem.

A ball thrown vertically into the air has the equation of motion h=48+32t16 t 2 . h=48+32t16 t 2 .

(a) How high is the ball at t=0 t=0 (the initial height of the ball)?

(b) How high is the ball at t=1 t=1 (after 1 second in the air)?

(c) When does the ball hit the ground? (Hint: Determine the appropriate value for h h then solve for t t .)

##### Solution

(a) 48 feet  (b) 64 feet  (c) t=3 (a) 48 feet  (b) 64 feet  (c) t=3

#### Exercise 15

A woman’s glasses accidently fall off her face while she is looking out of a window in a tall building. The equation relating h h , the height above the ground in feet, and t t , the time in seconds her glasses have been falling, is h=6416 t 2 . h=6416 t 2 .

(a) How high was the woman’s face when her glasses fell off?

(b) How many seconds after the glasses fell did they hit the ground?

### Sample Set B—Type Problems

#### Exercise 16

The length of a rectangle is 6 feet more than twice its width. The area is 8 square feet. Find the dimensions.

##### Solution

length=8;width=1 length=8;width=1

#### Exercise 17

The length of a rectangle is 18 inches more than three times its width. The area is 81 square inches. Find the dimensions.

#### Exercise 18

The length of a rectangle is two thirds its width. The area is 14 square meters. Find the dimensions.

##### Solution

width= 21 length= 2 3 21 width= 21 length= 2 3 21

#### Exercise 19

The length of a rectangle is four ninths its width. The area is 144 square feet. Find the dimensions.

#### Exercise 20

The area of a triangle is 14 square inches. The base is 3 inches longer than the height. Find both the length of the base and height.

b=7;h=4 b=7;h=4

#### Exercise 21

The area of a triangle is 34 square centimeters. The base is 1 cm longer than twice the height. Find both the length of the base and the height.

### Sample Set C—Type Problems

#### Exercise 22

The product of two consecutive integers is 72. Find them.

##### Solution

9,8or8,9 9,8or8,9

#### Exercise 23

The product of two consecutive negative integers is 42. Find them.

#### Exercise 24

The product of two consecutive odd integers is 143. Find them. (Hint: The quadratic equation is factorable, but the quadratic formula may be quicker.)

##### Solution

13,11or11,13 13,11or11,13

#### Exercise 25

The product of two consecutive even integers is 168. Find them.

#### Exercise 26

Three is added to an integer and that sum is doubled. When this result is multiplied by the original integer the product is 20. Find the integer.

n=2,5 n=2,5

#### Exercise 27

Four is added to three times an integer. When this sum and the original integer are multiplied, the product is 1. 1. Find the integer.

### Sample Set D—Type Problems

#### Exercise 28

A box with no top and a square base is to be made by cutting out 2-inch squares from each corner and folding up the sides of a piece of cardboard.The volume of the box is to be 25 cubic inches. What size should the piece of cardboard be?

##### Solution

4+ 12.5 inches 4+ 12.5 inches

#### Exercise 29

A box with no top and a square base is to made by cutting out 8-inch squares from each corner and folding up the sides of a piece of cardboard. The volume of the box is to be 124 cubic inches. What size should the piece of cardboard be?

### Sample Set E—Type Problems

#### Exercise 30

A study of the air quality in a particular city by an environmental group suggests that t t years from now the level of carbon monoxide, in parts per million, will be A=0.1 t 2 +0.1t+2.2. A=0.1 t 2 +0.1t+2.2.

(a) What is the level, in parts per million, of carbon monoxide in the air now?

(b) How many years from now will the level of carbon monoxide be at 3 parts per million?

##### Solution

(a) carbon monoxide now 2.2 2.2 parts per million
(b) 2.37years 2.37years

#### Exercise 31

A similar study to that of problem 21 suggests A=0.3 t 2 +0.25t+3.0. A=0.3 t 2 +0.25t+3.0.

(a) What is the level, in parts per million, of carbon monoxide in the air now?

(b) How many years from now will the level of carbon monoxide be at 3.1 parts per million?

### Sample Set F—Type Problems

#### Exercise 32

A contractor is to pour a concrete walkway around a wading pool that is 4 feet wide and 8 feet long. The area of the walkway and pool is to be 96 square feet. If the walkway is to be of uniform width, how wide should it be?

x=2 x=2

### Astrophysical Problem

#### Exercise 33

A very interesting application of quadratic equations is determining the length of a solar eclipse (the moon passing between the earth and sun). The length of a solar eclipse is found by solving the quadratic equation

( a+bt ) 2 + ( c+dt ) 2 = ( e+ft ) 2 ( a+bt ) 2 + ( c+dt ) 2 = ( e+ft ) 2

for t t . The letters a,b,c,d,e, a,b,c,d,e, and f f are constants that pertain to a particular eclipse. The equation is a quadratic equation in t t and can be solved by the quadratic formula (and definitely a calculator). Two values of t t will result. The length of the eclipse is just the difference of these t t -values.

The following constants are from a solar eclipse that occurred on August 3, 431 B.C.

a = 619 b = 1438 c = 912 d = 833 e = 1890.5 f = 2 a = 619 b = 1438 c = 912 d = 833 e = 1890.5 f = 2
Determine the length of this particular solar eclipse.

## Exercises For Review

### Exercise 34

((Reference)) Find the sum: 2x+10 x 2 +x2 + x+3 x 2 3x+2 . 2x+10 x 2 +x2 + x+3 x 2 3x+2 .

#### Solution

3x+14 ( x+2 )( x2 ) 3x+14 ( x+2 )( x2 )

### Exercise 35

((Reference)) Solve the fractional equation 4 x+12 + 3 x+3 = 4 x 2 +5x+6 . 4 x+12 + 3 x+3 = 4 x 2 +5x+6 .
(Hint: Check for extraneous solutions.)

### Exercise 36

((Reference)) One pipe can fill a tank in 120 seconds and another pipe can fill the same tank in 90 seconds. How long will it take both pipes working together to fill the tank?

51 3 7 51 3 7

### Exercise 37

((Reference)) Use the quadratic formula to solve 10 x 2 3x1=0. 10 x 2 3x1=0.

### Exercise 38

((Reference)) Use the quadratic formula to solve 4 x 2 3x=0. 4 x 2 3x=0.

#### Solution

x=0, 3 4 x=0, 3 4

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Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

##### Who can create a lens?

Any individual member, a community, or a respected organization.

##### What are tags?

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks

#### Module to:

My Favorites (?)

'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

#### Definition of a lens

##### Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

##### What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

##### Who can create a lens?

Any individual member, a community, or a respected organization.

##### What are tags?

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks