college algebra

1.5 - 1

10TH EDITION

LIAL

HORNSBY

SCHNEIDER

COLLEGE ALGEBRA

1.5 - 21.5 - 2

1.5Applications and Modeling with Quadratic Equations

Geometry ProblemsUsing the Pythagorean TheoremHeight of a Projected ObjectModeling with Quadratic Equations

1.5 - 31.5 - 3

Geometry Problems

Problem Solving When solving problems that lead to quadratic equations, we may get a solution that does not satisfy the physical constraints of the problem. For example if x represents a width and the two solutions of the quadratic equation are – 9 and 1, the value – 9 must be rejected since a width must be a positive number.

1.5 - 41.5 - 4

Example 1 SOLVING A PROBLEM INVOLVING THE VOLUME OF A BOX

A piece of machinery is capable of producing rectangular sheets of metal such that the length is three times the width. Equal-sized squares measuring 5 in. on a side can be cut from the corners so that their resulting piece of metal can be shaped into an open box by folding up the flaps. If specifications call for the volume of the box to be 1435 in.3, what should the dimensions of the original piece of metal be?

1.5 - 51.5 - 5


Solution

Step 1 Read the problem. We must find the dimensions of the original piece.

1.5 - 61.5 - 6


Step 2 Assign a variable. We know the length is three times the width, so let x = width (in inches) and thus 3x = the length.The box is formed by cutting 5 + 5 = 10 in. from both the length and the width.

Solution

x – 103x – 10

5

3x – 10 5

5 5

55 5

5 5

x – 10 x

3x

1.5 - 71.5 - 7


Solution

Step 3 Write an equation. The formula for the volume of a box is V = lwh.

Volume = length width height

1435 (3 10)x ( 10)x (5)

Note that the dimensions of the box must be positive numbers, so 3x – 10 and x – 10 must be greater than 0, which implies 10

3x 10x and

1.5 - 81.5 - 8


Solution

Step 3 Write an equation. The formula for volume of a box is V = lwh.

Volume = length width height

1435 (3 10)x ( 10)x (5)

These are both satisfied when x > 10.

1.5 - 91.5 - 9


Solution

Multiply.

Step 4 Solve the equation.

21435 15 200 500x x

20 15 200 935x x Subtract.

20 3 40 187x x Divide by 5.

0 (3 11)( 17)x x Factor.

1.5 - 101.5 - 10


Solution


0 (3 11)( 17)x x Factor.

3 11 0x or 17 0x Zero-factor property

113

x or 17x Solve.

A length cannot be negative.

1.5 - 111.5 - 11


Solution

Step 5 State the answer. Only 17 satisfies the restriction x > 10. Thus, the dimensions of the original piece should be 17 in. by 3(17) = 51 in.

1.5 - 121.5 - 12


Solution

Step 6 Check. The length of the bottom of the box is 51 – 2(5) = 41 in. The width is 17 – 2(5) = 7 in. The height is 5 in. (the amount cut on each corner), so the volume of the box is

341 7 5 1435 in. , as required.V lwh

1.5 - 131.5 - 13

Using Pythagorean Theorem

Pythagorean theorem from geometry is used in Example 2. The two sides that meet at the right angle are the legs and the side opposite the right angle is the hypotenuse.

Leg a

Leg b

Hypotenusec

1.5 - 141.5 - 14

Pythagorean Theorem

In a right triangle, the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse.

2 2 2a b c Leg a

Leg b

Hypotenusec

1.5 - 151.5 - 15

Example 2 SOLVING A PROBLEM USING THE PYTHAGOREAN THEOREM

Erik Van Erden finds a piece of property in the shape of a right triangle. He finds that the longer leg is 20 m longer than twice the length of the shorter leg. The hypotenuse is 10 m longer than the length of the longer leg. Find the lengths of the sides of the triangular lot.

1.5 - 161.5 - 16


Solution Step 1 Read the problem. We must find the lengths of the three sides.

Step 2 Assign a variable. Let s = the length of the shorter leg (in meters).Then 2s + 20 = length of the longer leg, and(2s + 20) + 10 or 2s + 30 = the length of the

hypotenuse.

s

2s + 20

2s + 30

1.5 - 171.5 - 17


Solution

Step 3 Write an equation.

2a 2b 2c

2s 2(2 20)s 2(2 30)s

Be sure to substitute

correctly here.

Pythagorean theorem

1.5 - 181.5 - 18


Solution

Remember the middle term

here.

Square the binomials.


2 2 2(4 80 400) 4 120 900s s s s s

Remember the middle term

here.

1.5 - 191.5 - 19


Solution

Standard form

or

2 2 2(4 80 400) 4 120 900s s s s s 2 40 500 0s s

( 50)( 10) 0s s Factor.

50 0s 10 0s Zero-factor property

50s or 10s Solve.


1.5 - 201.5 - 20


Solution

Step 5 State the answer. Since s represents a length, – 10 is not reasonable. The lengths of the sides of the triangular lot are 50 m, 2(50) + 20 = 120 m, and 2(50) + 30 = 130 m.

Step 6 Check. The lengths 50, 120, and 130 satisfy the words of the problem, and also satisfy the Pythagorean theorem.

1.5 - 211.5 - 21

Height of a Projected Object

If air resistance is neglected, the height s (in feet) of an object projected directly upward from an initial height of s0 feet, with initial velocity v0 feet per second is

20 016 ,s t v t s

where t is the number of seconds after the object is projected. The coefficient of t

2, – 16, is a constant based on the gravitational force of Earth. This constant varies on other surfaces, such as the moon and other planets.

1.5 - 221.5 - 22

Example 3 SOLVING A PROBLEM INVOLVING THE HEIGHT OF A PROJECTILE

If a projectile is shot vertically upward from the ground with an initial velocity of 100 ft per sec, neglecting air resistance, its height s (in feet) above the ground t seconds after projection is given by

216 100 .s t t

1.5 - 231.5 - 23


250 16 100 .t t

Solution

a. After how many seconds will it be 50 ft above the ground?

We must find the values of t so that height s is 50 ft. Let s = 50 in the given equation.

216 100 50 0t t Standard form

28 50 25 0t t Divide by 2.

1.5 - 241.5 - 24


Solution


28 50 25 0t t Divide by 2.

2( 50) ( 50) 4(8)(25)2(8)

t Quadratic formula

50 170016

t

1.5 - 251.5 - 25


Solution


50 170016

t

.55t or 5.70t

Use a calculator.

1.5 - 261.5 - 26


Solution


.55t or 5.70t

Both solutions are acceptable, since the projectile reaches 50 ft twice: once on the way up (after .55 sec) and once on the way down (after 5.70 sec).

1.5 - 271.5 - 27

b. How long will it take for the projectile to return to the ground?


Solution When the projectile returns to the ground, the height s will be 0 ft, so let s = 0 in the given equation. 20 16 100t t

0 4 (4 25)t t Factor.

4 0t or 4 25 0t Zero-factor property

0t 4 25t or

6.25t Solve.

1.5 - 281.5 - 28

b. When the projectile returns to the ground, the height s will be 0 ft, so let s = 0 in the given equation.


Solution

The first solution, 0, represents the time at which the projectile was on the ground prior to being launched, so it does not answer the question. The projectile will return to the ground 6.25 sec after it is launched.

1.5 - 291.5 - 29

Example 4

The bar graph shows sales of SUVs in the United States, in millions.The quadratic equation

2.00579 .2579 .9703S x x

ANALYZING SPORT UTILITY (SUV) VEHICLES

models sales of SUVs from 1992 to 2003, where S represents sales in millions, and x = 0 represents 1992, x = 1 represents 1993 and so on.

1.5 - 301.5 - 30

a. Use the model to determine sales in 2002 and 2003. Compare the results to the actual figures of 4.2 million and 4.4 million.

Example 4

2.00579 .2579 .9703S x x


Solution

Original model

2.00579( ) .2579( )10 1 9 30 . 70S For 2002, x = 10.

4.1 million

1.5 - 311.5 - 31


Example 4

2.00579 .2579 .9703S x x


Solution

Original model

2.00579( ) .2579( )11 1 9 31 . 70S For 2003, x = 11.

4.5 million

1.5 - 321.5 - 32


Example 4 ANALYZING SPORT UTILITY (SUV) VEHICLES

Solution

The prediction is .1 million less than the actual figure of 4.2 in 2002 and .1 million more than the actual prediction of 4.4 million in 2003.

1.5 - 331.5 - 33


Solution

23.5 .00579 .2579 .9703x x Let S = 3.5 in the original model.

20 .00579 .2579 2.5297x x Standard form

b. According to the model, in what year do sales reach 3.5 million? (Round down to the nearest year.) Is the result accurate?

1.5 - 341.5 - 34


Solution 20 .00579 .2579 2.5297x x Standard form

2.2579 (.2579) 4(.00579)( 2.5297)2(.00579)

x

Quadratic formula


1.5 - 351.5 - 35


Solution 2.2579 (.2579) 4(.00579)( 2.5297)

2(.00579)x

Quadratic formula

52.8x 8.3x or


1.5 - 361.5 - 36

Reject the negative solution and round 8.3 down to 8. The year 2000 corresponds to x = 8. Thus, according to the model, the number of SUVs reached 3.5 million in the year 2000. The model closely matches the graph, so it is accurate.


Solution


college algebra

Documents