-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 nnumber. The set of rational numbers is denoted by Q. Q = { a | a, b Z, b  0}

b

Definition

The set of positive rational nnumbers is denoted by Q+_.

Q+_{= {}a a_{| 0 and a, b , b  0}}

b b

Definition

The set of negative rational numbers is denoted by Q–.

Q–_{= {}a a_{| 0 and a, b , b  0}}

b 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.

In short, is a positive rational number if a and b are both positive integers or both nega-tive integers.

For example, are positive rational numbers, and denoted by 2 .

7 2 _and –2 7 –7   a 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.

In short, is a negative rational number if a is a positive integer and b is a negative integer, or if a is a negative integer and b is a positive integer.

For example, are negative rational numbers. We can write negative rational

numbers in three ways: –5 –5 5 .

4 4  –4 –5 5 and 4 –4   a b

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, ...}

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, ...}

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, ...}

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.

For example, are rational numbers.

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, b  and b  0}

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

Some rational numbers

-7 7 2 7 2 1 3 2 4 5 2 13 a b 8 2_{, , – , ,} 3 0 _and 17 3 5 7 7 1 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 Integers -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 Whole numbers -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 Natural numbers

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.

For example, = 0.324242424...

0.6 is a terminating decimal, and 0.324–– is a repeating decimal. There are some decimals which do not repeat or terminate.

For example, the decimals 0.1012001230001234000 ...

3.141592653 ... = 

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.

3 321

0.6, and 0.324

5 990 

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 nnumbers.

The set of real numbers is denoted by R.

R = Q Q

R = R+_{ {0}  R–}

R+ _{is the set of positive} real numbers

R– _{is the set of negative} real numbers

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.

The real numbers fill up the number line.

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 dia-gram on the left.

-5 -4 -3 -2 -1 0 1 2 3 4 5

-2.35 -0.5 0.6 e p

Some rational numbers

N  W  Z  Q  R Q  R Real NNumbers

1. Understanding Square Roots

Remember that we can write a a as a2_{. We call a}2_{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.

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.

Objectives

Definition

If a2_{= b then a is the square rroot 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 a22_{= b then a = ñb (a b, b  0).}

Here are the square roots of all the perfect squares from 1 to 100.

12_{= 1  ñ1 = 1 6}2_{= 36  ò36 = 6}

22_{= 4  ñ4 = 2 7}2_{= 49  ò49 = 7}

32_{= 9  ñ9 = 3 8}2_{= 64  ò64 = 8}

42_{= 16  ò16 = 4 9}2_{= 81  ò81 = 9}

52_{= 25  ò25 = 5 10}2_{= 100  ó100 = 10}

The equation x2 _{= 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

In other words, if x is a non-negative real number, then if x is a negative real number, then _x2 _{– .x}

2 _{, and} x x 2 _{| |} if 0. – if 0. x x x x x x    _{ }  

For example,

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

Note

x = ñ9 and x2_{= 9 have different meanings in the set of all real numbers.}

 ñ9 = = |3| = 3  If x2_{= 9 then x = 3 or x = –3.} 2 3 EXAMPLE

1

a. ò81 = 9 b. ñ1 = 1 c. ñ0 = 0 d. ò64 = 8 e. f. ó_{0.64 = 0.8} g. –ó100 = –10 h. –ó0.09 = –0.3 i. ò–4 is undefined j. _(–4)2 _{ 16 4} k. _–42 _{ –16 is undefined} 4 2 9  3

Evaluate each square root.

a. ò81 b. ñ1 c. ñ0 d. ò64 e. ñ9 f. ó0.64 g. –ó100 h. –ó0.09 i. ò–4 j. _(–4)2 k. _–42 Solution EXAMPLE

2

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. _{10000 100} 1225 35

Evaluate each square root.

a. ó100 b. ó121 c. ó144 d. ó169 e. ó225 f. ó361

g. ó400 h. ó625 i. 1225 j. ₁₀₀₀₀

EXAMPLE

3

a. ñ2ñ8 = ó28 = ò16 = 4 b. c. ò50ñ2 = ó502 = ó100 = 10 d. e. f. 10 90  10 90  900 30 576  36 16  36  16 6 4 24   25 1 25 1  25 5 7 7  7 7  49 7

Simplify each of the following.

a. ñ2 ñ8 b. ñ7 ñ7 c. ò50ñ2 d. ò25ñ1 e. ó576 f. ò10ò90

Solution

2. Properties of Square Roots

Property

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

Property

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

.

a a

b b 

Mathematics is a universal language.

b  0 óa . 6 = ña .ñ6 a  0 _

Note

If a 0 then

ña  ña = óa  a = _a2 _a.

For example, 2 2 25 16 25 16 5 4 20, 3 27 3 27 81 9, 36 36 6 ( 0), and 5 5 5 5 25 5. a a a a                   

For example, 24 24 4 2, and

6 6 1 1 1_. 49 49 7      If a > 0 then 1 1. a a a a= = =

EXAMPLE

4

a. b. c. d. e. f. g. h. i.          3 3 2 2  3 3 2 2 1 1 1 x y x y xy x y x y x y x y  _  _{    } 5 6 5 6 4 4 2 2 2 2 2 2 2 ( ) a b a b _{a b a b a b} ab ab       3 3 2 2 24 24 _{4 4 2} 6 6 a a _{a a a} a a   625 625 25 144 144 12  1  1 1 – – – 100 100 10  1  1 1 64 64 8   16 16 4 49 49 7    50 50 25 5 2 2 25 25 5 = 9 = 9 3

Simplify the expressions.

a. b. c. d. e. f. g. h. i.   3 3 x y x y  5 6 2 a b ab 3 24a 6a 625 144 1 – 100 1 64 16 49 50 2 25 9 Solution Property

For any real number a and n Z, (a  0).

Proof

n factors of ña n factors of a

           ( _a)n _{a a a} ... _{a a a a} ... _{a a}n  ( _a)n _an For example,        2 2 3 3 8 8 ( ) , ( 5) 5 125, ( 2) 2 256 16. a a a and

EXAMPLE

5

          2 4 2 6 4 4 2 6 2 2 2 2 2 3 2 2 2 3 ( 2) ( 5) – ( 5) – ( 2) 2 5 – 5 – 2 (2 ) (5 ) – 5 – (2 ) 2 5 – 5 – 2 4 25 – 5 – 8 16 Evaluate (ñ2)4_{+ (ñ5)}4_{– (ñ5)}2_{– (ñ2)}6_. Solution

3. Working with Pure and Mixed Radicals

Definition

A radical eexpression is an expression of the form

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

 2_{a a} radical sign radicand index

_ñ

n

a

. n_a Definition

A mixed rradical is a radical of the form

(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.

n

x a

Property

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

For example,                             2 2 2 2 2 2 2 2 8 4 2 2 2 2 2 2 2, 27 9 3 3 3 3 3 3 3, 32 16 2 4 2 4 2 4 2, and 50 25 2 5 2 5 2 5 2.     2 _and 2 _. a b a b a b a b

EXAMPLE

6

a. b.                 2 2 2 2 2 48 3 27 – 108 243 2 4 3 3 3 3 – 6 3 9 3 8 3 9 3 – 6 3 9 3 (8 9 – 6 9) 3 20 3                 8 2 32 18 72 98 2 2 8 2 3 2 6 2 7 2 2 (2 8 3 6 7) 6 2   _   _        _   _ 2 2 2 2 2 8 2 2 = 2 2 2 32 = 2 4 2 = 8 2 18 = 3 2 = 3 2 72 = 6 2 = 6 2 98 = 7 2 = 7 2 =

Simplify the expressions.

a. ñ8 + 2ò32 – ò18 + ò72 – ò98 b. 2ò48 + 3ò27 – ó108 + ó243 Solution EXAMPLE

7

a. b. c. d. e. _{x y} _{x y}2  2    10 10 10 10 100 10 1000  2    5 3 5 3 25 3 75  2    3 5 3 5 9 5 45  2  2    2 2 2 2 2 2 4 2 8

Write the numbers as pure radicals.

a. 2ñ2 b. 3ñ5 c. 5ñ3 d. 10ò10 e. xñy

Solution

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,

EXAMPLE

8

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 =

ó

252 +

ó

492 +

ó

812 = 5ñ2 + 7ñ2 + 9ñ2 = (5 + 7 + 9)ñ2 = 21ñ2 f. 5ñx – ò9x + ó64x = 5ñx – 3ñx + 8ñx = (5 – 3 + 8)ñx = 10ñx Perform the operations.

a. ñ3 + ñ3 b. 2ñ5 + ñ5 c. 3ñ6 + 4ñ6 d. 10ñ5 – 3ñ5

e. ò50 + ò98 + ó162 f. 5ñx – ò9x + ó64x

EXAMPLE

9

Compare the following numbers.

a. ñ7 ... 3 b. 3ñ5 ... 2ò10 c. 2ñ7 ... 3ñ3 d. –2ñ3 ... –3ñ2 Solution a. b. c. d.     2 2 –2 3... – 3 2 – 2 3... – 3 2 – 12 – 18 –2 3 –3 2     2 2 2 7...3 3 2 7... 3 3 28 27 2 7 3 3     2 2 3 5...2 10 3 5... 2 10 45 40 3 5 2 10   7...3 7... 9 7 9 7 3 Solution Property

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

1. 2.

Proof

1. In order to verify these expressions, suppose that t = òm + ñn.

t2_{= (òm + ñn)}2_{= (òm + ñn)  (òm + ñn)}

= (òm  ñn) + (òm  ñn) + (ñn  òm ) + (ñn  ñn) (by the distributive property)

= m + (òm  ñn) + (ñn  òm ) + n

= m + n + 2ómn (by the commutative property)

a b

 t2_{= a + 2ñb  t =}

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

 2 a b   – – 2 ( ) m n a b m n    2 m n a b

EXAMPLE

10

Simplify the expressions. Use the property to help you. a. b. c. d. e. 5 21 f. 2 3 6 – 4 2  6 32  5 2 6  3 2 2 Solution a. b. 2+1 2  1 3+2 3  2 c. 4+2 4  2 d. 4+2 4  2

e. We need a 2 in front of ò21 before we can use the property. Therefore, let us multi-ply the expression by

7+3 7  3 3+1 3  1 f. 2  2 3  4 2 3  3 1  3  1 2 2 2 2 2       2    2 (5 21)  10 2 21  7 3  7  3 5 21 5 21 2 2 2 2 2 2 2_. 2    2    6 – 4 2 6 – 2 2 2 6 – 2 2 2 6 – 2 8 4 – 2 2 – 2        6 32 6 2 8 4 2 2 2    5 2 6 3 2      3 2 2 2 1 2 1

Check Yourself 1

1. Simplify the expressions.

a. ñ2  ñ2 b. ñ8  ò32 c. ò3x  ó12x d. ñ2  ò18

e. f. g. h.

2. Evaluate the following.

a. (ñ3)2_{+ (ñ4)}4_{– (ñ5)}2_{– (ñ2)}4 _{b. (ña)}4_{+ (ñb)}2_{– (ñc)}6

3. Simplify the expressions.

a. ò18 b. ò50 c. ò48 d. ò20 e. 5ñ3 – 2ñ3 + ñ3 f. 2ñ2 + 3ñ2 – 4ñ2 g. ò50 – ò18 – ò32 h. ó12x + ó27x – ó48x 1 – 49   3 3 a b a b 12 3 32 2

4. Write each number as a pure radical.

a. 5ñ3 b. 3ñ5 c. 4ñ2 d. 2ñ5 e. añb

5. Perform the operations.

a. b. 5ñ2 – ñ8 c. ò27 – ò48 d.

6. Compare the numbers.

a. 3ñ5 and 2ò10 b. c. –2ñ5 and –3ñ3

7. Write each expression in its simplest form.

a. b. c. d. e. f. g. h. i.

Answers

1. a. 2 b.16 c.6x d.6 e.4 f.2 g. h. 2. a. 10 b.a2_{+ b – c}3 _{3. a.3ñ2 b.5ñ2} c. 4ñ3 d.2ñ5 e. 4ñ3 f.ñ2 g.–2ñ2 h.ò3x 4. a.ò75 b.ò45 c.ò32 d. ò20 e. 5. a.3ò30 b.3ñ2 c.–ñ3 d.3ñ3 6. a.3ñ5 > 2ò10 b. c.–2ñ5 > –3ñ3 7. a.ñ2 – 1 b.2 + ñ2 c.ñ5 – ñ2 d.ñ2 + 1 e.ñ5 – 2 f.2 – ñ3 g.4 h.6 i.2 1 1 2  3 2 a b 1 – 7 1 ab   3 – 5 3 5   ( 7 1) ( 8 – 28 )   ( 6 – 2) ( 8 2 12 ) 7 – 48 9 – 4 5  3 8 7 – 2 10  6 2 8 3 – 2 2 1 1 and 2 3  27 75 12 – 4 4 4  10 15 6 2 3 2

EXAMPLE

11

Simplify the following.

a. b. c. 1 1 9 16 1 6 6 72 16 16    4 21 13 9

Start from the radical on the ‘inside’ of the expression and move outwards.

a. Start with ñ9, on the inside, and work outwards.

b. c. 1 1 9  1 25  1 5 9 3 16 16 4 4 2             1 4 1 6 6 72 16 6 6 72 6 6 72 16 16 4 1 6 6 72 6 6 36 2 6 6 6 6 36 6 6 36 6                    4 21 13 9 4 21 13 3 4 21 16 4 21 4 4 25 4 5 9 3 Solution

EXAMPLE

12

a. Evaluate b. c. x x x... 5. Find .x  ... 7. Find . a a a a a 2 2 2 2 ... a. b. c.              2 2 5 ( ... ) 5 ... 25 5 25 20 x x x x x x x x x        2 2 7 ... 7 ( ... ) 7 ... 49 7 49 7 a a a a a a a a a a a a a a        2 2 2 2 Let 2 2 2 2 ... . ( 2 2 2 2 ... ) 2 2 2 2 2 ... 2 ( 2 2 2 ... ) x x x x x x x x  x x  2 x x    (simplify) x 2. Therefore, 2 2 2 2 ... 2. Solution

(take the square of both sides) (remove a square root)

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 = 24

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

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

EXAMPLE

13

Perform the operations.

a. ñ2(ñ5 + ñ3) b. ñ3(3ñ3 + 2ñ2) c. 2ñ5(ñ3 + ñ2 + 2ñ5 – ñ7)

EXAMPLE

16

Multiply and simplify.

a. 3 5  3 – 5 b. 2 2  2 – 2 c. a b a– b

EXAMPLE

15

Multiply and simplify.

a. (ñ2 + 1)  (ñ2 – 1) b. (ñ5 + ñ3)  (ñ5 – ñ3) c. (1 – 2ñ2)  (1 + 2ñ2)

d. (ña + 1)  (ña – 1) e. (ña + ñb)  (ña – ñb)

EXAMPLE

14

Multiply and simplify.

a. (ñ2 + ñ3)  (ñ2 + ñ3) b. (5 + ñ5)  (5 + ñ5) a. ñ2(ñ5 + ñ3) = ñ2  ñ5 + ñ2  ñ3 = ó_{2  5 +} ó_{2  3 = ò10 + ñ6} b. ñ3(3ñ3 + 2ñ2)= ñ3  3ñ3 + ñ3  2ñ2 = 3 ó_{3  3 + 2 }ó_{3  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 Solution a. (ñ2 + 1)  (ñ2 – 1) = ñ2  ñ2 – ñ2  1+1  ñ2 – 1  1 = (ñ2)2_{ 1}2_{= 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_{– 1}2_{= a – 1 (a  0)}

e. (ña + ñb)  (ña – ñb) = (ña)2_{– (ñb)}2_{= a – b (a, b  0)}

Solution a. b. c. _a _b _{a– b}  _a2_–b       2 2   2 2 2 – 2 (2 2) (2 – 2) 2 – ( 2) 4 – 2 2       2 2    3 5 3 – 5 (3 5) (3 – 5) 3 – ( 5) 9 – 5 4 2 Solution Solution 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

17

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 Solution

5. Rationalizing Denominators

Look at the numbers They are all fractions, and each fraction

has an irrational number as the denominator. In math, it is easier to work with fractions that have a rational number as the denominator.

1 3 10 19

, , , and .

5 2 12 13

Definition

Changing the denominator of a fraction from an irrational number to a rational number is called rationalizing tthe ddenominator 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 by a suitable factor. For example, if the fraction is in the form we multiply both the numerator and the denominator by ñb.

So, Note that and have the same value: they are

equivalent fractions.

Look at some more examples:

3 3 2 3 2 3 2 6 6_, 2 2 2 2 2 2 2 2 4 3 3 3 3 3 3 3 3, and 3 3 3 3 3 3 3 5 2 3 5 3 5 2 3 10 3 10 . 2 2 4 2 2 2 2 2 2 2 2                           ab b a b . a b a a b ab b b b b b b       , a b

a. b. c. d. 2 2 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 5) 5 – (2 5) 15 2 – 6 10 – 10 – 4 5 15 2 – 6 10 – 10 – 4 5 25 – 20 5            2 2 6 2 ( 6 2) (1 3) 6 6 3 2 1 2 3 1– 3 (1– 3) (1 3) 1 – ( 3) 6 18 2 6 6 3 2 2 6 2 6 4 2 _{– 6 – 2 2} 1– 3 –2 –2  _  _  _                   2 2 3 – 2 ( 3 – 2) (2 2 1) 3 2 2 3 1– 2 2 2 – 2 1 2 2 – 1 (2 2 – 1) (2 2 1) (2 2) – 1 2 6 3 – 2 2 2 – 2 2 6 3 – 4 – 2 8 – 1 7                2 2 5 5 3 2 2 5 (3 2 2) 5 3 5 2 2 3 – 2 2 3 – 2 2 3 2 2 (3 – 2 2)(3 2 2) 3 – (2 2) 3 5 2 10 3 5 2 10 3 5 2 10 9 – 8 1                   Solution 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.

For example, let us rationalize ñ3 – ñ2 is the conjugate of ñ3 + ñ2.

Therefore, we multiply the numerator and the denominator by ñ3 – ñ2 to rationalize the denominator. 2 2 1 3 – 2 1 ( 3 – 2) 3 – 2 3 – 2 3 – 2 3 – 2 3 – 2 1 3 2 3 – 2 ( 3 2) ( 3 – 2) ( 3) – ( 2)           1 . 3 2 (a + b)(a – b) = a2_{– b}2

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

Remark

EXAMPLE

18

Rationalize the denominators.

a. b. c. d. 3 2 – 2 5 2 5 6 2 1– 3  3 – 2 2 2 – 1 5 3 – 2 2

EXAMPLE

19

Rationalize the denominators to find the sum. 3 2 3 2 2  3 – 2 2 Solution 2 2 3 2 3 3 – 2 2 2 3 2 2 3 2 2 3 – 2 2 3 2 2 3 – 2 2 3 – 2 2 3 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) 3 – 2 6 6 4 7 – 6 6 – 7 3 – 8 –5 5                                      

Check Yourself 2

1. Rationalize the denominators and simplify.

a. b. c. d. e. f.

g. h. i. j.

2. Rationalize the denominators and simplify.

a. b. c. d. e. f.

g. h. i. j. k.

3. Rationalize the denominators and simplify.

a. b. c. d. e.

Answers

1. a. b. c.ñ2 d. –ò15 e. ò15 f. g. h. i. j.ab 2. a.ñ2 + 1 b.ñ6 – 2 c. d. e. f. g.5ñ2 – ò10 + 3 ñ5 – 3 h. i. j. k.–9ñ3 – 6ñ7 3. a. b.ñ2 c. d. e.17 3 – 3 6 –16 3 2 4  6 6 2 5 3 2 – a b ab a b   – – a ab a b 35 3 2  4 – 6 10 3 5 2  3 6 6 2  5 – 1 2 2 xy b a 10 20 30 4 15 3 21 7 3 2 3 – 3 1 1– 3 3 3  2 2 3 – 2 1 2 – 1 2 2 1 1 3 3  3 – 3 2 2 2 – 2 2 2 1 1 5 2 5 – 2 3 3 3 – 2 7 – a b a b  a a b 2 5 – 7 7 – 5 2 5 – 2 2 10 – 6 3 – 2 2 3 – 2 5 1 5 – 1  3 6 – 2 2 5 1 2 3 – 2 1 2 – 1 3 4 2 a b a b   3 3 2 x y x y   3 a b a  1 2 10 3 5 2 6 3 5 3 3 –5 5  1 2 2 5 3 3 7

EXERCISES

1

.1

44.. Write each number as a mixed radical.

a. ñ8 b. ò72 c. ó243

d. e. ó125 f. _{x y}3 2

1000

22.. Simplify the expressions.

a. ñ3  ñ3 b. ñ5  ñ5 c. ñ3  ò12 d. ñ3  ò27 e. ò2x  ò8x f. g. h. _{3 2x 4 18xy}2 3a 5a 6xy 24xy

11.. Evaluate the square roots.

a. ò36 b. ó100 c. –ó121

d. _16x2 e. _{25 y} 2 f. _{121 a} 4

33.. Simplify the expressions.

a. b. c. d. e. f. 3 3 3 12 xy x y 3 2 32 24 x y x 3 72 2 2 x x 108 27 72 8 50 2

55.. Perform the operations.

a. 3ñ3 + 2ñ3 b. 6ñ5 + ñ5 c. –5ñx + 5ñx d. ñ6 – 3ñ6 e. 3ò18 + 2ò72 f. ò80 – ó125 + ò45 g. ò75 + ó108 – ò48 + ò27 h. i. j. _{0.9 – 2.7  1.7  0.4} 2.25 – 2.89 1.44 3 3 9x  16x – 4x 25x

66.. Write each expression in its simplest form.

a. b. c. d. e. f. g. h. i. f. k. _{2 2 4  12} 2 8 – a  a a 4 15 – 4 – 15 3 8  3 – 8 7 – 4 3 5 24 8 – 2 7 11 96 8 2 12 5 – 2 6 3 2 2

1111.. Perform the operations. a. b. c. d. 4 2 _– 2 3 3 2 2 3 – 2 2 5 _– 2 2 3 – 11 2 3 11 3 2 3 1  3 – 1 1 1 3  2

1122.. Perform the operations.

a. b. c. 1 1 1 ... 1 2 1  3 2  4 3   100  99 1 2 2 11 72    5 2 – 5 – 2 5 – 1 

1100.. Rationalize the denominators.

a. b. c. d. e. f. g. h. i. j. k. l. 10 2 21 7 3   2 2 3 – 5 3 2 3 10 2 7 – 5 4 3 3 2 3 1 3 – 1  –2 3 2 1 1– 2 1 2 4 3  1 – 11 3 2 6  3 3 77.. Simplify the expressions.

a. b. c. d. 13 6 6 9 1 8 16 9 16 96     7 _–2 _{6 9} 3 9  32 21– 23 4

88.. Find x in each equation.

a. b.

c. 3x 3x 3x... 9

3 3 3 3 ... x

2 2 2 2 ... x

99.. Find the products.

a. ñ5  (ñ2 + ñ3) b. ñ7  (1 + ñ7) c. –ñ2  (ñ3 – ñ8 + 1) d. ñ2  (ñ8 + ò32) 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. 5 2 3  2 3 16 – 9 3 2 3 3  2 3 – 3 2 1  2 – 1

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.

Objectives

1. n

th

_Roots

In section 8.2 we studied exponential equations. For example, 2n

 2n= 2 is an exponential equation. Let us solve it. 2n_{ 2}n_{= 2}2n_{= 2 (a}n_{ a}m_{= a}n+m₎

22n_{= 2}1

2n = 1, n = If we substitute for n in the original equation we will get but we know that ñ2  ñ2 = 2. Therefore,

1 2 2 2  2  2. 1 1 2 2 2 2 2, 1 2 1 . 2

A. RATIONAL EXPONENTS

Definition

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

If an_{= b then a =} 1_n n _{. a is called the n}tthh_{root of b. It is denoted by} nn_ñb.

b  b

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

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

2_{ña = ña.}

Look at some examples of different roots:

52_{= 25 and 5 = ò25 ‘the square root of 25 is 5’,}

23_{= 8 and 2 =} 3_{ñ8 ‘the cube root of 8 is 2’,}

33_{= 27 and 3 =} 3_{ò27 ‘the cube root of 27 is 3’, and}

24_{= 16 and 2 =} 4_{ò16 ‘the fourth root of 16 is 2’.}

Let x R – {–1, 0, 1}. If xm_{= x}n _{then m = n.} Remark

2. Rational Exponents

Definition

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

For example, the numbers have rational exponents.

1 2 2 2 2 3 8 , 4 , and 2 m n . m n_bm _bn

EXAMPLE

20

Write the expressions in radical form.

a. b. c. d. e. f. 1 1 2 3 ( )a 1 4 4 ( )x 3 4 x 2 3 5 1 5 32 1 2 3 a. b. c. d. e. f. 1 1 3 2 3 ( )a  a 1 4 4 4 4 ( )x  x 3 4 3 4 x  x 2 3 2 3 3 5  5  25 1 5 5 32  32 1 2 2 3  3 3 Solution

EXAMPLE

21

Write the expressions with radical exponents.

a. b. 3 c. 7 _x14 d. 3_a2 e. 5 _ab2 64 5 ₅ 3 a. b. c. d. e. 1 1 2 5_ab2 _(ab2 5_{) a b}5_ 5 2 3 _a2 _a3 14 7 _x14 _x7 _x2 3 3 3_{64 4}3 _43 _41_4 5 5₃5 _35 _31 _3 Solution

EXAMPLE

22

Simplify the following.

a. b. c. d. e. 5 0 4 –16 3 –27 4 81 3₈ a. (8 > 0) b. (81 > 0) c. (–27 < 0 and 3 is odd)

d. is not a real number (–16 < 0 and 4 is even)

e. 50 = 0 4 –16 3 3 _–273_(–3)3 _(–3)3 _–3 4 4 4_{81 3}4 _34 _3 3 3 3_{8  2}3 _23 _2 Solution Rule

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

1. If b > 0 then nñb is a positive real number.

2. If b = 0 then nñb is zero.

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

Check Yourself 3

1. Write the expressions in radical form.

a. b. c. d.

e. f. g. h.

2. Write the expressions with rational exponents.

a. b. c. d. e. f. g. h.

3. Simplify the expressions.

a. ò16 b. c. d. e. f. g. h.

Answers

1. a.ñ2 b. c. d. e. f. g. h. 2. a. b. c. d. e. f.a2 _{g. h. 3. a.4 b.3 c.–4 d.5 e.2} f.–2 g.xy2 _h.2ab2 1 3 2 6 (x y ) 1 2 4 3 (x y ) 2 3 x 1 2 3 (xy ) 2 5 a 1 2 2 1 3 2 2 3_{(x y x )} 2 2 a b 6_x 2 3 6_{x y} 5_x3 3_a2 3_a 4_{16 a b} 4_ 8 2 4 x y 9 –512 6 64 4 625 3 –64 3 27 12 _{x y}6_ 4 3_{x y}2_ 4 4_a8 6 _x4 3_{x y} 2 5 _a2 4₄ 3 2 1 2 3 ((x y x ) ) 1 2 2 2 (a b ) 1 1 2 3 (x ) 1 2 3 6 (x y ) 3 5 x 2 3 a 1 3 a 1 2 2

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. if n is odd. if n is even. 2. 3. 4. 5. 6. 7. 8. m n m n a   a m n m n n m n m n m n m a a a b b b      m n m n m _an_b  _an _  _bm _ m n _{a b}n_ m n m n _a  _am and n n n n n n a b a b a b a   b n n n a a b b  n_an _bn_{a b} | | n_an _{ a} n_an _a

Look at the examples of each property.

1. a. b.

c. 4_164_{2 |2| 2}4 _{ } d. 4_{(3) |3| 3}4 _{ }

5_{32 }5₍₂₎5 _–2 3

Check Yourself 4

1. Simplify the expressions.

a. b. c. d. e.

2. Perform the operations.

a. ñ3  ò12 b. c. d.

3. Simplify the expressions.

a. b. c. d.

4. Simplify the expressions.

a. b. c. d.

5. Perform the operations.

a. _23₄ b. 3_{x }4_x3 c. 3_a2 _6_a3 d. 432316 4_{x y}5_ 6 32 4 3 81 3 40 3 3 3 4 5 4 x x 3 3 625 5 2 8 3_x3_{x y}2 2 5_x2 _5 _x3 3 3 3 9 3 ₃ 27x 7 –128 5_(–3)5 4_(–2)4 4 16 2. a. ñ5  ñ4 = ó54 = ò20 b. c. 3. a. b. c. 4. a. b. c. 5. a. b. 6. 7. a. b. 8. a. b. c. 4 _x8 _4 2 _x8 _8 _x8 _x 4 3 4 3 2 24 12    12  12 3 3 2 6 5 _  5 _ 5 3 2 3 2 2 6 3 3 2 3 a a a b b b     3 4 3 4 4 12 12 4 3 4 ₃ 3 2 2 2 16 27 3 3 3      2 3 2 3 3 3 2 6 3 2 6 6 5 3  5   3  5 3  125 9  1125 3 2 5_{3 }  ₅2 _6₂₅ 6 3 2 6_{4  2}2 _  ₂2 _3₂ 3 3 _{x y}3_ 6_{ z}2 _ 3_{(x y} 2 3_{)  z}2 _{ x y z}2 2 4 4 4 4 5 x x x x  x 3 3 3 3 3 2 3 2 3  8 3  24 4 2 2 4 4 4 4 4 4 4 4 4 64 64 _{16 2 2 2} 4 4 x x _{x x x x} x x        3 3 3 3 81 81 _{27 3} 3 3    8 8 4 2 2 2    5 _x5_{y }5_{x y} 3 3 3 3_{2  4}3_{2 4  8 2}3_{ 2}

6. Simplify the expressions.

a. b. c. d.

7. Write each expression in its simplest form.

a. b. c. d. e.

Answers

1. a.2 b.2 c.–3 d.–2 e.3x 2. a.6 b.3 c.x d. 3. a. b.5 c.x d. 4. a. b. c. d. 5. a. b. c. d. 6. a. b.₆ 9 c.20_a7 d.24_x7 7. a.6₃ b.20_x c.x d.x2 _e. 9₈₁ 2 12256 27 12 7 4 2 6 a a 12 x x 6 2 2 2 4 xy xy 4 2 2 3 3 3 3 2 5 3₉ 1 2 2 3 x y 3 3_{3 3} 4 _x16 3 3 3 _x27 4 5 x 3 3 6 ₄ 8 3 x x 4 ₃ 5 2 a a 3 6 2 3 4 4 3

C. RADICAL EQUATIONS

Definition

An equation that has a variable in a radicand is called a radical eequation.

For example, the equations ñx = 25, are radical equations.

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

1 3, and 2 – 1 3 5

x  x  x

EXAMPLE

23

Solve x 1 3.

(take the square of both sides)

2 2 ( 1) 3 1 9 8 x x x      Check: Therefore, 8 is a solution. ? ? 8 1 3 9 3 3 3     Solution EXAMPLE

24

Solve 3x 1 5. 2 2 ( 3 1) 5 3 1 25 3 24 8 x x x x       Check: Therefore, 8 is a solution. ? ? 3 8 1 5 25 5 5 5      Solution

EXAMPLE

25

Solve 2x  3 3 8. 2 2 2 3 3 8 2 3 8 – 3 2 3 5 ( 2 3) 5 2 3 25 2 22 11 x x x x x x x              Check: Therefore, 11 is a solution. ? ? ? 2 11 3 3 8 25 3 8 5 3 8 8 8          Solution EXAMPLE

26

_{Solve x}2_{12  x 6.} 2 2 2 2 ( x 12) (x 6) x    2 12 x   12 36 12 –24 –2 x x x     Check: Therefore, –2 is a solution. ? 2 ? ? (–2) 12 – 2 6 4 12 4 16 4 4 4        Solution EXAMPLE

27

Solve 3 3x 2 5. 3 3 3 ( 3 2) 5 3 2 125 3 123 41 x x x x       Check: Therefore, 41 is a solution. ? 3 ? 3 ? 3 3 41 2 5 123 2 5 125 5 5 5        Solution EXAMPLE

28

Solve 4x1 – 5 – 1 0.x  2 2 4 1 – 5 – 1 0 ( 4 1) ( 5 – 1) 4 1 5 – 1 2 x x x x x x x        Check: Therefore, 2 is a solution. ? ? ? 4 2 – 1 – 5 2 – 1 0 9 – 9 0 3 – 3 0 0 0       Solution

EXAMPLE

29

Solve 3x 1 3x 6 5. Check: Therefore, 1 is a solution. ? ? ? 3 1 1 – 3 1 6 5 4 9 5 2 3 5 5 5           2 2 2 2 3 1 5 – 3 6 ( 3 1) (5 – 3 6) 3 1 25 – 2 5 3 6 3 6 10 3 6 30 ( 3 6) 3 3 6 9 3 3 1 x x x x x x x x x x x x                       Solution

(take the square of both sides) (simplify)

Check Yourself 5

1. Solve each equation and check your answer.

a. ñx = 5 b. ñx + 1 = 3 c. óx+1 = 6 d. ñx = –3 e. f. g. h. i. j. k. l. m. n. o. p.

Answer

1. a.25 b.4 c.35 d. e.25 f.8 g.3 h. i.2 j. k. 6 l.3 m.26 n.2ñ2 o.1 p.1 4 1 2 17 5 4 4 2 – 1 x  x 3 3 3x 2 5x 3 2 x 3 1 3 x  2 _{1 2 –} x   x 3 – 2 5 x  x  4 – 3 – 2 – 2 0x x  2x  6 –x 5 1 – 3 0 2 x _ 3 – 5 4 6x   3 x1 – 1 8 2 – 1 7x 

EXERCISES

1

.2

22.. Write the expressions in radical form.

a. b. c. d. e. _{( )a}3 12 6 ab c x 3 – 2 3 ( ) 2 2 3 b 1 2 a

44.. Perform the operations.

a. b. c. d. e. 4 f. ( 2 – 6)( 2 3 ) 17 6 8 3 7 10 2 2 9 27    3 3 11– 3 4 – 27 3 6 2  3 6 3 6 3 2 108 

55.. Solve the equations.

a. b. c. ò7x = x d. e. f. g. h. i. j. k. l. m. n. o. ñ6x – x= 5 p. q. r. s. 1 – 1 2 2 – x 2 x  4 3 – 2 6 – 3 1 8 4 x x  3 _{3 – 6} 2 1 3 1 9 9 x x  1 2 1– 2 8 x  x  x x  6 2 3 x  16 – 2 x x 2 – 5x x– 4 3 7 – 6 4x  2 _{9 5} x   2 _{5 1} x   x 2 – 3 – 2 4x  5 – 1 3 7x   3 7x 6 5 3 5 – 4 6x  3 4x6 3 – 14 8x  2x 1 3

33.. Simplify the expressions.

a. b. c. d. e. f. g. h. i. j. k. l. m. n. o. p. 3 4 q. 4₂33_{2 2}4 5 3 9 3 3 3 5 4 3 6 21 7 3 3 2 250 162 2 4 4 3 27 3 3 2 4 3 3 2 a  a 3 _64x6 3 –27 4 (–5) 5 243 3 5 5 16 10 4 5_a2 3 3

11.. Write the expressions in exponential form and simplify if possible. a. ò21 b. c. d. e. f. g. 3 1 3 5 6 ( 3) 3 5 2 3_{a x}2 3 7 5 2 7

66.. Simplify the expressions.

a. 23  24 b. 53  52 c. 43  42 4 d. 26  2–4 22 e. 3x_{ 3}–2x_{ 3}3x+1 _{f. 2}x_{ 3}x_{ 5}x g. (x – y)2_{ (y – x)}3_{ (x – y)}–4 h. 298_{+ 2}98_{+ 2}98_{+ 2}98

1100.. Solve the equations for x. a. 4x+1_{= 2}3x – 4 _{b. 3}x – 4_{= 81} c. d. 8x+1_{= 4}2x – 3 e. (2x – 3)3_{= (x + 1)}3_{f. (2x – 4)}6_{= x}6 g. 5  23x – 1_{– 2}3x+1_{= 256} h. i. j. 1111.. 3a_{= 25 and 3}b_{= 5. Find} 1122.. 3x_{= 4. Find 9}2x_{+ 3}x+1_.

1133.. 2x_{= 3}y_{= a. Write (12)}xy_{in terms of a, x, and y.}

1144.. 2a – 3 and 4b + 7 are integers with 32a – 3_{= 5}4b + 7_.

Find a + b.

1155.. (x + 1)x2– 16 _{= 1. Find the sum of the possible}

values of x. 1166.. 444_{= 16}x_{. Find x.} 1177.. |x + y – 3| + |x – y – 1| = 0. Find x. 1188.. Simplify |x – 4| + 2|3 – x| if 3 < x < 4. . a b 3 –1 2 –4 3 9 _{((–3) )} 81 (243) x x  _  2 –1 2 3 4 – 3 4 – 128 4 x x x    12 12 12 81 4 4 4 x x x x x x   _   –2 1 3– 9 ( ) 27 x _ x

88.. Perform the operations.

a. (–2)5

 (–2)3 b.–32

 (–3)3 (–3)

c. (–4)2_{ (–2}2₎3_{ (–2}–2₎4 _d. e. (–a)7_{ (–a}4_{)  (–a)}–2 f. (–a–3₎–2_{ (–a}2₎–3_{ (–a}–1₎–1

3 –3 2 6 3 (–2 ) (–2 ) 1 (– ) 2 

99.. Simplify the expressions.

a. 2x – 1_{+ 2}x – 1_{+ 2}x – 1_{+ 2}x – 1 b. ax_{+ 2  a}x_{– 3a}x c. 3x+1_{+ 3}x – 1_{+ 3}x+2_{– 3}x – 2 d. e. f. g. (–1)101_{ (–1)}125_{ (–1)}100_{ (–1)}99_{ (–1)}49 h. ((–39) (–2) ) (–1)3₁₂₅ –4 0_{–127 126}2003 (–3) 6 2     13 11 9 12 10 3 3 – 3 3 3   –7 –5 2 –7 (2 3 ) 36  10 10 10 10 5 5 5 5 x x x x x x x x      

77.. Simplify the expressions.

a. b. c. d. e. f. 5 4 4 5 2 10  –3 –4 (–2) (–2) 1 1 x x a b   –1 – 2 x x a a 3 4 5 5 5 3 2 2

2222.. Find x + y. 42 – 42 – 42 – .... y 72 72 72 ... and x    2233.. Find a. 2a 2a 2a...  12 – 12 – 12 – ... 3. 2244..Simplify 427 11– 23  30 30 30 ... . 2255..4 _x2_–4_x2_–4 _x2_{–... 2.} Find x. 2266..Simplify 1 1 _... 1 1 _. 2 4 4 6  252 254 254  256

2277.. How many different natural number solutions does the equation ñx + ñy = ó300 have?

2288.. Find (ñ2 – 1)7_. (Hint:( 1 ) 2 1) 2 – 1   7 ( 2 1)  57122  57121. 3311.. Find x2_. (Hint: (a – b)3_{= a}3_{– 3ab(a – b) – b}3₎ 3 3 2 – – 2 4. x x  3300.. Evaluate (50 51 52 53) 1.    2299.. Find 15. 8 x 1 9 16 1 9 24 – 16 . x x x     

1199.. Find the sum of the possible values of x that satifsfy the equation |(|2x + 4|) – 2|= 4.

2200.. Find the product of the possible values of x that satisfy the equation |2x + 3| = |5 – 3x|.

2211.. Simplify the expressions.

a. b. c. d. e. f. g. h. i. j. if x < y < 0 k. l. m. 4 7  4 – 7 4 4 15 – 6  15 6 – 3 2.7 5.4 7.1 2 0.2 2_– 2 – x y x y x y x y     ( 6 20 ) ( 5 – 1) 2   4 12  4 – 12 5 5 7 – 2  2 7 2 2 12 16 3 4 18 3 29 – 16 3   2 2 5 – 1 5 1 2.89 6.25 – 1.96 72 48 54  3 4 2 (–3) – –8 16

CHAPTER REVIEW TEST

1

A

11.. Simplify (|(–(–2)3_{) – 2|) – [3 |–|(–3)}2_{– 3||].} A)–8 B)–12 C)8 D)28 22.. Calculate |1 – ñ2| + |1 + ñ2|. A)0 B)ñ2 C)2 D)2ñ2 99.. Simplify 4x+1_{ 8}x – 1_{ 16}x_. A)23x _B)24x _C)29x – 1 _D)28x+1 77.. a is a positive real number. Which one of the

following numbers is negative?

A)a–1 _B)a–2 _C)–a–3 _D)–(–a)3

44.. –1 < x < 3. Calculate |x + 1| + |x – 3|.

A)2x B)2x – 2 C)2 D)4

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

A)–3 B)–2 C)2 D)4

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

A)0 B)1 C)2 D)3

66.. |x – 2| = |x + 3|. Find the value of x which sat-isfies the equation.

A) B) C)1 D) 3 2 1 – 2 1 – 4 88.. Simplify A) B) 1 C)–8 D)8 8 1 – 8 –1 3 1 (– ) . 2       1100.. Simplify A)22m _{B) 2}m +1 _C)2m – 1 _D) 2 1 2 2 m m   4 4 . 2 2 2 2 m m m m m m    

1177.. Evaluate A)–ñ3 B)ñ3 C)ñ3 – 1 D)ñ3 + 1 4 – 12. 1188.. Evaluate A)1 B)ñ3 C)ñ3 – 1 D)ñ3 + 1 2 2 4  12 . 1144.. Evaluate 3ñ8 – 2ñ2 + ò32. A)ñ6 B)8ñ2 C)ò38 D)26ñ2 1166.. Evaluate (3ñ2 – ò12)  (ò18 + 2ñ3). A)36 B) 180 C)ñ6 D)6

1133.. a = 2x_{, b = 3}x_{, and c = 5}x_{. Express 90}x_{in terms}

of a, b, and c. A)(abc)2 _B)a2_bc2 _C)ab2_{c D)a}2_b4_c2 1111.. Find x. A)1 B)3 C)13 D)27 1 2 3 3 3 3 _27. 39 x _ x _ x  1122.. Find a. A)0 B)–1 C)–3 D)3 – 2 2 1 1 2 – 0. 2 a b a b     1155.. Evaluate A)–1 B) C)1 D) 2 3 1 2 2.25 – 2.89 1.44. 1199.. Evaluate A) B)ñ2 C) 9 D)3m+1 4 3 2 3 2 3 2 27 9 _. 12 4 m m m m   2200.. Evaluate A)7 B)2ñ7 C)5ñ7 D)5 5 5 . 7 – 2  7 2

CHAPTER REVIEW TEST

1

B

1100.. k = 1254_{ 64}2_{. How many digits are there in the}

number k? A)10 B)11 C)12 D)13 77.. 3x_{= 5, 2}y_{= 3, and 5}z_{= 0.125. Find x  y  z.} A)3 B) 10 C)–3 D)30 88.. 3x+1_{+ 2  3}x – 1_{= 33. Find x.} A)1 B)2 C)3 D)4 44.. What is half of 220_? A)210 _B)221 _C)219 _D) 10 1 2

55.. Find the simplest form of

A)3x _{B) C)3}2x _D)3x+1 x 1 3 3 3 3 . 9 9 9 x x x x x x     33.. Find |x|. A)2 B)4 C)8 D)16 3 (2 ) 2 3 16. ( ) x x 

99.. a = (52₎3_{, b = 5}(23)_{, and c = 5}(32)_{. Which statement}

is true? A) a > b > c B)a > c > b C)b > c > a D)c > b > a 11.. Evaluate A) –3 B)–3 C)1 D)–1 2 55 72 64 21 82 15 (–1) (–1 ) – (–1) _. (–1 ) – (–1) – (–1)  22.. Evaluate A)3 B) 1 C)2–1 _D)–6 3 –2 –1 –2 –1 2 4 . 2 – 3  66.. Evaluate A) B) C) D) 10 3 5 3 5 6 1 6 69 68 68 67 5 – 5 _. 5 5

1188.. Evaluate A)ñ5 B)2ñ5 C)2 D)1 3 5 – 3 – 5 . 2  1199.. Evaluate A)6 ñ8 B)2 C)8 D)3 ñ8 3 3 2 4 2 4 ... . 2200.. Find x. A)9 B)105 C)114 D)120 5 ₁ 3 7 _{2 3} 8 2. 4 x x    1133.. Simplify A)2ñ3 B)–1 C)1 D)3ñ3 12 48 – 27_. 75 – 2 3  1166.. Evaluate A)2ñ2 B)–2 C)–ñ2 D)4 1 1 . 2 2  2 – 2 1144.. Evaluate A)1 B)2 C)3 D)4 4 3 13 13 – 64 . 1111.. Find x. A) 2 B) 2 C)9 D)10 9 2 3 1 3 243 _{9 .} 81 x x x x    _ 1122.. Evaluate A) B) C) D) 2 3 3 2 41 4 41 16 9 1 1 . 16   1177.. Evaluate A) B)ñ6 – ñ3 C)ñ3 D) 3 2 3 15 11 2 30 – 8 2 15_. 3 – 2 2   1155.. Find n. A) B) C) D) 1 6 11 7 4 7 11 24 4_{x x x}3 2 _xn_.

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.

Objectives

1. What is Statistics?

Statistics is the science of collecting, organizing, summarizing and analyzing data, and draw-ing 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.

A. BASIC CONCEPTS

Definition

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 con-clusions.

For example, the information might be about:



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.

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.

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 car-ried 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 ttable.

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.

8 7 9 3 5

10 8 10 6 8

7 7 6 5 9

4 5 9 6 4

The following table shows the frequency distribution of these marks. It is a frequency distri-bution 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:

We can use the following formula to find the percentage of values in each class:

where

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  % f 100% n  

class frequency percent

A B O AB 4 7 9 5 16 % 28 % 36 % 20 % Total 225 Total 1100

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.

B. PRESENTING AND INTERPRETING DATA

A graph is a diagram

that relates two or more different types of information. Subject Mark Maths Physics Chemistry Biology Computer History Music 9 7 7 8 10 5 6 EXAMPLE

2

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.

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

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

Mathematics

Physics

Chemistry Biology Computer History

Music Lessons 0 1 2 3 4 5 6 7 8 9 10 Marks

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.

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

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

Mathematics

Physics

Chemistry Biology Computer History

Music Lessons 0 1 2 3 4 5 6 7 8 9 10 Marks Mathematics Physics

Chemistry Biology Computer History

Music Lessons 0 1 2 3 4 5 6 7 8 9 10 Marks

EXAMPLE

3

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. Year Production 1992 1993 1994 1995 1996 1997 1998 1999 2000 Car Production 110 659 133 006 99 326 74 862 65 007 91 326 88 506 125 026 140 159

Solution 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.

1992 1993 1994 1995 1996 1997 1998 Year 0 20 40 60 80 100 120 140 160 1999 2000 Number of cars (10 000) 1992 1993 1994 1995 1996 1997 1998 Year 0 20 40 60 80 100 120 140 160 1999 2000 Number of cars (10 000)

EXAMPLE

4

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.

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

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

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?

Turk

ey

USA

United

Kingdom Norway Australia Canada Germany

0 1 2 3 4 5 6 7 8

Share of education expenditures as a percentage of GDP in selected countries * 3.4% 7% 5.2% 6% 6% 6.4% 5.3%

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. 16 d.632000 e.1732

EXERCISES

2

.1

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

8 5 6 10 4 7 2 7 6 3 1 7

5 9 2 6 5 4 6 6 8 4 10 8

Construct a frequency distribution table for this data.

22.. A student’s expenses can be categorized as shown in the table.

Present this information in a bar graph.

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

Expenses Percent oof ttotal iincome. Food 30% Rent 27% Entertainment 13% Clothing 10% Books 15% Other 5%

33.. The following table shows the favorite sport chosen by each of forty students in a class.

Sport Number oof cclass mmembers Football 8 Basketball 5 Volleyball 7 Swiming 12 Wrestling 3 Karate 2 Judo 4

Present this information in a circle graph.

44.. The following table shows the amount of sea fish caught in Turkey in 2003.

Fish Quantity ((1000 ttons) Anchovy 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.

1999 2000 2001 2002 2003 100 000 200 000 300 000 400 000 500 000 600 000

700 000 Hazelnut production in Turkey (tons)

a. Estimate the total production for all five years.

b. Which year had the highest production?

c. Find the combined production for 2002 and 2003.

Definition

An equation that can be written in the form

ax2_{+ bx + c = 0, a 0}

is called a quadratic eequation.

EXAMPLE

1

Determine whether the following equations are quadratic or not.

a. x2_{+ 1 = 0 b. c. 2x}2_{– 3x = 5} d. x2_{– 2x}–1_{+ 3 = 0 e. (x – 1)(x + 2) = 0 f. (x – 2)x}2_{= 0} 2 1 _{– 2 +5 = 0} 2x x Equation 3x2_{+ 5x – 9 = 0} ñ2x2_{+ 5x = 0} (ñ3 + 1)x2_{= 0 ñ3 + 1} 3 ñ2 5 –9 1 – x + 3x2_{= 0 3 –1 1} a b c 5 0 –1 0 1 – x2_{= 0 –1 0 1} 0 0 2 ₁ – + = 0 2 3 4 x x 1 2 1 – 3 1 4 2 1 – + = 0 2 x 1 2

I’m sick of being an unknown

In the equation, a, b, and c are real nnumber ccoefficients and x is a variable. A quadratic equa-tion written in the form ax2_{+ bx + c = 0 is said to be in standard fform. Sometimes, a}

quad-ratic equation is also called a second ddegree eequation. For example,

x2_{+ 3x – 5 = 0, 2x}2_{– x – 1 = 0 and ñ2x}2_{– x + 3 = 0}

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,

3x2_{+ 5x = 0, 2x}2_{= 0 and x}2_{– 9 = 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.

1 2