OpenStax-CNX module: m21915
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∗
Quadratic Equations: Applications
Wade Ellis
Denny Burzynski
This work is produced by OpenStax-CNX and licensed under the
Creative Commons Attribution License 3.0†
Abstract
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 zerofactor 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 ve-step method. Objectives of this module: become more procient at using the ve-step
method for solving applied problems.
1 Overview
• The Five-Step Method
• Examples
2 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
for solving word problems.
ve-step method
Five-Step Method of Solving Word Problems
Step
Step
Step
Step
Step
1:
2:
3:
4:
5:
Let x (or some other letter) represent the unknown quantity.
Translate the verbal expression to mathematical symbols and form an equation.
Solve this equation.
Check the solution by substituting the result into the equation found in step 2.
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.
∗ Version
1.4: Jun 1, 2009 11:17 am -0500
† http://creativecommons.org/licenses/by/3.0/
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3 Sample Set A
Example 1
A producer of personal computer mouse covers determines that the number N of covers sold is
related to the price x of a cover by N = 35x − x2 . At what price should the producer price a mouse
cover in order to sell 216 of them?
Step 1 :
Let x = the price of a mouse cover.
Step 2 :
Since N is to be 216, the equation is
216 = 35x − x2
Step 3 :
216
=
35x − x2
Rewrite in standard form.
2
x − 35x + 216
=
0
Try factoring.
(x − 8) (x − 27)
=
0
x−8=0
or
x − 27 = 0
x=8
or
x = 27
Check these potential solutions.
Step 4 :
If x = 8,
If x = 27,
2
=
216 Is this correct?
35 · 27 − 272
=
216 Is this correct?
280 − 64
=
216
Is this correct?
945 − 729
=
216
Is this correct?
216
=
216
Yes, this is correct.
216
=
216
Yes, this is correct.
35 · 8 − 8
These solutions check.
Step 5 :
The computer mouse covers can be priced at either $8 or $27 in order to sell 216 of them.
4 Practice Set A
Exercise 1
(Solution on p. 13.)
A manufacturer of cloth personal computer dust covers notices that the number N of covers sold
is related to the price of covers by N = 30x − x2 . At what price should the manufacturer price the
covers in order to sell 216 of them?
Step 1:
Step 2:
Step 3:
Step 4:
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Step 5: In order to sell 216 covers, the manufacturer should price them at either
.
Exercise 2
or
(Solution on p. 13.)
It is estimated that t years from now the population of a particular city will be
P = t2 − 24t + 96, 000.
How many years from now will the population be 95,865?
Step 1:
Step 2:
Step 3:
Step 4:
Step 5:
5 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 =the width. Then, 2x + 4 = the length.
Step 2 :
The area of a rectangle is dened to be the length of the rectangle times the width of the rectangle. Thus,
x (2x + 4) = 30
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Step 3 :
4
x (2x + 4)
=
30
2x + 4x =
30
2
2
=
0
Divide each side by 2.
2
x + 2x − 15
=
0
Factor.
(x + 5) (x − 3)
=
0
2x + 4x − 30
Step 4 :
Step 5 :
x
= −5, 3
x
= −5
x
=
3
2x + 4 = 2 · 3 + 4
=
10
has no physical meaning so we disregard it. Check x = 3.
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.
Width = 3 inches and length = 10 inches.
6 Practice Set B
Exercise 3
(Solution on p. 13.)
The length of a rectangle is 3 feet more than twice its width. The area is 14 square feet. Find the
dimensions.
Exercise 4
(Solution on p. 13.)
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 = 21 b · h.
7 Sample Set C
Example 3
The product of two consecutive integers is 156. Find them.
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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 +x =
156
2
x2 + x − 156
=
0
(x − 12) (x − 13)
=
0
x = 12, − 13
This factorization may be hard to guess. We could also use the quadratic formula.
Step 4 :
x2 + x − 156 = 0
a = 1,
b = 1,
√
x =
=
=
x =
x+1
=
If x = 12 :
c = −156
12 (2
12
12 −4(1)(−156)
2(1)
√
−1± 1+624
2
−1±25
2
−1±25
2
12, − 13
Check 12, 13 and − 13, 12.
−1±
If x = −13
=
24
2
= 12 and −1−25
=
2
13, − 12
−26
2
−13 (−
= −13
Step 5 :
There are two solutions:
8 Practice Set C
Exercise 5
The product of two consecutive integers is 210. Find them.
Exercise 6
(Solution on p. 13.)
(Solution on p. 13.)
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.
9 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?
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−13 (−13
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Step 1:
Step 2:
6
Let x = the length (and width) of the piece of cardboard.
The volume of a rectangular box is
V = (length) (width) (height)
8 = (x − 4) (x − 4) 2
Step 3 :
8 = (x − 4) (x − 4) 2
8 = x2 − 8x + 16 2
8 = 2x2 − 16x + 32
2x2 − 16x + 24 = 0
Divide each side by 2.
x2 − 8x + 12 = 0
Factor.
(x − 6) (x − 2) = 0
x = 6, 2
x cannot equal 2 (the cut would go through the piece of cardboard). Check x = 6.
Step 4 :
Step 5 :
(6 − 4) (6 − 4) 2
=
8
Is this correct?
(2) (2) 2
=
8
Is this correct?
8
=
8 Yes, this is correct.
The piece of cardboard should be 6 inches by 6 inches.
10 Practice Set D
Exercise 7
(Solution on p. 13.)
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?
11 Sample Set E
Example 5
A study of the air quality in a particular city by an environmental group suggests that t years from
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now the level of carbon monoxide, in parts per million, in the air will be
A = 0.3t2 + 0.1t + 4.2
(a) What is the level, in parts per million, of carbon monoxide in the air now?
Since the equation A = 0.3t2 + 0.1t + 4.2 species the level t years from now, we have t = 0.
A = 0.3t2 + 0.1t + 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 2 :
8
=
0.3t2 + 0.1t + 4.2
Step 3 :
8
=
0.3t2 + 0.1t + 4.2
0
=
0.3t2 + 0.1t − 3.8
a
=
0.3, b = 0.1, c = −3.8
√
2
t
=
=
(0.1) −4(0.3)(−3.8)
2(0.3)
√
−0.1± 0.01+4.56
4.57
= −0.1±
0.6
0.6
−0.1±2.14
0.6
=
3.4 and − 3.73
This does not readily factor, so
we'll use the quadratic formula.
−0.1±
√
=
t
t = −3.73 has no physical meaning. Check t = 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.
12 Practice Set E
Exercise 8
(Solution on p. 13.)
A study of the air quality in a particular city by an environmental group suggests that t years from
now the level of carbon monoxide, in parts per million, in the air will be
A = 0.2t2 + 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.
13 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
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width, how wide should the contractor make it?
Step 1: Let 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
Step 3:
(20 + 2x) (40 + 2x) − 20 · 40
=
544
800 + 120x + 4x − 800
=
544
2
=
544
4x2 + 120x − 544
=
0
Divide each term by 4.
x2 + 30x − 136
=
0
Solve by factoring.
(x − 4) (x + 34)
=
0
2
120x + 4x
x−4=0
or
(This is dicult to factor so we may
wish to use the quadratic formula.)
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
Area of pool = (20)(40) = 800
Area of walkway = 1344 − 800 = 544
Yes, this is correct.
This solution checks.
Step 5: The contractor should make the walkway 4 feet wide.
14 Practice Set F
Exercise 9
(Solution on p. 13.)
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?
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15 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.
15.1 Sample Set AType Problems
Exercise 10
(Solution on p. 13.)
The manufacturer of electronic fuel injectors determines that the number N of injectors sold is
related to the price x per injector by N = 22x − x2 .At what price should the manufacturer price
the injectors so that 112 of them are sold?
Exercise 11
The owner of a stained-glass shop determines that the number N of pieces of a particular type of
glass sold in a month is related to the price xper piece by N = 21x − x2 . At what price should the
shop buyer price the glass so that 162 sell?
Exercise 12
(Solution on p. 13.)
It is estimated that t years from now the population of a certain city will be
P = t2 − 15t + 12, 036
(a) What is the population now?
(b) How many years from now will the population be 12,000?
Exercise 13
It is estimated that t years from now the population of a certain city will be
P = t2 − 16t + 24, 060
(a) What is the population now?
(b) How many years from now will the population be 24,000?
Exercise 14
(Solution on p. 13.)
If an object is thrown vertically upward, its height h, above the ground, in feet, after t seconds is
given by h = h0 + v0 t − 16t2 , where h0 is the initial height from which the object is thrown and v0
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 + 32t − 16t2 .
(a) How high is the ball at t = 0 (the initial height of the ball)?
(b) How high is the ball at t = 1 (after 1 second in the air)?
(c) When does the ball hit the ground? (Hint: Determine the appropriate value for hthen solve for
t .)
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Exercise 15
A woman's glasses accidently fall o her face while she is looking out of a window in a tall building.
The equation relating h, the height above the ground in feet, and t, the time in seconds her glasses
have been falling, is h = 64 − 16t2 .
(a) How high was the woman's face when her glasses fell o?
(b) How many seconds after the glasses fell did they hit the ground?
15.2 Sample Set BType Problems
Exercise 16
(Solution on p. 13.)
The length of a rectangle is 6 feet more than twice its width. The area is 8 square feet. Find the
dimensions.
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.
Exercise 19
The length of a rectangle is four ninths its width. The area is 144 square feet. Find the dimensions.
Exercise 20
(Solution on p. 13.)
(Solution on p. 13.)
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.
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.
15.3 Sample Set CType Problems
Exercise 22
The product of two consecutive integers is 72. Find them.
Exercise 23
The product of two consecutive negative integers is 42. Find them.
Exercise 24
(Solution on p. 13.)
(Solution on p. 13.)
The product of two consecutive odd integers is 143. Find them. (Hint: The quadratic equation is
factorable, but the quadratic formula may be quicker.)
Exercise 25
The product of two consecutive even integers is 168. Find them.
Exercise 26
(Solution on p. 13.)
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.
Exercise 27
Four is added to three times an integer. When this sum and the original integer are multiplied,
the product is −1. Find the integer.
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15.4 Sample Set DType Problems
Exercise 28
(Solution on p. 13.)
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?
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?
15.5 Sample Set EType Problems
Exercise 30
(Solution on p. 13.)
A study of the air quality in a particular city by an environmental group suggests that t years from
now the level of carbon monoxide, in parts per million, will be A = 0.1t2 + 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?
Exercise 31
A similar study to that of problem 21 suggests A = 0.3t2 + 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?
15.6 Sample Set FType Problems
Exercise 32
(Solution on p. 13.)
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?
15.7 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
2
2
(a + bt) + (c + dt) = (e + f t)
2
for t. The letters a, b, c, d, e, and f are constants that pertain to a particular eclipse. The equation
is a quadratic equation in t and can be solved by the quadratic formula (and denitely a calculator).
Two values of t will result. The length of the eclipse is just the dierence of these t-values.
The following constants are from a solar eclipse that occurred on August 3, 431 B.C.
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a =
c
=
12
−619
b
=
912
d
= −833
1438
e = 1890.5
f = −2
Determine the length of this particular solar eclipse.
16 Exercises For Review
Exercise 34
( here ) Find the sum: +
.
Exercise 35
( here ) Solve the fractional equation + =
.
(
Check for extraneous solutions.)
Exercise 36
( here ) One pipe can ll a tank in 120 seconds and another pipe can ll the same tank in 90
seconds. How long will it take both pipes working together to ll the tank?
Exercise 37
( here ) Use the quadratic formula to solve 10x − 3x − 1 = 0.
Exercise 38
( here ) Use the quadratic formula to solve 4x − 3x = 0.
(Solution on p. 13.)
1
2x+10
x2 +x−2
2
x+3
x2 −3x+2
4
x+12
3
x+3
4
x2 +5x+6
Hint:
(Solution on p. 13.)
3
4
2
(Solution on p. 13.)
5
1 "Rational Expressions:
2 "Rational Expressions:
3 "Rational Expressions:
4 "Quadratic Equations:
2
Adding and Subtracting Rational Expressions" <http://cnx.org/content/m21936/latest/>
Rational Equations" <http://cnx.org/content/m21951/latest/>
Applications" <http://cnx.org/content/m21938/latest/>
Solving Quadratic Equations Using the Quadratic Formula"
<http://cnx.org/content/m21927/latest/>
5 "Quadratic Equations: Solving Quadratic Equations Using the Quadratic Formula"
<http://cnx.org/content/m21927/latest/>
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OpenStax-CNX module: m21915
Solutions to Exercises in this Module
Solution to Exercise (p. 2)
12 or 18
Solution to Exercise (p. 3)
In 9 and 15 years, the population of the city will be 95,865.
Solution to Exercise (p. 4)
width = 2 feet, length = 7 feet
Solution to Exercise (p. 4)
height = 6 meters, base = 8 meters
Solution to Exercise (p. 5)
14 and 15, amd 14 and 15
Solution to Exercise (p. 5)
2
Solution to Exercise (p. 6)
10 in. by 10 in.; 2 by 2 is not physically possible.
Solution to Exercise (p. 7)
(a). 5.1 parts per million (b). 3.6 years
Solution to Exercise (p. 8)
3 ft wide
Solution to Exercise (p. 9)
$8 or $14
Solution to Exercise (p. 9)
(a) 12, 036 (b) 3 and 12 years from now
Solution to Exercise (p. 9)
(a) 48 feet (b) 64 feet (c) t = 3
Solution to Exercise (p. 10)
length = 8; width = 1
Solution√to Exercise (p.√ 10)
21
width = 21 length =
Solution to Exercise (p. 10)
b = 7; h = 4
Solution to Exercise (p. 10)
−9, −8 or 8, 9
Solution to Exercise (p. 10)
−13, −11 or 11, 13
Solution to Exercise (p. 10)
n = 2, −5
Solution
to Exercise (p. 11)
√
4 + 12.5 inches
Solution to Exercise (p. 11)
2
3
(a) carbon monoxide now 2.2 parts per million
(b) 2.37 years
Solution to Exercise (p. 11)
x=2
Solution to Exercise (p. 12)
Solution to Exercise (p. 12)
51
Solution to Exercise (p. 12)
3x+14
(x+2)(x−2)
3
7
x = 0, 34
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