Applications of Quadratic Equations and Functionstlment.nait.ca/.../A64-Applications_of_Quadratic_Equations.pdf · The sum of the squares of two numbers is 34, ... Applications of

LAST REVISED November, 2008

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Algebra Module A64

Applications of Quadratic Equations and Functions

Module A64 − Applications of Quadratic Equations and Functions

1

Introduction to Applications of Quadratic Equations Statement of Prerequisite Skills Complete all previous TLM modules before beginning this module.

Required Supporting Materials Access to the World Wide Web. Internet Explorer 5.5 or greater. Macromedia Flash Player.

Rationale Why is it important for you to learn this material? This module contains theory on solving a variety of word problems plus a section on sketching quadratic equations.

Learning Outcome When you complete this module you will be able to… Solve a variety of word problems and sketch quadratic equations.

Learning Objectives 1. Solve number problems using quadratics. 2. Solve geometric word problems using quadratics. 3. Solve motion problems using quadratics. 4. Graph quadratic equations.

Connection Activity A rectangular piece of tin 9 cm by 12 cm is to be made into an open box with base area 60 cm 2 by cutting out equal squares from the four corners and then bending up the edges. Find to the nearest millimetre the length of the side of the square cut from each corner.

http://www.microsoft.com/windows/ie/default.asp

http://www.macromedia.com/shockwave/download/download.cgi?P1_Prod_Version=ShockwaveFlash

Module A64 − Applications of Quadratic Equations and Functions 2

OBJECTIVE ONE When you complete this objective you will be able to… Solve number problems using quadratics.

Exploration Activity

EXAMPLE 1 Two positive numbers differ by 5 and their product is 234. Find the numbers. SOLUTION: First we must identify the unknown, therefore:

• Let x = larger number

• Now if x is the larger number then x – 5 = the smaller number Now since we know the product of the two numbers is 234 we write,

(x)(x – 5) = 234 Solving this as a quadratic we get:

x2 – 5x – 234 = 0 Factoring this quadratic we get:

(x – 18)(x +13) = 0 Solving for x we get: x = 18 x = –13 Check:

i) x = –13 is not a solution since the number must be positive. ii) x = 18

larger number = 18 smaller number = 13 product is 18 × 13 = 234 Therefore the solution of 18 and 13 checks.


EXAMPLE 2 One positive number exceeds 3 times another positive number by 5. The product of the two numbers is 68. Find the numbers. SOLUTION: First we must identify one of the unknowns. Therefore:

• Let x = the smaller number.

• The larger number is 3 × the smaller number plus 5 which gives us 3x + 5. We write the product of the two numbers as: (x) × (3x + 5) = 68 Rewrite this as a quadratic to get: 3x2 +5x – 68 = 0 Factoring this quadratic we get:

(3x + 17)(x – 4) = 0 Solving the quadratic gives us:

x = 4 and x = 173−

Check:

i) x = 173− is excluded because the number must be positive.

ii) x = 4

smaller number = 4 larger number is: 3 × 4 + 5 = 17 4 × 17 = 68

3

Module A64 − Applications of Quadratic Equations and Functions 4

Experiential Activity One 1. Two positive numbers differ by 5. Their product is 36. Find the numbers. 2. Two consecutive even integers have a product of 360. Find these integers. 3. The sum of the squares of two numbers is 34, the first number being one less than

twice the second number. Determine the numbers. Show Me 4. The sum of the squares of three consecutive integers is 50. Find the numbers.

Experiential Activity One Answers 1. 9 and 4 2. 18 and 20 3. 5 and 3; or –5.4 and –2.2 4. 3, 4, and 5; or –3, –4 and –5


OBJECTIVE TWO When you complete this objective you will be able to… Solve geometric word problems using quadratics.


EXAMPLE 1 The width of a certain box is 3 cm, and the height is 3 cm less than the length. If the volume of the box is 84 cm3, find the length and the height of the box. Remember that the formula for the volume of a box is: V = 1 • w • h SOLUTION: First we will draw a sketch of the box.

h

lw

Identify the unknowns.

• height of box = h and the length of the box = h + 3 • width of the box = 3 and the volume of the box = 84 • volume of a box = length • width • height

Since V = 1 • w • h therefore 84 = (h + 3)(3)(h) Rewrite this as a quadratic. 84 = 3h2 + 9h 3h2 + 9h – 84 = 0 Factor and solve: (3h + 21)(h – 4) = 0 3h + 21 = 0 h – 4 = 0 h = –7 and h = 4 but h = –7 is a negative value and has no practical significance for height.

therefore h = 4

5


Check: height = 4, length = 7, width = 3 v = 4 × 7 × 3 = 84 cm3

EXAMPLE 2 A uniform strip is mowed on the inside edges of 4 sides of a field 80 m by 100 m. A rectangular interior section of 6000 m2 remains unmowed. Find the width of the strip. SOLUTION: First we will let x be the width of the unmowed strip and draw a sketch of the field.

100 – 2x

80 – 2x

unmowed section

area = 6000 m2

x

100

80

Now make sure we understand the labels used on the diagram.

• The dimensions of the field are 100 by 80.

• The length of the unmowed strip is found by subtracting 100 – x – x because x is the width of the unmowed part and must be subtracted twice giving 100 – 2x.

• The width of the unmowed strip is 80 – 2x.

Area = length • width Area (unmowed) = (100 – 2x)(80 – 2x)

6


Given the unmowed area is 6000 m2, the equation is written as:

6000 = (100 – 2x)(80 – 2x) 6000 = 8000 – 200x – 160x + 4x2 4x2 – 360x + 2000 = 0

Solve for factors by using the quadratic formula.

2360 (360) 4(4)(2000)8

360 312.418

x

x

± −=

±=

x = 5.95 x = 84.05 Check:

i) if x = 84.05 m, the strip mowed would be wider than the field which means

this answer could not be correct.

ii) if x = 5.95 m

width of unmowed = 68.1m

length of unmowed = 88.1

area = 88.1 × 68.1 = 6000 m2

7


Experiential Activity Two 1. The length of a rectangle is 1 cm more than the width. If 3 cm is added to the length

and 1 cm to the width, the new area is twice the original area. Find the dimensions of the original rectangle. A = 1 • w

2. If the radius of a circle is doubled, its area increases by 192π m2. Find the radius of

the circle. 2A rπ= 3. A man designing a machine part finds that if he changes a square part to a rectangular

part by increasing one dimension of the square by 2 mm, the area becomes 50 mm2. Determine the side of the square part.

4. A metal cube expands when heated. If the volume changes by 6.00 mm3 and each

edge is 0.20 mm longer after the cube is heated, what was the original length of the edge of the cube? Show Me

Experiential Activity Two Answers 1. 4cm × 5cm 2. 8 m 3. 6.14 mm 4. 3.06175 mm

8


OBJECTIVE THREE When you complete this objective you will be able to… Solve motion problems using quadratics.


EXAMPLE 1 A pilot flies 600 km to city B. On the return trip he could reduce his flying time by 30 minutes, by increasing his speed by 40 km/h. Find his original speed.

SOLUTION:

First we must identify one of the unknown speeds therefore we let x = the slower speed.

Distance = rate × time. That is D = r × t . Solving for time we get Dtr

=

Therefore, time (going) = 600x

Since speed was increased by 40 km/h on the return trip, the faster speed = x + 40

Using the formula Dtr

= , 600kmtime back = ( 40)km/hx +

Writing the equation with respect to time we get:

Time going – Time back = ½ hour

600 600 140 2x x

− =+

To solve this equation we must rewrite it as a quadratic. To do this we multiply every term by (x)(x + 40)(2). This will eliminate the fractions from the equation.

(600)(x + 40)(2) – (600)(x)(2) = (x)(x + 40)

x2 + 40x – 48000 = 0

(x + 240)(x – 200) = 0

x = –240 x = 200

9


Check:

i) x = –240 not a solution since speed cannot be negative.

ii) x = 200

time going = 600200

= 3 hours time back = 600240

= 2.5 hours

10


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Experiential Activity Three 1. After flying 1200 km a pilot determines that he can make the return trip in 1 hour less

if he increases his average speed by 100 km/h. What was his original average speed? 2. Two men start at the same time from the same place and travel along roads that are at

right angles to each other. One man travels 4 km/h faster than the other, and at the end of 2 hours they are 40 km apart. Determine their rates of travel.

3. By increasing his average speed by 10 km/h a motorist could save 36 minutes in

traveling a distance of 120 km. Find his actual average speed. Note: It is unsafe to exceed the posted speed limit. Show Me

4. A man travels down a river that is 36 km long. The time required for him to go down

the river and return to his starting point is 8 hours. If the rate of his boat in still water is 12 km/h, what is the rate of the river current?

Experiential Activity Three Answers 1. 300 km/h 2. 12 km/h and 16 km/h 3. 40 km/h 4. 6 km/h


OBJECTIVE FOUR When you complete this objective you will be able to… Graph quadratic equations.

Exploration Activity All quadratic equations we have studied so far have contained one unknown. However, to draw a graph requires 2 variables: the independent variable x and the dependent variable y. We rewrite the standard equation

ax2 + bx + c = 0 as y = ax2 + bx + c Now we choose values for x (independent variable) and use these to find values for y (dependent variable). We plot these points on a coordinate grid and sketch the graph of the function.

EXAMPLE 1 Graph the curve y = x2 – l0x + 16. SOLUTION: 1. For the table of values, choose values of x from –8 to +8 x –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 y 160 135 112 81 16 –5 0

(a) you complete the table of values by continuing to substitute the x values into the given function.

(b) now we will plot these points on graph paper and sketch the graph. This graph is

called a PARABOLA.

-40

-20

0

20

40

60

80

100

120

140

160

-8 -6 -4 -2 0 2 4 6 8 10 12

y=x2-10x+16

The selection of these points did not give us a complete sketch of the graph. So choose more positive x values and complete the graph.

12


Plot the following quadratic functions.

1. y = 2x2 + x – 3 3. y = x2 – 3x – 2 2. y = 4x2 – 8x + 4 4. y = 2x2 + 2 – 3x We can sometimes make our task easier if we have some idea as to what the graph looks like before we start. The following 5 steps will help us predict the shape of the graph. y = x2 – l0x + 16 STEP 1 1. If a > 0, i.e. a is positive, then the graph will open upwards.

2. If a < 0, i.e. a is negative, then the graph will open downwards.

3. If a = 0, then the standard quadratic of y = ax2 + bx + c can be rewritten as y = 0x2 +

bx + c. When this is simplified we get y = bx + c which is a linear equation.

For example in the quadratic y = x2 – l0x + 16, a = 1 which is greater than 0 so the curve opens upwards.

13


STEP 2 y–intercept let x = 0 and substitute into the quadratic function to solve for the corresponding y value,

y = x2–l0x+ 16, if x = 0

then y = 16 Therefore y–intercept is x = 0, y = 16 STEP 3 x–intercept(s) Let y = 0 and solve for the corresponding x values x2 – l0x + 16 = 0 (factor and solve or use the quadratic formula if not

factorable) (x – 8)(x – 2) = 0 x = 8, x = 2 when y = 0, x = 8 (1st x intercept) and when y = 0, x = 2 (2nd x intercept) STEP 4 axis of symmetry Using the quadratic formula:

2 42

b b acxa

− ± −=

Rewrite the formula:

2 4

2b b acx

a− + −

= 2 4

2b b acx

a− − −

=

or

2 4

2 2b b axa a− −

= +c

2 42 2

b b acxa a− −

= −

The quantity 2

ba− represents a midpoint. We add

2 42

b aca− to find one root and subtract

2 42

b aa− c to find the other root.

14


Therefore 2

bxa−

= is called the axis of symmetry.

From y = x2– l0x + 16 we see a = l, b = –l0, and c = 16

Therefore ( 10) 52 2

bxa− − −

= = =

graphically:

x

y

x = 5

Axis of symmetry

STEP 5 Maximum and minimum points This depends on whether a > 0 or a < 0 a > 0 (minimum point); a < 0 (maximum point) y = x2 – 10x + 16 so a is a minimum since it is > 0 From the previous step, we see that x = 5 is the axis of symmetry. Therefore the maximum point or minimum point must lie on this line. If we substitute x = 5, into the original equation, this will give us the y-coordinate of the vertex. (In this case a minimum point)

y = x2 – 10x + 16 for x = 5, we get y = –9

Minimum point is found at: x = 5 and y = –9 i.e., (5, –9) Summary: a > 0 ............ opens up y – int ........... (0,16) x – int ........... (8,0) and (2,0) vertex ........... (5,−9) minimum

15


EXAMPLE 2 Let us apply these 5 steps and sketch the graph of the following function. unction. y = x2 – 8x + 12 y = x2 – 8x + 12 SOLUTION: SOLUTION: Step 1Step 1 a > 0, therefore the curve opens upward. Step 2 For the y intercept, let x = 0, we get y = 12. Step 3 For x intercepts, let y = 0. x2 – 8x + 12 = 0 (x – 6)(x – 2) = 0

x = 6, x = 2 Step 4 axis of symmetry

( )8

42 2

ba

− −−= =

Step 5 minimum point at x = 4

substitute into original equation and get y = –4. Now using these 4 points we sketch the following graph.

-10-8-6-4-202468

10121416

-8 -6 -4 -2 0 2 4 6 8 10 12

(0, 12)

(4, -4)

x = 4

x

y

16


17

Try the following: Sketch the graphs of the following quadratics. 1. y = x2 +7x + 6 7. y = x2 + 11x – 12 2. y = 2x2 – 7x + 5 8. y = 4x2 – 8x + 3 3. y = 6x2 – 35x + 6 9. y = 9x2 – 5x 4. y = 16x2 –24 x+ 9 10. y = 4x2 – 28x + 49 5. y = 18x2 + 7x 11. y = 9x2 – 55x + 6 6. y = x2 + x – 110 12. y = 2x2 + 39x – 20 Some quadratics cannot be plotted using the 5 steps just outlined so you will have to supplement the information with a table of values. The following example illustrates this.


EXAMPLE 3 Graph y = x2 – 10x + 25 SOLUTION: Let us first find the 5 points that we normally use when we plot a quadratic.

a > 0, therefore graph opens upward.

For y–intercept, we let x = 0 and substitute this value into the original quadratic. Therefore at x = 0, y = 25 and the y–intercept is at the point (0, 25)

For x–intercepts, let y = 0

(x – 5)(x – 5) = 0 x = 5, x = 5

For axis of symmetry

10 5

2 2bxa−

= = =

Minimum point will be found along the axis of symmetry. Therefore substitute

x = 5, into y = x2 – 10x + 25 25 – 50 + 25 = y therefore y = 0

We have located these points on the grid below.

5

25

20

15

10

2 4 1086

18


Note: The x intercepts, and minimum points are all the same. This will make it difficult to plot the graph unless we determine some additional points. Therefore we must make a table of values. Make a Table of Values to find additional values.

x 2 8 10 y 9 9 25

Plot these points and finish the graph.

EXAMPLE 4 Graph y = x2 + x + l SOLUTION: Let us first find the 5 points that we normally use when we plot a quadratic.

a > 0, opens up.

For y–intercept, we let x = 0 and substitute this value into the original quadratic. Therefore at x = 0, y = 1 and the y–intercept is at the point (0, l)

For x–intercepts, let y = 0. If you attempt to factor this quadratic you will

discover the roots are imaginary. Therefore the curve does not cross the x– axis.

For axis of symmetry

12 2

ba− −

=

Minimum point at 12

x −= , is found by substituting this value into y = x2 + x + 1

21 1 1

2 2− −⎛ ⎞ ⎛ ⎞+ + =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

34

19


We have located these points on the grid below.

2

1

1 2

y-interceptmin. pt.

x

y

Note: Since the roots are imaginary we have only two points for our sketch: the minimum point and the y–intercept. This will make it difficult to plot the graph unless we determine some additional points. Therefore we must make a table of values. Make a Table of Values to find additional values.

x –2 –1 1 2 y 3 1 3 6

Plot these points and finish the graph.

20


21

Experiential Activity Four Sketch the graph of the following quadratic functions. 1. y = x2 – 12x + 36 3. y = 2x2 + 4x + 3 Show Me 2. y = x2 – 4 4. y = x2 + 4

Practical Application Activity Complete the Applications of Quadratic Equations and Functions assignment in TLM.

Summary This module presented theory on solving a variety of word problems plus a section on sketching quadratic functions.


Appendix

Graphs of Quadratic Functions

-4

-2

0

2

4

f(x)

-4 -3 -2 -1 1 2 3 4

x

Graph 1

-8

-6

-4

-2

0

2

4

6

8f(x)

-8 -6 -4 -2 2 4 6 8

x

Graph 2

22


-10

-8

-6

-2

0

2

4

6f(x)

-8 -6 -4 -2 2 4 6 8

x

Graph 3

-10

-8

-6

-2

0

2

4

6 f(x)

-6 -4 -2 2 4 6x

Graph 4

23


-10

-5

5

10

f(x)

-10 -5 5 10

x

Graph 5

-4

-2

0

2

4

6

10

12f(x)

-8 -6 -4 -2 2 4 6 8

x

Graph 6

24


-6

-4

-2

0

2

6

8

10f(x)

-6 -4 -2 2 4 6

x

Graph 7

-10

-8

-4

-2

0

2

4 f(x)

-2 2 4 6

x

Graph 8

25


-10

-8

-6

-4

-2

0

2

4

f(x)

-8 -6 -4 -2 2 4 6 8 x

Graph 9

-6

-4

-2

0

2

6

8

10 f(x)

-6 -4 -2 2 4 6

x

Graph 10

26


-6

-4

-2

0

2

4

6

8

10

12

f(x)

-8 -6 -4 -2 2 4 6 8 10

x

Graph 11

-10

-8

-4

-2

0

2

4 f(x)

-6 -4 -2 2 4 6

x

-6

Graph 12

27


-4

-6

-4

-2

0

2

6

8

10 f(x)

-3 -2 -1 1 2 3 4

x

Graph 13

-10

-8

-6

-2

0

2

4

6 f(x)

-3 -2 -1 1 2 3 x

Graph 14

28


-10

-5

0

5

10

f(x)

-4 -3 -2 -1 1 2

x

Graph 15

-10

-5

0

5

10 f(x)

-2 -1 1 2 3x

Graph 16

29


-15

-5

0

5

f(x)

-3 -2 -1 1 2 x

Graph 17

-10

-8

-4

-2

0

2

4

f(x)

-3 -2 -1 1 2 3 x

Graph 18

30


-10

-5

0

5

10

f(x)

-4 -3 -2 -1 1 2 3 4

x

Graph 19

-5

0

5

15 f(x)

-1 1 2 3

x

Graph 20

31


-5

0

5

15

f(x)

-3 -2 -1 1 2

x

Graph 21

-10

-5

0

5

10 f(x)

-2 -1 1 2 x

Graph 22

32


-10

-5

0

5

10

f(x)

-4 -3 -2 -1 1 2 3 4

x

Graph 23

-5

0

5

10

f(x)

-2 -1 1 2

x

Graph 24

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Applications of Quadratic Equations and Functionstlment.nait.ca/.../A64-Applications_of_Quadratic_Equations.pdf · The sum of the squares of two numbers is 34, ... Applications of