section 7.1

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Section 7.1. Rational Exponents and Radicals. Find the n th root of a number, if it exists. A. OBJECTIVES. Evaluate expressions containing rational exponents. B. OBJECTIVES. Simplify expressions involving rational exponents. C. OBJECTIVES. DEFINITION. N TH ROOT. - PowerPoint PPT Presentation

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Section 7.1

Rational Exponents and Radicals

OBJECTIVES

A Find the nth root of a number, if it exists.

OBJECTIVES

B Evaluate expressions containing rational exponents.

OBJECTIVES

C Simplify expressions involving rational exponents.

DEFINITION

If a and x are real numbers and n is a positive integer:

x is an nth root of a if xn = a

NTH ROOT

DEFINITIONPRINCIPLE NTH ROOT

If n is a positive integer, then an

denotes the principle nth root of a

DEFINITIONRATIONAL EXPONENTS AND THEIR ROOTS

a1/ n = an

If n is a positive integer andan is a real number:

DEFINITIONRADICAL EXPRESSION WITH AN M/N EXPONENT

am/ n = ( an )m = amn

Provided m and n are positive integers and an

is a real number.

LAWS OF EXPONENTS

I. r s r+s a a = a

If r, s, and t are rational, and a and b are real:

II.

ar

as = ar–s

LAWS OF EXPONENTS

III. r s r s (a ) = a

IV. (ar bs )t = artbst

If r, s, and t are rational, and a and b are real:

Practice Test

Exercise #1

Chapter 7Section 7.1A

Find, if possible.

a. 3

–64

= – 4

3 3= –4

b. –36

It is not a real number.

Find, if possible.

Practice Test

Exercise #4

Chapter 7Section 7.1B

a. –27 –

23

2

3

1=

27–

Evaluate if possible.

23

1=

27–

2

1=

–3 =

19

b. 8– 2

3

= 1

823

= 1

83

2

=

1

22 =

14

Evaluate, if possible.

Section 7.2

Simplifying Radicals

OBJECTIVES

A Simplify radical expressions.

OBJECTIVES

B Rationalize the denominator of a fraction.

OBJECTIVES

C Reduce the order of a radical expression.

DEFINITIONnTH ROOT

( 0)1/n na = a a

when n is a positive integer.

LAWS

I. ( 0)n na = a a

For Simplifying Radical Expressions

( 0)Product ruII. le n n n a, b ab = a b

LAWSFor Simplifying Radical Expressions

( 0 > 0)Quotient rulIII. e n

nn

a ,b a a

= b b

LAWSFor Simplifying Radical Expressions

DEFINITION

ann = | a |

ann

n is an even positive integer

DEFINITION ann

n is an odd positive integer

ann = a

Practice Test

Exercise #7b

Chapter 7Section 7.2A

Simplify.

44 –b. x

= – x

44

= – x

= x

Practice Test

Exercise #8b

Chapter 7Section 7.2A

Simplify.

b. 3

54a4b12

= 3ab4 3 2a

= 3

27 • 2 • a3 • a • b12

= 3

33 a3 b12 • 2a

Practice Test

Exercise #9b

Chapter 7Section 7.2A

Simplify.

b. 3 5

x6

= 3

53

x6

=

35

x 2

Section 7.3

Operations with Radicals

OBJECTIVES

A Add and subtract similar radical expressions.

OBJECTIVES

B Multiply and divide radical expressions.

OBJECTIVES

C Rationalize the denominators of radical expressions involving sums or differences.

DEFINITION

Radical expressions with the same index and the same radicand.

LIKE RADICAL EXPRESSIONS

DEFINITION

The expressions a + b and a – b are conjugates of each other.

CONJUGATE

Practice Test

Exercise #14

Chapter 7Section 7.3A

Perform the indicated operations.

a. 32 + 98

= 16 • 2 + 49 • 2

= 4 2 + 7 2

= 11 2

b. 112 – 28

= 16 • 7 – 4 • 7

= 4 7 – 2 7

= 2 7

Perform the indicated operations.

Practice Test

Exercise #16b

Chapter 7Section 7.3B

Perform the indicated operations.

= 3

3x •3

9x 2 –3

3x •3

16x

= 3

27x3 –3

48x 2

= 3x –3

8 • 6x 2

= 3x – 23

6x 2

333 2b. 3 9 – 16

x x x

Practice Test

Exercise #18b

Chapter 7Section 7.3B

Find the product.

– 6 b + 3 6. 3

–= 6 + 3 6 3

2 2–= 6 3

= 6 – 3

(a + b)(a – b) = a2 – b2

= 3

Practice Test

Exercise #20

Chapter 7Section 7.3C

Rationalize the denominator.

2

x – 5

=

2

x – 5 •

x + 5

x + 5

=

2 ( x + 5)

x – 25

=

2 x + 10x – 25

Section 7.4

Solving Equations Containing Radicals

OBJECTIVES

A Solve equations involving radicals.

OBJECTIVES

B Solve applications requiring the solution of radical equations.

DEFINITIONPOWER RULE OF EQUATIONS

All solutions of the equation P = Q are solutions of the equation Pn = Qn , where n is a natural number.

TO SOLVE EQUATIONS CONTAINING RADICALS

PROCEDURE

• Isolate

• Raise

• Simplify

• Repeat

• Solve

• Check

Practice Test

Exercise #21

Chapter 7Section 7.4A

There is no real number solution

because x + 2 ° negative number.

Solve.

a. x + 2 = – 2

When solving by squaring both sides:x + 2 = 4

x = 2

However, the solution x = 2 does not check.

Solve.

b. x + 2 = x – 10

2 2 + 2 = – 10x x

x + 2 = x 2 – 20x + 100

0 = x 2 – 21x + 98

0 = – 14 – 7x x

x – 14 = 0 oror x – 7 = 0

x = 14 x = 7

Solve. x = 14 x = 7

Check:

9 ° – 3

7 + 2 = ?

7 – 10

14 + 2 = ?

14 – 10

16 = 4

The only solution is x = 14 .

b. x + 2 = x – 10

x = 7,

x = 14,

Practice Test

Exercise #22

Chapter 7Section 7.4A

x – 3 – x = – 3

x – 3 = x 2 – 6x + 9

x – 3 = x – 3

2 2

– 3 = – 3x x

–3 = x 2 – 7x + 9

Solve.

0 = – 4 – 3x x

x – 4 = 0 oror x – 3 = 0

0 = x 2 – 7x + 12

–3 = x 2 – 7x + 9

x = 4 x = 3

x – 3 – x = – 3

Solve.

x = 4 x = 3

Check:

–3 = – 3

x = 4, 4 – 3 – 4 = ?

– 3

The solutions are x = 4 or x = 3 .

x = 3, 3 – 3 –3 = ?

– 3

–3 = – 3

x – 3 – x = – 3

Solve.

Section 7.5

Complex Numbers

OBJECTIVES

A Write the square root of a negative integer in terms of i.

OBJECTIVES

B Add and subtract complex numbers.

OBJECTIVES

C Multiply and divide complex numbers.

OBJECTIVES

D Find powers of i.

DEFINITIONCOMPLEX NUMBER

a + bi

If a and b are real numbers, the following is a complex number:

Real part Imaginary part

RULESADDING AND SUBTRACTING COMPLEX NUMBERS

(a + bi) + (c + di) = (a + c) + (b +d)i

(a + bi) – (c + di) = (a – c) + (b – d)i

PROCEDUREDIVIDING ONE COMPLEX NUMBER BY ANOTHER

Multiply the numerator and the denominator by the conjugate of the denominator.

Practice Test

Exercise #26b

Chapter 7Section 7.5A

Write in terms of i.

b. –98 = –1 • 49 • 2

= i • 7 2

= 7i 2

Practice Test

Exercise #27b

Chapter 7Section 7.5B

Find.

b. 5 + 2 – 7 – 8i i

= 5 + 2i – 7 + 8i

= – 2 + 10i

= 5 – 7 + 2i + 8i

Practice Test

Exercise #29b

Chapter 7Section 7.5C

Find.

b.

4 – 3i3 – 5i

3 + 4 – 3 =

3

5

3 + 5 – 5

i i

i i

2 2

4 3 + 4 5 + –3 3 + –3 5 =

3 – 5

i i i i

i

Find.

b.

4 – 3i3 – 5i

2 2

4 3 + 4 5 + –3 3 + –3 5 =

3 – 5

i i i i

i

=

12 + 20i – 9i – 15i 2

9 – 25i 2

Find.

b.

4 – 3i3 – 5i

=

12 + 20i – 9i – 15i 2

9 – 25i 2

=

12 + 11i + 159 + 25

=

27 + 11i34

= 2734

+ 1134

i

Practice Test

Exercise #30b

Chapter 7Section 7.5D

Write the answer as 1, -1, i or -i.

b. i – 9

=

1

i 9 =

1

i8 • i

= 1

i 4 2

• i

=

1i

= 1

1 • i =

1 • ii • i

=

i

i 2 =

i–1

= – i

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