-Algebra Class 8 (Zambak)-Zambak Publishing

A. RATIONAL NUMBERS: THE SET Q 1. Understanding Rational Numbers Definition

The ratio of an integer to a non-zero integer is called a rational number. The set of rational a numbers is denoted by Q. Q = { | a, b  Z, b  0} b

2. The Set of Positive Rational Numbers If a rational number represents a point on the number line on the right side of zero, then it is called a positive rational number. a is a positive rational number if a and b are both positive integers or both negab tive integers. 2 –2 2 and For example, are positive rational numbers, and denoted by . 7 –7 7

In short,

Definition

The set of positive rational numbers is denoted by Q+. Q+ = {

a a |  0 and a, b  , b  0} b b

3. The Set of Negative Rational Numbers If a rational number represents a point on the number line on the left side of zero, then it is called a negative rational number. a In short, is a negative rational number if a is a positive integer and b is a negative integer, b or if a is a negative integer and b is a positive integer. –5 5 are negative rational numbers. We can write negative rational and 4 –4 5 –5 5 numbers in three ways: –   . 4 4 –4

For example,

Definition

The set of negative rational numbers is denoted by Q–. a a Q– = { |  0 and a, b  , b  0} b b 10

Algebra 8

A. THE SET OF REAL NUMBERS 1. Understanding Real Numbers In algebra we use many different sets of numbers. For example, we use the natural numbers to express quantities of whole objects that we can count, such as the number of students in a class, or the number of books on a shelf. The set of natural numbers is denoted by N. N = {1, 2, 3, 4, 5, ...} -6

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

Natural numbers

The set of whole numbers is the set of natural numbers together with zero. It is denoted by W. W = {0, 1, 2, 3, 4, 5, ...} -6

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

Whole numbers

The set of integers is the set of natural numbers, together with zero and the negatives of the natural numbers. It is denoted by Z. Z = {..., –5, –4, –3, –2, –1, 0, 1, 2, 3, 4, 5, ...} -6

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

Integers

We use integers to express temperatures below zero, distances above and below sea level, and increases and decreases in stock prices, etc. For example, we can write ten degrees Celsius below zero as –10°C. To express ratios between numbers, and parts of wholes, we use rational numbers. 8 2 3 0 17 For example, , , – , , and are rational numbers. 3 5 7 7 1 The set of rational numbers is the set of numbers that can be written as the quotient of two integers. It is denoted by Q. Q={

a | a, b   and b  0} b 7 2 -7

-6

-5

-4

1 2 -3

-2

-1

2 3 0

5 4 1

2

13 2 3

4

5

6

7

Some rational numbers

Radicals

11

We can write every rational number as a repeating or terminating decimal. Conversely, we can write any repeating or terminating decimal as a rational number. 3 321  0.6, and  0.324 = 0.324242424... 5 990 –– 0.6 is a terminating decimal, and 0.324 is a repeating decimal.

For example,

There are some decimals which do not repeat or terminate. For example, the decimals

0.1012001230001234000 ... 3.141592653 ... = 

R = R+  {0}  R– R+ is the set of positive real numbers R– is the set of negative real numbers

2.71828 ...

=e

1.4142135 ...

= ñ2

do not terminate and do not repeat. Therefore, we cannot write these decimals as rational numbers. We say that they are irrational.

Definition

A number whose decimal form does not repeat or terminate is called an irrational number. The set of irrational numbers is denoted by Q or I. Definition

The union of the set of rational numbers and the set of irrational numbers forms the set of all decimals. This union is called the set of real numbers. The set of real numbers is denoted by R. R = Q  Q For every real number there is a point on the number line. In other words, there is a one-to-one correspondence between the real numbers and the points on the number line.

Real Numbers

-2.35 -5

-4

-3

-2

-0.5

0.6

-1 0 1 2 Some rational numbers

e

p 3

4

5

The real numbers fill up the number line.

NWZQR Q  R

12

We can summarize the relationship between the different sets of numbers that we have described in a diagram. As we know, the set of natural numbers is a subset of the set of whole numbers, the set of whole numbers is a subset of the set of integers, the set of integers is a subset of the set of rational numbers, and the set of rational numbers is a subset of the set of real numbers. This relationship is shown by the diagram on the left. Algebra 8

1. Understanding Square Roots Remember that we can write a  a as a2. We call a2 the square of a, and multiplying a number by itself is called squaring the number. The inverse operation of squaring a number is called finding the square root of the number. Objectives

After studying this section you will be able to: 1. Understand the concepts of square root and radical number. 2. Use the properties of square roots to simplify expressions. 3. Find the product of square roots. 4. Rationalize the denominator of a fraction containing square roots. Definition

If a2 = b then a is the square root of b (a  0, b  0). We use the symbol ñ to denote the square root of a number. ñb is read as ‘the square root of b’. So if a2 = b then a = ñb (a  b, b  0). Here are the square roots of all the perfect squares from 1 to 100. 

ñ1 = 1

2 =4



ñ4 = 2

32 = 9



ñ9 = 3

2



ò16 = 4

2



ò25 = 5

12 = 1 2

4 = 16 5 = 25



ò36 = 6

7 = 49



ò49 = 7

82 = 64



ò64 = 8



ò81 = 9



ó100 = 10

62 = 36 2

2

9 = 81 2

10 = 100

2

The equation x = 9 can be stated as the question, ‘What number multiplied itself is 9?’ There are two such numbers, 3 and –3. Rule

If x  R then

 x if x  0. x2 | x |   – x if x  0.

In other words, if x is a non-negative real number, then if x is a negative real number, then Radicals

x2  x, and

x2  – x. 13

For example,

32  3, ( 32  9  3), and (–3)2  –(–3)  3 ( (–3) 2  9  3).

We can conclude that the square root of any real number will always be greater than or equal to zero. ò–9 is undefined. Negative numbers have no square root because the square of any real number cannot be negative. ò–9  3, since 32 is 9, not (–9). ò–9  –3, since (–3)2 is 9, not (–9).

Note x = ñ9 and x2 = 9 have different meanings in the set of all real numbers.  ñ9 = 32 = |3| = 3  If x2 = 9 then x = 3 or x = –3.

EXAMPLE

1

Solution

EXAMPLE

2

Solution

Evaluate each square root. a. ò81

b. ñ1

g. –ó100

h. –ó0.09 i. ò–4

d. ò64

b. ñ1 = 1

d. ò64 = 8

e.

g. –ó100 = –10

h. –ó0.09 = –0.3

j.

k.

(–4)2  16  4

e. ñ9

(–4)2

j.

a. ò81 = 9

k.

f. ó0.64

–4 2

c. ñ0 = 0

4 2  9 3

f. ó0.64 = 0.8 i. ò–4 is undefined

–42  –16 is undefined

Evaluate each square root. a. ó100

b. ó121

c. ó144

d. ó169

g. ó400

h. ó625

i.

j.

1225

e. ó225

f. ó361

10000

a. ó100 = 10

b. ó121 = 11

c. ó144 = 12

d. ó169 = 13

e. ó225 = 15

f. ó361 = 19

g. ó400 = 20

h. ó625 = 25

i.

j. 14

c. ñ0

1225  35

10000  100 Algebra 8

2. Properties of Square Roots Property

For any real number a and b, where a  0, and b  0, ñañb = óab. For example,

25  16  25  16  5  4  20, 3  27  3  27  81  9, 36 a2  36 a2  6 a ( a  0), and 5  5  5  5  25  5.

Note

b0

If a  0 then

a0

ña  ña = óa  a =

óa . 6 = 

ña . ñ6

a2  a.

Mathematics is a universal language.

3

Simplify each of the following.

Solution

a. ñ2ñ8 = ó28 = ò16 = 4

b.

7  7  7  7  49  7

c. ò50ñ2 = ó502 = ó100 = 10

d.

25  1  25  1  25  5

EXAMPLE

a. ñ2 ñ8

b. ñ7 ñ7 c. ò50ñ2

d. ò25ñ1

e.

576  36  16  36  16  6  4  24

f.

10  90  10  90  900  30

e. ó576

f. ò10ò90

Property

For any real numbers a and b, where a  0, and b > 0, a b

If a > 0 then a a

=

a a

Radicals

= 1 = 1.

For example,

24 6



1  49



a . b

24  4  2, and 6 1 49



1 . 7 15

EXAMPLE

4

Simplify the expressions. a.

g.

Solution

25 9

b.

24a 3

h.

6a

50

c.

2 a5  b6 ab

1 64

d.

e. –

1 100

f.

x3  y3

a.

25 25 5 = = 9 3 9

b.

c.

16 16 4   49 49 7

d.

1  64

f.

625 625 25   144 144 12

e. – g.

h.

i.

1 1 1 – – 100 10 100 24a3

a5  b6 ab2

xy 3

3

x y

50 2



50  25  5 2

1 64



1 8

24 a3  4  a2  4  a2  2a 6a



6a

625 144

xy

i.

2

16 49



a5  b6  a4  b4  ( a2  b2 )2  a2 b2 ab2



xy  x3  y3

1  x2  y2

1 2

2

x  y



1 xy

Property

For any real number a and n  Z, ( a )n  an

(a  0).

Proof ( a )n  a  a  a  ... 

a

n factors of ña

a  a  a  ... a 

an

n factors of a

For example, ( a )2  a2  a, ( 5)3  53  125, and ( 2 )8  2 8  256  16. 16

Algebra 8

EXAMPLE

5

Solution

Evaluate (ñ2)4 + (ñ5)4 – (ñ5)2 – (ñ2)6. ( 2 )2  ( 5)4 – ( 5)2 – ( 2 ) 6  2 4  5 4 – 5 2 – 2 6  (22 )2  (52 )2 – 52 – (23 )2  22  52 – 5 – 2 3  4  25 – 5 – 8  16

3. Working with Pure and Mixed Radicals Definition

A radical expression is an expression of the form index

n

a. radical sign

ña n

radicand

Square roots have index 2. However, we usually write square roots in their shorter form, ña: 2

a a

Definition

A mixed radical is a radical of the form x  n a (x  Q, x  {–1, 0, 1})

For example, 3ñ2, 6ñ7, and 9ó115 are mixed radicals. ò55, ò99, and ò27 are not mixed radicals. We say that they are pure radicals. We can convert between mixed and pure radical numbers to simplify radical expressions. Property

For any real numbers a and b, where a  0 and b  0, a2  b  a b and a b 

For example,

a 2  b.

8  4  2  2 2  2  2 2  2  2 2, 27  9  3  32  3  32  3  3 3, 32  16  2  4 2  2  4 2  2  4 2, and 50  25  2  5 2  2  5 2  2  5 2.

Radicals

17

EXAMPLE

6

Simplify the expressions. a. ñ8 + 2ò32 – ò18 + ò72 – ò98 b. 2ò48 + 3ò27 – ó108 + ó243

Solution

a.

8 = 22  2 = 2 2 2 32 = 2 4 2  2 = 8 18 = 3  2 = 3 2 2

2

72 = 6  2 = 6 2 98 = 7  2 = 7 2 2

   2       



8  2 32  18  72  98  2 2  8 2  3 2  6 2 7 2  2  (2  8  3  6  7)  6 2

b. 2 48  3 27 – 108  243  2 4 2  3  3 3 2  3 –

6 2  3  9 2 3

 8 3  9 3 – 6 3 9 3  (8  9 – 6  9) 3  20 3 EXAMPLE

7

Solution

Write the numbers as pure radicals. a. 2ñ2

b. 3ñ5

c. 5ñ3

a. 2 2  2 2  2  2 2  2  b. 3 5  32  5  9  5  c. 5 3  52  3 

25  3 

4 2 

d. 10ò10

e. xñy

8

45 75

d. 10 10  10 2  10  100  10  1000 e. x y  x2 y Property

For any non-zero real numbers a, b, c, and x, añx + bñx – cñx = (a + b – c)ñx .

Note ña + ñb  óa+b For example, ñ9 + ò16 = 3 + 4 = 7, but ó9 + 16 = ò25 = 5. 18

Algebra 8

EXAMPLE

8

Perform the operations. a. ñ3 + ñ3

b. 2ñ5 + ñ5

c. 3ñ6 + 4ñ6

e. ò50 + ò98 + ó162 Solution

d. 10ñ5 – 3ñ5

f. 5ñx – ò9x + ó64x

a. ñ3 + ñ3 = (1 + 1)ñ3 = 2ñ3 b. 2ñ5 + ñ5 = (2 + 1)ñ5 = 3ñ5 c. 3ñ6 + 4ñ6 = (3 + 4)ñ6 = 7ñ6 d. 10ñ5 – 3ñ5 = (10 – 3)ñ5 = 7ñ5 e. ò50 + ò98 + ó162 =

ó252 + ó492 + ó812 = 5ñ2 + 7ñ2 + 9ñ2

= (5 + 7 + 9)ñ2 = 21ñ2 f. 5ñx – ò9x + ó64x = 5ñx – 3ñx + 8ñx = (5 – 3 + 8)ñx = 10ñx EXAMPLE

9

Solution

Compare the following numbers. a. ñ7 ... 3

b. 3ñ5 ... 2ò10

a.

b. 3 5...2 10

7...3

c. 2ñ7 ... 3ñ3

d. –2ñ3 ... –3ñ2

c. 2 7...3 3

d. –2 3... – 3 2

7... 9

32  5... 2 2  10

22  7... 32  3

– 22  3... – 32  2

7 9

45  40

28  27

– 12  – 18

2 7 3 3

–2 3  –3 2

7 3

3 5  2 10

Property

Let a, b, m, and n be four real numbers, satisfying a = m + n and b = m  n. Then, 1.

2.

m  n  a2 b

m – n  a–2 b

( m  n)

Proof 1. In order to verify these expressions, suppose that t = òm + ñn. t2 = (ò m + ñn)2 = (ò m + ñn)  (ò m + ñn) = (ò m  ñn) + (òm  ñn) + (ñn  òm ) + (ñn  ñn) (by the distributive property) = m + (ò m  ñn) + (ñn  ò m ) + n = m + n + 2ómn a

(by the commutative property)

b

 t = a + 2ñb  t = a  2 b 2

2. We can prove the second part in the same way. Try it yourself. Radicals

19

EXAMPLE

10

Solution

Simplify the expressions. Use the property to help you. a.

32 2

b.

52 6

e.

5  21

f.

2 3

a.

3  2 2  2  1  2 1 2+1

c.

c.

6  32

b.

52 6  3  2

21

3+2

d.

6–4 2

32

6  32  6  2 8  4  2  2  2 4+2

d.

42

6 – 4 2  6 – 2  2 2  6 – 2 2 2  2  6 – 2 8  4 – 2 2 – 2 4+2

42

e. We need a 2 in front of ò21 before we can use the property. Therefore, let us multi2

ply the expression by

2

. 7+3

5  21 

2 2

 5  21  3+1

2

f.

2

 2 3 

2  (5  21) 2



73

10  2 21 2



7 3 2



7 3  2 2

31

42 3 2



3 1 2

3 1  2 2



Check Yourself 1 1. Simplify the expressions. a. ñ2  ñ2 e.

32

b. ñ8  ò32 f.

12

2 2. Evaluate the following.

3

a. (ñ3)2 + (ñ4)4 – (ñ5)2 – (ñ2)4

c. ò3x  ó12x g.

a b 3

a b

3

d. ñ2  ò18 h. –

1 49

b. (ña)4 + (ñb)2 – (ñc)6

3. Simplify the expressions. a. ò18

b. ò50

f. 2ñ2 + 3ñ2 – 4ñ2 20

c. ò48

d. ò20

g. ò50 – ò18 – ò32

e. 5ñ3 – 2ñ3 + ñ3 h. ó12x + ó27x – ó48x Algebra 8

4. Write each number as a pure radical. a. 5ñ3

b. 3ñ5

c. 4ñ2

d. 2ñ5

e. añb

5. Perform the operations. a. 6

10 15 2 3 2

b. 5ñ2 – ñ8

c. ò27 – ò48

d.

27 75 12  – 4 4 4

6. Compare the numbers. a. 3ñ5 and 2ò10

b.

1 and 2

1 3

c. –2ñ5 and –3ñ3

7. Write each expression in its simplest form. a.

3–2 2

b.

c.

62 8

g. ( 6 – 2 )  ( 8  2 12 )

7 – 2 10

d.

3  8 e.

h. ( 7  1)  ( 8 – 28 )

i.

f.

9–4 5

7 – 48

3– 5  3 5

Answers 1. a. 2 b. 16 c. 6x d. 6 e. 4 f. 2 g. c. 4ñ3 d. 2ñ5 e. 4ñ3 f. ñ2

1 ab

h. –

1 7

2. a. 10 b. a2 + b – c3 3. a. 3ñ2 b. 5ñ2

g. –2ñ2 h. ò3 x 4. a. ò75 b. ò45 c. ò32 d. ò20 e. a2  b

1 1 c. –2ñ5 > –3ñ3 7. a. ñ2 – 1  2 3 b. 2 + ñ2 c. ñ5 – ñ2 d. ñ2 + 1 e. ñ5 – 2 f. 2 – ñ3 g. 4 h. 6 i. 2

5. a. 3ò30 b. 3ñ2 c. –ñ3 d. 3ñ3 6. a. 3ñ5 > 2ò10 b.

EXAMPLE

11

Simplify the following. a.

Solution

4  21  13  9

b.

6 6 72

1 16 16

c.

1 1

9 16

Start from the radical on the ‘inside’ of the expression and move outwards. a. Start with ñ9, on the inside, and work outwards. 4  21  13  9  4  21  13  3  4  21  16  4  21  4  4  25  4  5  9  3

b.

6 6 72

1 4 1  6 6 72 16  6 6 72 16 16 4  6 6 72 

1  6 6 36 2

 6 6  6  6 36  6  6  36  6

c. Radicals

1 1

9 25 5 9 3  1  1   16 16 4 4 2 21

EXAMPLE

12

a. Evaluate c.

Solution

a a a a ...  7. Find a.

b.

2 2 2 2 ...

x  x  x  ...  5. Find x.

a. Let x  2 2 2 2 ... . x2  ( 2 2 2 2 ... )2

(take the square of both sides)

x2  2 2 2 2 2 ... 

(remove a square root)

x

x 2x

( 2 2 2 ...  x)

2

x  x 2x  x x

(simplify)

 x  2. Therefore,

b.

2 2 2 2 ...  2.

a a a a ...  7 ( a a a a ... )2  7 2

( x  x  x  ... )2  5 2

c.

x  x  x  x  ...  25   5

a a a a ...  49    7

a  7  49

x  5  25 x  20

a7

4. Multiplying Square Roots To multiply expressions containing square roots, we used the product property of square roots: ña  ñb = óa  b. We can also use the distributive property of multiplication over addition and subtraction to simplify the products of expressions that contain radicals. For example, 2ñ8  3ñ2 = 2  3  ñ8  ñ2 = 6ò16 = 64

Multiply the rational part by the rational part and the radical part by the radical part.

= 24 ñ2  (ñ3 + 2ñ2) = ñ2  ñ3 + ñ2  2  ñ2 = ñ6 + 2  ñ2  ñ2 = ñ6 + 2  2 = ñ6 + 4 22

Algebra 8

EXAMPLE

13

Solution

Perform the operations. a. ñ2(ñ5 + ñ3)

b. ñ3(3ñ3 + 2ñ2)

c. 2ñ5(ñ3 + ñ2 + 2ñ5 – ñ7)

a. ñ2(ñ5 + ñ3) = ñ2  ñ5 + ñ2  ñ3 = ó2  5 + ó2  3 = ò10 + ñ6 b. ñ3(3ñ3 + 2ñ2)= ñ3  3ñ3 + ñ3  2ñ2 = 3  ó3  3 + 2  ó3  2 = 3  3 + 2  ñ6 = 9 + 2ñ6 c. 2ñ5(ñ3 + ñ2 + 2ñ5 – ñ7)= 2ñ5  ñ3 + 2ñ5  ñ2 + 2ñ5  2ñ5 – 2ñ5  ñ7 = 2ò15 + 2ò10 + 4ò25 – 2ò35 = 2ò15 + 2ò10 + 20 – 2ò35

EXAMPLE

14

Solution

Multiply and simplify. a. (ñ2 + ñ3)  (ñ2 + ñ3)

b. (5 + ñ5)  (5 + ñ5)

a. (ñ2 + ñ3)  (ñ2 + ñ3) = ñ2  ñ2 + ñ2  ñ3 + ñ3  ñ2 + ñ3  ñ3 = ñ4 + ñ6 + ñ6 + ñ9 = 2 + 2ñ6 + 3 = 5 + 2ñ6 b. (5 + ñ5)  (5 + ñ5)= 52 + 2  5  ñ5 + (ñ5)2 = 25 + 10ñ5 + 5 = 30 + 10ñ5

EXAMPLE

15

Solution

Multiply and simplify. a. (ñ2 + 1)  (ñ2 – 1)

b. (ñ5 + ñ3)  (ñ5 – ñ3)

d. (ña + 1)  (ña – 1)

e. (ña + ñb)  (ña – ñb)

c. (1 – 2ñ2)  (1 + 2ñ2)

a. (ñ2 + 1)  (ñ2 – 1) = ñ2  ñ2 – ñ2  1+1  ñ2 – 1  1 = (ñ2)2  12 = 2 – 1 = 1 b. (ñ5 + ñ3)  (ñ5 – ñ3) = (ñ5)2 – (ñ3)2 = 5 – 3 = 2 c. (1 – 2ñ2)  (1 + 2ñ2) = 12 – (2ñ2)2 = 1 – 4  2 = 1 – 8 = –7 d. (ña + 1)  (ña – 1) = (ña)2 – 12 = a – 1 (a  0) e. (ña + ñb)  (ña – ñb) = (ña)2 – (ñb)2 = a – b (a, b  0)

EXAMPLE

16

Solution

Radicals

Multiply and simplify. a.

3 5  3– 5

b.

a.

3  5  3 – 5  (3  5)  (3 – 5)  3 2 – ( 5) 2  9 – 5  4  2

b.

2  2  2 – 2  (2  2 )  (2 – 2 )  2 2 – ( 2 ) 2  4 – 2  2

c.

a b  a– b 

2 2  2– 2

c.

a b  a– b

a2 – b 23

EXAMPLE

17

Solution

Multiply and simplify. a. (ñ3 + ñ2)  (ñ5 – 1)

b. (ñ5 + ñ3)  (ñ7 + ñ2)

c. (2ñ3 + 1)  (ñ5 + 1)

d. (3ñ2 – 2)  (ñ5 – ñ3)

a. (ñ3 + ñ2)  (ñ5 – 1)= (ñ3  ñ5) – (ñ3  1) + (ñ2  ñ5) – (ñ2  1) = ò15 – ñ3 + ò10 – ñ2 b. (ñ5 + ñ3)  (ñ7 + ñ2) = (ñ5  ñ7) + (ñ5  ñ2) + (ñ3  ñ7) + (ñ3  ñ2) = ò35 + ò10 + ò21 + ñ6 c. (2ñ3 + 1)  (ñ5 + 1) = (2ñ3  ñ5) + (2ñ3  1) + (1  ñ5) + 1 = 2ò15 + 2ñ3 + ñ5 + 1 d. (3ñ2 – 2)  (ñ5 – ñ3)= (3ñ2  ñ5) – (3ñ2  ñ3) – (2ñ5 + 2ñ3) = 3ò10 – 3ñ6 – 2ñ5 + 2ñ3

5. Rationalizing Denominators 1

Look at the numbers

3

10

19

. They are all fractions, and each fraction 5 2 12 13 has an irrational number as the denominator. In math, it is easier to work with fractions that ,

,

, and

have a rational number as the denominator. Definition

Changing the denominator of a fraction from an irrational number to a rational number is called rationalizing the denominator of the fraction. Rationalizing the denominator does not change the value of the original fraction. To rationalize the denominator, we multiply the numerator and denominator of the fraction a by a suitable factor. For example, if the fraction is in the form , we multiply both the b numerator and the denominator by ñb. a

So,

b

a



b

b



b

ab



bb

a

ab . Note that b



b

and

ab have the same value: they are b

equivalent fractions. Look at some more examples: 3 2 3 3

3

2 2

3 

2



2



3 5

24

3





3 5 2 2



2 3 3 

 2 2

3 2 2 2

3 3 3 3 





3  2 2  2



6 4



6 , 2

3 3  3, and 3

3 52 2 2 2



3  10 3 10 .  22 4 Algebra 8

Definition

An expression with exactly two terms is called a binomial expression. Two binomial expressions whose first terms are equal and last terms are opposite are called conjugates, i.e. a + b and a – b are conjugates. If a  0 and b  0, then the binomials xña + yñb and xña – yñb are conjugates. We can use conjugates to rationalize denominators that contain radical expressions. 1 For example, let us rationalize . ñ3 – ñ2 is the conjugate of ñ3 + ñ2. 3 2 Therefore, we multiply the numerator and the denominator by ñ3 – ñ2 to rationalize the denominator. 1 3 2



3– 2 3– 2



1  ( 3 – 2) ( 3  2 ) ( 3 – 2 )



3– 2 2

( 3) – ( 2 )



2

3– 2 3– 2   3– 2 3–2 1

Remark

(a + b)(a – b) = a2 – b2 (ña + ñb)(ña – ñb) = a – b where a  0 and b  0. EXAMPLE

18

Rationalize the denominators. a.

Solution

a.

b.

5 3–2 2 5 3–2 2

3– 2 2 2 –1



6 2 1– 3

Radicals

3 2 –2 52 5

3–2 2

2 2 –1 

32 2 3 2 2



6 2

d.

1– 3

5  (3  2 2 ) (3 – 2 2 )(3  2 2 )



3 2 –2 52 5

5 3  5 2 2 3 2 – (2 2 ) 2

3 5  2 10 3 5  2 10   3 5  2 10 9–8 1



( 3 – 2 ) (2 2  1) 3  2 2  3  1 – 2  2 2 – 2 1   (2 2 – 1) (2 2  1) (2 2 ) 2 – 12

 

d.

5

c.





c.

3– 2

b.

2 6  3–2 2 2 – 2 8 –1



2 6  3–4– 2 7

( 6  2 ) (1  3) 6  6  3  2 1  2  3   (1 – 3) (1  3) 12 – ( 3) 2 6  18  2  6 6  3 2  2  6 2 6 4 2   – 6 –2 2 1– 3 –2 –2



(3 2 – 2) (5 – 2 5) 3 2  5 – 3 2  2 5 – 2 5 – 2 2 5   (5  2 5) (5 – 2 5) 5 2 – (2 5) 2



15 2 – 6 10 – 10 – 4 5 15 2 – 6 10 – 10 – 4 5  25 – 20 5 25

EXAMPLE

19

Rationalize the denominators to find the sum. 3



3 2 2 3

Solution



3 2 2

2 3–2 2  3 3–2 2   2 3 2 2       3 – 2 2  3 2 2 3–2 2   3–2 2 3 2 2  2

  

( 3  ( 3 – 2 2 ))  ( 2  ( 3  2 2 )) ( 3  2 2)  ( 3 – 2 2) ( 3  3) – ( 3  2 2 )  ( 2  3) ( 2  2 2 ) ( 3)2 – (2 2 )2 3–2 6  6 4 7– 6 6 –7   3–8 –5 5

Check Yourself 2 1. Rationalize the denominators and simplify. a.

g.

3 7

5

b.

1

3

a  b

h.

2 10

c. 2

d. –5 

2 x  y

i.

a3

1 2

e.

5

3 5

3 5

f.

3

2 6

a 3  b4

j.

x3  y3

3

a  b2

2. Rationalize the denominators and simplify. a.

g.

1

2

b.

2 –1 2 5–2

3– 2 2 5– 7

h.

2 10 – 6

c.

i.

7– 5

2

3

d.

5 1 a

6 –2

a b

j.

a b

a– b

5 1

e.

f.

5 –1

3– 2 2 3– 2

3

k.

3 3–2 7

3. Rationalize the denominators and simplify. a. d.

1 5 2 2 2 1

–

1



b.

5– 2 2 2 –1



3

e.

2 2

2 2– 2 3



2 2 2

–

2

3 1 1– 3

1

c.



3 3



1 3– 3

3 3 3

Answers 1. a.

15 21 b. 3 7

b. ñ6 – 2 c. i. 26

30 4

c. ñ2 d. –ò15 e. ò15 f.

5 –1 3 6 6 d. 2 2

e.

3 5 2

f.

g.

4– 6 10

10 20

h.

b a

i.

2 xy

j. ab 2. a. ñ2 + 1

g. 5ñ2 – ò10 + 3 ñ5 – 3 h.

6 –16  3 2 a – ab a  b  2 ab 2 5 j. k. –9ñ3 – 6ñ7 3. a. b. ñ2 c. d. 6 4 3 a–b a–b

35  3 2

e. 17 3 – 3 6 Algebra 8

EXERCISES

1 .1

1. Evaluate the square roots. a. ò36 16x2

d.

5. Perform the operations.

b. ó100 e.

c. –ó121

25  y2

f.

121  a4

a. 3ñ3 + 2ñ3

b. 6ñ5 + ñ5

c. –5ñx + 5ñx

d. ñ6 – 3ñ6

e. 3ò18 + 2ò72

f. ò80 – ó125 + ò45

g. ò75 + ó108 – ò48 + ò27

2. Simplify the expressions. a. ñ3  ñ3

b. ñ5  ñ5

c. ñ3  ò12

d. ñ3  ò27

e. ò2x  ò8x

f.

g.

h. 3 2 x  4  18 xy2

3a  5a

6 xy  24xy

h.

9x3  16 x3 – 4 x 25 x

i.

2.25 – 2.89  1.44

j.

0.9 – 2.7  1.7  0.4

3. Simplify the expressions. 50

a.

d.

2

72 x3 2 2x

b.

e.

72 8 32 x3 y2 24 x

c.

f.

108

6. Write each expression in its simplest form.

27

3xy3

a.

32 2

b.

5–2 6

c.

8  2 12

d.

11  96

e.

8–2 7

f.

5  24

g.

7–4 3

h.

3 8  3– 8

i.

4  15 – 4 – 15

f.

a  2  8a – a

k.

2  2 4  12

3

12 x y

4. Write each number as a mixed radical. a. ñ8

b. ò72

c. ó243

d.

e. ó125

f.

Radicals

1000

x3 y2

27

7. Simplify the expressions. a.

32  21 – 23  4

b.

7 2  –  6 9 3 9

c.

d.

8  16 

10. Rationalize the denominators. a.

3 3

d. 1  2 4 3

1  9  16 96

g. j.

13  6  6  9

3 2

b.

6

1

e.

3 1 3 –1 3

4 3 3 2

2 2

k.

2 3

f.

1– 2

h.

c. –

3– 5

i. l.

1 11 –2 3 2

10 2 7 –5 10  2 21 7 3

8. Find x in each equation. b. 3 3 3 3 ...  x

a.

2 2 2 2 ...  x

c.

3x  3x  3x  ...  9

9. Find the products. a. ñ5  (ñ2 + ñ3) b. ñ7  (1 + ñ7) c. –ñ2  (ñ3 – ñ8 + 1)

11. Perform the operations. a. b. c. d.

d. ñ2  (ñ8 + ò32)

1 3

1



2

3 3 1

2



3 –1

5 2 3 – 11

4 2 3 2 2

–

2 2 3  11

2 3

–

3–2 2

e. ñ6  (2ñ3 + 3ñ2) f. (3 + ñ5)  (3 – ñ5) g. (2ñ2 – 3)  (2ñ2 + 3) h. (2ñ3+2)  (2ñ3 – 2) i. (ò12 + ñ8)  (ñ3 – ñ2) j. (–ò12 + 2ñ2)  (ñ2 + ñ3) k. l. m. 28

2 1 

2 –1

12. Perform the operations. a.

5–2

5 –1

1 2

b.

2  11  72

2 33  2 3–3 5  2 3  2  3  16 – 9 3

5 2 –

c.

1 2 1



1 3 2



1 4 3

 ... 

1 100  99 Algebra 8

Objectives

After studying this section you will be able to: 1. Understand the concepts of nth root and rational exponent. 2. Write numbers in radical or rational exponent form. 3. Understand the properties of expressions with rational exponents. 4. Use the properties of rational exponents to solve problems.

A. RATIONAL EXPONENTS 1. nth Roots In section 8.2 we studied exponential equations. For example, 2n  2n = 2 is an exponential equation. Let us solve it. 2n  2n = 22n = 2

(an  am = an+m)

22n = 21 2n = 1, n =

1

1

1 1 for n in the original equation we will get 2 2  2 2  2, . If we substitute 2 2 1

but we know that ñ2  ñ2 = 2. Therefore, 2 2  2  2 2. Remark

Let x  R – {–1, 0, 1}. If xm = xn then m = n. Definition

For any natural number n and a, b  R. 1

n If an = b then a = b n  b. a is called the nth root of b. It is denoted by ñb. n

In the expression ña, a is called the radicand and n is called the index. n

Remember that we do not usually write the index for square roots: ña = ña.

2

Look at some examples of different roots: 52 = 25 and 5 = ò25

‘the square root of 25 is 5’,

2 = 8 and 2 = ñ8

‘the cube root of 8 is 2’,

3 = 27 and 3 = ò27

‘the cube root of 27 is 3’, and

2 = 16 and 2 = ò16

‘the fourth root of 16 is 2’.

3 3 4

Radicals

3

3 4

29

2. Rational Exponents Definition

If m and n are positive integers (n > 1), and b is a non-negative real number, n

m is called a rational exponent. n 2

1

m

bm  b n .

2

For example, the numbers 8 3 , 4 2 , and 2 2 have rational exponents. EXAMPLE

20

Write the expressions in radical form.

Solution

b. 32 5

1

1

Solution

f.

1

( a 3 )2

2

c. 5 3  3 52  3 25

b. 32 5  5 32

3

21

e. ( x4 ) 4

1

a. 3 2  2 3  3

1

1

d. x 4

c. 5 3

d. x 4  4 x3 EXAMPLE

3

2

1

1

a. 3 2

1

4

e. ( x4 )4  x4

f.

1

3

( a 3 )2 

a

Write the expressions with radical exponents. 3

a.

5

35

a.

5

35  3 5  31  3

d.

3

a2  a 3

b.

c.

64

5

7

3

d.

x14

a2

3

3

b.

3

64  43  4 3  41  4

e.

5

ab2  ( ab 2 )5  a 5  b 5

2

1

5

e.

1

ab2 7

c.

14

x14  x 7  x2

2

Rule

If b is any real number and n is a positive integer (n > 1): 1. If b > 0 then ñb is a positive real number. n

2. If b = 0 then ñb is zero. n

3. If b < 0 and n is odd, then ñb is a negative real number. n

4. If b < 0 and n is even, then ñb is not a real number. n

EXAMPLE

22

Solution

30

Simplify the following. b.

4

a.

3

8

a.

3

8  23  2 3  2

3

3

c.

81

3

(8 > 0) 3 3

–27

4

d. b.

4

e.

–16 4

4

81  34  3 4  3

c.

3

–27  3 (–3)3  (–3)  –3

(–27 < 0 and 3 is odd)

d.

4

–16 is not a real number

(–16 < 0 and 4 is even)

e.

5

0 =0

5

0

(81 > 0)

Algebra 8

Check Yourself 3 1. Write the expressions in radical form. 1

1

2

3

a. 2 2

b. a 3

c. a 3

d. x 5

1

1

e. ( x2  y3 )6

1

1

1

g. ( a2  b2 ) 2

( x2 )3

f.

h. (( x  y)  x2 ) 3

2. Write the expressions with rational exponents. 3

a.

4

b.

2

4

5

c.

a2

3

d.

x  y2

e.

625 e.

6

6

4

x4 f.

a8

g.

–512

g.

3

x2  y4 h.

12

x6  y4

3. Simplify the expressions. a. ò16

3

b.

3

27 c.

–64

d.

4

9

64 f.

x2  y4 h. 4 16  a4  b8

Answers 1. a. ñ2 b. 3 a c. 3 a2 d. 5 x3 e. 6 x2  y3 1

1

2

b. 2 2

c. a 5

d. ( xy2 )3

f. 6 x g. a2  b2 1

2

h. 3 ( x  y)x2

1

e. x 3 f. a2 g. ( x2  y4 )3 h. ( x3 y2 )6

1

2. a. 2 3

3. a. 4 b. 3 c. –4 d. 5 e. 2

f. –2 g. xy2 h. 2ab2

B. PROPERTIES OF RADICALS Property

For all real numbers a and b, where a > 0 and b > 0, and for any integer m and n, where m > 1 and n > 1: 1.

2.

an  a

if n is odd.

5.

n

a

n

an | a |

if n is even.

6.

m

a b

n

a  b n a b

m

a

n

3.

n m

n

n

n

a



b n

n

7.

a b

4. a  b 

n

8. an  b

and

n

an b  a 

n

n

am

n

b

m n

mn



mn

a 

mn

an bm

mn

an 

 mn

mn

bm 

m n

an  bm

an bm

a

b

Look at the examples of each property. 1.

Radicals

3

b.

5

32  5 (2)5  –2

4

d.

4

(3)4 | 3| 3

a.

3

27  33  3

c.

4

16  2 4 | 2| 2

31

2.

a. ñ5  ñ4 = ó54 = ò20 5

c. 3.

8

8  4 2 2



2 4

4.

64 x2 4

4x

4

4

b. x  x 

5.

4

3

3



3

81 3  27  3 3

x4  x 

4

3

24

x5 3

6

4  22  3

6.

81

23  2

x3  y6  z2  3 ( x  y2 )3  z2  x  y2 z2

6

a.

3

3

64 x2 4 4 4 4  16 x  4 2 4  x  2 4  x  2 x 4x

3

3

3

2  4  324 8

b.

a. 2  3  3 2 3  3  3 8  3 

c.

3

3

x5 y5 xy

a. c.

b.

5 3 3

23

34

2



7.

a.

8.

a.

3

5 

c.

4

x8 

4

3

34

53 

24 33

3 2

32

3

b.

22  2

2 3

 12

32 

6

5 3  32 

6

125  9 

b.

6

x8 

8

3

3 2

52  25

32

a2

6

1125 3

2 4 12 16  27 33

5 5

4 2

6

5

a b

b.

4 3



32

b3

12 

6

4  3 2

a2 b3

12 

24

12

x8  x

Check Yourself 4 1. Simplify the expressions. a.

4

b.

16

4

(–2)4

c.

5

(–3)5

d.

7

–128

c.

5

x 2  x3

5

d.

3

x  3 x2 y2

4

x5

e.

3

27x3

2. Perform the operations. a. ñ3  ò12

b.

3

3

3 9

3. Simplify the expressions. 3

2

a.

b.

8

625 3

5

c.

4

d.

x

3 3

3

d.

4

x5  y6

d.

4

32  16

4. Simplify the expressions. a.

3

b.

40

3

81

c.

4

32

c.

3

a2  a3

5. Perform the operations. a. 32

3

2 4

b.

3

4

x  x3

6

3

Algebra 8

6. Simplify the expressions. 3

a.

3

4

4

b.

3

4

6

c.

2

5

a3

d.

a2

6

x4

8

x3

7. Write each expression in its simplest form. 3

a.

b.

3

5 4

x

c.

3 3 3

x27

d.

4

x16

3

e.

3

3 3

Answers 1. a. 2 b. 2 c. –3 d. –2 e. 3x 2. a. 6 b. 3 c. x d. x  3 y2 4. a. 2 3 5 b. 3 3 3 c. 2 4 2 6. a. 12

256 27

b.

6

9 2

c. 20 a7

d. xy 4 xy2 d. 24 x7

5. a. 2 6 2 7. a. 6 3

3. a.

1 2

b. 5 c. x d. 3 9

b. x 12 x c. a 6 a

d. 4 

12

27

b. 20 x c. x d. x2 e. 9 81

C. RADICAL EQUATIONS Definition

An equation that has a variable in a radicand is called a radical equation. For example, the equations ñx = 25,

x  1  3, and

2 x – 1  3 x  5 are radical equations.

Let us look at some examples of how to solve radical equations. EXAMPLE

23

Solution

Solve

x  1  3.

( x  1)2  32 (take the square of both sides)

Check:

x 1 9

?

8 13 ?

9 3

x8

33

Therefore, 8 is a solution. EXAMPLE

24

Solution

Solve

3x  1  5.

( 3x  1)2  52 3x  1  25 3x  24 x8

Radicals

Check:

?

3  8  15 ?

25  5 55

Therefore, 8 is a solution. 33

EXAMPLE

25

Solution

2 x  3  3  8.

Solve

2x  3  3  8

Check:

?

2  11  3  3  8

2x  3  8 – 3

?

25  3  8

2x  3  5

?

( 2 x  3)2  5 2

5  3 8

2 x  3  25

88

2 x  22

Therefore, 11 is a solution.

x  11 EXAMPLE

26

Solution

Solve

x2  12  x  6.

( x2  12 )2  ( x  6)2

Check:

?

(–2)2  12  – 2  6 ?

x2  12  x2  12 x  36

4  12  4

12 x  –24

?

16  4

x  –2

44

Therefore, –2 is a solution. EXAMPLE

27

Solution

Solve

3

3x  2  5.

3

( 3x  2 )3  5 3

Check:

3x  2  125

3

?

3  41  2  5 3

3x  123

?

123  2  5 3

x  41

?

125  5 5 5

Therefore, 41 is a solution. EXAMPLE

28

Solution

Solve

4x  1 – 5x – 1  0.

4x  1 – 5x – 1  0 ( 4 x  1)2  ( 5 x – 1) 2 4x  1  5 x – 1 x2

Check:

?

4  2 – 1 – 5  2 – 1 0 ?

9 – 9 0 ?

3 – 3 0 0 0

Therefore, 2 is a solution. 34

Algebra 8

EXAMPLE

29

3x  1  3x  6  5.

Solve

3x  1  5 – 3 x  6

Solution

(take the square of both sides)

( 3x  1)2  (5 – 3 x  6 ) 2

3x  1  25 – 2  5  3 x  6  3 x  6

(simplify)

10  3x  6  30 ( 3 x  6 )2  3 2 3x  6  9 3x  3 x1

Check:

?

3  1  1 – 3  1  6 5 ?

4  9 5 ?

2  35 5 5

Therefore, 1 is a solution.

Check Yourself 5 1. Solve each equation and check your answer. a. ñx = 5

b. ñx + 1 = 3

d. ñx = –3

e.

2x – 1  7

f.

3 x 1 –1  8

g.

3x – 5  4  6

h.

5x  1 –30 2

i.

2x  6 – x

j.

4x – 3 – 2 x – 2  0

k.

x3  x–2 5

l.

x2  1  2 – x

n.

x 2

o.

m.

3

x 1  3

p.

4

x  2x – 1

3

c. óx+1 = 6

3

3

3x  2  5x

4

Answer 1. a. 25 b. 4 c. 35 d.  e. 25 f. 8 g. 3 h. Radicals

1 17 i. 2 j. k. 6 l. 3 m. 26 n. 2ñ2 o. 1 p. 1 2 5 4 35

EXERCISES

1 .2

1. Write the expressions in exponential form and simplify if possible. a. ò21 3 5

e.

7

b. f.

2

2

7

c.

5

( 3)6

g.

3

3

c. ò7x = x

1 3

3 –3 c. ( ) 2 2

2

1

b. b 3

d.

3

4x  6

5x – 4  6

f.

3

7x  6  5

g.

5x – 1  3  7

h.

2x – 3 – 2  4

i.

x2  5  x  1

j.

x2  9  5

7x – 6  4

l.

2x – 5  x – 4

x  x  16 – 2

n.

6  x

e.

d. x

3 2

ab c

3. Simplify the expressions. 3

3

b.

4 5

d.

3

5 5

e.

5

g.

3

–27

h.

3

j.

3

3

2 4 3

m.

p.

k.

2

3

250

3

4

q.

c.

1016

243

f.

(–5)4

i.

a3  a2

4

3  27 21

6

7

n.

3 9 3

a2

4 3

4

2

3 3

2

o.

4

2

r.

3 3

s.

c. e.

3

4

4 3



1 9

8x – 2  6

1 4x – 3

1

1

2– x

–

2 x

 2

6

11 – 3 4 – 27

d.

2

17  6 8

f.

3

9

2 x1

6. Simplify the expressions. 3

2 3 6

108

33x – 6

5

b.

6

3

x  1 2 x  x 1– 2 x  8 3

q.

2

5

3

3 2

p.

162

l.

2

o. ñ6x – x= 5

3

4. Perform the operations. a.

3

m.

64x6

4

3

1

e. ( a )6

k.

a.

3x – 14  8

b.

3

2. Write the expressions in radical form. a. a 2

2x  1  3

a. 2

d. a x

5

5. Solve the equations.

7 3 10  2 9 27

( 2 – 6 )( 2  3 )

a. 23  24

b. 53  52

c. 43  42  4

d. 26  2–4  22

e. 3x  3–2x  33x+1

f. 2x  3x  5x

g. (x – y)2  (y – x)3  (x – y)–4 h. 298 + 298 + 298 + 298

36

Algebra 8

7. Simplify the expressions. a.

25 23

b.

x 1

d.

10. Solve the equations for x.

53 54

c. –3

a b x 1

e.

5

(–2) (–2)–4

f.

a. 4x+1 = 23x – 4

ax –1 ax – 2 5 2 104

4

c. 9x –2  (

b. 3x – 4 = 81

1 3– x ) 27

d. 8x+1 = 42x – 3

e. (2x – 3)3 = (x + 1)3 f. (2x – 4)6 = x6 g. 5  23x – 1 – 23x+1 = 256

8. Perform the operations. a. (–2)5  (–2)3

b. –32  (–3)3  (–3)

c. (–4)2  (–22)3  (–2–2)4

d.

(–2 3 )–3  (–2 2 )6 1 (– )3 2

e. (–a)7  (–a4)  (–a)–2

h.

12 x  12 x  12 x  81 4x  4x  4x

i.

3  4x – 3  4 x –1 –

j.

3  93 x –1  ((–3)2 ) –4 81  (243) x

22 x  128 4

11. 3a = 25 and 3b = 5. Find

a . b

12. 3x = 4. Find 92x + 3x+1.

f. (–a–3)–2  (–a2)–3  (–a–1)–1

13. 2x = 3y = a. Write (12)xy in terms of a, x, and y. 14. 2a – 3 and 4b + 7 are integers with 32a – 3 = 54b + 7. Find a + b.

9. Simplify the expressions. 2

15. (x + 1)x

a. 2x – 1 + 2x – 1 + 2x – 1 + 2x – 1

– 16

= 1. Find the sum of the possible

values of x.

b. ax + 2  ax – 3ax c. 3x+1 + 3x – 1 + 3x+2 – 3x – 2

4

16. 44 = 16x. Find x.

10 x  10 x  10 x  10 x d. 5x  5x  5x  5x

e.

(2 –7  3–5 )2 36 –7

f.

313  311 – 39 312  310

g. (–1)101  (–1)125  (–1)100  (–1)99  (–1)49 h.

3

–4 0

((–39)  (–2) )  (–1) (–3)125  6 –127  2126

Radicals

17. |x + y – 3| + |x – y – 1| = 0. Find x. 18. Simplify |x – 4| + 2|3 – x| if 3 < x < 4.

2003

37

19. Find the sum of the possible values of x that satifsfy the equation |(|2x + 4|) – 2| = 4.

22 . x  72  72  72  ... and y  42 – 42 – 42 – ... . Find x + y.

20. Find the product of the possible values of x that satisfy the equation |2x + 3| = |5 – 3x|.

23 . 2a  2a  2a  ...  12 – 12 – 12 – ...  3. Find a.

21. Simplify the expressions. 3

a.

4

(–3)2 – –8  16

24 .Simplify

72  48

b.

2.89  6.25 – 1.96 2

d.

5 –1

2



1

4

2 4

1



4 6

 ... 

1 252  254



1 254  256

27. How many different natural number solutions

12 2  16 2

5

g.

7– 2

h.

does the equation ñx + ñy = ó300 have? 5



2 7

28. ( 2  1)7  57122  57121. Find (ñ2 – 1)7.

4  12  4 – 12

(Hint: (

1 2 –1

)  2 1 )

( 6  20 )  ( 5 – 1) 2

x y x – y  2

k.

38

4

3

f.

m.

4

18  3  29 – 16

e.

l.

4

25 . x2 – x2 – x2 –...  2. Find x. 26 .Simplify

5 1 3

j.

3

27  11 – 2  30  30  30 ... .

54

c.

i.

4

2

xy if x < y < 0 x–y

2.7  5.4  7.1  2  0.2 4

15 – 6 

4

15  6 – 3

4 7  4– 7

29.

1

15 1 9 16  9   24 – 16  . Find x  . 8 x x x

30. Evaluate

31.

3

(50  51  52  53)  1.

3

x  2 – x – 2  4. Find x2.

(Hint: (a – b)3 = a3 – 3ab(a – b) – b3) Algebra 8

.

CHAPTER REVIEW TEST

1A

1. Simplify (|(–(–2)3) – 2|) – [3  |–|(–3)2 – 3||]. A) –8

B) –12

C) 8

D) 28

6. |x – 2| = |x + 3|. Find the value of x which satisfies the equation. A) –

2. Calculate |1 – ñ2| + |1 + ñ2|. A) 0

B) ñ2

C) 2

1 4

D) 2ñ2

3. How many elements are there in the solution set of |x – 2| + 1 = 0? B) 1

C) 2

D) 3

B) 2x – 2

C) 2

C) 1

D)

D) 4

5. ||x + 1| – 5| = 3. Find the sum of the possible

following numbers is negative? B) a–2

1

C) –a–3

D) –(–a)3

–1

8. Simplify (– )3  .  2  A) –

1 8

B)

1 8

C) –8

D) 8

A) 23x

10. Simplify

B) 24x

C) 29x – 1

D) 28x+1

4m  4m . 2m  2m  2m  2m

values of x. A) –3 Chapter Review Test 1A

3 2

9. Simplify 4x+1  8x – 1  16x.

4. –1 < x < 3. Calculate |x + 1| + |x – 3|. A) 2x

1 2

7. a is a positive real number. Which one of the A) a–1

A) 0

B) –

2 m 1

B) –2

C) 2

D) 4

A) 22m

B) 2m +1

C) 2m – 1

D) 2 m  2 39

11.

16. Evaluate (3ñ2 – ò12)  (ò18 + 2ñ3).

3x 1  3x  2  3x  3  27. Find x. 39

A) 1

12. 2 a – b  2 –

B) 3

1 2 2 a  b 1

A) 0

C) 13

D) 27

17. Evaluate

 0. Find a.

B) –1

C) –3

D) 3

13. a = 2x, b = 3x, and c = 5x. Express 90x in terms of a, b, and c. A) (abc)

2

15. Evaluate A) –1 40

A) –ñ3

18. Evaluate A) 1

2

B) a bc

2

2

C) ab c

2

B) 8ñ2

D) 26ñ2

2.25 – 2.89  1.44.

B)

1 2

C) 1

D) 6

4 – 12.

B) ñ3

C) ñ3 – 1

D) ñ3 + 1

2  2 4  12 .

B) ñ3

C) ñ3 – 1

D) ñ3 + 1

D) a b c

19. Evaluate

C) ò38

C) ñ6

B) 180

4 2

14. Evaluate 3ñ8 – 2ñ2 + ò32. A) ñ6

A) 36

A)

3 2

20. Evaluate D)

2 3

A) 7

27 m3  9 m2 . 12 m3  4 m2

B) ñ2

5 7– 2



B) 2ñ7

9 4

C)

5 7 2

D) 3m+1

.

C) 5ñ7

D) 5 Algebra 8

CHAPTER REVIEW TEST 1. Evaluate A) –

(–1)55  (–172 ) – (–1) 64 . (–121 ) – (–1)82 – (–1)15

3 2

2. Evaluate

1B

B) –3

6. Evaluate

C) 1

D) –1

2 –2  4–1 . 2 –2 – 3–1

1 B) 3

A) 3

A)

569 – 568 . 568  567

1 6

5 6

B)

C)

5 3

D)

10 3

7. 3x = 5, 2y = 3, and 5z = 0.125. Find x  y  z. A) 3 C) 2

–1

B) 10

C) –3

D) 30

C) 3

D) 4

D) –6

8. 3x+1 + 2  3x – 1 = 33. Find x. 3

3.

x(2 )  16. Find |x|. ( x2 )3

A) 2

B) 4

A) 1 C) 8

B) 2

D) 16

3

2

9. a = (52)3, b = 5(2 ), and c = 5(3 ). Which statement is true?

4. What is half of 220? A) 210

B) 221

5. Find the simplest form of A) 3x

B)

Chapter Review Test 1B

1 3x

C) 219

D)

1 210

3x  3x  3x . 9x  9x  9x

C) 32x

A) a > b > c

B) a > c > b

C) b > c > a

D) c > b > a

10. k = 1254  642. How many digits are there in the number k? D) 3x+1

A) 10

B) 11

C) 12

D) 13 41

11.

32 x  243x  9x  3. Find x. 81x1

A)

2 9

B) 2

41 A) 16

12  48 – 27 75 – 2 3

A) 2ñ3

15.

4

B) –1

4

C) 1

3

B) 2

19. Evaluate C) 3

D) 4

A) ñ8 6

3

11 24

B)

4 7

20. C)

11 7

D)

1 6

5

8 x 1

3 7

42 x  3

A) 9

.

B) –2

C) –ñ2

11  2 30 – 8  2 15 3–2 2

D) 4

.

B) ñ6 – ñ3 D)

3 5 – 3– 5 2

A) ñ5

D) 3ñ3

13  13 – 64 .

2 –2

C) ñ3

18. Evaluate

.

2 2

1



3  15

A)

2 D) 3

x x2 x  xn . Find n.

A) 42

3 C) 2

1

A) 2ñ2

D) 10

17. Evaluate

41 B) 4

13. Simplify

A) 1

C) 9

9 1 1 . 16

12. Evaluate

14. Evaluate

16. Evaluate

B) 2ñ5

3

3 2

.

C) 2

D) 1

3

2 4 2 4 ... .

B) 2

C) 8

D) ñ8

C) 114

D) 120

3

 2. Find x.

B) 105

Algebra 8

Objectives

After studying this section you will be able to: 1. Define statistics as a branch of mathematics and state the activities it involves. 2. Describe some different methods of collecting data. 3. Present and interpret data by using graphs. 4. Describe and find four measures of central tendency: mean, median, mode, and range.

A. BASIC CONCEPTS 1. What is Statistics? Statistics is the science of collecting, organizing, summarizing and analyzing data, and drawing conclusions from this data. In every field, from the humanities to the physical sciences, research information and the ways in which it is collected and measured can be inaccurate. Statistics is the discipline that evaluates the reliability of numerical information, called data. We use statistics to describe what is happening, and to make projections concerning what will happen in the future. Statistics show the results of our experience. Many different people such as economists, engineers, geographers, biologists, physicists, meteorologists and managers use statistics in their work.

Definition

statistics Statistics is a branch of mathematics which deals with the collection, analysis, interpretation, and representation of masses of numerical data. The word statistics comes from the Latin word statisticus, meaning ‘of the state’. The steps of statistical analysis involve collecting information, evaluating it, and drawing conclusions. For example, the information might be about:

    44

what teenagers prefer to eat for breakfast; the population of a city over a certain period; the quality of drinking water in different countries of the world; the number of items produced in a factory. Algebra 8

The study of statistics can be divided into two main areas: descriptive statistics and inferential statistics. Descriptive statistics involves collecting, organizing, summarizing, and presenting data. Inferential statistics involves drawing conclusions or predicting results based on the data collected.

2. Collecting Data We can collect data in many different ways.

a. Questionnaires A questionnaire is a list of questions about a given topic. It is usually printed on a piece of paper so that the answers can be recorded. For example, suppose you want to find out about the television viewing habits of teachers. You could prepare a list of questions such as:

 Do you watch television every day?  Do you watch television: in the morning? in the evening?

 What is your favourite television program?  etc. Some questions will have a yes or no answer. Other questions might ask a person to choose an answer from a list, or to give a free answer. When you are writing a questionnaire, keep the following points in mind:

1. A questionnaire should not be too long. 2. It should contain all the questions needed to cover the subject you are studying. 3. The questions should be easy to understand. 4. Most questions should only require a ‘Yes/No’ answer, a tick in a box or a circle round a choice.

In the example of a study about teachers’ television viewing habits, we only need to ask the questions to teachers. Teachers form the population for our study. A more precise population could be all the teachers in your country, or all the teachers in your school.

Statistics and Graphics

45

b. Sampling A sample is a group of subjects selected from a population. Suppose the population for our study about television is all the teachers in a particular city. Obviously it will be very Xdifficult to interview every teacher in the city individually. Instead we could choose a smaller group of teachers to interview, for example, five teachers from each school. These teachers will be the sample for our study. We could say that the habits of the teachers in this sample are probably the same as the habits of all the teachers in the city.

population

sample

A sample is a subset of a population.

The process of choosing a sample from a population is called sampling. The process of choosing a sample from a population is called sampling. When we sample a population, we need to make sure that the sample is an accurate one. For example, if we are choosing five teachers from each school to represent all the teachers in a city, we will need to make sure that the sample includes teachers of different ages in different parts of the city. When we have chosen an accurate sample for our study, we can collect the data we need and apply statistical methods to make statements about the whole population.

c. Surveys One of the most common method of collecting data is the use of surveys. Surveys can be carried out using a variety of methods. Three of the most common methods are the telephone survey, the mailed questionnaire, and the personal interview.

3. Summarizing Data In order to describe a situation, draw conclusions, or make predictions about events, a researcher must organize the data in a meaningful way. One convenient way of organizing the data is by using a frequency distribution table. A frequency distribution table consists of two rows or columns. One row or column shows the data values (x) and the other shows the frequency of each value (f). The frequency of a value is the number of times it occurs in the data set. For example, imagine that 25 students took a math test and received the following marks.

46

8

7

9

3

5

10

8

10

6

8

7

7

6

5

9

4

5

9

6

4

9

3

8

8

6 Algebra 8

The following table shows the frequency distribution of these marks. It is a frequency distribution table. mark

(x)

1

2

3

4

5

6

7

8

9

10

frequency

(f)

0

0

2

2

3

4

3

5

4

2

Note The sample size is the number of elements in a sample. It is denoted by n. We can see from the table that the frequency of 7 is 3 and the frequency of 8 is 5. The sum of the frequencies is equal to the total number of marks (25). The number of students took test is called the sample size (n). In this example the sample size is 25. The sum of the frequencies and the sample size are the same. EXAMPLE

1

Twenty-five students were given a blood test to determine their blood type. The data set was as follows: A

B

AB

B

AB

A

O

O

AB

A

B

O

O

O

B

AB

A

O

B

O

O

B

AB

B

O

Construct a frequency distribution table of the data and find the percentage of each blood type. Solution

There are four blood types: A, B, O, and AB. These types will be used as the classes for the distribution. The frequency distribution table is: class

frequency

percent

A

4

16 %

B

7

28 %

O

9

36 %

AB

5

20 %

Total 25

Total 100

We can use the following formula to find the percentage of values in each class: f %   100% where n f = frequency, and n = total number of values (25). For example, in the class for type A blood, the percentage is 4  100%  16%. 25

Statistics and Graphics

47

B. PRESENTING AND INTERPRETING DATA A graph is a diagram that relates two or more different types of information.

When we have collected, recorded and summarized our data, we have to present it in a form that people can easily understand. Graphs are an easy way of displaying data. There are three kinds of graph: a line graph, a bar graph, and a circle graph (also called a pie chart).

1. Bar Graph The most common type of graph is the bar graph (also called a histogram). A bar graph uses rectangular bars to represent data. The length of each bar in the graph shows the frequency or size of a cooresponding data value.

48

Mark

Maths

9

Physics

7

Chemistry

7

Biology

8

Computer

10

History

5

Music

6

Music

History

Computer

10 9 8 7 6 5 4 3 2 1 0 Mathematics

Then we can draw bars to show the marks for each subject.

Subject

Biology

We begin by drawing a vertical scale to show the marks and a horizontal scale to show the subjects.

Chemistry

Solution

The following table shows the marks that a student received at the end of the year in different school subjects. Draw a vertical bar graph for the data in table.

Physics

2

Marks

EXAMPLE

Lessons

Algebra 8

2. Line Graph We can make a line graph (also called a broken-line graph) by drawing line segments to join the tops of the bars in a bar graph.

Music

History

Computer

Biology

Chemistry

Physics

Music

History

10 9 8 7 6 5 4 3 2 1 0 Mathematics

Lessons

Computer

Biology

Chemistry

Physics

Marks

10 9 8 7 6 5 4 3 2 1 0 Mathematics

Marks

For example, look at the line graph of the data from Example 5.2.

Lessons

To draw the line graph, we mark the middle point of the top of each bar and join up the points with straight lines.

EXAMPLE

3

Statistics and Graphics

The following table shows the number of cars produced by a Turkish car company between 1992 and 2000. Draw a bar graph and a line graph of the data in this table.

Car Production Year

Production

1992

110 659

1993

133 006

1994

99 326

1995

74 862

1996

65 007

1997

91 326

1998

88 506

1999

125 026

2000

140 159

49

160 140 120 100 80 60 40 20 1997

1998

1999

2000

1997

1998

1999

2000

1996

1995

1994

1993

0 1992

Number of cars (10 000)

Year 160 140 120 100 80 60 40 20 1996

1995

1994

1993

0 1992

First we need to choose the axes. Let us put the years along the horizontal axis and the production along the vertical axis of the graph. It will be difficult to show large numbers such as 133 006 on the vertical axis. Instead, we can choose a different unit for the vertical axis, for example: one unit on the axis means 10 000 cars. We write this information when we label the axis.

Number of cars (10 000)

Solution

Year

6

3

6%

6.4% 5.3%

5.2%

5 4

6%

3.4%

2 1

Germany

0 Canada

a. Which country spent the largest percentage of its GDP on education?

7%

7

Australia

domestic product. Look at the graph and answer the questions.

Share of education expenditures as a percentage of GDP in selected countries *

Norway

The graph below shows the amount of money that seven different countries spend on education in 2003, as a percentage of each country’s gross

8

United Kingdom

The gross domestic product (GDP) of a country is the total value of new goods and services that the country produces in a given year.

USA

4

Turkey

EXAMPLE

* Source: Education at a Glance 2003 - OECD indicators

b. Which country spent the smallest percentage of its GDP on education? c. Find the percentage difference between the countries which spent the largest and smallest percentage of their GDP on education. d. Which countries spent the same percentage of their GDP on education? 50

Algebra 8

Solution

a. The USA spent the largest percentage (7% of its GDP). b. Turkey spent the smallest percentage (3.4% of its GDP). c. 7 – 3.4 = 3.6% d. Norway and Australia spent the same percentage: both countries gave 6% of their GDP.

Practice Problems 1. The bar graph below compares different causes of death in the United States for the year 1999. Look at the graph and answer the questions. Comparative causes of annual deaths in the United States (1999)*

Cause * Source: World Health Organization

a. What was the most common cause of death? b. What was the least common cause of death? c. What is the ratio of the number of deaths caused by smoking to the number of deaths caused by alcohol? d. How many deaths are shown in the graph? e. On avarage, how many people died per day from each canse in 1999? (Hint: There were 365 days in 1999.)

Answers 1. a. Smoking b. Drug Induced c. Statistics and Graphics

16 3

d. 632000 e. 1732

51

EXERCISES

2 .1

1. The set of quiz scores in a class is as follows. 8

5

6 10 4

7

2

7 6

3

5

9

2

4

6

6 8

4 10 8

6

5

1

7

Construct a frequency distribution table for this data.

2. A student’s expenses can be categorized as shown in the table. Expenses Food Rent

4. The following table shows the amount of sea fish caught in Turkey in 2003. Fish Anchovy

Quantity (1000 tons) 416

Horse Mackerel

295

Scad

16

Gray mullet

12

Blue fish

11

Pilchard

11

Whiting

12

Hake

8

Other

32

Source: Turkey’s Statistical Yearbook 2004

Present this information in a circle graph.

Percent of total income. 30% 27%

Entertainment

13%

Clothing

10%

Books

15%

Other

5%

Present this information in a bar graph.

5. The following bar graph shows the hazelnut production in Turkey from 1999 to 2003. Use the graph to answer the questions. 700 000

Hazelnut production in Turkey (tons)

600 000 500 000

100 000

2003

200 000 2002

Number of class members 8

2001

Sport Football

300 000

2000

chosen by each of forty students in a class.

400 000

1999

3. The following table shows the favorite sport

Basketball

5

Volleyball

7

a. Estimate the total production for all five years.

Swiming

12

Wrestling

3

b. Which year had the highest production?

Karate

2

Judo

4

Present this information in a circle graph. 52

c. Find the combined production for 2002 and 2003. d. Draw a broken line graph of the data. Algebra 8

Definition

An equation that can be written in the form a0

ax2 + bx + c = 0, is called a quadratic equation. I’m sick of being an unknown

In the equation, a, b, and c are real number coefficients and x is a variable. A quadratic equation written in the form ax2 + bx + c = 0 is said to be in standard form. Sometimes, a quadratic equation is also called a second degree equation. For example,

1 x+3=0 2 are all quadratic equations. By the definition of a quadratic equation, a cannot be zero. However b or c or both may be zero. For instance, 2x2 = 0 and x2 – 9 = 0 3x2 + 5x = 0,

2x2 – x – 1 = 0 and ñ2x2 –

x2 + 3x – 5 = 0,

are also quadratic equations. We can see that quadratic equations are formed by second-degree polynomials. Polynomials of a different degree do not form quadratic equations. Let us look at the coefficients a, b, and c of some quadratic equations. Equation

a

b

c

3x2 + 5x – 9 = 0

3

5

–9

1 – x + 3x2 = 0

3

–1

1

ñ2x2 + 5x = 0

ñ2

5

0

–1

0

1 2

1 – x2 = 0

–1

0

1

(ñ3 + 1)x2 = 0

ñ3 + 1

0

0

x2 x 1 – + =0 2 3 4

1 2

– x2 +

EXAMPLE

54

1

1 =0 2

–

1 3

1 4

Determine whether the following equations are quadratic or not. 1 2 x – 2 x +5 = 0 2

a. x2 + 1 = 0

b.

d. x2 – 2x–1 + 3 = 0

e. (x – 1)(x + 2) = 0

c. 2x2 – 3x = 5 f. (x – 2)x2 = 0 Algebra 8

Solution

To the best of our knowledge, the origin of the term ‘quadratic’ is Latin. It is derived from quadratus which is the past participle of quadrare which means ‘to make square’. From this it is clear that part of the word is connected to the Latin word for ‘four’: it refers to squaring, and a square is a regular four-sided figure.

a, b, c, and e are quadratic equations. Equation d is not quadratic, since the power of x is –1, which does not meet the requirements for a quadratic. Equation f is a third degree equation, so it is not quadratic. To solve a quadratic equation we must find the values of the unknown x which make the left-hand and right-hand sides equal. Such values are called the solutions or roots of the quadratic equation. A number of techniques are available to help us obtain a solution to any quadratic equation.

A. SOLVING EQUATIONS OF THE FORM ax2 = 0 We have seen that the solutions (or roots) of a quadratic equation are the values of x that make the two sides of the equation equal. Let us find the roots of the simple quadratic ax2 = 0. ax2 = 0 (a  0) x2 = 0 xx=0 x = 0 or x = 0 x1 = x2 = 0 We can see that this equation has two equal roots. When the roots of a quadratic equation are the same, we say that the equation has a double root.

If A  B = 0, then A = 0 or B = 0.

EXAMPLE

2

Solution

3 Solve the equation  x2  0. 2 3  x2  0 2 x2  0 x1  x2  0

Quadratic Equations

55

B. SOLVING EQUATIONS OF THE FORM ax2 + bx = 0 Let us look at the solution of this more complex quadratic. ax2 + bx = 0 x(ax + b) = 0 x = 0 or ax + b = 0, so b x1 = 0 and x2 = – . a This kind of quadratic equation has two roots and one of them is always zero. EXAMPLE

3

Solution

Solve the equations. a. x2 + x = 0

b. –4x2 + 5x = 0

c. ñ3x2 – 2x = 0

a. x2 + x = 0

b. –4x2 + 5x = 0

c. ñ3x2 – 2x = 0

x(x + 1) = 0

x(–4x + 5) = 0

x(ñ3x – 2) = 0

x = 0 or x + 1 = 0

x = 0 or –4x + 5 = 0

x = 0 or ñ3x – 2 = 0

x1 = 0 or x2 = –1

x1 = 0 or x2 =

5 4

x1 = 0 or x2 =

2 3

=

2 3 3

C. SOLVING EQUATIONS OF THE FORM ax2 + c = 0 Look at the calculation. ax2 + c = 0,

c0

2

ax  c x2  

c a

x 

c a

Here the sign of

c is important. a

c > 0, the equation has no real solution, because we cannot find the square root of a a negative number.

If

c < 0, the equation has two real solutions. These roots are symmetric, i.e. they are the a same numeral with opposite signs.

If

Note All positive real numbers have two square roots. One root is the positive square root and the other root is the negative square root, i.e. if a2 = b and a is a positive real number, then a = ñb. 56

Algebra 8

EXAMPLE

4

Solution

Solve the equations. a. 3x2 – 27 = 0

b. 2x2 + 6 = 0

c. 7 – 4x2 = 2

a. 3x2 – 27 = 0

b. 2x2 + 6 = 0

c. 7 – 4x2 = 2

3x2 = 27

2x2 = –6

x2 = 9

x2 = –3

x = 3

4x2 = 5 x2 =

no real solution (x2 cannot be negative)

x=

Check Yourself 1

5 4

5 5 = 4 2

Solve the equations. 1. –3x2 = 0

2. 5x2 – 20x = 0

3. 7x2 + 35 = 0

4. 2x2 – 8 = 0

Answers 1. 0 2. 0, 4 3. no real solution 4. 2 FIND THE MISTAKE! Let a and b be two arbitrary numbers such that a  b. Then (a – b)2 = a2 – 2ab + b2 = b2 – 2ab + a2 (a – b)2 = (b – a)2 a–b=b–a 2a = 2b a = b. Can you find the mistake in this working?

D. SOLVING EQUATIONS OF THE FORM ax2 + bx + c = 0 There are three basic methods for solving a quadratic equation of the form ax2 + bx + c = 0: 1. factoring, 2. completing the square, and 3. using the quadratic formula.

1. Factoring If we can write ax2 + bx + c = 0 as the product of two linear factors, then we can easily solve the equation. To solve a quadratic equation by factoring, follow the steps. 1. Write the equation in standard form, ax2 + bx + c = 0, a  0. 2. Factor the left side of the equation. Quadratic Equations

57

3. Apply the zero product property, that is, set each factor equal to zero. 4. Solve each equation to obtain the roots. EXAMPLE

5

Solve by factoring. a. x2 + 3x + 2 = 0

Solution

b. 6x2 – 19x – 7 = 0

a. x2 + 3x + 2 = 0

c. 2x2 = x + 3

b. 6x2 – 19x – 7 = 0 (2x – 7)(3x + 1) = 0 2x – 7 = 0 or 3x + 1 = 0 7 1 x1 = , x2 = – 2 3

(x + 2)(x + 1) = 0 x + 2 = 0 or x + 1 = 0 x1 = –2, x2 = –1 c. 2x2 = x + 3 2x2 – x – 3 = 0 (x + 1)(2x – 3) = 0 x + 1 = 0 or 2x – 3 = 0 3 x1 = –1, x2 = 2

Note When you are solving an equation, do not divide both sides by an expression containing the variable for which you are solving. You may be dividing by zero. For example, to solve x2 – 2x = 0, do not divide both sides by x, because x may be zero and you will also lose one of the solutions.

Check Yourself 2 Solve the following equations. 1. 3x2 = 5x + 2 Answers 2 1 1.  , 2 2. 2 , 5 3

2. (5x – 1)(x + 2) = x + 2

3. 4x(x + 1) = 3

3 1 3.  , 2 2

2. Completing the Square The idea behind this method is to adjust the left side of the equation ax2 + bx + c = 0 so that it becomes a perfect square, that is, the square of a first-degree polynomial. Expressions in the form x2 + 2xy + y2 and x2 – 2xy + y2 are perfect square polynomials. Numbers whose square roots are integers or quotients of integers are perfect squares.

58

For example, x2 + 6x + 9 and x2 – 4x + 4 are perfect squares, because x2 + 6x + 9 = (x + 3)2 and x2 – 4x + 4 = (x – 2)2. To write a quadratic equation as a perfect square, follow the steps, Algebra 8

1. Make sure a = 1 in the quadratic. If it isn’t 1, divide each term by a.

x+2 x

1

1

2. Rewrite the equation so that the constant term is alone on one side of the equation. 3. Take half of the coefficient of the x term and square it.

x

x+2

4. Add this number to both sides of the equation.

1

5. Factor the left-hand side into a perfect square.

1

2

6. Solve for x by using the square root property. x2 + 4x

Let us look at some examples of completing the square.

(x + 2)2 = x2 + 4x + 4

Start _____________

Add _____________

Result ___________________________

x2 + 4x

4

x2 + 4x + 4 = (x + 2)2

x2 + 12x

36

x2 + 12x + 36 = (x + 6)2

x2 – 6x

9

x2 – 6x + 9 = (x – 3)2

x2 + x

1 4

x2 + x +

2

4

1 1 = (x + )2 2 4

Note The expression x2 + 2bx is equivalent to (x + b)2 – b2. EXAMPLE

6

Solution

Solve by completing the square. a. x2 + 6x – 7 = 0

b. 2x2 – 4x + 1 = 0

a. x2 + 6x – 7 = 0

b. 2 x2 – 4x+ 1= 0 1 x2 – 2 x+ = 0 2 1 x2 – 2 x = – 2 1 x2 – 2 x+ 1= – + 1 2 1 (x – 1)2 = 2 1 x – 1=  2

x2 + 6x = 7 x2 + 6x + 9 = 7 + 9 (x + 3)2 = 16 x + 3 = ò16 x1 = –7,

x2 = 1

x1 =1 – Quadratic Equations

2 , 2

x2 =1+

2 2 59

Check Yourself 3 Solve the equations. 1. x2 + 8x – 3 = 0

2. 2x2 – 5x – 3 = 0

3. 2x2 – 2 = 4x

Answers 1. –4  ò19

2. 3, 

1 2

3. 1  ñ2

3. The Quadratic Formula The final method will work on any quadratic equation. Therefore, we can use it when the other easier methods fail or are not easy to apply. Look at the derivation of the quadratic formula. ax2 + bx+ c = 0

(a  0, a, b, c  )

b c x2 + x+ = 0 a a

(divide both sides by a)

b c x2 + x = – a a

(try to get a perfect square)

b b c b x2 + x+( )2 = – +( )2 2a 2a a a

(add (

(x+

b 2 ) to both sides) 2a

b 2 4ac+ b2 ) = 2a 4a2

x+

b b 2  4ac b 2  4ac  = = 2a 2a 4a2

x=

b2  4ac b  b 2  4ac –b  = , 4a2 2a 2a

x1 

b+ b 2  4ac b – b 2  4ac , x2  2a 2a

QUADRATIC FORMULA If ax2 + bx + c = 0, a  0, then x =

EXAMPLE

60

7

– b  b 2 – 4 ac 2a

Solve x2 + 2x – 8 = 0 using the quadratic formula. Algebra 8

Solution Make sure that you write a quadratic equation in standard form before you identify the values a, b, and c.

EXAMPLE

8

Solution

First we identify the coefficients a, b, and c. For this equation a = 1, b = 2, and c = –8. Let us substitute the values of a, b, and c into the quadratic formula. x= x1 =

9

Solution

2  6 4 = = 2, 2 2

x2 =

2  6 –8 = = –4 2 2

Solve 3x2 + 2x – 4 = 0. For this equation a = 3, b = 2, and c = –4. x=

EXAMPLE

2 b  b 2  4ac 2  2  4  1  (–8) 2  36 = = 2a 2 1 2

2  b  b2  4ac 2  2  4  3  (–4) 2  52 = = 2a 23 6

x1 =

2 + 52 2 + 2 13 1+ 13 = = 6 6 3

x2 =

2  52 2  2 13 1  13 = = 6 6 3

Find the real solutions of the equation 9 +

3 2  = 0 , x  0. x x2

3 2  = 0 , is not a quadratic equation. However, x x2 we can make it quadratic by multiplying each side by x2, since x  0. The result is

In its present form, the equation 9 +

9x2 + 3x – 2 = 0, x  0. Now, a = 9, b = 3, and c = –2. b2  4ac = 32  4  9  (–2)  81 x1 =

3  81 12 2 = = 29 18 3

x2 =

3 + 81 6 1 = = 18 18 3

Check Yourself 4 Solve the equations. 2. x2 

1. 4x2 + 3x – 1 = 0

5 = 3 x 2

3. 2x2 – 4x = 5

Answers 1. 1, Quadratic Equations

1 4

2.

3  19 2

3.

2  14 2 61

4. Discriminant of a Quadratic Equation discriminant of a quadratic equation

Definition

The quantity b2 – 4ac is called the discriminant of a quadratic equation. The discriminant tells us whether the equation has real solutions, and also tells us how many roots of an equation exist. The discriminant is denoted by  (delta). x=

b   , 2a

 = b2  4ac

For a quadratic equation ax2 + bx + c = 0, the value of  determines the number of real roots. 1. If  > 0, there are two distinct real roots. 2. If  = 0, there is one real root (a double root). 3. If  < 0, there is no real root. Use the discriminant to check the number of roots before you solve a quadratic equation. EXAMPLE

10

Solution

Solve x2 + 6x + 7 = 0. First, find .  = b2 – 4ac = 36 – 4  7  1 = 8 So  is positive. x1 =

6 + 8 = 3 + 2, 2

x2 =

6  8 = 3  2 2

We can see that  > 0 and there are two real roots. EXAMPLE

11

Solution

Solve x2 – 4x + 4 = 0. First, check .  = b2 – 4ac = 16 – 4  4  1 = 0 So  = 0 and x1 =

4 0 , 2

x2 =

4+ 0 . 2

Hence x1 = x2 = 2. We can see that = 0 and there is only one real root (a double root). EXAMPLE

12

Solution

Solve x2 – 2x + 5 = 0.  = b2 – 4ac = 4 – 4  1  5 = –16  is negative, so there is no real root.

62

Algebra 8

EXAMPLE

13

Solution

For which values of k does the equation 3x2 – 4x + k = 0 have no real solution? If there is no real solution, the discriminant must be negative. Therefore,  = (–4)2  4  3  k = 16 12 k < 0 . 16  12 k < 0, 16 < 12 k, 16 < k, 12

4 k> . 3

So the equation has no real solution for k >

EXAMPLE

14

Solution

4 . 3

For what values of m does the equation x2 + 3mx – 5m – 1 = 0 have a double root? A quadratic equation has a double root if its discriminant is 0. Therefore,  = (3m)2 – 4  1  (–5m – 1) = 0. 9m2 + 20m + 4 = 0 (9m + 2)(m + 2) = 0 9m + 2 = 0 or m + 2 = 0. So m1 =

EXAMPLE

15

Solution

2 and m2 = 2 . 9

The equation mx2 + (2m + 1)x + m – 1 = 0 has two real roots. Find m. If the quadratic equation has two real roots, then its discriminant is positive.  = b2  4ac = (2m+ 1)2  4m(m 1) = 4m2 + 4m+ 1  4m 2 + 4m = 8 m+ 1  = 8 m+ 1 > 0 8 m > 1 m>

–1 8

Therefore, the equation has two real roots if m > Quadratic Equations

–1 . 8 63

EXAMPLE

16

Solution

Prove that (a2 + b2)x2 + 2(a + b)x + 2 = 0 has no real root if a and b are unequal.

 = 4(a + b)2 – 4(a2 + b2)  2 = 4(a2 + 2ab + b2) – 8(a2 + b2) = 4a2 + 8ab + 4b2 – 8a2 – 8b2 = –4a2 + 8ab – 4b2 = –4(a – b)2 But (a – b)2 is always non-negative, and it is also non-zero, since a  b. So  < 0. Thus the equation (a2 + b2)x2 + 2(a + b)x + 2 = 0 has no real roots if a and b are unequal.

EXAMPLE

17

Solution

From each corner of a square piece of sheet metal, a man removes a square of side 3 cm. He turns up the edges to form an open box. If the box holds 48 cm3, what are the dimensions of the piece of sheet metal? Let x be the length of a side of the square. 3

x–6

3 3

3

x–6

x–6

3

3 3

x–6

3

Since the volume of the box is 48 cm3, we have 3(x – 6)2 = 48 (x – 6)2 = 16 x – 6 = 4 x1 = 10, x2 = 2. We discard the solution x = 2 since length cannot be negative. So the sheet of metal is 10 cm by 10 cm. 64

Algebra 8

EXAMPLE

18

Solution

A ladder 5 m long is leaning against a house. The distance from the bottom of the ladder to the house is 1 m less than the distance from the top of the ladder to the ground. How far is the bottom of the ladder from the house?

Let us use x to represent the distance from the top of the ladder to the ground. The ladder then forms the right triangle shown in the diagram. By using the Pythagorean Theorem, we get the equation x2 + (x – 1)2 = 52 x2 + x2 – 2x + 1 = 25 2x2 – 2x – 24 = 0

5

x

x2 – x – 12 = 0 (x + 3)(x – 4) = 0 x = –3 or x = 4.

x–1

Since the length cannot be negative, x = 4. So the distance from the bottom of the ladder to the house is 3 m. EXAMPLE

19

Solution

A motorboat heads upstream a distance of 48 km on a river whose current is running at 3 km per hour (km/h). Then the motorboat returns. The trip upstream and back takes 12 hours. Assuming that the motorboat maintained a constant speed relative to the water, what was its speed? We use x to represent the constant speed of the motorboat relative to the water. Then the true speed going upstream is x – 3 km/h, and the true speed going downstream is x + 3 km/h. Since Distance = Velocity  Time, we can write Time = Distance / Velocity. Therefore, the boat takes travel upstream and downstream.

48 hours to x–3

48 km

48 hours to travel x+ 3 x – 3 km/h

Since the total time is 12 hours, 48 48 + = 12; x  3. x – 3 x+3

x + 3 km/h

Multiply both sides by (x – 3)(x + 3): Quadratic Equations

65

48(x – 3) + 48(x + 3) = 12(x – 3)(x + 3) x2 – 8x – 9 = 0 (x – 9)(x + 1) = 0 x = 9 or x = –1. Since the speed cannot be negative, the speed of the boat is 9 km/h.

Check Yourself 5 1. Evaluate the discriminant of the equation 2x2 – x – 3 = 0, and describe the roots. 2. The sum of two numbers is 10 and sum of their squares is 68. Find the numbers. 3. For which value(s) of m does the equation mx2 + mx – 1 = 0 have a double root? Answers 1.  = 25, two real roots

2. 2, 8

3. –4

M AT H F U N A mathematician, a physicist, and an engineer were traveling through Scotland when they saw a black sheep through the window of the train. ‘Aha,’ said the engineer, ‘I see that Scottish sheep are black.’ ‘Hmm,’ said the physicist, ‘You mean that some Scottish sheep are black. We haven’t seen all the sheep yet.’ ‘No,’ said the mathematician, ‘All we know is that there is at least one sheep in Scotland, and that at least one side of that one sheep is black! We haven’t seen the other side of the sheep yet.’

66

Algebra 8

T HE G OLDEN R ATIO The Golden Ratio appears again and again in art, architecture, music and nature. Its origins go back to the days of the ancient Greeks, who thought that a rectangle with sides in the Golden Ratio, called a golden rectangle, exhibited the most aesthetically pleasing proportion. The use of the Golden Ratio has been of interest to artists and architects since before the building of the Parthenon in Greece in the fifth century B.C. The rectangle drawn around the Parthenon with its upper triangular structure intact, as shown in Figure 1, is a golden rectangle. 1

1

x–1

1

P

x Figure 1

Figure 2

Figure 2 shows the dimensions of a golden rectangle. The ratio of length to width in the rectangle is x to 1. A square P with sides of 1 unit has been marked, leaving a smaller rectangle Q. For the smaller rectangle Q, the ratio of length to width is 1 to x – 1. In order for the larger rectangle to be a golden rectangle, the two ratios need to be equal, creating a proportion: x 1 = 1 x–1 When we apply cross multiplication to this proportion, we obtain a quadratic equation: x 1   x(x –1) = 1 ; x2 – x = 1 ; x2 – x – 1 = 0. 1 x 1 Since we cannot factor this equation, we apply the quadratic formula. The solutions are x1 

1 5 1 5 and x2  . 2 2

Because x represents the length of a rectangle, the negative solution is discarded 1 5  1.618034 . This is the Golden Ratio. 2 The Golden Ratio occurs in nature as well as in art. For example, in sunflowers,

and the positive solution is

the ratio of the number of clockwise spirals to the number of counterclockwise spirals approximates the Golden Ratio.

Q

EXERCISES

3 .1

A. Solving Equations of the Form ax2 = 0 1. Solve the equations. a. ñ2x2 = 0

b. (ñ3 – 2)x2 = 0

1 c.  x2 = 0 5

2

d. 0.07x = 0

B. Solving Equations of the Form ax2 + bx = 0 2. Solve the equations. 2

D.Solving Equations of the Form ax2 + bx + c = 0 9. Solve by factoring. a. x2 – x = 0

b. –3x2 + x = 0

c. x2 – 49 = 0

d. x2 – 25 = 0

e. x2 – x – 2 = 0

f. x2 + 3x + 2 = 0

g. x2 – 2x + 1 = 0

h. 6x2 + x – 15 = 0

i. 10x2 – 19x + 6 = 0

j. 12x2 – 5x – 2 = 0

10. Solve by factoring. 2

a. 2x + 5x = 0

b. –7x + 3x = 0

c. 3x2 – 8x = 0

d.

2 x  6 x2 = 0 3

a. x2 + (a + 1)x + a = 0 b. (x + 1)2 – 2(x + 1)(x – 3) + (x – 3)2 = 0 c. 2ax2 + (5b – 2a)x – 5b = 0

11. Solve by completing the square.

3. Solve (x + 2)2 = 2(x + 2). 4. Solve (2x – 1)(x + 3) = –3. 5. Solve (x + 1) – 3(x + 1) = 0.

a. x2 – 4x – 1 = 0

b. x2 + 4x = 3

c. x2 – 6x – 13 = 0

d. 3x2 – 2x + 4 = 0

e. 4x2 + 8x + 15 = 0

f. 2x2 + 7x + 11 = 0

2

g.

C. Solving Equations of the Form ax2 + c = 0 6. Solve the equations. a. x2 – 16 = 0

b. 7x2 + 3 = 0

c. 9x2 – 25 = 0

d. 0.6x2 – 15 = 0

e. –5x2 + 4 = 0

f.

2 x2  =0 25 8

7. Solve the equations. a. (x – 2)2 – 9 = 0 2

c. (x + 1) = 5

2

d. (1 – x) – 1 = 8

8. Solve (4x + 1)(x – 1) = (x – 1)(x + 1) – 3(x – 3). 68

h. 3x2 + x =

1 2

12. Solve by using the quadratic formula. a. x2 – 4x + 2 = 0

b. 2x2 + x – 1 = 0

c. 4x2 + 12x + 9 = 0

d. 3x2 – 5x + 1 = 0

e. 5x2 + 4x + 7 = 0

f. 3x2 – 7 = 2 – 2x

g. x2 + x + 1 = 0 h. 25x2 + 40x + 16 = 0 i. (2x – 3)2 = 11x – 19 j.

b. (x + 3)2 + 7 = 0

1 2 1 x  3x + = 0 2 2

3 2 1 1 x  x  =0 4 4 2

2 k. 4x  1 = x(10 x  9) 3

l.

x2  1 =11( x +1) 2 Algebra 8

13. Solve by using the quadratic formula. a. 4  b.

x2 – (4k + 2)x + 7k + 2 = 0 form a perfect square?

1 2 =0  x x2

1 1 2 + = x 1 x 1 3 2

2 2 c. (x  3)  (x  2) = 1  x 16 4 2

d.

17. For which values of k does the equation

18. The longest side of a right triangle is 6 cm less than twice the length of the medium side. The shortest side is 6 cm. Find the length of the two other sides.

1 1 5 + = x 1 x  4 4

14. Use the discriminant to determine whether each quadratic equation has two real solutions, a double root, or no real solution, without solving the equation. a. 2x2 + 3x + 1 = 0

19. A wire that is 32 cm long is cut into two pieces, 

and each piece is bent to form a square. The total area enclosed by the two squares is 34 cm2. Find the length of each piece of wire.

b. x2 + 5x – 6 = 0

c. 4x2 + 12x + 9 = 0 d. 25x2 – 20x + 4 = 0 e. x2 + 4x + 7 = 0

f. 2x2 – x + 2 = 0

Mixed Problems 20. a, b, c, and d are four consecutive even natural 15. Consider the equation ax2 + 3x + 10 = 0. For



which values of a does the equation have

numbers. The sum of a and c is one fifth of the product of b and d. Find a, b, c, and d.

a. two distinct real roots? b. one double root? c. no solution?

21. Two squares have sides (x + 6) cm and (2x + 1) 

cm respectively. The sum of their areas is 697 cm2. Find the areas of the squares.

16. For which values of m does the equation 2x2 + 2x + m + 4 = 0 have a. two real solutions? b. one solution? c. no real solutions? Quadratic Equations

22. A year ago, a father was eight times as old as his 

son. Now his age is the square of his son’s age. How old are they now? 69

We have seen that the roots of an equation depend on its coefficients. Therefore, there exist certain relations between the coefficients and the roots of an equation. In this section we will consider the relations between the roots and the coefficients a, b and c of a quadratic equation. We know that the roots of the quadratic equation ax2 + bx + c = 0, a  0 are x1 = François Viète (or Vieta) 1540-1603, French mathematician. Vieta was a founder of modern algebra, who introduced the use of letters as algebraic symbols and correlated algebra with geometry and trigonometry. Vieta presented methods for solving equations of second, third and fourth degree. He knew the connection between the positive roots of equations and the coefficients of the different powers of the unknown quantity. The word ‘coefficient’ is actually due to Vieta. When Vieta applied numerical methods to solve equations, he used methods which were similar to those used by earlier Arabic mathematicians.

b   b +  and x2 = . 2a 2a

Let us use these formulas above to find the sum of the roots of a quadratic equation. x1 + x2 =

b   b+  + 2a 2a

=

b   + (  b+  ) 2a

=

2b b = 2a a

Therefore,

x1 + x2 = -

b . a

We can use the same expressions for x1 and x2 to find the product of the roots of a quadratic equation.  b     b+   (  b   )  (  b+  ) x1  x2 =      = 4a2  2a   2a  =

b2  (b2  4ac ) 4ac c = 2= 4a2 4a a

Therefore,

x1  x2 =

c a

.

These relations were discovered by François Vieta, a French mathematician, and so they are together called Vieta’s theorem. 70

Algebra 8

Vieta’s theorem

Theorem

Let x1 and x2 be the roots of the quadratic equation of the form ax2 + bx + c = 0, a  0. Then b x1 + x2 =  , a

c x1  x2 = . a

By using Vieta’s theorem we can now find the sum and product of the roots of a quadratic equation without calculating the roots.

Note By using Vieta’s theorem we can also see the following. 1.

x + x2 – b / a b 1 1 (sum of the reciprocals of the roots) + = 1 = = x1 x2 x1x2 c/a c 2

c b2  2ac  b  2. x12 + x22 = (x1 + x2 )2  2x1x2 =    2  = (sum of the squares of the roots) a a2  a 

3. x13 + x23 = (x1 + x2 )3  3x1x2 (x1 + x2 ) = EXAMPLE

20

Solution

Derive a formula for the difference of the roots of a quadratic equation in standard form. Let us consider the two differences. First, x1  x2 =

b   b+  b   + b   2     = = = . 2a 2a 2a 2a a

Similarly, x2  x1 =

 . a

So we can say that x1  x2 = EXAMPLE

21

Solution

3abc  b 3 (sum of the cubes of the roots) a3

 . a

Find the sum and product of the roots of the given equations, without solving the equations. a. 2x2 + 6x + 5 = 0

b. x2 – 3x – 5 = 0

a. a = 2, b = 6 and c = 5

b. a = 1, b = –3 and c = –5

b 6 x1 + x2 =  = = 3 a 2 x1  x2 =

c 5 = a 2

b 3 x1 + x2 =  = = 3 a 1 x1  x2 =

c –5 = = –5 a 1

Note The quadratic equation 2x2 + 6x + 5 = 0 has no real root. However, by Vieta’s theorem, the 5 sum of the roots is –3 and the product of the roots is , which are real numbers. Can you 2 say why? Quadratic Equations

71

Check Yourself 6 1. Find the sum and the product of the roots of the following equations, using Vieta’s theorem. 7 a. 3x2 + 5x – 1 = 0 b. x2 – 4x + = 0 c. –x2 + 7x – 1 = 0 2 2. For the previous equations, find a. 1 + 1 . b. x12 + x22. c. x13 + x23. x1 x2 Answers 5 1 7 1. a.  , – b. 4, c. 7, 1 2. a. 5 b. 9 c. 322 3 3 2 EXAMPLE

22

Solution

x1 and x2 are the non-negative roots of the equation 3x2 + 2mx + 1 = 0. Given x1 = 3x2, find x1, x2, and m. x1 = 3x2 x1  x2 =

c 1 = ; a 3

1 3x2  x2 = ; 3

x1 = 3x2 then x1 = 3  x1 + x2 = 

2m ; 3

1 x2 2 = ; 9

x2 = 

1 since x2 > 0; 3

x2 =

1 3

1 =1 3

1 2m 1+ =  ; 3 3

4 2m = 3 3

m = 2 EXAMPLE

23

Solution

x1 and x2 are the roots of the equation x2 – 3x + 1 = 0. Find the value of

Since x1 + x2 = –

x1 x2

+

x2 x1

.

(–3) 1 = 3 > 0 and x1  x2 = =1> 0, both x1 and x2 are positive. 1 1

x1 x + 2 =k x2 x1 k2 =

x12 2 x1x2 x2 2 + + x2 x1 x1x2

k2 =

x13 + x2 3 + 2 x1x2 x1x2

9 + 27 1 k = + 2 1 = 20 1 2

k1 = 2 5 , k2 = 2 5 k= 2 5 72

(why?) Algebra 8

EXAMPLE

24

Solution

x2 + x – 6m = 0 and x2 – 2mx + 3 = 0 have a common root. Find m. Let x1 and x2 be the roots of x2 + x – 6m = 0 and x1 and x3 be the roots of x2 – 2mx + 3 = 0, then  x1 + x2 = – 1 (1)   x1 + x3 = 2m

and (2)

x1x2 = – 6m .  x1x3 = 3

From (1) we get x2 – x3 = –1 – 2m. From (2) we get

x2 = –2m ; x2 = –2mx3. x3

From (1) and (2), 2 mx3  x3 = 1  2m x3 (2 m+ 1) = 1+ 2m x3 =

1+ 2 m = 1. 1+ 2 m

If we substitute 1 for x in the second equation, we have 4 – 2m = 0, so m = 2.

Note We can apply Vieta's theorem to a cubic polynomial equation. Let x1, x2 and x3 be the roots of the equation ax3 + bx2 + cx + d = 0. Then, b 1. x1 + x2 + x3 =  . a

c 2. x1x2 + x1x3 + x2 x3 = . a

d 3. x1x2 x3 =  . a

Check Yourself 7 1. One of the roots of the equation 2x2 – mx + 8 = 0 is 3 more than the other root. Find m. 2. x1 and x2 are the roots of the equation x2 + x – m + 1 = 0. x1 – x2 = 5 is given. Find m, x1, and x2. Answers 1. 10 2. 7, 2, –3

Mathematics is one component of any plan for liberal education. Mother of all the sciences, it is a builder of the imagination, a weaver of patterns of sheer thought, an intuitive dreamer, a poet. The study of mathematics cannot be replaced by any other activity. Quadratic Equations

73

EXERCISES

3 .2

1. Find the sum and the product of the roots of each

6. One of the roots of the equation

equation, without solving it.

3 6x2 + 13x + n2 + 2n – 2 = 0 is – . Find n and 2 the other root of the equation.

a. 10x2 – 11x – 12 = 0 b.

7 2 9 5 x + x – =0 8 7 3

c. (x – 2)(3x – 4) = 13 d. x + 7 = (2x – 1)(3x – 2) 2

7. Let x1 and x2 be the roots of the equation

2

e. (4x + 3) = (3x + 1)

4x2 + 5x = 0. Find x12x2 + x1x22.

f. x2 = ñ2(3x – ñ2x)

2. The sum of the roots of the equation

8. Consider the equation

(5k + 2)x2 + 7kx – 8k = 0 is 3. Find the product

(m + 2n)x2 – (m + 2n)x + m – n = 0.

of the roots.

If the arithmetic and the geometric means of the roots of the equation are equal, find the relation between m and n.

3. The product of the roots of the equation (4m2 – 1)x2 + (2m + 1)x + 2m – 1 = 0 is 5. Find m.

9. Let x1 and x2 be the roots of the equation x2 – 8x + 5k = 0. x1 = 2x2 – 1 is given. Find k.

4. Find the sum of the squares of the roots of the following quadratic equations. a. 2x2 + 5x + 1 = 0 b. –x2 + 7x + 2 = 0 c. 3(x + 2)(x – 1) = 4(x – 2) – 1

10. Let x1 and x2 be the roots of the equation 

x2 + (m + 1)x + m + 2 = 0. 2x1 + 3x2 = –13 is given. Find m.

d. 4x(3 – 4x) = x – 1

5. One of the roots of the equation 4 20x2 + (2m2 + 2m + 1)x + 20 = 0 is – . Find 5 m and the other root of the equation. 74

11. Let x1 and x2 be the roots of the equation (k + 1)x2 – kx + k – 4 = 0.

6 6 + = 30 is given. x1 x2

Find k. Algebra 8

12. Consider the equation x2 – (3k + 1)x + 8 = 0. 2 1

2 2

x + x = 20 is given. Find k.

13. Consider the equation x2 + (m + 1)x – m = 0. 2 1

2 2

x + x = 13 is given. Find m.

18. Let p and q be the roots of the equation 

2x2 – 5x + p2 + q2 = 0. Find the discriminant of the equation.

19. Find x12 + 

1 2

x1

, if x1 is the root of the equation

2

1 1    x +   6  x +  +9 = 0. x x  

14. Let x1 and x2 be the roots of the equation kx2 + (k – 1)x – 2 + k = 0. 2 2 3 + = is given. Find k. x1 + 3 x2 + 3 2

20. Consider the equation x3 + cx + 1 = 0. 

x1 =

1 1 is given. Find the roots of the + x2 x3

equation and c.

15. Consider the equation x2 + (k – 2)x + k – 6 = 0. If the equation has two negative roots, find all the possible values of k.

16. Find two positive consecutive numbers such that the sum of their squares is 85.

21. The roots of the equation x2 – 5x + p = 0 are also



the roots of the equation x3 + qx + 30 = 0. Find p + q.

22. The sum of the two roots of the equation 2x3 – x2 – 7x – 3 = 0 is 1. Find the roots of the equation.

23. From each corner of a square piece of sheet 17. Find the product of the roots of the equation 

1 1 1 1  , a + b  0. + = a b a+b+ x x

Quadratic Equations

metal, a man removes a square of side 2 cm. He turns up the edges to form an open box which holds 24 cm3. What are the dimensions of the piece of sheet metal? 75

Let x1 and x2 be the roots of the equation ax2 + bx + c = 0. b c We can write the equation ax2 + bx + c = 0 as x2 + x + = 0. a a b   x1 + x2 =  a , and  We know that  c   x1  x2 = a .

So we can write the equation again as x2 – (x1 + x2)x + (x1  x1) = 0. In other words, if we know the sum and product of the roots of a quadratic equation then we can write the equation as x2 – Sx + P = 0 where S = x1 + x2 and P = x1  x2. This means that we can derive (find) a quadratic equation if we know its roots.

Note If the roots of a quadratic equation are x1 and x2, then (x – x1)  (x – x2) = 0. EXAMPLE

25

Solution

Find a quadratic equation whose roots are –1 and 5. x1 = –1 and x2 = 5 S = x1 + x2 = –1 + 5 = 4 P = x1  x2 = (–1)  5 = –5 Hence, the equation is x2 – Sx + P = 0 x2 – 4x – 5 = 0.

EXAMPLE

26

Solution

Find a quadratic equation whose only root is x1 = x2 =

1 . 3

1 3

1 1 2 S = x1 + x2 = + = 3 3 3 1 P = x1  x2 = . 9 Hence, the equation is x2  76

2 1 x+ =0, or 9 x2 6 x+1 =0. 3 9 Algebra 8

EXAMPLE

27

Solution

Find the equation whose roots are 2 + ñ3 and 2 – ñ3. x1 = 2 + ñ3, x2 = 2 – ñ3 S = x1 + x2 = 4 P = x1  x2 = (2 + ñ3)  (2 – ñ3) = 1 Hence, the equation is x2 – 4x + 1 = 0.

EXAMPLE

28

Solution

Find the equation whose roots are 1 more than the roots of x2 – 3x – 4 = 0. Let the roots of x2 – 3x – 4 = 0 be x1 and x2. Let the roots of the equation we are looking for be x3 and x4. x1 + x2 = 3,

x1  x2 = –4

x3 = x1 + 1,

x4 = x2 + 1

S = x3 + x4 = (x1 + 1) + (x2 + 1) = x1 + x2 + 2 = 3 + 2 = 5 P = x3  x4 = (x1 + 1)  (x2 + 1) = x1  x2 + x1 + x2 + 1 = (–4) + (3) + 1 = 0 Hence the equation is x2 – 5x = 0. EXAMPLE

29

Solution

The equation x2 + (k + 4)x + k – 4 = 0 has symmetric roots. Find them. x1 = t, x2 = –t (symmetric roots) x1 + x2 =  t +(  t) =

b a

( k+ 4) 1

0 = k  4 k = 4

x1  x2 =

c a

t  ( t) =

k 4 1

t2 = 8 t2 = 8

t = 2ñ2 Therefore, x1 = 2ñ2, x2 = –2ñ2.

Check Yourself 8 1. Find a quadratic equation whose roots are –2 and 3. 2. The roots of the equation x2 – x – 2 = 0 are x1 and x2. Find the equation whose roots are x3 and x4, where x3 = 2x1 + 1 and x4 = 2x2 + 1. Answers 1. x2 – x – 6 = 0 2. x2 – 4x – 5 = 0 Mathematics has beauty but not everyone sees it. Quadratic Equations

77

EXERCISES

3 .3

1. Find the equation with the given roots.

6. The roots of the equation x2 – kx – 3k + 1 = 0 are two more than the roots of the equation x2 – (k – 4)x – 4k = 0. Find k.

1 3 , 2 2

a. –1, 1

b.

c. 0, 4

d. 2 + ñ2, 2 – ñ2

e. ñ3 – ñ2, ñ3 + ñ2

f.

1 1 , p2 q 2

7. Find three consecutive integers a, b, c, such that a2 + b2 + c2 = 110.

8. Find the number that is 3 more than twice its 2. Find the quadratic equation whose sum of the

square root.

roots is –2 and product of the roots is 5.

9. A swimming pool can be filled by two pipes x1 and x2. Find the equation whose roots are x3 and x4, such that x3 = 2x1 and x4 = 2x2.

together in six hours. If the larger pipe alone takes five hours less than the smaller pipe alone to fill the pool, find the time in which each pipe alone would fill the pool.

4. Find the equation whose roots are 2 less than the

10. Alex can do a job in one hour less than Jane. If

3. The roots of the equation x2 – 4x – 3 = 0 are

roots of the equation 2x2 – 6x + 58 = 0.

5. The roots of each given equation are x1 and x2.

6 5 hours. How long would it take each person working

Alex and Jane work together the job takes

alone?

Write a new equation with roots x3 and x4. a. x2 – 2x – 3 = 0

b. x2 + 5x + 4 = 0

x3 = x1 – 1

x3 = 2x1 + 1

x4 = x2 – 1

x4 = 2x2 + 1

c. 6x2 – 9x – 6 = 0 x3 = x1 – x2 x4 = x2 – x1

11. A man completed a job for $156. It took him seven hours longer than he expected and so he earned $14 an hour less than he anticipated. How long did the man expect the job to take?

d. x2  2 x+ 1 = 0 x3 = x4 =

x12 x2 2 2

x x1

12. The sum of the numerator and denominator of a certain positive fraction is 11. If 1 is added to both the numerator and the denominator, the fraction 3 is increased by . Find the fraction. 56

e. x2 – 2mx + 3m – 2 = 0 f. mx2 – 2mx – 1 = 0

78

x3 = x1 – 2x2

x3 = 2x1 – x2

x4 = x2 – 2x1

x4 = 2x2 – x1

13. The area of an isosceles right triangle is 81 m2. Find the perimeter of the triangle. Algebra 8

14. The distance between two cities A and B is

19. Two painters working together can paint the front

140 km. A car driving from A to B left at the same time as a car driving from B to A. The cars met after one hour, then the first car reached city B 35 minutes later than the second car reached city A. Find the speed of each car.

of a house in 16 hours. One of the painters alone can finish this job in 24 hours less time than the other painter alone. How much time does each worker need to do this job alone?



20. A motorcyclist traveled at a constant speed for 60 km. If he had gone 10 km/h faster, he would have shortened his traveling time by one hour. Find the speed of the motorcyclist.

15. An aeroplane traveled a distance of 400 km at an average speed of x km/h. Write down an expression for the time taken. On the return journey, the speed increased by 40 km/h. Write down an expression for the time for the return journey. If the return journey took 30 minutes less than the outward journey, write down an equation in x and solve it.

21. The area of a triangle is 36 m2. The length of the base is twice the height. Find the length of the base and the height of the triangle.

22. A jeweler wishes to mix an alloy of 25% silver with another alloy of 40% silver. How much of each should he use to produce 60 kg of an alloy which is 30% silver?

16. A car drove from one city to another and returned 

by a different route. The outward journey was 48 km and the return journey was 8 km shorter. The speed of the car on the return journey increased by 4 km/h. The return journey took one hour less time. Find the speed of the car on the outward journey.

23. One of two pipes can fill a pool 24 hours faster 

than the other one. The slower pipe filled the pool for eight hours, then the other pipe was opened. The pipes filled together for twenty hours and 2 of the pool. Find the time that each pipe 3 requires to fill the pool alone.

filled

17. A group of women plan to share equally in the $14,000 cost of a boat. At the last minute three of the women decide not to pay. This raises the share for each of the remaining women by $1500. How many women were in the original group?

18. A car drove from a city A to a city B. The distance between the two cities is 350 km. After 200 km, the speed of the car decreased by 20 km/h. The total trip took 5 hours. Find the speed of the car in the first part of the journey. Quadratic Equations

24. A worker can clean a pool in four hours less time than it takes another worker. If the men work 8 together the job takes hours. How long would 3 it take each man working alone?

25. The length of a rectangle is 2 m more than its width and the area is 48 m2. Find the length and width of the rectangle. 79

A. WRITING EQUATIONS IN QUADRATIC FORM standard form of an equation

Definition

An equation is in standard form if the only term on the right-hand side of the equation is zero. For example, the equations 6x2 + 2x – 3 = 0 and x4 – 5 = 0 are both in standard form. The equation 6x2 + 2x = 3 and x4 = 5 are not in standard form. Certain equations that are not quadratic can be expressed in quadratic form using substitutions. These equations can be recognized because when they are written in standard form, the exponent of the variable in one term is half the exponent of variable in the other term. For example, we can write standard form equations such as x4 + 17x2 + 72 = 0 2x8 + 4x4 = 0 x – ñx – 12 = 0 as quadratic equations, because the exponent of the first variable is twice the exponent of the second variable. Look at the steps to write an equation as a quadratic. 1. Let t be a variable term with the half exponent. 2. Substitute t in all the terms with the variable. 3. Solve for t. 4. Back substitute for the original variable. EXAMPLE

30

Solution 1

Solve x4 – 13x2 + 36 = 0. The equation x4 – 13x2 + 36 = 0 is not a quadratic equation but we can write it as (x2)2 – 13x2 + 36 = 0. For this reason, it is a quadratic in x2. Let x2 = t. First we solve for t, then solve the resulting equations for x. (x2)2 – 13x2 + 36 = 0 x2 = t, so t2 – 13t + 36 = 0. By factoring, (t – 4)(t – 9) = 0 t = 4 or t = 9. Since t = x2 x2 = 4 x = 2

80

or

x2 = 9 x = 3

. Algebra 8

Solution 2

Alternatively, we can directly factorize the equation. x4 – 13x2 + 36 = 0 (x2 – 4)(x2 – 9) = 0 x2 = 4 x = 2

or

x2 = 9 x = 3

In both solutions, the roots of the given equation are –3, –2, 2, 3. EXAMPLE

31

Solution

Solve (5x – 1)2 + 4(5x – 1) – 5 = 0. For the equation (5x – 1)2 + 4(5x – 1) – 5 = 0, we let t = 5x – 1 so that t2 = (5x – 1)2. Then the original equation becomes t2 + 4t – 5 = 0. First solve for t: (t + 5)(t – 1) = 0, so t = –5 or t = 1. Now, solve for x. 5x – 1 = –5 5x = –4 x1 =

or

–4 5

5x – 1 = 1 5x = 2 x2 =

2 5

Hence, the roots of the equation are EXAMPLE

32

Solution

–4 2 and . 5 5

Solve (a2 – a)2 – 2(a2 – a) = 0. Let t = a2 – a. The equation becomes t2 – 2t = 0 t(t – 2) = 0 t = 0 or t = 2. Now, solve for a. 1. a2 – a = 0 a(a – 1) = 0 a = 0 or a = 1 2. a2 – a = 2 a2 – a – 2 = 0 (a – 2)(a + 1) = 0 a = 2 or a = –1 Hence, the roots of the equation are –1, 0, 1, 2.

Quadratic Equations

81

EXAMPLE

33

Solution

4

1 2  Solve  x   + 5 = 2 x2 + 2 . x x  First x2  0, so x  0. Let x 

1 = t. x

How can we write the right-hand side in terms of t? Let’s take the square of t: 2

1 1  t2 =  x   = x2 + 2  2, so x x  1 x2 + 2 = t 2 + 2. Now, let us multiply both sides of the equation by 2: x 2 2 x2 + 2 = 2t2 + 4. x Then we have the equation t4 +5 = 2t2 + 4 t4  2t 2 +1= 0 (t2  1)2  0 t2  1= 0 t =1 or t = 1. 1 Since t = x  , x 1 1 x  =1 or x  = 1. x x Let us multiply the equations by x: x2  x  1= 0 or x2 + x 1= 0. By using the quadratic formula, x1 =

1 5 1+ 5 1+ 5 1  5 , x2 = , x3 = , x4 = . 2 2 2 2

Check Yourself 9 Solve the equations. 1. (x + 15)2 – 3(x + 15) – 18 = 0. 2. (x + 5)4 – 6(x + 5)2 – 7 = 0. 3. x – 9ñx + 14 = 0. Answers 1. –18, –9 2. –5 ñ7 3. 4, 49 82

Algebra 8

B. EQUATIONS INVOLVING PRODUCTS AND QUOTIENTS If the product of two or more numbers is zero, then at least one of the factors must be zero. If the quotient of a division is zero, then the dividend must be zero.

Note 1. P  Q = 0 if and only if P = 0 or Q = 0, where P = P(x) and Q = Q(x). 2.

EXAMPLE

34

Solution

P = 0 if and only if P = 0 and Q  0. Q

Solve (x2 – 1)(x2 – 2x – 8) = 0. Try to factorize each part if possible: (x – 1)(x + 1)(x – 4)(x + 2) = 0. If the product is zero then at least one of the factors is zero. x–1=0;

x=1

x+1=0;

x = –1

x–4=0;

x=4

x+2=0;

x = –2

Thus, the roots of the equation are –2, –1, 1, 4.

EXAMPLE

35

Solution

Solve (x2 – 4x + 10)(x2 – 5x + 2) = 0. We cannot factorize the parts. Let us try to solve x2 – 4x + 10 = 0 and x2 – 5x + 2 = 0 as two different quadratic equations. x2 – 4x + 10 = 0;  = 16 – 4  1  10 = –24 < 0. Therefore, this equation has no real root. x2 – 5x + 2 = 0 ;

 = 25 – 4  1  2 = 17

This time, by the quadratic formula, the roots are x1 = Quadratic Equations

5  17 5+ 17 , x2 = . 2 2 83

EXAMPLE

36

Solution

Solve x3 – x2 – 4x + 4 = 0. First try to factorize the expression. x3 – x2 – 4x + 4 = x2(x – 1) – 4(x – 1) = (x – 1)(x2 – 4) = (x – 1)(x – 2)(x + 2). Now, the question becomes: solve (x – 1)(x – 2)(x + 2) = 0. So the solution is x–1=0;

x1 = 1

x–2=0;

x2 = 2

x+2=0;

EXAMPLE

37

Solution

Solve

x3 = –2.

x2  5x  6 = 0. x+ 2

The denominator of a fraction cannot be equal to zero, so x + 2  0, x  –2. We need to solve x2 – 5x – 6 = 0. x2 – 5x – 6 = (x – 6)(x + 1) (x – 6)(x + 1) = 0 x1 = –1 or x2 = 6 Since these roots are not equal to –2, x1 = –1 and x2 = 6 are both solutions to the equation.

EXAMPLE

38

Solution

Solve

x2 + 7 x  8 = 0. x2  3x+ 2

First solve x2 – 3x + 2 0: x  2, x 1. x2 + 7x – 8 = (x – 1) (x + 8) = 0 So x1 = 1 and x2 = –8 are the roots of the numerator. We know that x  1, so the only solution is –8.

Note It is very important to check the roots of the numerator to see whether they make the denominator zero or not. We can do this either by substituting the roots of the numerator in the denominator, or by finding the roots of the denominator directly and checking whether they are common or not. 84

Algebra 8

EXAMPLE

39

Solution

Solve

x  7 x+ 3  = 1. x  4 x+ 4

x – 4  0, and x + 4  0, so x  4, and x  –4 (since the denominators cannot be zero). x  7 x+ 3   1 = 0. x  4 x+ 4

Now we have

Let us make the denominators common: (x  7)( x+ 4)  ( x+ 3)( x  4)  ( x2 16) = 0. (x+ 4)(x  1)  x2  2 x =0 (x  4)(x+ 4)

–x2 – 2x = 0 ; –x(x + 2) = 0 ; x1 = 0 and x2 = – 2.

EXAMPLE

40

Solution

Solve

4 x+ 1 x2  5  = . x+ 1 x  1 x2  1

4 x+ 1 x2  5 is not a quadratic equation. However,  = x+ 1 x  1 x2  1 we can make it quadratic by multiplying each side by x2 – 1 since (x2 – 1  0 ; x  1). The result is

In its present form, the equation

4(x – 1) – (x + 1)2 = x2 – 5 4x – 4 – x2 – 2x – 1 = x2 – 5 –2x2 + 2x = 0 –2x(x – 1) = 0 x1 = 0 or x2 = 1, but x  1, so x = 0 is the only possible solution.

EXAMPLE

41

Solution

2

5x  x  Solve  + 6 = 0.   x+ 2  x+ 2 

x + 2  0 ; x  –2 Let t =

x , then the equation becomes x+ 2

t2 – 5t + 6 = 0 (t – 3)(t – 2) = 0 t1 = 2 or t2 = 3. Quadratic Equations

85

Now, solve for x: x =2 x+ 2

or

x =3 x+ 2

x = 2x+ 4

x = 3 x+ 6

x2 = 4

x1 = 3.

So the roots are –4 and –3.

Check Yourself 10 1. Solve

9 + 2 x = 4. 4x+ 3

2. Solve

x+ 1 x  3 12  = . x  2 x+ 2 x2  4

Answers 1 3 1.  ; 4 2

2. no real solution

C. EQUATIONS INVOLVING RADICALS When the variable in an equation occurs in a square root, cube root, and so on, that is, when it occurs in a radical, the equation is called a radical equation. For example, the equations x+ 1 = 2 and

3

x – 1 – 5 = 0 are radical equations. Sometimes a suitable operation will

change a radical equation to an equation that is linear or quadratic. To solve a radical equation, follow the procedure. 1. Isolate the radicals Isolating a radical means putting the radical on one side of the equation and everything else on the other side, using inverse operations. If there are two radicals in the equation, isolate one of the radicals. 2. Get rid of the radical sign Raise both sides of the equation to a power equal to the index of the isolated radical. 3. If there is still a radical sign left, repeat steps 1 and 2. 4. Solve the remaining equation 86

Algebra 8

5. Check for extraneous solutions When you solve a radical equation, extra solutions may come up when you raise both sides to an even power. These extra solutions are called extraneous solutions. In radical equations you check for extraneous solutions by putting the values you found into the original equation. If the left side of the equation does not equal the right side then you have an extraneous solution.

Note 1. If a value is an extraneous solution, it is not a solution to the original problem. 2. It is very important to check your results in the original equation. In many equations, one of the results may not satisfy the original equation. However, sometimes it is possible that all results that you have found will be acceptable. EXAMPLE

42

Solution

Solve the radical equation

2 x+ 5 = 7.

Here the radicand is already alone; we do not need to isolate it. So take the square of both sides: 2x + 5 = 49 ; 2x = 44 ; x = 22. Now let us check to see if x = 22 is an extraneous solution: 2 x+ 5 = 7 2  22 + 5 = 7 49 = 7 7 = 7.

Since the last statement is true, x = 22 is not an extraneous solution. Therefore, there is one solution to this radical equation, x = 22. EXAMPLE

43

Solution

Solve

2 x  5 + x = 4.

First we isolate the radical: 2 x  5 = 4  x.

Now take the squares of both sides to eliminate the square root: 2x – 5 = 16 – 8x + x2. The new equation is x2 – 10x + 21 = 0 (x – 3)  (x – 7) = 0 x = 3 or x = 7. Let us check the results in the original equation: Quadratic Equations

87

x=7;

2 .7  5 + 7 = 4

x=3;

2  3  5 +3 = 4

9 +7 = 4

1+ 3 = 4

3+7 = 4

4=4

10 = 4

This is true, so x = 3 is a solution.

This is false, so 7 is an extraneous solution. Hence, the only solution to the equation is x = 3. EXAMPLE

44

Solution

Solve

10 x+ 56  2 x +8 = 4 .

In this question there are two radical expressions. We can isolate only one expression, so it is better to isolate the more complex one. So we have 10 x+ 56 = 2 x+ 8 + 4.

Take the squares of both sides: 10 x+ 56 = (2 x+ 8) + 2  4  2 x+ 8 +16 .

This is a new equation involving radical expressions. Follow the same steps again to isolate the second radical. 8 x+ 32 = 8  2 x+ 8 x+ 4 = 2 x+ 8 x2 + 8 x+ 16 = 2 x+ 8 x2 + 6 x+ 8 = 0 (x+ 4)  (x+ 2) = 0

x = –4 ,

x = –2

Now, check these results in the original equation. x = –4 ;

2  (–4) + 8  10 (–4) +56 = 4 0  16 = 4 4 = 4

This is true! x = –2 ;

2  (–2) + 8  10 (–2) + 56 = 4 4  36 = 4 2  6 = 4 4 = 4

This is true! Hence, both –4 and –2 are solutions to the equation. 88

Algebra 8

EXAMPLE

45

Solution

Solve the equation 5 + 3 x+ 3 = 3.

5 + 3 x+ 3 = 3 3

x+ 3 = 2 (by taking the cube of both sides)

x+ 3 = 8

x = –11. We do not need to check for extraneous solutions because this is an odd power. Therefore, –11 is the only solution to the equation.

EXAMPLE

46

Solve the equation

4x+ 1 + x+ 2 = 10 x+ 5 .

( 4 x+1 + x+ 2 )2 = ( 10 x+ 5) 2

Solution

4x+ 1+ 2 (4x+ 1)(x+ 2) + x+ 2 = 10 x+ 5 2 (4 x+1)( x+ 2) = 5 x+ 2 (2 (4 x+1)(x+ 2))2  (5 x+ 2)2 4(4 x+1)(x+ 2)  25 x2 + 20 x+ 4 16 x2 + 36 x+ 8 = 25 x2 + 20 x+ 4 9x2  16 x  4 = 0 (9 x+ 2)( x  2) = 0 x= 

2 or x = 2 9

Now, check these results in the orijinal equation. x= 

2 ; 9

x=2 ;

8 2 20 1 16 5  +1 +  + 2 =  + 5 = + = ; 9 9 9 9 9 3 8 +1 + 2 + 2 = 20 + 5 ;

5 =5

Both statements are true, so both x =  Quadratic Equations

5 5 = 3 3

2 and x =2 are solutions. 9 89

Check Yourself 11 Solve the equations. 2x  4 = x  2

1.

5x+ 3 = 4

4.

3x2  2 x+15 + 3x2  2 x+ 8 = 7

2.

3.

x+ 2  4 x+ 8 = –3

Answers 1 13 1. 2. 2, 4 3. 7 4.  , 1 3 5

D. EQUATIONS INVOLVING AN ABSOLUTE VALUE On the real number line, the absolute value of x is the distance from the origin to the point x. For example, there are two points whose distance from the origin is three units, –3 and 3. So the equation |x| = 3 has two solutions, 3 and –3. Let us first remember the mathematical definition of absolute value. absolute value of a function

Definition

For all real numbers x,  f x  , for f x   0  f ( x) =   – f x  , for f x  < 0.

We can use this information to begin solving equations involving one or more absolute values. EXAMPLE

47

Solution

The absolute value of a number is never negative. |a|  0

EXAMPLE

48

Solution 1

Solve the equation |x – 2| = 5. Case 1 __________________

Case 2 _________________

x–20; x2 x–2=5 x=7

x–2 0. So, D(m) = (3, ).

B. COMPOSITION OF FUNCTIONS Aside from the four basic operations there is a very important way of combining functions to get a new function. We call this a composition. This operation is illustrated as follows: Here we see two different functions f and g. Their common point is that the range of one is the domain of the other. Because of that fact the domain of the function g is linked with the range of the function f. Consider x = –2: g(–2) = 3 f(g(–2)) = f(3) = 0

f(g(x)) means g is applied first, f is applied second.

Read f(g(x)) as “f of g of x”.

In other words, –2 finally maps to 0. Note that to find the final mapping we first use g and then f. If we name that final mapping as h, then h(–2) = 0 where h(x) = f(g(x)). That final mapping h is another function and is the composition of f and g as shown below on the right:

f

g Range of g –2 0 1 3 Domain of g

Range of f

2

–1

3

0

5

1

Domain of f f(g(x)) h –2 0 1 3

Domain of h

–1 0 1 Range of h

As an algebraic example if f(x) = ñx and g(x) = 3x + 1, we know that f(x) is the rule “take the square root of the number x” and g(x) is the rule “multiply the number x by 3 and add 1”. In that case f(g(x)) means you must “first multiply the number x by 3 and add 1, then take the square root of the result” so we can define h( x) = f ( g( x)) = g( x) = 3x +1. Here we say that h(x) is a composition of f(x) and g(x), that is h(x) is a composite function. The domain of f(g(x)) is the set of all x in the domain of g such that g(x) is in the domain of f.

Functions

157

Composite functions are composed of “bridges” that link the domain of “first” with the range of “last”.

composition

Definition

f(g(x)) is defined as composition of f and g. It is also denoted by (f  g)(x) or simply f  g. fog

g x

f g(x)

f(g(x))

Arrow diagram for (f o g)(x) EXAMPLE

54

Solution 1

Given that f(x) = x2 + 3x, g(x) = x – 4, find f(g(x)) and g(f(x)). f(g(x)) = g(x)2 + 3g(x) = (x – 4)2 + 3(x – 4) = x2 – 8x + 16 + 3x – 12 = x2 – 5x + 4 g(f(x)) = f(x) – 4 = (x2 + 3x) – 4 = x2 + 3x – 4 So f(g(x)) = x2 – 5x + 4 and g(f(x)) = x2 + 3x – 4. What can you say about f(g(x)) and g(f(x))? Is it true that they are the same?

Solution 2

f(g(x)) = f(x – 4) = (x – 4)2 + 3(x – 4) = x2 – 5x + 4 g(f(x)) = g(x2 + 3x) = (x2 + 3x) – 4 = x2 + 3x – 4

Note In general, f(g(x))  g(f(x)). 158

Algebra 8

EXAMPLE

55

Let f(x) = x + 1, g(x) = x3. Find: a. (f  g)(2) b. (g  f)(1) c. (f  f)(5) d. (g  g)(–1)

Solution 1

a. (f  g)(x) = f(g(x)) = g(x) + 1 = x3 + 1 (f  g)(2) = 23 + 1 = 9 b. (g  f)(x) = g(f(x)) = (f(x))3 = (x + 1)3 3

(g  f)(1) = (1 + 1) = 8 f  g means g is applied first, f is applied second and in general f  g  g  f.

Composite functions are formed of functions that are inside another function.

c. (f  f)(x) = f(f(x)) = f(x) + 1 = x + 2 (f  f)(5) = 5 + 2 = 7 d. (g  g)(x) = g(g(x)) = (g(x))3 = x9 (g  g)(–1) = (–1)9 = –1

Solution 2

Note that these values can be calculated without finding the formula. a. In case of ( f  g)(2) we may first calculate g(2) = 8 and then f(8) = 9 to find the answer since ( f  g) (2) = f(g(2)) = f(8) = 9. b. (g  f )(1) = g( f(1)) = g(2) = 8 c. ( f  f )(5) = f( f(5)) = f(6) = 7 d. (g  g)(–1) = g(g(–1)) = g(–1) = –1

EXAMPLE

56

Solution 1

Let f ( x) =

x , g(x) = x5 and h(x) = x2 + x. Find f  g  h. x 1

Here we have a composition of three functions. ( f  g  h)( x) = f ( g( h( x))) =

Solution 2 Functions

g( h( x)) ( h( x))5 ( x2 + x)5 = = 2 . 5 g( h( x)) – 1 ( h( x)) – 1 ( x + x)5  1

( f  g  h)(x) = f(g(h(x))) = f(g(x2 + x)) = f((x2 + x)5) =

( x2 + x)5 . ( x2 + x)5 – 1 159

EXAMPLE

57

Solution

Write the function h( x) = 5x + 3 as a composition of two functions. Let us read what the formula tells us to do: “first add 3 to five times the number x and then take the square root of all”. Let f(x) = 5x + 3 and g(x) = ñx. Note that f is applied first, g is applied second. So h(x) = (g  f)(x). The composite function form is never unique. For example, consider the previous example: If n is any nonzero integer, we could choose f(x) = (5x + 3)1/n and g(x) = xn/2. Thus, there are an infinite amount of composite function forms. Generally, our aim is to choose a formula such that the expression for the function is simple, as in the above example.

EXAMPLE

58

Solution

Given f(x) = 3x + 1 and g(x) = 2x – m such that (f  g)(x) = (g  f )(x), find g(

1 ). 10

( f  g)(x) = f(g(x)) = 3  g(x) + 1 = 3(2x – m) + 1 = 6x – 3m + 1 (g  f )(x) = g( f(x)) = 2  f(x) – m = 2(3x + 1) – m = 6x + 2 – m

1 Since ( f  g)(x) = (g  f )(x), 6x – 3m + 1 = 6x + 2 – m which gives m = – . 2 1 1 7 So we have g(x) = 2 x + . Therefore, g( ) = . 2 10 10 EXAMPLE

59

Solution

If the rule for function f  g  h is “first take the square root of a number plus twenty-five, second divide the new number plus 2 by itself, third take the cube of the newest number” and p(x) = 3x2 + 6x. Find (f  g  h)(x) and (h  p  f)(–2). First of all let us find formulae for f, g and h. Since f  g  h means h is applied first, g second and f third we have h( x) = x + 25, g( x) =

x+ 2 , f ( x) = x 3. x 3

 x + 25 + 2  According to the rule, ( f  g  h)(x) =  .  x + 25   Now let us find (h  p  f)(–2):

(h  p  f )(–2) = h(p( f(–2))) = h(p(–8)) = h(144) = 13. 160

Algebra 8

EXAMPLE

60

Solution

Given that f(g(x)) = 4x – 1 and g(x) = x + 2, find f(x). Since f(g(x)) = 4x – 1 and g(x) = x + 2 we have f(x + 2) = 4x – 1. Let x + 2 = a. Then x = a – 2. Substituting x = a – 2 in f(x + 2) = 4x – 1 we get f(a – 2 + 2) = 4(a – 2) – 1 or f(a) = 4a – 9. Since this is just a matter of notation in the final formula instead of a we choose x to get f(x) = 4x – 9.

Check Yourself 13 x +1 and g(x) = x, find f + g, f – g, fg, f / g. 2x  4 2. Given that f(x) = x2 and g(x) = 3x + 4, find (f  g)(1) and (g  f)(1).

1. Given that f ( x) =

3. Write the function h(x) = (2x – 7)5 as a composition of two functions such that h(x) = (f  g)(x). Answers 2 2 2 1. 2 x  3x +1 , 2 x +5 x +1 , x + x , x2+1 2x  4 2x  4 2x  4 2x 4x

Functions

2. 49, 7 3. f(x) = x5, g(x) = 2x – 7

161

C. INVERSE OF A FUNCTION 1. One-to-one Functions Consider the two functions represented by a map as shown below: f

g

·1

·4

·1

·2

·5

·2

·3

·6

·3

·4 ·5

Each number in the domain of f takes a different value from the range. But it is not the same for the function g since 1 and 2 take the same value in the range. We name a function one-to-one if each element in the domain corresponds to exactly one element from the one and g is not one-to-one. range. So here f is one-tto-o To understand whether a given formula belongs to a one-to-one function or not we need an algebraic tool. The following definition for one-to-one function meets our needs: one-tto-o one function

Definition

one if, for each x1  x2 in its domain, f(x1)  f(x2). A function f is one-tto-o EXAMPLE

61

Which one of the following is a one-to-one function? a. f(x) = 2x – 1

Solution

b. f(x) = x2

Let us use the definition for one-to-one function to find out whether they support the definition or not: a. Let x1  x2. Then f(x1) = 2x1 – 1 and f(x2) = 2x2 – 1. Let’s assume that f(x1) = f(x2), so 2x1 – 1 = 2x2 – 1 which means x1 = x2. But this is wrong since we let x1  x2! So our assumption that f(x1) = f(x2) is also wrong. That means when x1  x2, we have f(x1)  f(x2). Therefore, the function is one-to-one. b. Clearly, if we choose –2  2 we get f(–2) = f(2) = 4. Although we can find an infinite amount of such examples, just one of them is enough to decide that function is not one-to-one.

162

Algebra 8

We know that the graph of a relation is a function if any vertical line one if crosses the curve at most once. Similarly, a function is one-tto-o any horizontal line crosses the curve at most once. For example: y

y

x

x

Graph of a function which is not one-to-one

Graph of a one-to-one function

HORIZONTAL LINE TEST FOR THE ONE-TO-ONE FUNCTIONS A function is one-to-one if, and only if, no horizontal line crosses its graph at more than one point. Note that in the previous example we could have found the answers without using the definition but just by plotting their graphs and applying the horizontal line test. EXAMPLE

62

Which one of the following graphs belong to one-to-one functions? a.

b.

y

x

Solution

c.

y

x

x

y

y

y

x

y

x

x

a. This graph does not belong to a one-to-one function. b. This graph does not belong to a function, so we can not talk about one-to-one function. c. This graph belongs to a one-to-one function.

Note A function that is only increasing or decreasing in its domain is one-to-one. Functions

163

Check Yourself 14 3x  1 is a one-to-one function. 5 2. Draw a graph which belongs to a one-to-one function.

1. Prove that f ( x) =

3. Draw a graph which does not belong to a one-to-one function. Answers 1. Compare f(x1) and f(x2) when x1  x2. 2. Consider a function which is either increasing or decreasing. 3. Consider a function which is sometimes increasing and sometimes decreasing.

2. Definition for The Inverse of a Function As mentioned earlier in this book, many mathematical relations can be modelled as functions. For example,

C = f(r) = 2r

V = g(a) = a3

The circumference of a circle is a function of the radius r.

The volume of a cube is a function of the edge a.

In many cases we are interested in reversing this correspondence determined by a function:

r  m C  

C 2

The radius of a circle is a function of the circumference C.

a  n V   3 V

The edge of a cube is a function of the volume V.

As illustrated above, reversing the relation between two quantities produces a new function. 164

Algebra 8

Recall that a function is a relation that assigns to each element in the domain exactly one element from the range.

f x0

y0

Domain

Range

In the figure above we see an arrow diagram for function f. So “Using the rule f we say that x0 becomes y0”. We symbolize this as “f(x0) = y0”.

–1

Read f (x) as “f inverse of x”.

Now consider this question: Using the rule f what gives us y0? Of course the answer is x0. So “Using the rule f to get y0 we need x0”. We need a new notation so that “x0” will be our answer. And we symbolize this fact as “f –1(y0) = x0”.

1 . f ( x) 1 The reciprocal is f ( x)

f –1(x) doesn’t mean

f –1

written as (f(x))–1.

x0

y0

Range

Domain

Note The domain of a function is the same as the range of its inverse. The range of a function is the same as the domain of its inverse.

The domain and the range change their places in an inverse of a function just as the server and the opponent change their places in a table tennis game.

Functions

165

EXAMPLE

63

Given the function f = {(0, 2), (–1, 4), (4, 6), (5, 5)}, a. find f –1.

Solution

b. find f –1(4).

a. Interchange the x and y-coordinates of each ordered pair of f to find f –1: f

–1

= {(2, 0), (4, –1), (6, 4), (5, 5)}

b. To find the value of f –1(4), notice that f –1(4) is the second coordinate when the first coordinate is 4 in the function f –1. So f –1(4) = –1. EXAMPLE

64

Solution

If f(x) = 2x – 3, find f –1(5). If we substitute 5 in place of x we find f(5) = 2  5 – 3 = 7 but not f –1(5)! Finding f –1(5) means finding the x value for which f(x) = 2x – 3 gives 5 as a result. 2x – 3 = 5, so x = 4 or f –1(5) = 4. If f –1 also supports the definition for a function, we will call f –1 as the inverse of f. But this is not always possible. For example, consider the relation below: · -3

f

· -1

· -2 · -1

·0

·1

·1 ·2

·3

Domain

Range

Clearly, this is a function since each element is assigned to exactly one element. Now let’s consider its inverse: · -3

f –1

· -1 ·0 ·1

· -2 · -1 ·1

·2

·3

Range

Domain

As we see f –1(1) = –3 or f –1(1) = 0 is not possible for a function (just one element must be assigned for each element in the domain). So although f is a function, f –1 is not a function. That means the inverse of f doesn’t exist.

Note It is not true that every function has an inverse. For the inverse of a function to be defined, it is necessary that different elements in the domain always give different values. As we know such functions are called one-to-one functions. 166

Algebra 8

As another example let us consider the following function graph of which is given below: y y = f(x)

x

To find the inverse, we use the same procedures that we used for relations. Drawing the reflection with respect to y = x we get the following picture (below left): y

y

y = f –1(x)

y = f –1(x)

y = f (x) x

x

y=x

We know that a set of points in the coordinate plane is the graph of a function if, and only if, no vertical line crosses the graph at more than one point. But that rule doesn’t hold for f –1 (look at the picture above right). That means the inverse of f doesn’t exist. Note that the inverse we draw will be a function if f has no horizontal line crossing the graph at more than one point, that is if f is one-to-one. We know that the range of a function becomes the domain of its inverse. We also know that no element from the domain of a function must be left unassigned. Sometimes the range of a function is explicitly so that it is larger than the function’s its real range, that is some elements of the range are not used. In that case when we talk about the inverse some elements in the domain of inverse will be unassigned. This will result in an absence of the inverse function. For example, if f(x) = x + 2 such that f:   , although the range seems to be  it is in fact  (Note that when we put any integer in x + 2, we always get other integers). To guarantee that we will not face such functions we must be sure that any element in the range is assigned by an element from the domain. Such functions are called onto functions. CRITERIA FOR EXISTENCE OF INVERSE OF A FUNCTION A function has an inverse if, and only if, it is one-to-one and onto. If a function is given by only formula or graph (where no explicit range is given), it is naturally an onto function. So there is no need to think about this detail to define its inverse. Functions

167

Now we can give the formal definition for the inverse of a function: inverse function

Definition

Let f be a one-to-one and onto function with the domain A and the range B. Then its inverse function f –1 has the domain B and the range A such that f(x) = y  f –1(y) = x. By definition the inverse function f –1 undoes what f does. That is, if we take x, apply f, and then apply f –1, we arrive back at x where we started. Similarly, f undoes what f –1 does. That is why f and f –1 are the inverses of each other.

Note f(f –1(x)) = x and f –1(f(x)) = x. EXAMPLE

65

Solution

Prove that f(x) =x3 and g(x) = 3 x are inverses of each other. When we find f(g(x)) = ( 3 x )3 = x, we can see that f(x) and g(x) are the inverses of each other. Note also that g(f(x)) = x.

EXAMPLE

66

Solution

If f(x) = x + 2 and f –1(x) = x – a, find a. We know that f(f –1(x)) = x. So f(x – a) = x or (x – a) + 2 = x. That gives a = 2.

3. Finding The Inverse of a Function Given the graph of a function, to find the inverse we use the following procedure: 1. Verify that the function is one-to-one by applying the horizontal line test. 2. Take symmetry of the graph of the function with respect to the line y = x. EXAMPLE

67

Answer the following using the graph on the right:

y

(7,11) y = f(x)

a. Does f have an inverse? Why?

(6,6)

b. Draw the graph of the inverse function f –1. c. Find f –1(-6), f –1(-3), f –1(0), f –1(6), f –1(11).

(0,-3)

(4,0)

x

–1

d. Find the domain and the range of f . Solution

(-10,-6)

a. Applying the horizontal line test we can see that f is one-to-one. So it has an inverse y

x

168

Algebra 8

b. We draw the line y = x and reflect f with respect to it to draw the graph of f –1: y (11,7) (0,4)

(6,6)

–1

y = f (x)

(-3,0) x

(-6,-10)

c. f –1(6) = –10, f –1(–3) = 0, f –1(0) = 4, f –1(6) = 6, f –1(11) = 7. d. f –1 can take any x between –6 and 11, inclusive, so the domain is [–6, 11]. The value y of f –1 can be any number x between –10 and 7, inclusive, so the range is [–10, 7].

Given the formula for a function, to find the inverse we use the following procedure: 1. Verify that the function is one-to-one. 2. Solve the equation y = f(x) for x and interchange x and y in the end.

EXAMPLE

68

Given that f(x) = 2x + 5, answer the following: a. Does f have an inverse? Why? b. Find the formula of the inverse function f –1. c. Find f –1(19), f –1(1). d. Find the domain and the range of f –1.

Solution

a. f is a linear function which is always increasing. That means it is one-to-one and so it has an inverse. y5 . b. Let y = 2x + 5. Then x = 2 Interchanging x and y we have the inverse function as y =

x5 x5 or f –1(x) = . 2 2

1 5 x5 19  5 1 = 2. we get f 1(19) = = 7 and f (1) = 2 2 2 Note that we could also solve equations 2x + 5 = 19 and 2x + 5 = 1 to find f –1(19) and

c. Using the formula f 1( x) =

f –1(1), respectively. d. Clearly, D( f –1) =  and E( f –1) = . Functions

169

FINDING THE INVERSE OF A FUNCTION  To plot the graph of the inverse of a function, reflect the graph of the function with respect to the line y = x.  To find the formula for the inverse of a function, solve the equation y = f(x) for x and interchange x and y. EXAMPLE

69

If possible draw the graphs of inverses for the following functions: a.

b.

y

y y = g(x)

(0,4)

y = f(x)

(-6,0)

Solution

(-3,3)

(2,3)

(0,0)

x

x

a. Clearly (horizontal line test), f is one-to-one. So the inverse is the symmetry of f with respect to the line y = x: y y = f –1(x)

y = f (x) (0,4) (-6,0)

(4,0)

x

(0,-6)

b. g is not one-to-one (horizontal line test). We cannot draw the graph of the inverse function since it doesn’t exist. EXAMPLE

70

Given that the following one-to-one functions, find their inverses. a. f(x) = 4 – 3x

Solution

b.

f ( x) =

2x – 6 3

c. g(x) = x3 – 5

Since we know that the functions are one-to-one, we can find their inverses directly a. Let y = 4 – 3x. Then x = b. Let y =

4– y 4– x 4– x . Interchanging x and y: y = or f –1( x) = . 3 3 3

2x – 6 3y +6 3x +6 3 x +6 . Then x = . Interchanging x and y: y = or f –1( x) = . 3 2 2 2

c. Let y = x3 – 5. Then x = 3 y +5. Interchanging x and y: y = 3 x +5 or g –1( x) = 3 x +5. 170

Algebra 8

EXAMPLE

71

If possible find the inverses of the following functions: a.

Solution

f ( x) =

2x + 4 3x  5

a. We can verify that f ( x) =

b.

f ( x) = 3 2 x  1+7

c. f(x) = x2 – 2x

2x + 4 is one-to-one as follows: 3x  5

Let x1  x2. Then assume that f(x1) = f(x2). So, 2 x1 + 4 2 x2 + 4 = 3x1  5 3x2  5 6 x1x2  10 x1 +12 x2  20 = 6 x1x2 +12 x1 10 x2 20 22 x2 = 22 x1 x2 = x1.

But this is not correct since we let x1  x2. This means that if x1  x2, then f(x1)  f(x2). So f is one-to-one. Now let’s find its inverse: f ( x) = y =

2x + 4 3x  5

3xy  5y = 2x + 4 3xy  2 x = 4+5 y x(3y  2) = 4+5 y x=

4+5 y 3y  2

Interchanging x and y we have y =

4+5 x 4+5 x or f 1( x) = . 3x  2 3x  2

b. We leave the verification of the fact that f is one-to-one to student and proceed to finding its inverse. y = 3 2 x  1+7 y  7 = 3 2x 1 ( y  7)3 = 2 x  1 ( y  7)3 +1 =x 2 ( x  7)3 +1 ( x  7) 3 +1 or f 1( x) = . 2 2 c. Choosing x = 0 and x = 2 we can realize that f will give the same value. So f is not one-to-one. Proof of this using definition is left to the student in exercises. As a result f does not have an inverse.

Interchanging x and y we have y =

Functions

171

Check Yourself 15 1. If f(2) = 3, f(5) = 6, f(6) = –1, find f –1(–1), f –1(3), f –1(6). 2. If possible draw the inverse of f graph of which is given on the right. 3. Find the inverse of f ( x) =

y

(-3,2)

2x + 3 . 7

(0,0)

Answers 1. 6, 2, 5

y = f(x)

x (1,-3)

y

2. (–3, 1)

3. f 1( x) =

(0, 0)

x

7x  3 2

(2, –3)

The functions which are not one-to-one can have their domains restricted so that they become one-to-one. As a result their inverses will then be functions.

EXAMPLE

72

Given that f(x) = x2 + 1, a. find the largest possible domain so that the inverse exists. b. find f –1(x) using the new domain.

Solution

a. The inverse exists if we have a one-to-one function. Note that for each value of x and its negative we have the same value of y. So, if we ignore the negative x values, we will have a one-to-one function. That means to have an inverse, domain must be restricted to [0, ). b. Let y = x2 + 1 and we will try to solve the equation for x: x2 = y – 1, so x =  y – 1. If we interchange x and y we have y =  x – 1. Note that the range of the inverse function is the domain of the original function, that is [0, ). So y = x – 1. Therefore, 1 the inverse function is f ( x) = x – 1.

172

Algebra 8

EXAMPLE

73

Solution

Prove that f(x) = x2 – 2x – 2 has an inverse if D( f ) = [1, ) and then find its inverse. f(x) = x2 – 2x – 2 = x2 – 2x + 1 – 3 = (x – 1)2 – 3. Here (x – 1)2 is equal to the same number for x – 1 = x0 and x – 1 = –x0 for any x0 > 0. This fact prevents f(x) = (x – 1)2 – 3 from being a one-to-one function. But if we guarantee that x – 1 is never negative, then the function will be one-to-one. And that is possible when x – 1  0, x  [1, ). To find the inverse, we solve y = (x – 1)2 – 3 for x: y = ( x  1)2  3 y + 3 = ( x  1)2 y + 3 =| x  1| Since x  1  0 we have

y+ 3 = x 1 or

y+ 3 +1= x.

Interchanging x and y we have f 1( x) = x + 3 +1.

EXAMPLE

74

Solution

 x +1  –1 3 If f 1   = x , find f (2) + f(8). x  

Let us find f –1(2). We don’t have the formula for f –1(x) but for f 1(

x +1 ). The expression x

inside the brackets must be equal to 2 since we are looking for f –1(2). When we solve

x +1 = 2, we find x = 1. x

x +1 ) = f 1(2) =13  1. x 1 x +1 ) = x3 Finding f(8) is simple. Note that f is inverse of f –1. So if we need f(8), the result of f ( x x +1 x +1 must be equal to 8, that is f 1( will be f(8) for the corresponding x value. ) = 8, where x x

That means when x = 1, f 1(

So, x3 = 8, x = 2 and

x +1 2+1 3 = = . x 2 2

3 5 Finally, f 1(2)+ f (8) =1+ = . 2 2 Functions

173

EXAMPLE

75

Using the graph below find (f  g–1  f –1)(5) + (f  g  f)(–3). y y = f(x)

1 1

x y = g(x)

Solution

Note that we don’t have the graph of any inverse function and we don’t need them at all. To find the value of a function just find the y-value for the given x-value on the graph. To find the value of an inverse function just find the x-value for the given y-value on the graph. (f  g–1  f –1)(5) = f(g–1(f –1(5))) = f(g–1(4)) = f(0) = 2 (f  g  f)(–3) = f(g(f(–3))) = f(g(0)) = f(4) = 5 So (f  g–1  f –1)(5) + (f  g  f)(–3) = 2 + 5 = 7.

EXAMPLE

76

Solution

If (f  g)(x) = 5x – 2 and f(x) = 4x – 3, find g–1(x). (f  g)(x) = f(g(x)) = 4  g(x) – 3 = 5x – 2 Solving the last equation for g(x) we have g( x) = To find the inverse, we solve y =

5x +1 4y  1 for x: x = 4 5

1 Interchanging x and y we have g ( x) =

EXAMPLE

77

Solution

4x  1 . 5

Given f(x) = 2x – 7, find f –1(4x + 1). Let us find f –1(x): f ( x) = 2 x – 7 = y y +7 x= 2 x +7 f –1( x) = 2 Now let us find f –1(4x + 1): f –1( 4x +1) =

174

5x +1 4

(4 x +1)+7 = 2x+ 4 2 Algebra 8

EXAMPLE

78

Given that f(x) = 3x + 1 and g(x) = 2x + 2, find a. (f  g)–1

Solution

b. g–1  f –1

a. Let us find f  g. y = (f  g)(x) = f(g(x)) = 3(2x + 2) + 1 = 6x + 7 To find the inverse, x = (f  g)–1(x) =

y–7 6

x–7 . 6

b. Let us find f –1 and g–1. y = f(x) = 3x + 1 To find the inverse, x = f –1(x) =

y –1 3

x –1 . 3

y = g(x) = 2x + 2. To find the inverse, x =

y–2 2

x–2 . 2 x–1 –2 x–7 (g–1  f –1)(x) = g–1( f –1(x)) = 3 = 2 6

g–1(x) =

Note that we have (f  g)–1(x) = (g–1  f –1)(x). Is it always correct? Why?

Check Yourself 16 1. Given that f(x) = x2 – 4, restrict the domain so that the inverse exists and find its formula. 2. Given that f(2x + 1) = x – 5, find f –1(0). 3. Given that f(g(x)) = 2x + 1 and g(x) = x + 4, find f –1(x). 4. Given that f(x) =

x –1 and g(x) = 3x, find (f  g)–1 and g–1  f –1. 4

Answers 1. [0,  ), Functions

x+ 4

2. 11 3.

x +7 2

4.

4x +1 4 x +1 , 3 3 175

EXERCISES

4 .3 5. Find the required values using the given data:

A. Basic Operations 1. Find f + g, f – g, fg, f / g for the following functions.

a. f ( x) =

2

a. f(x) = x + 1, g(x) = x – 1

3x +5 x+ 2 and g( x) = , x +1 x 1

( f  g  f )(0) = ?

b. f(x) = x3 + 3x2, g(x) = x2 + 5x

 x +1  2 b. f(2x + 1) = 3x – 1 and g   = x +1,  x 1 ( f  g)(–1) = ?

c. f ( x) = x + 3, g( x) = x + 2 d. f ( x) = x  1, g( x) = x +1

c. f (3x  1) =

2. If

1 f ( x) = 3x + 4, g( x) = x 2 + x, h( x) = , x

( f  g)(1) = ?

find

2 x +1 if x > 2 d. f ( x) =  and if x  2  x  2

the following functions.   b.  f  h  ( 1) g  

a. (f + g)(3)

if x > 0  x g( x) =  if x  0  x ( f  g  f )(–2) = ?

c. (hg – f)(4)

B. Composition of Functions

6. If f(x) = x3 – 3x2 + 3x – 1 and g( x) =

3. Find f  f, f  g, g  f, g  g for the following

solve ( f  g)(x) = 0.

functions. a. f ( x) =

x +1 , g( x) = x2 + x x 1

(g  f )(x) = 3f(x) – 1,

b. f(x) = x – 2x + 1, g(x) = x + 1

find a such that ( f + g)(a) = 19.

8. If f ( x) =



4. Express h(x) in terms of f(x) and g(x) such that h(x) = (f  g)(x).

C. Inverse of a Function functions or not. a. {(x, 1), (y, 1), (z, 2), (k, 3)}

7

 x4 b. h( x) =    5 

b. {(1, 1), (2, 2), (3, 3), (4, 1)} 3

2

c. h(x) = (2x – x) – (2x – x) + 4 176

  1 x , find  f  f  ...  f  ( x).    1+ x 2004 times  

9. State whether the following relations are one-to-one

a. h( x) = 2 x  4

2

2 x +1 , x+ 3

7. If ( f  g)(x) = 5g(x) + 4 and

2

2x + 3 x+ 3 c. f ( x) = , g( x) = x 3x  2

x+ 3 x2 and g( x+1) = , 2x  1 x 1

c. {(a, b), (b, a), (c, d), (d, c)} Algebra 8

10. State whether the following functions are one-to-one

13. Plot the graphs of inverse functions for the

or not.

following functions, if possible:

a. f(x) = x2 – 1, D( f ) = {0, 1, 2, 3}

a.

b.

y

y

(4,5)

b. f(x) = x2, D( f ) = +

(0,4)

c. f(x) = x2 d. f(x) = 3x + 5 e. f ( x) =

x

x 1 x2

x

(2,-2)

(-8,-3)

c.

y

(-5,-5)

(5,7)

d.

11. State whether the following graphs belong to b.

y

(0,3)

x

(0,0)

(2,-3)

(-6,-3)

one-to-one functions or not. a.

y

(-6,7) (-5,4)

(4,1) (-2,-1)

(5,-5)

e.

y

(6,-5)

f.

y

y (0,5)

(-5,3)

x

x

x

x

(4,-2)

x

(6,-5)

c.

y

14. Find the required values using the given data for the functions with the inverse.

x

2 x +5 1 , f (3) = ? 5 b. f –1(x) = x2 – 3, D(f –1) = [0, ), f(–3)= ?

a. f ( x) =

12. Find the inverses of the following one-to-one functions: 3x + 4 x 1

a. f(x) = 6 – 4x

b.

f ( x) =

c. f(x) = 4x3 – 5

d.

f ( x) = 3  5 x  1+9

Functions

c. 2 f ( x)  1=

f ( x)+1 1 , f (2) = ? x2

d. f (3x +1) =

4 x  1 1 , f (3) = ? x2

3  e. f(x2 + x + 1) = x – 2, D( f ) =  ,   , 4  f –1(0) = ? 177

15. Find the following using the given data for

18. Find the following using the graph of functions 

functions with inverses:

y = f –1(x) y

 2 x +1  1 a. f   = x +1, f ( x) = ?  x 1 

b. f ( x  1) =

below:

3x  2 1 , f ( x) = ? x + 21

1

 3x  5  1 c. f   = x, f (2 x) = ?  x +1 

1

Mixed Problems

a. (g  f –1)(–2) + ( f –1  g  f )(2) b. ( f –1  g)(2) + ( f  g)(3)

16. Find the following using the given data for 

functions with inverses: a. f(x) = 3x + 1, g(x) = 2x + 3, (f  g–1)(x) = ?

19. Prove that 

1 x+ 2 b. f ( x) = , g( x) = , ( g 1  f )( x) = ? x 2 x +1

c. f 1( x) =

2 x +1 1 3x +1 , g ( x) = , ( f 1  g1 )( x) = ? x3 x 2

d. f 1( x) =

x  2 1 2 , g ( x) = , ( f  g)1( x) = ? x x2

f. g( x) =

3x +1 , ( f  g)( x) = 2 x+ 3, f 1( x) = ? x 1

g. g( x) =

2 x +1 , ( g 1  f )( x) = 2 x  1, f 1( x) = ? x

17. Find the required values using the given data for functions with inverses:

20. If f and g are even functions such that f(0) = g(4) = –2, g(0) = f(4) = 4, f(–2) = g(–2) = 0, find

a. f(x) = x + 1, g(x) = 3x – 4, ( f

b. f(x) = x – 2, g(x) = 2x + 1, ( f  g–1)(3) = ? x+ 3 , g( x) = 6x +1, ( f  g)1(2) = ? c. f ( x) = x2

2x + 3  1  2 x +5  d. f   = x, g   = 3x  2,  x +1   x+ 3  ( f  g )(3) = ? 178

21. Prove that if f and g are even, then f + g, f – g, fg, 

f / g are also even.

22. Prove that if f and g are odd, then f + g, f – g are odd and fg is even.

 g)(2) = ?

2

1

( g  f )(4)+( f  g)(–4)+( g  f )(–4) . ( f  g)(2)+( g  f )(2)+( f  g)(4)



–1

a. f(x) = x2 – 2x is not one-to-one. b. f(x) = 2x3 + 4 is one-to-one.

e. f(x) = 2x + 3, (g  f )(x) = 4x – 5, g(x) = ?



x y = g(x)

23. Find the inverses of the following functions: 

a. f(x) = x2 + 6x, D( f ) = (–, –3] b. f(x) = x2 – 8x + 5, D( f ) = [4, )

24. Prove that (g  f )–1 = f –1  g–1.



Algebra 8

Cryptography is the art of secret writing. Its aim is to protect information by transforming it into a coded language which unwanted eyes are unable to use. This transformation, however, must be done in a reversible way so that individuals intended to view the information may do so. Some of the classical methods of cryptography are as follows (However, none of them are used today, because they are considered either insecure or impractical).

Simple Substitution: When we encode the message MATH IS COOL , it becomes NBUIAJTADPPM. Here, the rule is that every letter is substituted by the letter that follows it. That is, A by B, B by C, “space” by A, etc.

Mixing: Take a key sequence consisting of the first few natural numbers in mixed order, for example 3, 5, 1, 2, 4. The rule is that the first letter will move to the 3rd place, the second to 5th place, the third to 1st place, the fourth to 2nd place, the fifth to 4th place, and so on for the next five letters. So MATH IS COOL becomes THM A CIOS OL. Note that the “space” character is also treated as a letter.

Method of Vigenere: Pair each letter with a number with its order in the alphabet as follows: A B C D E F G H I J K L M N O P Q R S T U V W X Y Z 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 Functions

179

Our message MATH IS COOL becomes 13 1 20 8 27 9 19 27 3 15 15 12 Choose a key word, for example, FUNCTION which is 6 21 14 3 20 9 15 14. Now, add 6 to the first letter of our encoded message, 21 to the second, 14 to the third, and so on. The new message will be read as 19 22 34 11 47 18 34 41 9 36 29 15. If the number is more than 27, divide it by 27 and take the remainding number. Finally, the message will become SVGKTRGNIIBO. Find the pair for these numbers from the table above. The whole procedure is summarized in the table below. 13 1 20 8 27 9 19 27 3 15 15 12 6 21 14 3 20 9 15 14 6 21 14 3

convert the message using table. convert the keyword and write it as the same length as the original message.

19 22 34 11 47 18 34 41 9 36 29 15

add the top two rows.

19 22 7 11 20 18 7 14 9

divide the previous row by 27 and write the remainding number.

9

2 15

Note that all of these methods are functions that are defined from a limited domain to a range. We can invent even more complicated methods by applying a few of them together for the same message (Think about the composition of functions!). Here it is very important to know the rules. To decode a crypted message we just apply the rule in reverse order (Think about the inverse of a function!). Below is an encoded message. We first applied the Method of Vigenere with the key word CAT (each letter is paired by its order in the alphabet as “space” being the 27th letter), and then we applied the simple substitution.

GQHKTVXWFEVCSPM Try to decode it! What will happen if the rule is a function that has no inverse? What happens if we assign the same number to two different letters? If a few of the rules are applied in order, by which rule should we start the decoding procedure? Try to develop your own method of encryption. Algebra 8 180

Definition

quadratic function A function f(x) is a quadratic function if y = f(x) = ax2 + bx + c,

a0

where a, b, and c are real numbers. For example, f(x) = 2x2 + 3x + 1, y = x2 – 1, y = –ñ2x2 and f(x) = 2x – x2 are all quadratic functions. y = 3x – 4 is not a quadratic function, because a = 0. The domain of a quadratic function is the set of all real numbers. The graph of a quadratic function f(x) = ax2 + bx + c is called a parabola. a>0 y

a0

a 0, the parabola opens upward.

When a < 0, the parabola opens downward.

The point V in the parabolas above is called the vertex of the parabola. The vertex is the lowest or the highest point of the parabola. The vertical line drawn through the vertex is called the axis of symmetry. It divides the curve into two symmetrical halves. The points x1 and x2 are called the x-iintercepts of the parabola. They are the zeros of the function. The point c is called the y-iintercept of the parabola. 182

Algebra 8

A. GRAPHING y = ax2 How can we find the graph of the quadratic function y = ax2? If we have the function, we can plot the graph by making a table of values. To find the values, we substitute different values of x into the equation to obtain the corresponding y values. These x and y values provide the coordinates for points which we can plot to form the shape of the graph. Let us graph the function y = ax2. If a > 0, we get the table of ordered pairs opposite. Then we plot the points

x

–

–2

–1

0

1

2

+

y

+

4a

a

0

a

4a

+

on a graph and draw a parabola through them. y 4a 3a 2a a -2

-1

1

x

2

y = ax2, a > 0

We can see that the vertex of the parabola is at the origin (0, 0), and the axis of symmetry lies along the y-axis (the line x = 0). If a < 0, we get a different set of ordered pairs.

x

–

–2

–1

0

1

2

+

y

–

4a

a

0

a

4a

–

y -2

-1

a

1

2

x

2a 3a 4a

Quadratic Functions

183

1

EXAMPLE

Sketch the graphs of the functions. a. y = x2 , y = 2x2 and y = 1 x2 2 a.

Solution

As |a| increases, the parabola becomes narrower. As |a| decreases, the parabola becomes wider.

x

–

–2

–1

0

1

2



y = x2



4

1

0

1

4



y = 2x2



8

2

0

2

8



1 2 x 2



2

1 2

0

1 2

2



y= y

ax2 x2

b. y =  x2 , y = 2x2 and y =  1 x2 2

1 x2 2 x2 2x2

y 4

a>1

3

x y

2

x2 ax2

1 -4

-3

-2

x

-1

1

–

–2

–1

0

1

2



y = –x2

–

–4

–1

0

–1

–4

–

y = –2x2

–

–8

–2

0

–2

–8

–

–

–2

–

1 2

0

–

1 2

–2

–

2

3

4

x

00

a 0, the vertex is the minimum value and k = –4 is the minimum value of the function.

b. h = 

b 4 = =2 2a 2  (–1)

k = f(h) = –22 + 4  2 + 5 = 9 So the vertex is V(2, 9). The equation of the axis of symmetry is x = 2. Since a = –1 < 0, the vertex is the maximum value and k = 9 is the maximum value of the function. 186

Algebra 8

c. h = 

b 3 3 = = 2a 2 1 2 2

5  3  3 k = f (h) =    + 3     + 1 =  4  2  2  3 5 So the vertex point is V   ,   .  2 4

3 The equation of the axis of symmetry is x =  . 2

Since a = 1 > 0, the vertex is the minumum value and y =  the function. EXAMPLE

3

Solution

5 is the minimum value of 4

The line 3x + 1 = 0 is the axis of symmetry of the parabola y = –2x2 + mx – 1. Find the maximum value of the function. 3x + 1 = 0 ; x = 

1 is the equation of axis of symmetry. 3

1 b m 4 h=  =  = ; m=  3 2a 2  (–2) 3

Since a = –2 < 0, the vertex is the maximum value. 2

4 1 7  1 k = f (h) =  2         1 =  3 3 9  3

So k = 

EXAMPLE

4

Solution

7 is the maximum value. 9

The minimum value of the function y = x2 – 8x + n is –4. What is the value of n? Given k = –4 and h =  k = f (h) ;

b 8 = = 4 ; h = 4. 2a 2 1

 4 = 42  8 4 + n ;

n = 12

Check Yourself 2 Find the vertex of each parabola. 1. y = x2 – 10x + 20

2. y = 1 x2 + x – 8 2

3. y = 4 – x2

4. y = 4x + x2

Answers 1. (5, –5) 2. (–1, –8.5) Quadratic Functions

3. (0, 4)

4. (–2, –4) 187

2. Intercepts Consider the function y = f(x) = ax2 + bx + c,

a  0.

If x = 0, then y = c. The point (0, c) is called the y-intercept of the graph. We mean that the parabola intersects the y-axis at this point. If y = 0, then ax2 + bx + c = 0. Let x1 and x2 be roots of this equation. The points (x1, 0) and (x2, 0) are called the x-intercepts of the graph. We mean that the parabola intersects the x-axis at these points. There are three possibilities for the x-intercepts. 1. If  > 0, the parabola intersects the x-axis at two distinct points. 2. If  = 0, the parabola is tangent to the x-axis. 3. If  < 0, the parabola does not intersect the x-axis. D0

a>0

D=0 D 0, the graph has a minimum point and the parabola opens upward. When a < 0, the graph has a maximum point and the parabola opens downward. 2. Find the coordinates of the vertex point, V(h, k). 3. Find the x- and y-intercepts. 4. Sketch the graph of the parabola.

EXAMPLE

7

Solution

Graph the functions. a. y = x2 – 3x – 10

b. y = –2x2 – 8x

d. y = –3x2 + 12

e. y = x2 – 2ñ2x + 2

c. y = x2 + 4x + 5

a. a = 1 > 0, so the parabola opens upward. h= 

3 b = , 2a 2

k = f (h) = (

y

3 2

-2

3 2 3 49 )  3   10 =  ; 2 2 4

3 49 ) is the vertex point. V( , – 2 4

x = 0 ; y = –10, (0, –10) is the y-intercept.

x

5

f(x)=x2 – 3x – 10 -10 49 4

V

y = 0 ; x2 – 3x – 10 = 0 ; x = –2, x = 5 ; (–2, 0) and (5, 0) are the x-intercepts. 190

Algebra 8

b. a = –2 < 0, so the parabola opens downward. h= 

V

8 = 2, 2  (–2)

k = f (h) = 2(–2)  8 (–2) = 8 ; 2

V(–2, 8)

y 8

f(x)=–2x2 – 8x

x = 0 ; y = 0, (0, 0) is the y-intercept. y = 0 ; –2x2 – 8x = 0 ; x = 0, x = –4 ; (–4, 0) and (0, 0) are the x-intercepts.

-4

c. a = 1 > 0, so the parabola opens upward. h= 

y

4 = 2, 2 1

5

k = f (h) = (–2)2 + 4 (–2) 2 + 5 = 1 ;

x

-2

f(x)=x2+4x+5

V(–2, 1)

x = 0 ; y = 5, (0, 5) is the y-intercept. 2

y = 0 ; x + 4x + 5 = 0 ;

1

V

x

-2

 = 42 – 4  1  5 = –4 < 0 ; the parabola does not cut the x-axis. d. a = –3 < 0, so the parabola opens downward. h= 

y

0 = 0, 2(  3)

k = f (h) = 3(0)2 +12 = 12 ;

V 12

V(0, 12) f(x)=–3x2+12

x = 0 ; y = 12, (0, 12) is the y-intercept. y = 0 ; –3x2 + 12 = 0 ; x = 2 ; (–2, 0) and (2, 0) are the x-intercepts.

2

-2

e. a = 1 > 0, so the parabola opens upward. h= 

y f(x)=x2 – 2ñ2x+2

2 2 = 2 2 1

k = f (h) = ( 2 )2  2 2  2 + 2 = 0 ;

x

V( 2, 0)

x = 0 ; y = 2, (0, 2) is the y-intercept.

2 V ñ2

x

2

y = 0 ; x – 2ñ2 + 2 = 0 ; x1 = x2 = ñ2 ; the parabola is tangent to the x-axis at the point (ñ2, 0). Quadratic Functions

191

EXAMPLE

8

Solution

Sketch the graph of the function y = –x2 – 2x + 8 for |x|  3. Find the range of the function. a = –1 < 0, so the parabola opens downward. h= 

y

2 = 1, 2  (–1)

9 8

k = f (h) = (–1)2  2 (–1) + 8 = 9 ;

V(–1, 9)

5

x = 0 ; y = 8, (0, 8) is the y-intercept. y = 0 ; –x2 – 2x + 8 = 0 ; x1 = –4, x2 = 2 ; (–4, 0) and (2, 0) are the x-intercepts. |x|  3 ; –3  x  3 x = –3 ; y = –9 + 6 + 8 = 5, so (–3, 5) is on the parabola. x = 3 ; y = –9 – 6 + 8 = –7, so (3, –7) is on the parabola. So the range of the function is [–7, 9].

EXAMPLE

9

Solution

3

-4 2

-3 -1

x

-7

Find the values of p for which px2 + 4x + p is greater than zero for all real values of x.

px2 + 4x + p > 0 for all real numbers x.

y = px2+ 4x+p

This is possible only if p > 0 and  < 0. (1)

p>0 2

2

4 – 4p < 0 (2)

x

(1) p > 0 (2) 42 – 4p2 < 0 ; 42 – 4p2 = 0 ; p = 2 p

–¥

–2

0

2

¥

(1)

–

–

+

+

(2)

–

+

+

–

system

So p  (2, ).

Check Yourself 4 Sketch the graph of each function. 1. y = x2 – 5x + 4 192

2. y = –x2 + 4x – 3

3. y = 4x2 – 20x + 25

4. y = 3x2 – 2x Algebra 8

Answers 1. y

2.

3.

y

4

1

3

4.

y

y

25

x

2/3

-3 4 x

1

5/2

x

x

4. Shifting Graphs Let y = f(x) be a function. Vertical Shifting y = f (x ) + k

k > 0 shift the graph of y = f(x) k units upward. (see Figure 1)  k < 0 shift the graph of y = f(x) |k| units downward. (see Figure 1)

Horizontal Shifting y = f (x+ h)

 h > 0 shift the graph of y = f(x) h units to the left. (see Figure 2)   h < 0 shift the graph of y = f(x) |h| units to the right. (see Figure 2)

Reflection y = –f(x)

Reflect the graph of y = f(x) in the x-axis. (see Figure 3)

Horizontal and Vertical Shifting y = f(x + h) + k Shift the graph of y = f(x) |k| units upward or downward and |h| units to the left or to the right. k>0

y

y

x2+k

h>0

x2 (x – h)2

(x+h)2

x2 k

x2 – k

0

x

y

-h

0

h

x

x2

x

0

-k –x2

Figure 1 EXAMPLE

10

Quadratic Functions

Figure 2

Figure 3

Sketch the graph of the each function using the shifting method. a. y = x2 + 3

b. y = x2 – 3

c. y = (x – 3)2

e. y = –(x – 4)2

f. y = (x + 2)2 – 3

g. y = –(x – 2)2 + 1

d. y = (x + 3)2

193

Solution

a.

y

b.

x2+3

c.

y

x2

d.

y

(x+3)2

x2 (x – 3)2

x2

y x2

x2 – 3

3 x

0

0

x

0

x

3

0

-3

x

-3

e.

f.

y

g.

y

y

(x+2)2 x2

x2

x2

(x+2)2 – 3

4

1

x

0

-2

x

0

x

2

-3

–(x – 4)2

–x2

0

–(x – 2)2+1

y

Check Yourself 5 Sketch the graph of the each function by shifting. 1. y = x2 + 4 Answers 1. y

2. y = –(x + 1)2 2.

x2+4

3. y = (x – 2)2 – 2 3.

y -1

x2

x

–(x+1)2

y

–x2

4.

(x–2)2

y 1

x

-1

4

x2

4. y = –(x + 2)2 + 1

(x–2)2–2

4 2 -2

-2 –(x+2)2+1

2

-3

x

-4

x –(x+2)2

–x2

5. Parabolas with Absolute Value (Optional) EXAMPLE

194

11

Sketch the graph of each function. a. y = |x2 – 3x + 2|

b. y = –x2 + 2|x| + 3

c. y = x|x + 2|

d. y = |2x2 – 12x + 16| + 2 Algebra 8

Solution

a. First graph the function y = x2 – 3x + 2.

y

Then, take above the part of the graph which is below

2

the x axis, symmetric with respect to x-axis. a = 1 > 0, so the parabola opens upward. h= 

3 3 = , 2 1 2

f(x)=|x2 – 3x+2| 1 1/4 3/2

1

-1/4

2

x

2

3 1  3 k = f (h) =    3  + 2 =  ; 2 4 2 1 3 V  ,   is the vertex point. 4 2

x = 0 ; y = 2, (0, 2) is the y-intercept y = 0 ; x2 – 3x + 2 = 0 ; x = 1, x = 2 ; (1, 0) and (2, 0) are the x-intercepts. b. y = –x2 + 2|x| + 3 Case 1 ____________________________________________________________________ If x  0, |x| = x ; y = –x2 + 2x + 3 a = –1 < 0, so the parabola opens downward. h= 

2 = 1, 2  (–1)

k = f (1) = 12 + 2 1+ 3 = 4 ;

V(1, 4) is the vertex point.

x = 0 ; y = 3, (0, 3) is the y-intercept. y = 0 ; –x2 + 2x + 3 = 0 ; x = –1, x = 3 ; (–1, 0) and (3, 0) are the x-intercepts. Case 2 ____________________________________________________________________ If x < 0, |x| = –x ; y = –x2 – 2x + 3

y 4

a = –1 < 0, so the parabola opens downward.

3

2 h=  = 1, 2  (–1) k = f (–1) = (–1) 2  2 (–1) + 3 = 4 ;

f(x)=–x2+2|x|+3

-3

-1

1

x

3

V(–1, 4) is the vertex point.

x = 0 ; y = 3, (0, 3) is the y-intercept. y = 0 ; –x2 – 2x + 3 = 0 ; x = –3, x = 1 ; (–3, 0) and (1, 0) are the x-intercepts. Quadratic Functions

195

c. y = x|x + 2| Case 1 ________________________________________________________________________ If x + 2  0, x  –2, |x + 2| = x + 2 ; y = x2 + 2x a = 1 > 0, so the parabola opens upward. h= 

2 = 1, k = f (–1) = (–1) 2 + 2 (–1) = 1 ; 2 1

V(–1, –1)is the vertex point.

x = 0 ; y = 0, (0, 0) is the y-intercept. y = 0 ; x2 + 2x = 0 ; x = –2, x = 0 ; (–2, 0) and (0, 0) are the x-intercepts. Case 2 ________________________________________________________________________ If x + 2 < 0 ; x < –2, |x + 2| = –(x + 2) ;

y

2

y = –x – 2x 2

a = 1 > 0, so the parabola opens upward.

f(x)=x|x+2|

1

2 h=  = 1, 2  (–1)

-3

-2

-1

x

-1

k = f (–1)  (–1) 2  2 (–1)  1 ;

V(–1, 1) is the vertex point. x = 0 ; y = 0, (0, 0) is the y-intercept. y = 0 ; –x2 – 2x = 0 ; x = –2, x = 0 ; (–2, 0) and (0, 0) are the x-intercepts. d. y = |2x2 – 12x + 16| + 2

y 2

First graph the function y = |2x – 12x + 16|. Then shift this graph 2 units up.

18

f(x)=|2x2 – 12x+16|+2

16

a = 2 > 0, so the parabola opens upward. h= 

y=|2x2 – 12x+16|

12 = 3, 22

k = f (h) = 2(3)2  12  3 +16 = 2 ; V(3, –2)is the vertex point.

x = 0 ; y = 16, (0, 16) is the y-intercept.

4

y = 0 ; 2x2 – 12x + 16 = 0 ;

2

x = 2, x = 4 ; (2, 0) and (4, 0) are the x-intercepts.

196

2 3 4 -2

x

Algebra 8

Check Yourself 6 Sketch the graph of each function. 1. y = |x2 – 6x + 5|

2.

y = –x|x| + 4

Answers 1. y

2.

y

5 4

4 2 1

3

5

x

x

C. EQUATION OF A PARABOLA We have learned how to construct a parabola if we are given its equation. But if we are given the graph of a parabola, how can we write its equation? There are three different approaches to finding the equation of a parabola. The approach we use depends on the information we know. 1. If we know the vertex point of the parabola, V(h, k) then we use the formula y = a(x – h)2 + k to write the equation of the parabola. We need to know another point on the parabola to write the equation. 2. If we know the x-intercepts of the parabola then we use the formula y = a(x – x1)(x – x2) to write the equation of the parabola. We need to know another point on the parabola to write the equation. 3. If we know any three points on the parabola, then we use the formula y = ax2 + bx + c to write the equation of the parabola. EXAMPLE

12

Write the equation of each parabola. a.

b.

y 2

3

x -3

Quadratic Functions

y

y=g(x)

4

y=f(x) 2

c.

y

y=h(x)

2 1

x

3

4

x

197

Solution

a. The vertex point of the parabola is V(2, 2), so y = a(x – 2)2 + 2. Also, the parabola passes through the origin (0, 0), so this point satisfies the equation of the parabola. 0 = a(0  2)2 + 2 ; 0 = 4 a+ 2 ;

a= 

1 2

1 y =  (x  2)2 + 2 2 1 y =  x2 + 2 x 2

This is the equation of the parabola. b. Let y = ax2 + bx + c. Since the points (0, 3), (–3, 3) and (1, 4) are on the graph of the parabola, these points satisfy the equation of the parabola. (0, 3), 3 = c

 b  b2  4ac 2a

(–3, 3), 3 = 9 a  3 b+ 3 ; (1, 4), 4 = a+ b+ 3 ;

9a 3b = 0 a+ b = 1

9a  3b = 0 (1)  (2)  a+ b = 1 From (1) and (2),

1 3 a= , b= . 4 4

So the equation is y =

1 2 3 x + x+3 . 4 4

c. Let y = a(x – 3)(x – 4) (0, 2) is on the parabola. 2 = a(0  3)(0  4) ;

a=

1 6

1 y = (x  3)( x  4) 6 So the equation is y = 198

1 2 7 x  x+2 . 6 6 Algebra 8

EXAMPLE

13

x1 and x2 are the roots of the parabola below. Find the value of x1 + x2. y y=g(x)

3

-6 x1

x2 2

x

-2

Solution

g(2) = g(–6) = 3

y B

A

3

C

y=g(x)

AB = BC, so the x-coordinate of the vertex point is h=

x2 2

-6 x1

x

-2

6 + 2 = 2. 2

h =

x + x2 b ; 1 = 2 ; 2a 2

x1 + x2 = 4.

T

EXAMPLE

14

Solution

A parabolic concrete bridge support needs to pass through the points (–50, 0), (0, 30), and (50, 0). Write the equation of the parabola formed by the bridge.

Let y = ax2 + bx + c. Since the points (–50, 0), (0, 30), and (50, 0) are on the graph of the parabola, these points satisfy the equation of the parabola. (0, 30), c = 30 (–50, 0), 0 = 2500a – 50b + 30 ;

250a – 5b = –3

(1)

(50, 0), 0 = 2500a + 50b + 30 ;

250a + 5b = –3

(2)

From (1) and (2), a=

–3 , b = 0. 250

So the equation is y = –

Quadratic Functions

3 2 x +30 . 250 199

EXAMPLE

15

The shape of the famous “Gateway to the West” arch, which is the spirit of the City of St. Lois can be modeled by a parabola. The equation for the parabola is y = –

1 2 21 x + x. 150 5

a. Sketch the graph of the arch's equation on a coordinate axis. b. What is the distance from one side of the arch to the other? Solution

a. a = –

1 < 0, so the parabola opens downward. 150 21 5

h= –

2 ( –

1 ) 150

y 661.5

V

= 315,

1 21 315 2 +  315 150 5 = 661 .5 V(315, 661.5)

k = f (315) = –

630 315

x

x = 0 ; y = 0 (0, 0) is the y-intercept. 1 2 21 x + x=0 ; 150 5 x = 0, x = 630 ; (0, 0) and (630, 0) are the x-intercepts. y=0 ; –

b. From the x-intercepts, we can see that the distance along the bridge is 630 m. EXAMPLE

16

Solution

A farmer has 120 m of fencing. He wants to put a fence around three sides of a rectangular plot of land, with the side of a barn forming the fourth side. Find the maximum area the farmer can enclose. What dimensions give this area? Let x represent the width of the plot. Then, since there are 120 m of fencing, x + x + length = 120 ; length = 120 – 2x. The area of the plot is A = (120 – 2x)  x = 120x – 2x2. We can see that a < 0, so the graph of the function of the area has its maximum value at the vertex point. To make the area as large as possible, let us therefore find the vertex of the graph of the function A = –2x2 + 120x. –b 120 =– = 30. This is the width of the plot at its maximum The x-coordinate of the vertex is 2a –4 area. Now we have the solution: A = –2  302 + 120  30 = 1800 m2 and the dimensions of the plot are 30 m by (120 – 2  30)m, i.e. 30 m by 60 m.

200

Algebra 8

EXAMPLE

17

Solution

Find the maximum vertical distance d between the parabola and the line in the figure. The distance between the vertex point of the parabola and the midpoint of the AB line segment in the figure will be the maximum distance.

y

b –4 = =1 So h = – 2 a –4

3

k = f (1) = 5 ,

5 4

f(x)=x – 2 d

2

V (1, 5).

B

1

Let us find the coordinates of point A and point B.

-2

-1

2

–2 x + 4x+ 3 = x – 2

-1

1

3

2

4

5

x

-2

2 x2 – 3x – 5 = 0 A

5 (2 x – 5)(x+1) = 0 ; x = , x = –1 2

-3 -4

5 1 A(–1, –3) and B( , ) 2 2

A

5 1 –1+ –3 + 2, 2 ) = ( 3 , –5 ) M( 2 2 4 4 So d = (

f(x)=–2x2+4x+3

M

B ( 5 , 1) 2 2

(–1, –3)

3 –5 1 625 626 626 – 1)2 +( – 5) 2 = + = = . 4 4 16 16 16 4

Check Yourself 7 Write the equation of each parabola. 1.

2.

y

y

3. 3

y

4

3

x

-1

3

2

-3

x

x

-2

-4

-1

Answers 1. y =

1 2 2 x – x–1 3 3

2. y = –

1 2 3 x + x–3 4 4

3. y = –

1 2 x – x+ 2 4

The things of this world cannot be made known without a knowledge of mathematics.

Quadratic Functions

201

P ARABOLIC R

EFLECTORS

Parabolic curves are used in the design of lighting systems, telescopes, and radar antennas, mainly because of the reflective property you can see in the figures below. axis

axis

F

Figure a . A reflecting telescope: light rays parallel to the axis are concentrated at the focus.

Figure b . A parabolic flashlight: a light source at the focus sends out beams of light parallel to the axis.

Figure a shows the application of the reflective property of a parabola to create a reflecting telescope. The eyepiece of the telescope is placed at the focus F of a parabolic mirror. Light enters the telescope in rays that are parallel to the axis of the parabola. We know from physics that when light is reflected, the angle of incidence equals the angle of reflection. Hence, the parallel rays of light strike the parabolic mirror so that they all reflect through the focus, which means that all the parallel rays are concentrated at the eyepiece. This maximizes the light-gathering ability of the mirror. Flashlights and automobile headlights (see Figure b) simply reverse this process. A light source is placed at the focus of a parabolic mirror. The light rays strike the mirror with an angle of incidence equal to the angle of reflection, and each ray is reflected along a path parallel to the axis. As a result, the light emits a light beam of parallel rays.

Radar utilizes both of these properties. First, a pulse is transmitted from the focus to a parabolic surface. As with a reflecting telescope, parallel pulses are transmitted in this way. The reflected pulses then strike the parabolic surface and are sent back to be received at the focus.

202

Algebra 8

5 .1

EXERCISES

6. The equation of the axis of symmetry of the

A. Graphing y = ax2 1. Graph each set of functions in the same plane. a. y = 2x2, y – 4x2 = 0, 2x2 = 3y

parabola y = 2x2 + (m + 1)x – 4 is 4x – 3 = 0. Find the minimum value of the function.

b. y = –3x2, y + 4x2 = 0, –3x2 = 2y c. y = 3x2, 3y = x2, 2y + x2 = 0, y + 2x2 = 0

7. – 2. Determine whether each point lies on the graph

4 is the minimum value of the function 3

y = 3x2 + 2x – n. Find the value of n.

of the function 4y + 5x2 = 0. 45   a.  3, –  4  

4  b.  , 1 5 

c. (–2, –5)

d. (4, 10)

8. The vertex point of the parabola y = –3x2 – 2(2k + 5)x + 4 is on the y-axis.

3. If the parabola y = px2 passes through the following

Find k.

points, find p. a. (–3, 3)

 4 b.  2,   5

c. (3, –7)

4  d.  , –2  5 

2

4. Graph the function y =  x if x  4. 2

9. Find the x- and and y-intercepts of each function. a. y = –x2 + 4x

b. y = (x – 3)2

c. y = 8(x + 7)2 + 4

d. y = x2 – 8

e. y = 2x2 + 5x + 3

f. y = –x2 – 4x – 5

g. y = 16x2 + 24x + 9 h. y = –

B. Graphing y = ax2 + bx + c

2 2 x –4 3

5. Find the coordinates of the vertex point of each function. Write the equation of the axis of symmetry and determine the maximum or minimum value of the function. a. y =

3 2 x 4

c. y = 3x2 + 2

d. y = –x2 – 1

e. y = –2x2 + 5x

f. y = 0, 7x2 + 0,8x

g. y = 3x2 – 4x + 3

h. y = –x2 + x + 1

i. y = 3(x + 1)(x – 4) j. y = –2(x + 4)2 k. y = –(x + 3) – 5 Quadratic Functions

Find the values of m for which the parabola a. does not cut the x-axis.

b. y = –3x2

2

10. The function y = –x2 + (m – 2)x + m – 3 is given.

2

l. y = (x – 1) + 3

b. is tangent to the x-axis. c. cuts the x axis at two distinct points.

11. The parabola y = mx2 – (m + 1)x + 2m – 1 passes through the point A(2, 9). Find the value of m. 203

12. Sketch the graph of each function.

22. Sketch the graph of each function by shifting.

a. y = 3x2 – 9

b. y = 2x2 + 5

a. y = x2 + 4

b. y = x2 – 2

c. y = –3(x + 4)2

d. y = 2(x + 7)2

c. y = –x2 – 2

d. y = –(x + 5)2

e. y = x2 – 2x – 3

f. y = –x2 – 4x

e. y = (x – 4)2

f. y = –(x – 3)2

g. y = x2 – 4x + 1

h. y = x2 – 5

g. y = –(x + 1)2 – 2

h. y = (x – 2)2 + 3

i. y = (x – 3)2 + 4

j. y = (x + 3)(4 – x)

13. Sketch the graph of the function y = x2 – 2x – 8 if –1 < x < 5.

23. Sketch the graph of each function. a. y = |x2 – 4x + 3|

b. y = –2x2 + 4|x| + 1

c. y = x|x – 2|

d. y = |x2 – 6x + 8| – 1

14. Sketch the graph of the function y = –2x2 + 4x + 1 if |x – 1| < 2.

C. Equation of a Parabola 15. Find the minimum and the maximum values of the function y = x + 4x + 7 if x  [–4, 1]. 2

24. Write the equation of each parabola. a.

b.

y

y

3 2

16. Find the minimum or the maximum value of the

1

function if m + n = 16.

5

x

1

6 x

17. Find the minimum or the maximum value of the function y = m2 + n if m – 2 = n.

c.

d.

y

y 4 3

3

18. The parabola y = 2ax + bx – 3b passes through 2

the points (1, –3) and (2, –5). Find the values of a and b.

19. Write the equation of the parabola which is symmetric to y = 3(x + 1)2 with respect to the line x = 2.

20. Given f(x) = –3(x – 1)2 + 2, show that f(x)  f(1).

-3

e.

x

f.

y 3

x

-1

y

5

3

x -3

-3

3

x

21. f(x) = x2 + 3mx + 4m and f(x)  0 are given. Find m. 204

-8

Algebra 8

25. Write the equation of the parabola through each

30. The figure shows the graph 

set of points.

y

of the function

a. A(–1, 0),

B(0, 4),

C(1, 2)

y = mx2 – 7mx + 5.

b. A(0, –4),

B(2, 0),

C(4, –4)

If |AB| = 3, find m.

A

B

x

26. Write the equation of the parabola through each vertex point V and point A. a. V(0, –4), A(3, 5)

b. V(2, 0), A(4, 4)

31. Find the area of the

 1 15  c. V(–1, –4), A(2, 5) d. V   ,   , A(0, –4) 4   2



y

trapezoid OABC in the figure.

5 2

y=–2x2+6x C

B A

O

27. In the figure,

y

3  |OB| = |AO|. Find c.

y=–x2 – 4x – c A

O

B x

32. Find the area of the 

y

|OB| = 5  |AO| and V is the vertex point of the parabola. Find k.

k

C

V

O

A

A

x

2

B O

x

2



29. The figure shows the

y

graph of the function y = ax2 + bx + c. Find the value of

B

2

33. Find the area 

y

square OABC in the figure. 24

28. In the figure, 

-5

1

x

y

of the rectangle ABCD in the figure.

D

C y=x2 – 6x+5

O

A

B

x

a + b + c. Quadratic Functions

x

205

39. The figure shows the

Mixed Problems



34. x1 and x2 are the x-intercepts of the function 

y = mx2 – 4x – 2m – 2. If x1  x1 < 0, find the possible values of m.

y

graph of the function y = –x2 + bx + 1. Find the value of a + b + c + d.

y=–x2+bx+1

5 c

a

2

x

d

35. The perimeter of a rectangle is 32 cm. Find its 

maximum possible area.

40. In the figure, 36. The function y = 5x – 3x + k is tangent to the 2





line y = 2. Find k.

y

the area of the triangle ABC is 6 cm2. Find k.

y=k(x2+5x+4) C A

B O

x

37. 3 is the minimum value of the function 

y = mx2 – 2mx + 2m + 1. Find m.

38. The figure shows the

y

graph of the function

6



41. In the figure, the 

y = –x2 + bx + c. of the point A.

2 A

206

B

maximum value of the 49 function is . 4 Find the area of the

Find the coordinates x

y C A -2

B O

5

x

triangle ABC. Algebra 8

An equation is a statement that says two expressions are equal. For example, a = b is an equation. An inequality is a statement that says two expressions may or may not be equal. For example, a > b, a  b, and a < b are all inequalities. An equation usually has a finite number of solutions, but an inequality may have an infinite number of solutions. We can show the solutions of an inequality as an interval and as a graph. Look at the following examples of intervals and their graphs. Interval

Inequality

Graph

the interval (–, a)

xa

a

the interval [a, )

xa

a

For any two real numbers a and b, exactly one of the following is true:

the open interval (a, b)

a –2b. The order of the inequality is reversed. EXAMPLE

2

Solution

Solve the inequality 2x + 3  5x and graph this solution. 2x + 3  5x –3x  –3 1x

1

Therefore, x  [1, ), or

EXAMPLE

3

Solution

.

Solve the inequality 2 < 3x – 1  8 and graph this solution. 2 < 3x – 1  8 3 < 3x  9 1 0, the following statements are true.

Property

1. |x|  k means –k  x  k. 2. |x| > k means x < –k or x > k. EXAMPLE

4

Solution

Solve the inequality |2x – 1| < 5 and graph this solution. |2x – 1| < 5 –5 < 2x – 1 < 5 –4 < 2x < 6 –2 < x < 3 Therefore, x  (–2, 3), or

Quadratic Inequalities

–2

3

. 209

Check Yourself 1 Solve the inequalities. 1. –x + 7 > 2x + 1 2. 11  x+ 3  17 4 3. |2x – 1| + 2  7 Answers 1. x < 2 2. –47  x  65 3. –2  x  3 FIND THE MISTAKE! a>4 4a > 16 4a – a2 > 16 – a2 a(4 – a) > (4 – a)(4 + a) a>4+a 0>4 Can you find the mistake in this working?

Sign Chart In general, to solve a linear inequality such as ax + b > 0 or ax + b  0 we need to know the sign of the polynomial ax + b, a  0. Look at the steps. First we find the zero of the polynomial: b ax+ b = 0 ; x =  . a

Then we construct a sign chart. x ax+b

x0= –

–¥ ax+b has sign opposite to a

b a

¥ ax+b has the same sign as a

This sign chart shows us: b  if x    ,   , the sign of the polynomial is opposite to the sign of a, a    b  if x    ,   , the sign of the polynomial is the same as the sign of a.  a  210

Algebra 8

EXAMPLE

5

Solution

Solve the inequality 3x – 2  0. 2 Find the zero: 3 x  2 = 0 ; x = . 3 Draw the sign chart: x

x0=

–¥

2 3

–

3x – 2

¥

+

2  If x   ,  , then 3 x2 is negative. 3  2  If x   ,   , then 3 x2 is positive. 3  2  2  Since 3 x – 2  0, x is in the interval  ,   , i. e. x   ,  . 3 3     EXAMPLE

6

Solution

Solve the inequality 6 – 2x  0 by using a sign chart. 6 – 2x = 0 ; x = 3 x 6 – 2x

–¥

3

+

¥

–

Since 6 – 2x  0, the sign is positive. So x  (–, 3].

Check Yourself 2 Solve the inequalities by using a sign chart. x+ 14 x  12 1. 5 – 2x < 0 2.  3 6 8 Answers 5 1. x > 2. x  –20 3. x  –10 2

3. 6x + (x – 2)(x + 2)  (x + 4)2

Mathematics, of course, is not the only cornerstone of opportunity in today’s world. Reading is even more fundamental as a basis for learning and for life. What is different today is the great increase in the importance of mathematics to so many areas of education, citizenship, and careers.

Quadratic Inequalities

211

A rectangular room has dimensions 30 m  12 m  12 m. A spider is in the horizontal center of one end wall, one unit away from the ceiling. A fly is in the horizontal center of the opposite wall, one unit away from the floor. What is the shortest distance the spider needs to travel (without leaving a surface) to get to the fly (which remains stationary)?

EXERCISES

6 .1

1. Write each inequality using interval notation. a. x > –2

b. x  6

c. x < –7

d. x  4

e. 2  x < 4

f. –1 < x  0

g. –5 < x <

1 2

h. 0 ,5  x 

a. –2x + 3 < 4

3 2

2. Determine the sign of each polynomial. a. 4x + 1

b. –3x – 5

x 7 c.  + 2 3

d. ñ3x – 6

212

3. Solve and graph the inequalities.

b.

x x x x +   2 3 4 5

c. 

3x x + 5 < 1+ 2 2

d. 

x+ 3 x  2 2x   4 2 3

e. (3x + 1)2 – (x + 2)(4x – 1) > 5(x – 1)2 + 6x f. (a + 3)x – 5  1

(a > –3)

g. (a4 + 4)x + 3 > 0 (a  ) Algebra 8

Definition

quadratic inequality A quadratic inequality is an inequality that can be written in one of the forms ax2 + bx + c > 0,

ax2 + bx + c < 0,

ax2 + bx + c  0,

ax2 + bx + c  0,

for the real numbers a, b, and c, a  0.

We will never be equals.

We have seen how to solve linear inequalities such as 2(x + 3) + 3  5(x – 1). But how do we solve quadratic inequalities such as x2 – 5x  –6? First we write the inequality in standard form, leaving only zero on the right side: x2 – 5x + 6  0.

In this example, we are looking for values of x that will make the quadratic on the left side greater than or equal to zero.

Note If  and  are used in the inequality, then remember that the zeros of the polynomial are included in the solution set. First, let us find the zeros of the polynomial x2 – 5x + 6: (x – 2)(x – 3) = 0 positivepositive=positive positivenegative=negative negativenegative=positive

x = 2 or x = 3. Then we construct a sign chart for each linear factor of the polynomial, and their product. x

–¥

3

2

¥

x–2

–

+

+

x–3

–

–

+

+

–

+

(x – 2)(x – 3)

Since x2 – 5x + 6  0, we need to take the positive intervals. Therefore, the solution is the union of intervals (–, 2] and [3, –), i.e. x  (–, 2]  [3, ). Quadratic Inequalities

213

We can also construct the sign chart in one step. The zeros of the polynomial divide the real number line into three intervals, (–, 2], [2, 3] and [3, ). We know that the polynomial has constant sign in each of these three intervals. If we select a test number in each interval and evaluate the polynomial at that number, then the sign of the polynomial at this test number must be the sign for the whole interval. Let us try testing each interval in our problem. Choose a number from each interval, and substitute for x in the original inequality. For example, we could choose the numbers 1, 2.5, and 4. Test number

1

2.5

4

Value of the polynomial

2

–0.25

5

Sign of the polynomial

+

–

+

With this information, we can draw the sign chart. x x2

–¥

3

2

+

– 5x + 6

–

¥

+

This is the same as the last line of the previous chart. We can also use the discriminant of a quadratic to help complete the sign chart. We have already seen that the discriminant  of a quadratic equation ax2 + bx + c = 0 tells us it the equation has two real roots ( > 0), one double root ( = 0), or no real solutions ( < 0). If  < 0, then the polynomial ax2 + bx + c always has the same sign as a. x

–¥

¥

ax2 + bx + c

same sign as a

If  = 0, then of the polynomial ax2 + bx + c has the same sign as a but we must consider the zero of the polynomial. x

x1=x2= –

–¥

ax2 + bx + c

b 2a

same sign as a

¥ same sign as a

If  > 0, the polynomial ax2 + bx + c has the opposite sign to a between the zeros of the polynomial and the same sign as a in other intervals. x ax2 + bx + c

214

–¥

x2

x1 same sign as a

opposite sign to a

¥ same sign as a

Algebra 8

EXAMPLE

7

Solution

Solve the inequality –x2 – 4 < –5x. –x2 – 4 < –5x –x2 + 5x – 4 < 0 (–x + 4)(x – 1) = 0 x = 4 or x = 1 x

–¥

–x2 + 5x – 4

4

1

–

¥

+

–

Therefore, x  (–, 1)  (4, ). EXAMPLE

8

Solve the inequalities. a. 3x + 4  x2

Solution

c. 9x2 – 12x + 4  0

b. x(x + 2) > 35

d. x2 

2x  1 0, there are two real roots, and m  (–, –2)  (0, ).

EXAMPLE

10

Solve the inequalities. a. (3 – x)(x3 – 2x2 – 8x)(x2 + 3) < 0

216

b.

(3x+ 2)(x  5) 0 x(x – 1)(x2 + x +1)

c.

(x2  x  6)(x  1)17 0 x8 (1  x2 )55 Algebra 8

Solution

a. First we find all the zeros of the polynomials, then we determine the sign for each polynomial and multiply the signs of each polynomial. (3 – x)(x3 – 2x2 – 8x)(x2 + 3) = (3 – x)x(x2 – 2x – 8)(x2 + 3) 3–x=0; x=3 x=0 x2 – 2x – 8 = 0 ; x = –2 or x = 4 x2 + 3 = 0 ; no real solution. x

(3 –

x)x(x2

–¥

–2

0

3

4

¥

3–x

+

+

+

–

–

x

–

–

+

+

+

x2 – 2x – 8

+

–

–

–

+

x2 + 3

+

+

+

+

+

–

+

–

+

–

– 2x –

8)(x2

+ 3)

We need a value less than zero, so x  (–, –2)  (0, 3)  (4, ).

positive = positive positive positive = negative negative negative = positive negative

b. First we find all the zeros of the polynomials. The equality part of the original inequality is satisfied for these zeros and they must be included in the final solution set. On the other hand, since division by zero is never allowed, the zeros of x4 – x must not be included in the solution set. (3x+ 2)(x  5) x(x  1)(x2 + x+ 1) 2 3x+ 2 = 0 ; x =  3 x5=0 ; x=5 x=0 x  1= 0 ; x = 1 x2 + x+ 1 = 0; no real solution. x

–

–¥

2 3

0

1

5

¥

3x + 2

–

+

+

+

+

x–5

–

–

–

–

+

x

–

–

+

+

+

x–1

–

–

–

+

+

+x+1

+

+

+

+

+

(3x + 2)(x – 5)

+

–

+

–

+

x2

x4 – x

2  We need a value greater than or equal to zero, so x    ,    (0 , 1)  [5 , ). 3  Quadratic Inequalities

217

c.

(x2  x  6)(x  1)17 0 x8 (1  x2 )55

x2 – x – 6 = 0 ; x = –2 or x = 3 (x – 1)17 = 0 ; x – 1 = 0 ; x = 1 x8 = 0 ; x = 0 (double root) (1 – x2)55 = 0 ; 1 – x2 = 0 ; x = 1 or x = –1 (1 is also a double root) If the power is an odd number, you can ignore it when you calculate the zero. If the power is an even number, consider it just as 2 when you calculate the zero.

x x2 – x – 6

–¥

–2

–1

0

1

3

¥

+

–

–

–

–

+

(x – 1)17

–

–

–

–

+

+

x8

+

+

+

+

+

+

(1 – x2)55

–

–

+

+

–

–

+

–

+

+

+

–

(x2 – x – 6)(x – 1)17 x8(1 – x2)55

We need a value less than or equal to zero, so x  [–2, –1)  [3, ). The signs of M  N and M , N  0 are the same. N

Let us summarize the key steps to solving any inequality. 1. Write the polynomial inequality in standard form. 2. Find all zeros of the polynomial(s). 3. Determine the character of the roots. 4. Determine the sign of the coefficient of leading term of the polynomial(s). 5. Construct a sign chart. 6. In the sign chart, from right to left start with the sign of the coefficient of the leading term a. After each root change the sign, but if there is a double root do not change the sign. Let us solve Example 10c in another way.

EXAMPLE

11

Solution

Solve the inequality

(x2  x  6)(x  1)17  0. x8 (1  x2 )55

x2 – x – 6 = 0 ; x = –2 or x = 3 (x – 1)17 = 0 ; x – 1 = 0 ; x = 1 x8 = 0 ; x = 0 (double root) (1 – x2)55 = 0 ; 1 – x2 = 0 ; x = 1 or x = –1 (1 is also a double root)

218

Algebra 8

x (x2

– x – 6)(x –

–¥

1)17

–2

–1

+

x8(1 – x2)55

0

–

1

+

3

+

¥

+

–

Therefore, x  [–2, –1)  [3, ). EXAMPLE

12

Solution

Solve the inequality (x2 – 2x – 8)(x2 + x – 12)  0. Find the zeros of the polynomials. x2 – 2x – 8 = 0 ; x = 4 or x = –2 x2 + x – 12 = 0 ; x = –4 or x = 3 x

–¥

(x2 – 2x – 8)(x2 + x – 12)

–4

+

–2

3

–

4

+

¥

–

+

So, x  [–4, –2]  [3, 4]. EXAMPLE

13

Solution

Find the domain of the function y =

x+ 2 3x  12 x2

.

3x – 12x2 > 0 3x – 12x2 = 0 x = 0 or x =

1 4 x

3x –

–¥

1 4

0

12x2

–

¥

+

–

 1 So x   0 ,  .  4 EXAMPLE

14

Solution

Solve the inequality (x – 1)2(x – 2)3(x – 3)4(x – 4)5  0. (x – 1)2 = 0 ; x – 1 = 0 ; x = 1 (double root) (x – 2)3 = 0 ; x – 2 = 0 ; x = 2 (x – 3)4 = 0 ; x – 3 = 0 ; x = 3 (double root) (x – 4)5 = 0 ; x – 4 = 0 ; x = 4 x

–¥

(x – 1)2(x – 2)3(x – 3)4(x – 4)5

1

+

2

+

3

–

4

–

¥

+

So x  (–, 2]  {3}  [4, ]. Quadratic Inequalities

219

EXAMPLE

15

Solution

x4  3x3 + 2x2 < 0. x3  5x2

Solve the inequality x4  3x3 + 2x2 0 x2 – 7x – 8 = 0 ; x = –1, x = 8 (2) x2 – 4x + 3 > 0 x2 – 4x + 3 = 0 ; x = 1, x = 3 We need to find values so that both polynomials are greater than zero. Let us check the chart. x

–¥

–1

1

3

8

¥

(1)

+

–

–

–

+

(2)

+

+

–

+

+

system

We can see that both polynomials are greater than zero when x  (–, –1)  (8, ). This is the intersection of the solutions. EXAMPLE

22

Solution

 x2 + x  4 x2 (2)

(1) x < 0

(1) x  0

EXAMPLE

If

f ( x)  g( x) then

x2 + x  2 > x.

2

(2) x + x – 2 = 0 ;

 g( x) < 0   f ( x)  0

(2) x – 2 > 0 ;

x1 = –2, x2 = 1

x – 2 = 0, x = 2

or  g( x)  0   f ( x)  g2 (x)

x

Case 1

–¥

–2

0

1

2

¥

(1)

–

–

+

+

+

(2)

+

–

–

+

+

(1)

–

–

+

+

+

(2)

–

–

–

–

+

system

Case 2

system

So x  (–, –2]  (2, ).

EXAMPLE

26

Solution

Solve the inequality (x2 + x + 1)2 – 4(x2 + x + 1) + 3 < 0. For the inequality, we let t = x2 + x + 1. Then the original inequality becomes t2 – 4t + 3 < 0. First let us solve the inequality for t. t2 – 4t + 3 = 0 ; t = 1, t = 3 x

–¥

t2 – 4t + 3

3

1

+

–

¥

+

So 1 < t < 3. Now solve for x. 1 < x2 + x + 1 < 3 0 < x2 + x < 2, which gives us the system of inequalities  x2 + x > 0 (1) .   x2 + x  2 < 0 (2) Quadratic Inequalities

227

(1) x2 + x > 0 x2 + x = 0 ; x = 0, x = –1 (2) x2 + x – 2 < 0 x2 + x – 2 = 0 ; x = –2, x = 1 x

–¥

–2

–1

0

1

¥

(1)

+

+

–

+

+

(2)

+

–

–

–

+

system

So x  (–2, –1)  (0, 1).

27

Solve the inequality |x2 – 3x + 2|  2x – x2.

Solution

Case 1 ________________________________________

EXAMPLE

 x2  3x+ 2  0 (1)   x2  3x+ 2  2 x  x2 (2)

(1) x2 – 3x + 2  0 x2 – 3x + 2 = 0 ; x = 1, x = 2 (2) x2 – 3x + 2  2x – x2 ; 2x2 – 5x + 2  0 1 2x2 – 5x + 2 = 0 ; x = , x = 2 2 Case 2 _______________________________________  x2  3x+ 2 < 0 (1)    (x2  3x+ 2)  2 x  x2 (2)

(1) x2 – 3x + 2 < 0 x2 – 3x + 2 = 0 ; x = 1, x = 2 (2) –(x2 – 3x + 2)  2x – x2 –x2 + 3x – 2  2x – x2 x–20 x–2=0; x=2 228

Algebra 8

x

Case 1

1 2

–¥

1

2

¥

(1)

+

+

–

+

(2)

+

–

–

+

(1)

+

+

–

+

(2)

–

–

–

+

system

Case 2

system

1  So x   , 1  (1, 2)  {2}, i.e. 2  1  x   , 2 . 2 

Check Yourself 4 Solve the systems.  x2 – 4x  0 1.   x – 3  0

 x2  4x+ 3 > 0 2.   x2 (x  7)2 > 0

3.

x2  x > 1+ x

Answers  1. [4, ) 2. x  0, x  7, (–, 1)  (3, ) 3. ( ,  ) 3

In the center of a square pond whose side measures 10 m grows a plant whose top reaches 1 m above the water level. If we pull the plant toward the bank, its top becomes even with the waters surface. What is the depth of the pond and the length of the plant?

Do not worry about your difficulties in mathematics. I can assure you that mine are still greater. Einstein Quadratic Inequalities

229

EXERCISES

6 .3

1. Solve the inequality systems.

4. Solve the inequalities.

 x+ 2 > 0 a.  2 x  3 < 0

 x  1 > 0 b.   x2  2 x  3 < 0

 x2 + 3x  10 > 0 c.  2 x2 + 11x  6 < 0

x  2  x+ 1  0  d.   x2  1  x  3 > 0 2 x2 + 2 < 5x   x2  x

 x2  4x+ 3 < 0 e.  2 x  4 < 0

f.

 x+ 2  x+ 1 > 1  g.   x  x  2 > 2

x 1 0 x+ 3

e.

x+ 1  x  2  1

f.

x+2

x2 – x – 7

d. |x – 4| + |2x + 6| > 10

b. 2x2 + mx2 – 2x + m – 1 > (1 + m)x2 – 4x – m

e.

x3 x2  5x+ 6

2

f. |2x + 1| – |5x – 2|  1 g. 230

x2  x  12 x3

c. (m + 2)x2 – 3x + m – 2 < 0

 2x

d. x2 – 6x + 4m2 > 7

9. Solve the inequality



(x2 + 6x + 14)2 – 9  (x2 + 6x + 15) + 9 < 0. Algebra 8

CHAPTER REVIEW TEST

6A

1. What is the solution of the inequality

5. What is the solution of the inequality

2 – 3(1 – x) < x + 1.8? A) x > 1.4

B) x < 0.4

D) x > –0.7

C) x > 0.7

E) x < 1.4

1 1 < ? x+ 2 x  1

A) x < 0 or x > 2

B) –2 < x < 1

C) x < –2 or x > 1

D) 0 < x < 2

E) –2 < x < 2

2. What is the solution of the inequality x2 + 5x – 14  0? A) [–2, 7]

B) [–7, 2]

D) [2, 7)

C) (–7, 2)

E) (–2, 7]

6. What is the greatest integer value of x which satisfies the inequality x2 – 13x + 36 < 0? A) 13

3. What is the solution of the inequality

B) 7

C) –8

D) 8

E) –13

x2  4 > 0? x2  1

A) (–, –2)  (1, )

B) (–, –2)  (2, )

C) (–, –1)  (1, )

D)  – {–1, 1}

E) (–, –2)  (–1, 1)  (2, )

7. What is the solution of the inequality 1 1 1 + + < 0? x x 2 x3

4. What is the solution of the inequality

A) (–, 0)

B) (0, )

C) 

D) 

E) {0}

(2  3x)(3 x  3)  0? 3x +1  1 2 A)   ,   [1, )  3 3

1  2   B)  ;     ; 1 3  3  

2  C)    (1, + ) 3

 1 2 D)   ,   (1, + )  3 3

1  2   E)  ;     ; 1 3 3   Chapter Review Test 6A

8. How many integer values of x are there which satisfy the inequality |x + 2| 

2 x +7 ?

A) 1

D) 5

B) 3

C) 4

E) 7 231

9. What is the solution of the inequality

13. What is the domain of the function

(1+ x2 )2001  (1  x)2002  0? ( x + 2)2003  (3  x)2005

A) x > –2

y=

B) x < –2

D) x > 3

C) x < 3

E) –2 < x < 3

x2

?

20  8 x  x2

A) [–1, 7]

B) [–10, 2]

D) [5, 4)

C) (–10, 2)

E) (–2, 10]

14. What is the solution of the inequality system 10. The equation 2x(ax – 4) – x + 6 = 0 has no real 2

roots. What is the smallest possible integer value of a? A) –1

B) 2

C) 3

D) 4

E) 5

11. What is the solution of the inequality

B) (–2, –1)

D) (–5, –1)  (1, )

A) (2, 3)

C) (1, 3)

E) (–1, 1)  (5, )

D) [3, 2)

12. What is the sum of the integer values of x which x2  8x +7 < 0? ( x + 2)2

A) 32

C) 24

232

B) 28

D) 20

C) (–2, 3)

E) (–3, –2)

 x < 2 ?   ( x  3)( x  2)  0

A) (–2, 2) D) [3, 2)

satisfy the inequality

B) [2, 3]

15. What is the solution of the inequality system

( x  3)( x + 2) 1?

A) (–2, 2) E) 16

D) (–, 2)

B) (1, )

C) (–2, –3) E) (2, ) Algebra 8

6B

CHAPTER REVIEW TEST 1. What is the solution of the inequality

5. What is the solution of the inequality

x +1 x  1  > x? 2 3

A) x > 1

x2 > x +1?

B) x < –3 D) x < 1

C) x < 3

1  B)  1,   2 

A) (–2, 4)

E) x > 6

C) (–, 2)

1  E)  ,   2 

D) [2, )

2. What is the solution of the inequality (x – 6)(x + 3)  2 – 2x? A) [–5, 4]

6. What is the smallest possible integer value of x

B) [–4, 5]

D) [–5, 4)

C) (–4, 2)

E) (–4, 5]

3. What is the solution of the inequality x2 + x 0 x+ 3 x+ 7 4 is given. x1 x2

What are the possible values of m? A) (–, )

E) 16

D) (3, 12)

B) [2, ]

C) (–, –3) E) (0, 12) Algebra 8

EXERCISES

1 .1

1. a. 6 b. 10 c. –11 d. 4x e. 5y f. 11a2 c. 2 d. 3x e.

y 2 3xy f. 3 2x

2. a. 3 b. 5 c. 6 d. 9 e. 4x f. 12xy g. aò15 h. 72xy

4. a. 2ñ2 b. 6ñ2 c. 9ñ3 d. 10ò10 e. 5ñ5 f. xyñx 5. a. 5ñ3 b. 7ñ5 c. 0 d. –2ñ6

e. 21ñ2 f. 2ñ5 g. 10ñ3 h. –13xñx i. 1 j.

4 3

6. a. ñ2 + 1 b. ñ3 – ñ2 c. ñ2 + ñ6 d. 2ñ2 + ñ3 e. ñ7 – 1

f. ñ3 + ñ2 g. 2 – ñ3 h. 2ñ2 i. ñ6 j. ñ2 k. ñ3 + 1

7. a. 6 b. 2 c. 4 d. 4

6 3 12

– 11 d. 11

11. a.

2 33 2 6

e. –1 – ñ2 f. 2ñ3 – 4 g. 2 + ñ3 h. 22 5 3 –1 c. 6ñ3 + 7ò11 d. 5 2

b.

EXERCISES 1

1. a. 212 b. 7 c. 5 7 d. xa 3

1

6

e. 215 f. 3 5 g. 3

c. 2 d.

5 4

e. ñ2 + 1 f. –2

l. 7 m. 9 n. 9 o. ñ6 + 1 p. 16 q. –1 r. 7. a. 4 b.

1 5

d. 2x e. 81 f. 13. ax+2y

a c. a d. ( )x 1 e. –2 f. 5 b 89 30

14. –

1 4

g. 1 h. 18 15. –2

28. 57122 – 57121

Answers to Exercises

j. 3 – ñ6 k. ñ5 + 1 l. 1

12. a. ñ2 b. 1 c. 9

–

1 3

2 c. ( )3 3

2. a. ña b. 3 b2

d. c xab

e. 4 a

18 7

j. 2 k. 3 l. 9 m.

1 5

n. 12 3 o.

16. 128

3. a. 6 3

b. 10 a

3

s. 2 6. a. 27 b. 55 c. 46 d. 24 e. 32x+1 f. 30x g. y – x h. 2100

8. a. 256 b. –729 c. –4 d. 64 e. a9 f. a

9. a. 2x+1 b. 0 c. 110  3x – 2

4 10. a. 6 b. 8 c. 5 d. 9 e. 4 f. { , 4} g. 3 h. 4 i. 3 j. –3 3

17. 2

18. x – 2

19. –4

20.

22 l. 0 m. ò14 22. 15 23. 0 3 19 29. 30. 2651 31. –121 8

f. 20 g. 2ñ7 h. 2ñ3 i. 2ñ2 j. 0 k. 27. 9

2 70  5 10 3

3 2 2 3 6

27 p. 24 313 q. 24 231 4. a. 1 5 39 17 5. a. 4 b. 26 c. 0 d. 54 e. 44 f. 17 g. h. i. 2 j. {–4, 4} k. 10 2 5

c. 100 d. ñ5 e. 3 f. 5 g. –3 h. 2x i. a2  6 a b. 6 24  33

i.

10. a. ñ3 b.

1 .2 2

1

12 3 – 8 23

9. a. ò15 + ò10

8. a. 2 b. 9 c. 24

b. 7 + ñ7 c. 4 – ñ6 – ñ2 d. 12 e. 6(ñ3 + ñ2) f. 4 g. –1 h. 8 i. 2 j. –2 k. 1 l. ñ3 m. ò13 c.

3. a. 5 b. 3

16 5

24. 3

11. 2

12. 268

21. a. 7 b. 8ñ3 c. 2.8 d. ñ5 e. 3 25. 3ñ2

26. 8 –

2 2

235

EXERCISES

2.

frequency (f)

1 2 1 3 3 5 3 3 1 2

3.

Percent

judo karate 10% 5%

35 30

wrestling 7%

25

football 20% basketball 12%

20 15

swiming 29%

10

volleyball 17%

Other

Expenses

5. a. 2705000 tons b. 2001 c. 1080000 tons

Other

EXERCISES

5 d. 5 e.  2 5 3 5

b. 0,

500 000 400 000 300 000 200 000 100 000

3 .1

1. a. 0 b. 0 c. 0 d. 0 2. a. 0,  c. 

600 000

2003

Pichard Whiting Hake

Hazelnut production in Turkey (tons)

2002

Blue fish

700 000

2001

d.

Scad Gray mullet

2000

Anchocy Horse mackerel

1999

4.

Books

0

Clothing

5 Food

1 2 3 4 5 6 7 8 9 10

Entertaiment

scores (x)

Rent

1.

2 .1

f. 

5 8 3 1 b. 0, c. 0, d. 0, 2 3 7 9 4 5

3. a. 0, –2 4. 0, 

5 2

5. –1, 2 6. a. 4 b. no real solution

7. a. –1,5 b. no real solution c. (–1 – ñ5), (–1 + ñ5) d. –2, 4 8. ñ3

5 3 1 2 2 3 1 c. 7 d. 5 e. –1, 2 f. –1, –2 g. 1 h.  , i. , j.  , 3 2 4 3 5 2 3

9. a. 0, 1

10. a. –a, –1 b. no real solution c. 

5b ,1 2a

11. a. (2 + ñ5), (2 – ñ5), b. (ñ7 – 2), (–ñ7 – 2) c. (3  ò22) d. no real solution e. no real solution f. no real solution 3 1 g. 3  2ñ2 h. –1  7 12. a. 2  ñ2 b. 1, c. x   d. 5  13 e. no real root f. 1  2 7 g. no real root 2 2 6 3 6 4 1 7 8 2 , 1 l. –1, 23 14. a. two real roots h. x   i. , 4 j.  , 1 k. 13. a. 1  33 b. 3  73 c. 1, 5 d. , 5 5 26 4 5 3 8 4 9 b. two real roots c. one double root d. one double root e. no real solution f. no real solution 15. a. a  40 9 7 7 7 9 1 b. a  c. a  16. a. m   b. m   c. m   17. – , 1 18. 8 cm, 10 cm 40 2 2 2 40 4

19. 12 cm, 20 cm 236

20. 4, 6, 8, 10

21. 256 cm2, 441 cm2

22. 7, 49 Answers to Exercises

EXERCISES

72 40 4 5 11 6 10 5 18 8 24 c. e.  , f. (3ñ2 – 2), 0 2. ,  b.  ,  ,  d. ,  49 21 3 6 10 5 3 3 7 7 7

1. a.

153 256

d. 10.

3 .2

5. m1 = –5, m2 = 4, x2  

14 ,4 3

19. 7

11. 5

20. c = 2, 1,

7 5 12.  , 3 3

1 5 2

EXERCISES

5 4

6. n1 = –4, n2 = 2, x2  

13. –6, 2 21. –13

14. 1

15. k > 6

22.  1 , 1  13 2 2

2 3

7. 0

16. 6, 7

3. 

2 5

4. a.

8. m = 2n 17. a  b

17 21 b. 53 c.  9 4

9. k = 3

18. 0

23. 4 + 2ñ3 cm

3 .3

1. a. x2 – 1 = 0 b. 4x2 – 8x + 3 = 0 c. x2 – 4x = 0 d. x2 – 4x + 2 = 0 e. x2 – 2ñ3x + 1 = 0 f. p2q2x2 – (p2+q2)x+1=0 2. x2 + 2x + 5 = 0 3. x2 – 8x – 12 = 0 4. x2 + x + 27 = 0 5. a. x2 – 4 = 0 b. x2 + 8x + 7 = 0 c. 4x2 – 25 = 0 d. x2 – 2x + 1 = 0 e. x2 + 2mx – 8m2 + 27m – 18 = 0 f. mx2 – 2mx – 8m – 9 = 0

6. 5

4 13. 18(ñ2 + 1) 7 14. from A to B 60 km/h, from B to A 80 km/h 15. 160 km/h 16. 16 km/h 17. 7 women 18. 80 km/h 19. 24 hours, 48 hours 20. 20 km/h 21. height = 6 m, base = 12 m 22. 40 kg 25%, 20 kg 40% 23. 60 hours, 84 hours 24. 4 hours, 8 hours 25. 6 m, 8 m

7. –7, –6, –5 or 5, 6, 7 8. 9 9. 15 hours, 10 hours 10. 3 hours, 2 hours 11. 6 hours 12.

EXERCISES

3 .4

1. a. 2, 3 b. ñ3 c. ò10, 2ñ3 d. –5 ñ7 e. –5

1 3 5 f. 79 g. 4 h. 2 ñ3 i. 2 ñ2 2. a. b. (1 ò10), –1, 3 2 2

c. –2, –1, 0, 1 d. 1, 2 e. –6, 1

7 6  42 1 3. a. 0,  , 1, 5 b. –10, (–4  ñ2), 2 c. 2 d. 5 e. f.  5, , 1 g. no solution 3 3 4

h. 1, 2, 4 i. 0 j. –3 ò15

4. a. 2 b.

11 – 21 c. –7 d. 1, 2 e. 12 f. 3 g. –1, 2 h. 0 i. –1, 7 j. no solution 2

8 5. a. 7, 8 b. 2 c. 0 d. 0, 2 e. 0 f. no solution g. – , 1 h. 2 3

g. 3 – ñ7, 2 h. 2 i. no solution d. –1, 4 Answers to Exercises

7. a. 0 b. 

6. a. –1 b.

7 , 3 c. 2 d. –1, 5 e. –1, 3 f. no solution 3

2 1 3 3 c. –2, 0 d. 2 e. – , , 2 f. x  (– , ] 8. a. 1 b. 0, 3 c. 3 2 2 4 2

1

237

EXERCISES

3 .5

5 7 1. a. (–6, –18), (18, 6) b. ( , – ) c. (–3, 6), (10, –7) d. (10, 15), (15, 10) e. (4, –1), (–4, 1) f. (3, 3), (–3, –3) 2 2 g. (–3, –4) h. (0, –5), (1, –4) 3 2. a. (–4, –4), (–6, –2) b. (0, 0), ( , 3) c. (0, 0) d. (1, 0) e. (2, 4), (4, 2), (ò22 – 4, –ò22 – 4), (–ò22 – 4, ò22 – 4) 2 7 7 7 ,– )(– , 2 2 2

f. (0, –2), (2, 0) g. (0, ñ2), (0, –ñ2) h. (2, 1), (–2, –1), (

EXERCISES 1. a. x = –1, y = 4

4 .1

b. x = 1, y = 0

2. a. {3}

b. 1

3. a. (–1, 1], (–3, 2) b. (2, 5), (–3, )

e. {3} f. 2

7 ) 2

c. {1, 3, 5, 7, 8, 9, 10}

d. {1, 2, 3, 4, 5, 7, 8, 9, 10}

4. a. A = {1, 2}, B = {a, b, c} b. A = {1, 2}, B = [2, 4]

5. a. {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}, {(–1, –1), (–1, 1), (1, –1), (1, 1)}, {(1, –1), (1, 1), (2, –1), (2, 1), (3, –1), (3, 1)}, {(–1, 1), (–1, 2), (–1, 3), (1, 1), (1, 2), (1, 3)} b. {(December, December), (December, January), (December, February), (January, December), (January, January), (January, February), (February, December), (February, January), (February, February)}, {(2, 2), (2, 5), (2, 6), (5, 2), (5, 5), (5, 6), (6, 2), (6, 5), (6, 6)}, {(December, 2), (December, 5), (December, 6), (January, 2), (January, 5), (January, 6), (February, 2), (February, 5), (February, 6)}, {(2, December), (2, January), (2, February), (5, December), (5, January), (5, February), (6, December), (6, January), (6, February)} 6. a.

b.

B 6

1 2 3 · · · ·

1 2 3 5

· · · ·

b c d e

1

A

b.

· · · · ·

1 2 3 4 5

· · · · ·

4

0 1 2 3 4

A

7. a. {(0, 0), (1, 1), (2, 4), (3, 9)}

B 4

4 3 2

4

8. a.

c.

B

b. {(–2, –5/3), (–1, –1), (0, –1/3), (1, 1/3), (2, 1), (3, 5/3)}

2 -2

3

9. a.

A

b.

y 3 2 1

7

1 2

d.

y

y

1 1

x

10. a. {(December, winter), (September, fall), 1 (June, summer), (March, spring)}, {December, September, June, March}, {winter, fall, summer, spring} b. x = 2y – 3, [–43, 65], [–20, 34] c. x = y2 + 2, [2, 11], [–3, 3] 238

c.

y

x

1 -1

3 x

Answers to Exercises

x

11. a.

b.

y

c.

y

y

(0, 4) (2, 0)

x

(-2, 2)

12.

x

(0, 0)

1

(4, 2)

(2, 2)

(0, -3)

y

(0, 3)

(0, 0)

x

(4, -1)

x

-1

13. {(a, a), (a, b), (b, b), (b, c), (c, a), (c, b), (c, c), (c, d), (d, a), (d, d)} 14. 84

y

17.

5 4

15. use the fact that n(A) + n(B) includes n(A  B) twice. 16. {(1, a, blue), (1, a, red), (1, b, blue), (1, b, red), (2, a, blue), (2, a, red), (2, b, blue), (2, b, red), (3, a, blue), (3, a, red), (3, b, blue), (3, b, red)}, use three coordinate axes that are perpendicular to each other in the space

1 -1

3

x

-5

4 .2

EXERCISES

x 2. a. f(x) = 5x + 2 b. f(x) = x2 – 2x c. f ( x) = + 3x3 2 3. a. multiply by 2, then subtract 4 b. add 1, then take the square root of all c. divide by three times itself increased by 1

1. a. function b. not c. not

+ 2 3 x2  1 4. a. –3, 33, 2x2 + 5x, –2x2 + 5x, 2x – 5ñx, 2x2 + 4hx – 5x + 2h2 – 5h b.  1 , 11 , 9 x2  1 ,  6 x , , 2 18 9x + 2 x2 + 2 x4 + 2

6a + 3b  1 2 4a + 4ab + b2 + 2

6. a.

b.

y -2 -1

2 2 4 4 2 2 c. – , undefined, 4u3 +1 , 3 x +22 x + 2 , 6u +1 2 , 6u + 2 4u +1 2 3 8u + 4u x + 3x +5x + 3 u + 2 u u + 4u + 4u

x -1 -1

3 1 2

d.

y

y

3 2 1

3 2

1 2 -1

c.

y

x

7. a. 1

1

4

9

x

–1/2

x

5. 

8 3

7 4 d. 21; –3 ; – 7 b. 6, –1; –6 c. 2;  3 5

8. a. function b. not c. function d. function e. not f. not

-4

c. (–, 4)  (4, ) d. (–, –3)  (–3, 3)  (3, ) e. [2, ) f. (–, –3)  (–3, 0)  (0, ) 5 10   10 1  3   3 5 g. [5, 6)  (6, ) h. [5, 10] i.  ,1 j. [1, 1)  1,  k. 3,    ,   10. a.   ,  b. [4, 1)  ( 1, 3 ) 3   3 2 4 5     2    c. (–1, 1)  (1, ) d. (–, –2)  (4, ) e. [–1, 2] f. [1, 2]  {0} g. (0, 1)  (1, 3) h. [–5, –1)  (1, 5] i. (–, –1]  [11, ) 9. a. 

b. 

2 2 11. a. f ( x) = 3  x b. f ( x) =  x  4x c. f ( x) = x  3  ( x + 2)(1  x) 2 x + 2x x5 b. –4, 4 c.  13. a. odd b. neither c. even d. neither e. even f. even g. odd h. odd i. neither j. odd

j. [–5, –2]  [2, 3)  (3, 5]

12. a. – 4 3 14. a. even b. odd c. even d. odd e. neither f. even Answers to Exercises

239

15. a.

(-4,6)

y

b.

(4,6)

c.

y

x

x (-2,-2)

(-4,6)

(-5,-6)

y

(-7,4)

(0,0)

x

y

(4,4)

(0,2) (-5,0) (-2,0)

(0,4)

(-2,-2)

x

(7,-4) (4,-6)

(5,0)

(-4,-4)

(0,-4)

(7,-4)

x

2

(0,-2)

(5,-6)

y

(-5,6)

(2,2)

(-7,4)

(2,0)

(-5,0)

(7,-4)

(-7,-4)

(2,-2)

y

(7,2) (-2,0)

(0,0)

16. a. (2,4)

(-7,2)

(7,4)

(-7,4)

y

(-2,4)

(0,4) (4,4)

(-4,4)

b.

(2,4)

x

y

(7,2)

(2,0) (5,0) (0,-2)

(-7,-2)

5

x -5

2

9 11 x

6

(-2,-4)

c.

(5,-6)

17. consider f(x2) – f(x1) where x1 < x2 on the given interval 18. 3 19. –2 20. a. ,  b. increasing on (–, –2] and [5, ), decreasing on [–2, 5] c. –6, 3, 6; 3 d. –4, 5, 8 e. (–, –6)  (3, 6)

y

345

7

x

21. a. (-, –10)  (–10, 10)  (10, ), [–7, ) b. –5, 5 c. 2 d. undefined e. [–5, 0)  (0, 5] f. (-, –10) (10, ) g. (–10, –6]  {0}  [6, 10) h. even i. increasing on [3, 6] and (10, ), decreasing on (–, –10) and [–6, –3], constant on (–10, –6], [-3, 0), (0, 3] and [6, 10) j. –7,  k. [–3, 0)  (0, 3] l. (–6, –3)  (3, 6) 22. 0 23. d(t) = 15t 24. a. F(x) = 1.8x + 32 b. 1.8

 10t if 0  t  90 225h2  if 0  h  2 25. a. V(t) = 10t b. V( h) =  c. h(t ) =  15 1200 h  1500 if 2  h  3 2+ t  90 if 90  t  210  120 26. until his twenties the person gained weight and kept his weight constant until his forties, then he lost weight

because of a diet or an illness and after his mid-forties he started to gain weight again. 27. he left home at 8 a.m. and 1 during the day visited four places, probably for a sale, and at 6 p.m. left for home. 28. a. (–, 4] b. [3, ) c.  0,  6   1  d. (–, 1) (1, ) 29. 100 30. 0, 2, –1 31. –8 32. decreasing on  ,   , increasing on   1 ,    5 5    2 13 2x 3 33. –x – 3x 34. 35. 12 36. 9 3x

EXERCISES

4 .3

1. a. x2 + x,  x2 + x + 2, x3 + x2  x 1, x5 +8 x4 +15 x3 , 240

1 with domain  \ {–1, 1} x 1

x2 + 3x with domain  \ {–5, 0} x +5

b.

x3 + 4 x2 +5 x, x3 + 2 x2  5 x,

c. x + 3+ x + 2, x + 3  x + 2, ( x + 3) x+ 2,

x+ 3 x+ 2

Answers to Exercises

d.

x  1+ x +1,

x  1  x +1,

2 x2 + 2 x , x4 + 2 x3 + 2 x2 + x x2  2 x +1

e. one-to-one 13. a.

8. x

2 2. a. 25 b. undefined c. –11 3. a. x, x 2 + x +1 , x +x 1 x +1 7 x +6 11 x 5 x + 3 10 x  3 b. x4 – 4x3 + 4x2, x2, x2 – 2x + 2, x + 2 c. , , , 2 x + 3 x + 3 4x +9 13  3 x

x4 41 c. f(x) = x3 – x + 4, g(x) = 2x2 – x 5. a. b. –1 5 11 9. a. not b. not c. one-to-one 10. a. one-to-one b. one-to-one c. not d. one-to-one

4. a. f(x) = ñx, g(x) = 2x – 4 c. 4 d. 9 6. 2 7. 2

x 1

x 1  x +1,

b. f ( x) = x7 , g( x) =

11. a. one-to-one b. one-to-one c. not

y (5, 4)

b. inverse function c. does not exist

12. a.

6x 4

x+ 4 x3

b.

d.

y (1, 4)

e.

x

(-2, 4) (3, 0)

x

(0, 0)

(4, -5)

f. inverse function does not exist

14. a. 5 b. 0 c. 3 d. –14 e. 7 15. a.

y (-5, 6)

(-3, -6)

(-3, -8)

5

 x  9  +1 d.    3 

x +5 4

y

(-3, 2) (-1, -2)

3

(-5, 6)

(7, 5)

(-2, 2)

x

c.

(7, -6)

x

(3, -5)

2x  1 2x 22 x  1 2x  1 6x  5 3x  7 b. c. 16. a. b. c. x d. e. 2x – 11 2x 3x  2 3x x2 2 x +1 2

3x  7 x 1 g. 17. a. 1 b. –1 c. 1 d.  17 18. a. 1 b. –2 19. a. find two different x-values that give the same x5 2x  4 16 y-value b. assume that for x1  x2, f(x1)  f(x2) 20. –2 21. Compare the functions when the argument is x and –x

f.

22. Compare the functions when the argument is x and –x

23. a.  x +9  3 b. x +11+ 4

24. Try to compose

each of the sides with g  f

EXERCISES

5 .1

y

1. a.

b.

4

-1

0

3 2

-3 -4

2x2=3y 1

2. a. yes b. no c. yes d. no

Answers to Exercises

-

y=2x2

2 3

y

c.

1 x

y – 4x2=0 2

y 0

-1

3

y=3x2

1 3

3y=x2

2y=–3x2 y=–3x2

y+4x2=0

-1 -

x

3. a.

1 1 7 25 b. c. – d. – 3 5 9 8

1 2

-2

1

x 2y+x2=0 y+2x2=0

241

4. a.

y -4 -3 -2 -1 0 1 2 3 4 1 x 2 -2 -

5. a. V(0, 0), x = 0, ymin = 0 b. V(0, 0), x = 0, ymax = 0 c. V(0, 2), x = 0, ymin = 2 d. V(0, –1), x = 0, ymax = –1 6 6 6 6 5 25 5 25 e. V( , f. V( , ), x = , ymin = ), x = , ymax = 7 5 7 5 4 8 4 8

9 2

2 5 2 5 g. V( , ), x = , ymin = 3 3 3 3

1 5 1 5 h. V( , ), x = , ymax = 2 4 2 4

3 75 3 75 i. V( , – ), x = , ymin = – j. V(–4, 0), x = –4, ymax = 0 2 4 2 4 5 41 k. V(–3, –5), x = –3, ymax = –5 l. V(1, 3), x = 1, ymin = 3 6. – 7. 1 8. – 9. a. (0, 0), (4, 0) 2 8 3 3 b. (3, 0), (0, 9) c. (0, 396) d. (–2ñ2, 0), (2ñ2, 0), (0, –8) e. (– , 0), (–1, 0), (0, 3) f. (0, –5) g. (– , 0), (0, 9) 2 4 h. (0, –4) 10. a. m  (–2ñ2, 2ñ2) b. m = 2ñ2 c. m  (–, -8

–2ñ2)  (2ñ2, ) 11. 3 y

12. a.

y

b. ñ3 x

-ñ3

c.

7

y x

x

y

h.

-48

i.

1 2+ñ2 x

4

3 x

-3 -4

y

ñ5 x

-ñ5 -5

-2

x

x y

y

j.

13.

12

13

2 – ñ2 -3

y

f.

1

-1

98

-7

y

y

e.

-4 1

-1

y

d.

5

-9

g.

-4

7

f(A)= [–9, 7)

-1 1 4

-2 3

x

-3

4

4 5

x

x

-5 -8 -9

y

14.

15. a. ymax = 12, ymin = 3

3 -1 1

3

x

16. ymax = 64

17. ymin = –

9 4

1 18. a = – , b =1 2

19. y = 3(x – 5)2 20. –3(x – 1)2 + 2  2; –3(x – 1)2  0; 3(x – 1)2  0 21. m  [0,

16 ] 9

-5

242

Answers to Exercises

y

22. a.

y=x2+4

y

b.

y=x2 y=x2 – 2

y=x2 4

-ñ2

y 16

-25

y=–x2 – 2

g.

3

y=(x – 4)2

-1

y

y=–x2

4

y=–(x+1)2

3)2

y

y

b.

c.

3

y

d.

1

3

1 2 3

x

-1

1

x

y

1

2

x

-1 2 3 4

x

-1

24. a. y = –(x – 1)2 + 3 b. y = f. y = –

1 (x – 3)(x + 3) 3

x

2

7

1

1

y=(x – 2)2

7

y=x2

y=–(x+1)2 – 2

23. a.

y=(x – 2)2+3

x -2

y=–(x –

y

h. -1

-9 y=–x2

y=–x2

y=–(x+5)2

x

x

4

x

y=–x2

y

f.

-5

x

-2

-2

y=x2

y

d.

x

ñ2

x

e.

y

c.

1 1 x(x – 5) c. y = (x + 3)2 d. y = (x + 1)2 + 3 e. y = –x2 + 4x – 3 6 3

25. a. y = –3x2 + x + 4 b. y = –x2 + 4x – 4

c. y = (x + 1)2 – 4 d. y = –x2 – x – 4 27. –12 28. 33. 10 square units 34. m  (–, –1)  (0, )

18 5

29. 0 30.

1 2

26. a. y = x2 – 4 b. y = (x – 2)2 31.

35. 64 square units 36.

25 4

32.

9 square units 4

49 37. 2 38. (–3, 0) 39. 9 40. 1 20

41. 35 square units

Answers to Exercises

243

EXERCISES

6 .1

1. a. (–2, ) b. (–, 6] c. (–, –7) d. [4, ) e. [2, 4) f. (–1, 0] g. (–5, x

2. a.

–¥

x

+

–¥

x

6 f. x  ; a+ 3

x

x

1. a. x2 –

–¥

5x+4

x

d.

–¥

1/4

c. (–, 0)  (2, )

–1/2 x

14/3

+ 0

b. x  0;

e. x > 2 ; 3

–

x

2/3 x

–3 a4+4

–3 g. x > 4 ; a +4

4

1

–

2. a. (4, 5) b. (0,

–

x

–¥

x 7 - + 2 3

x

6 .2

+

–16x2+8x – 1

x

c.

+

3

d. x  3;

6/(a+3)

EXERCISES

-5/3

1 3. a. x > – ; 2

+ 2

c. x > 2;

–¥

–3x – 5

6/ñ3

–

ñ3x – 6

x

b.

–

4x + 1

d.

-1/4

1 3 ) h. [0.5, ] 2 2

–

x

b. 2x2+x

+

3/2

-2

+

–6

e. –

–¥

–

+

2x2

–

–

–

–

+ + + + + +

– 3x+4

x

f.

x –4x2+10x – 25

x

c.

12x2+4ñ3x+1

4 7 4 4 f. ( , 1) g. (–, –4)  (2, ) ) c. (– , – ]  [0, ) d.  e. x = – 7 6 3 5

4. (–, –2ñ3 – 1)  (2ñ3 – 1, )

5. (3 – 2ñ2, 3 + 2ñ2)

–¥

–ñ3/6

+

+

3. a. (0, 2) b. 0, 2

6. {–6, –5, –4, –3, –2, –1, 0 1, 2}

1 3 2 2 5 7. a. (0, ), x  1 b. (– , – )  ( , 2) c. x  –2, (– , –1)  ( , 3) d. (– , )  (1, ) 3 2 3 3 2 1 3 7 e. (–4, –3)  (–2, – 1)  ( , 3) f. (–1, 5) g. (– , )  ( , ) h. (–8, –1) i.  – {2} j. (–, –7]  (–1, 0)  (0, 1]  (3, ) 2 2 3

k. (– ,

– 6 ]  [–1, 0)  [1, 2

6 4 1 11 ] l. (– , 1)  ( , 2) m. (–, 2) – {–4, –2} n. [0, ] o. (– , 1)  ( , ) 3 3 5 2

1 p. (–, –2)  (–2, –1)  (–1, 0) q. [– , 0]  [1, ) 2

3 8. a. (0, 1) b. (0, 1) c.  d. [ , 2) 2

9. a.  – {7, –7}

b. (0, 16)  (16, ) c. (–, –2)  [2, ) d. [2, 3]  [–1, 1] 10. a. (0, ñ3 – 1)  (–4, –3)  (–ñ3 – 1, –2)  (1, 2) 11. x  0, (–2, 1) 244

Answers to Exercises

EXERCISES 1. a. (–2, c. (–8, –

6 .3

3 ) b. (1, 3) c. (–6, –5) d. (–1, 1) e. (1, 2) f. [1, 2) g. (2, 4) h. (0, 1) 2

13 )  (0, 5) 2

2. a. (0, 1) b. (–1, 1)  (3, 5)

3 5 3. a. (– , 1]  [ ,  ) b. (– , – )  (3, ) c. (–, –5)  (–1, ) d. (–, –4)  (0, ) 2 3

3 2 2 13 5 3 e. [ , 2) f. [ , ] g. (–, 3) 4. a. (3, ) b. [ , 4) c. [4, ) d. (–3, 1) e. [3, ) f. (9, ) 5. a. [3, ) b. [–2, – ] 2 7 3 5 10 2

c. [2, 3] d. [–15, 7. [1, 5)  (10, )

TEST 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

B D A D B B C C C C

TEST 1. 2. 3. 4. 5. 6. 7. 8.

E B E A C D A D

13 – 5 417 ) e. [3, ) 64 2

TEST B B C B C D C D A B

6A 9. 10. 11. 12. 13. 14. 15. 16.

Answers to Exercises

– 5 –1 , 2

13 + 3 – 11 – 1 11+1 ] b. [–1, 3] c. [– 6, ) ( , 2 2 2

5 8. a. (2, ) b. (1, ) c. (– , – ) d. (–, –2)  (2, ) 2

1A 11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

6. a. [

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

C D B C B D C B D D

TEST E B D D C B A B

1. 2. 3. 4. 5. 6. 7. 8.

D B A C E B E C

1B 11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

TEST D C C B A C C D B C

1. 2. 3. 4. 5. 6. 7. 8.

C E C B E D C C

9. (–5, –1)

3A 9. 10. 11. 12. 13. 14. 15. 16.

TEST B B D A E C B D

1. 2. 3. 4. 5. 6. 7. 8.

B D E A D B A C

3B 9. 10. 11. 12. 13. 14. 15.

A E B D B B S

6B 9. 10. 11. 12. 13. 14. 15. 16.

C D A B E C D D 245

6]

246

Answers to Exercises

Answers to Exercises

247

248

Answers to Exercises