Tải bản đầy đủ - 0 (trang)
4: Products and Quotients Involving Radicals

# 4: Products and Quotients Involving Radicals

Tải bản đầy đủ - 0trang

398

Chapter 9 • Roots and Radicals

(d) 2x(2x ϩ 2y) ϭ 2x 2x ϩ 2x 2y

ϭ 2x2 ϩ 2xy

ϭ x ϩ 2xy

3

3

3

3

3

3

(e) 2

2( 2

4ϩ 2

10) ϭ 2

8ϩ 2

20 ϭ 2 ϩ 2

20

The distributive property plays a central role when we find the product of two binomials.

For example, (x ϩ 2)(x ϩ 3) ϭ x(x ϩ 3) ϩ 2(x ϩ 3) ϭ x2 ϩ 3x ϩ 2x ϩ 6 ϭ x2 ϩ 5x ϩ 6.

We can find the product of two binomial expressions involving radicals in a similar fashion.

Classroom Example

Multiply and simplify:

(a) ( 27 ϩ 22)( 26 ϩ 210)

(b) ( 25 Ϫ 2)(25 ϩ 8)

EXAMPLE 3

Multiply and simplify:

(a) (23 ϩ 25)(22 ϩ 26)

(b) 127 Ϫ 32127 ϩ 62

Solution

(a) (23 ϩ 25)(22 ϩ 26) ϭ 23(22 ϩ 26) ϩ 25(22 ϩ 26)

ϭ 2322 ϩ 2326 ϩ 2522 ϩ 2526

ϭ 26 ϩ 218 ϩ 210 ϩ 230

ϭ 26 ϩ 322 ϩ 210 ϩ 230

(b) (27 Ϫ 3)( 27 ϩ 6) ϭ 27( 27 ϩ 6) Ϫ 3(27 ϩ 6)

ϭ 27 27 ϩ 627 Ϫ 327 Ϫ 18

ϭ 7 ϩ 627 Ϫ 327 Ϫ 18

ϭ Ϫ11 ϩ 327

If the binomials are of the form (a ϩ b)(a Ϫ b), then we can use the multiplication pattern

(a ϩ b)(a Ϫ b) ϭ a2 Ϫ b2.

Classroom Example

Multiply and simplify:

(a) ( 212 ϩ 5)(212 Ϫ 5)

(b) (7 Ϫ 23)(7 ϩ 23)

(c) (26 ϩ 210)(26 Ϫ 210)

EXAMPLE 4

Multiply and simplify:

(a) (26 ϩ 2)(26 Ϫ 2)

(b) (3 Ϫ 25)(3 ϩ 25)

(c) (28 ϩ 25)(28 Ϫ 25)

Solution

(a) (26 ϩ 2)( 26 Ϫ 2) ϭ (26) 2 Ϫ 22 ϭ 6 Ϫ 4 ϭ 2

(b) (3 Ϫ 25)(3 ϩ 25) ϭ 32 Ϫ (25) 2 ϭ 9 Ϫ 5 ϭ 4

(c) (28 ϩ 25)( 28 Ϫ 25) ϭ ( 28) 2 Ϫ ( 25) 2 ϭ 8 Ϫ 5 ϭ 3

Note that in each part of Example 4, the final product contains no radicals. This happens

whenever we multiply expressions such as 2a ϩ 2b and 2a Ϫ 2b, where a and b are

rational numbers.

(2a ϩ 2b)(2a Ϫ 2b) ϭ ( 2a) 2 Ϫ ( 2b) 2 ϭ a Ϫ b

9.4 • Products and Quotients Involving Radicals

399

Expressions such as 28 ϩ 25 and 28 Ϫ 25 are called conjugates of each other.

Likewise, 26 ϩ 2 and 26 Ϫ 2 are conjugates, as are 3 Ϫ 25 and 3 ϩ 25. Now let’s see

how we can use conjugates to rationalize denominators.

Classroom Example

2

.

Simplify

27 ϩ 23

EXAMPLE 5

Simplify

4

25 ϩ 22

.

Solution

To simplify the expression, the denominator needs to be a rational number. Let’s multiply the

numerator and denominator by 25 Ϫ 22, which is the conjugate of the denominator.

4

25 ϩ 22

ϭ

ϭ

4

25 ϩ 22

и

25 Ϫ 22

25 Ϫ 22

25 Ϫ 22

25 Ϫ 22

is merely a form of 1

4( 25 Ϫ 22)

(25 ϩ 22)( 25 Ϫ 22)

ϭ

4(25 Ϫ 22)

5Ϫ2

ϭ

4( 25 Ϫ 22)

425 Ϫ 422

or

3

3

The next four examples further illustrate the process of rationalizing and simplifying

expressions that contain binomial denominators.

Classroom Example

Rationalize the denominator and

25

simplify

.

214 Ϫ 3

EXAMPLE 6

Rationalize the denominator and simplify

Solution

23

26 Ϫ 2

.

To simplify, we want to multiply numerator and denominator by the conjugate of the denominator, which is 26 ϩ 2.

23

26 Ϫ 2

ϭ

ϭ

Classroom Example

Rationalize the denominator and

20

simplify

.

225 ϩ 210

23

26 Ϫ 2

и

26 ϩ 2

26 ϩ 2

23(26 ϩ 2)

(26 Ϫ 2)(26 ϩ 2)

ϭ

218 ϩ 223

6Ϫ4

ϭ

322 ϩ 223

2

EXAMPLE 7

218 ϭ 2922 ϭ 322

Rationalize the denominator and simplify

14

223 ϩ 25

.

400

Chapter 9 • Roots and Radicals

Solution

To simplify, we want to multiply numerator and denominator by the conjugate of the denominator, which is 223 Ϫ 25.

14

223 ϩ 25

ϭ

ϭ

ϭ

14

223 ϩ 25

и

223 Ϫ 25

223 Ϫ 25

14(223 Ϫ 25)

(2 23 ϩ 25)(2 23 Ϫ 25)

14(2 23 Ϫ 25)

14(223 Ϫ 25)

ϭ

12 Ϫ 5

7

ϭ 2(223 Ϫ 25)    or    423 Ϫ 225

Classroom Example

Rationalize the denominator and

2a Ϫ 4

simplify

.

2a ϩ 5

EXAMPLE 8

Rationalize the denominator and simplify

Solution

2x ϩ 2

2x Ϫ 3

.

To rationalize the denominator, we need to multiply the numerator and denominator by

1x ϩ 3, which is the conjugate of the denominator.

2x ϩ 2

2x Ϫ 3

ϭ

ϭ

Classroom Example

Rationalize the denominator and

6 ϩ 27

.

simplify

27 Ϫ 9

2x ϩ 2

2x Ϫ 3

и

2x ϩ 3

2x ϩ 3

( 2x ϩ 2)(2x ϩ 3)

( 2x Ϫ 3)(2x ϩ 3)

ϭ

x ϩ 22x ϩ 32x ϩ 6

xϪ9

ϭ

x ϩ 52x ϩ 6

xϪ9

EXAMPLE 9

Rationalize the denominator and simplify

Solution

3 ϩ 22

22 Ϫ 6

.

To change the denominator to a rational number, we can multiply the numerator and denominator by 12 ϩ 6, which is the conjugate of the denominator.

3 ϩ 22

22 Ϫ 6

ϭ

ϭ

3 ϩ 22

22 Ϫ 6

и

22 ϩ 6

22 ϩ 6

(3 ϩ 22)( 22 ϩ 6)

( 22 Ϫ 6)( 22 ϩ 6)

ϭ

322 ϩ 18 ϩ 2 ϩ 622

2 Ϫ 36

ϭ

922 ϩ 20

Ϫ34

ϭϪ

922 ϩ 20

34

a

a

ϭϪ

Ϫb

b

9.4 • Products and Quotients Involving Radicals

401

Concept Quiz 9.4

For Problems 1–10, answer true or false.

n

n

n

1. The property 2x 2y ϭ 2xy can be used to express the product of two radicals as

2. The product of two radicals always results in an expression that has a radical even

after simplifying.

3. The conjugate of 5 ϩ 23 is -5 Ϫ 23.

4. The product of (2 Ϫ 27) and (2 ϩ 27) is a rational number.

5. To rationalize the denominator for the expression

28 ϩ 212

6.

22

22

7.

28 ϩ 212

225

4 Ϫ 25

, we would multiply by

25

25

.

ϭ 2 ϩ 26.

ϭ

1

2 ϩ 26

8. The product of (5 ϩ 23) and ( -5 Ϫ 23) is Ϫ28.

9. The product of (223 ϩ 25) and (223 Ϫ 25) is 7.

3

3

3

10. ( 28)( 23) ϭ 223

Problem Set 9.4

For Problems 1– 20, multiply and simplify where possible.

(Objective 1)

1. 27 25

2. 23 210

3. 26 28

4. 26 212

5. 25 210

6. 22 212

3

3

7. 29 26

3

3

8. 29 29

9. 28 212

30. 322(325 Ϫ 227)

31. ( 22 ϩ 6) ( 22 ϩ 9)

32. ( 23 ϩ 4) ( 23 ϩ 7)

33. ( 26 Ϫ 5) ( 26 ϩ 3)

34. ( 27 Ϫ 6) ( 27 ϩ 1)

10. 212 220

11. (3 23) (527)

3

29. 423( 22 Ϫ 225)

35. ( 23 ϩ 26)( 26 ϩ 28)

3

12. (5210)(723)

36. ( 22 ϩ 26)( 28 ϩ 25)

13. (Ϫ26)(524)

14. (523)(Ϫ6210)

37. (5 ϩ 210) (5 Ϫ 210)

15. (326) (426)

16. (227)(5 27)

38. ( 211 Ϫ 2) ( 211 ϩ 2)

17. (522) (4212)

18. (322)(2 227)

39. (322 Ϫ 23)(322 ϩ 23)

19. (4 23) (2215)

20. (727)(2 214)

40. (526 Ϫ 22)(526 ϩ 22)

3

3

For Problems 21– 42, find the products by applying the

form. (Objective 1)

41. (523 ϩ 226)(523 Ϫ 226)

42. (425 ϩ 527)(425 Ϫ 527)

21. 22( 23 ϩ 25)

22. 23( 25 ϩ 27)

23. 26( 22 Ϫ 5)

24. 28( 23 Ϫ 4)

For Problems 43– 54, find each product and express your

nonnegative real numbers. (Objective 1)

25. 22( 24 ϩ 210)

3

3

3

26. 23( 29 Ϫ 28)

43. 2xy 2x

44. 2x 2 y 2y

27. 212( 26 Ϫ 28)

28. 215( 23 Ϫ 25)

3

3

45. 225x 2 25x

3

3

46. 29x 23x

3

3

3

402

Chapter 9 • Roots and Radicals

47. (42a)(32ab)

59.

48. (52a)(62a)

49. 22x( 23x Ϫ 26y)

61.

50. 26x( 22x Ϫ 24y)

51. ( 2x ϩ 5)( 2x Ϫ 3)

63.

52. (3 ϩ 2x) (7 Ϫ 2x)

53. ( 2x ϩ 7)( 2x Ϫ 7)

65.

54. ( 2x ϩ 9)( 2x Ϫ 9)

For Problems 55–70, rationalize the denominators and simplify. All variables represent positive real numbers.

(Objective 2)

55.

57.

67.

69.

3

56.

22 ϩ 4

8

58.

26 Ϫ 2

5

2

25 ϩ 23

10

2 Ϫ 323

4

2x Ϫ 2

2x

2x ϩ 3

2a ϩ 2

2a Ϫ 5

2 ϩ 23

3 Ϫ 22

60.

62.

64.

66.

68.

70.

3

26 ϩ 25

5

322 Ϫ 4

7

2x ϩ 4

2y

2y Ϫ 6

2a Ϫ 3

2a ϩ 1

3 Ϫ 25

4 ϩ 28

23 ϩ 7

10

3 Ϫ 27

Thoughts Into Words

71. Explain how the distributive property has been used in

this chapter.

72. How would you help someone rationalize the denom24

inator and simplify the expression

?

212 ϩ 28

Further Investigations

that you obtained when you rationalized the denominators.

evaluate each expression. Then evaluate the result

2. False

3. False

4. True

1. True

9.5

5. False

6. True

7. False

8. False

9. True

10. True

OBJECTIVES

1

Find the solution sets for radical equations

2

Check the solutions for radical equations

3

Here are some examples of radical equations that involve one variable.

2x ϭ 3

22x ϩ 1 ϭ 5

23x ϩ 4 ϭ - 4

25s Ϫ 2 ϭ 22s ϩ 19

22y Ϫ 4 ϭ y Ϫ 2

2x ϩ 6 ϭ x

403

In order to solve such equations we need the following property of equality.

Property 9.4

For real numbers a and b, if a ϭ b, then

a2 ϭ b2

Property 9.4 states that we can square both sides of an equation. However, squaring both

sides of an equation sometimes produces results that do not satisfy the original equation. Let’s

consider two examples to illustrate the point.

Classroom Example

Solve 2m ϭ 5.

EXAMPLE 1

Solve 2x ϭ 3.

Solution

2x ϭ 3

( 2x) 2 ϭ 32

Square both sides

xϭ9

Since 29 ϭ 3, the solution set is 596.

Classroom Example

Solve 2m ϭ Ϫ4.

EXAMPLE 2

Solve 2x ϭ - 3.

Solution

2x ϭ Ϫ3

( 2x) 2 ϭ (Ϫ3) 2

xϭ9

Square both sides

Because 29 ϶ Ϫ3, 9 is not a solution, and the solution set is л.

In general, squaring both sides of an equation produces an equation that has all of the

solutions of the original equation, but it may also have some extra solutions that do not satisfy

the original equation. Such extra solutions are called extraneous solutions or extraneous

roots. Therefore, when using the “squaring” property (Property 9.4), you must check each

potential solution in the original equation.

We now consider some examples to demonstrate different situations that arise when solving radical equations.

Classroom Example

Solve 27x Ϫ 5 ϭ 3.

EXAMPLE 3

Solve 22x ϩ 1 ϭ 5.

Solution

22x ϩ 1 ϭ 5

( 22x ϩ 1) 2 ϭ 52

2x ϩ 1 ϭ 25

2x ϭ 24

x ϭ 12

Square both sides

404

Chapter 9 • Roots and Radicals

✔ Check

22x ϩ 1 ϭ 5

22(12) ϩ 1 ՘ 5

224 ϩ 1 ՘ 5

225 ՘ 5

5ϭ5

The solution set is 5126.

Classroom Example

Solve 26x ϩ 4 ϭ - 8.

EXAMPLE 4

Solve 23x ϩ 4 ϭ -4.

Solution

23x ϩ 4 ϭ Ϫ4

( 23x ϩ 4) 2 ϭ (Ϫ4) 2

3x ϩ 4 ϭ 16

3x ϭ 12

xϭ4

Square both sides

✔ Check

23x ϩ 4 ϭ Ϫ4

23(4) ϩ 4 ՘ Ϫ4

216 ՘ Ϫ4

4 ϶ Ϫ4

Since 4 does not check (4 is an extraneous root), the equation 23x ϩ 4 ϭ Ϫ4 has no real

number solutions. The solution set is л.

Classroom Example

Solve 327x ϩ 1 ϭ 4.

EXAMPLE 5

Solve 322y ϩ 1 ϭ 5.

Solution

322y ϩ 1 ϭ 5

5

22y ϩ 1 ϭ

3

5 2

( 22y ϩ 1) 2 ϭ a b

3

25

2y ϩ 1 ϭ

9

25

2y ϭ

Ϫ1

9

25

9

2y ϭ

Ϫ

9

9

16

2y ϭ

9

1

1 16

(2y) ϭ a b

2

2 9

8

9

Divided both sides by 3

Multiplied both sides by

1

2

✔ Check

322y ϩ 1 ϭ 5

3

B

3

8

2a b ϩ 1 ՘ 5

9

16

9

ϩ ՘5

B9

9

25 ՘

5

B9

5

3a b ՘ 5

3

5ϭ5

3

8

The solution set is e f .

9

Classroom Example

Solve 22x ϩ 11 ϭ 27x Ϫ 9

EXAMPLE 6

Solve 25s Ϫ 2 ϭ 22s ϩ 19.

Solution

25s Ϫ 2 ϭ 22s ϩ 19

( 25s Ϫ 2) 2 ϭ ( 22s ϩ 19) 2

5s Ϫ 2 ϭ 2s ϩ 19

Square both sides

3s ϭ 21

sϭ7

✔ Check

25s Ϫ 2 ϭ 22s ϩ 19

25(7) Ϫ 2 ՘ 22(7) ϩ 19

233 ϭ 233

The solution set is 576.

Classroom Example

Solve 23x Ϫ 14 ϭ x Ϫ 4.

EXAMPLE 7

Solve 22y Ϫ 4 ϭ y Ϫ 2.

Solution

22y Ϫ 4 ϭ y Ϫ 2

(22y Ϫ 4) 2 ϭ (y Ϫ 2) 2

Square both sides

2y Ϫ 4 ϭ y Ϫ 4y ϩ 4

2

0 ϭ y2 Ϫ 6y ϩ 8

0 ϭ (y Ϫ 4) (y Ϫ 2)

yϪ4ϭ0

or

yϪ2ϭ0

yϭ4

or

yϭ2

Factor the right side

Remember the property: ab ϭ 0 if and only

if a ϭ 0 or b ϭ 0

405

406

Chapter 9 • Roots and Radicals

✔ Check

if y ‫ ؍‬4

if y ‫ ؍‬2

22y Ϫ 4 ϭ y Ϫ 2       22y Ϫ 4 ϭ y Ϫ 2

22(4) Ϫ 4 ՘ 4 Ϫ 2       22(2) Ϫ 4 ՘ 2 Ϫ 2

24 ՘ 2

20 ՘ 0

2 ϭ 2

The solution set is 52, 46.

Classroom Example

Solve 22m ϩ 4 ϭ m.

EXAMPLE 8

0ϭ0

Solve 2x ϩ 6 ϭ x.

Solution

2x ϩ 6 ϭ x

2x ϭ x Ϫ 6

( 2x) 2 ϭ (x Ϫ 6) 2

x ϭ x2 Ϫ 12x ϩ 36

0 ϭ x2 Ϫ 13x ϩ 36

0 ϭ (x Ϫ 4)(x Ϫ 9)

xϪ4ϭ0

or

xϪ9ϭ0

xϭ4

or

xϭ9

We added Ϫ6 to both sides so that the term with

the radical is alone on one side of the equation

Apply ab ϭ 0 if and only if a ϭ 0 or b ϭ 0

✔ Check

if x ‫ ؍‬4

if x ‫ ؍‬9

2x ϩ 6 ϭ x       2x ϩ 6 ϭ x

24 ϩ 6 ՘ 4       29 ϩ 6 ՘ 9

2 ϩ 6 ՘ 4       3 ϩ 6 ՘ 9

8 ϶ 4

9ϭ9

The solution set is 596.

Note in Example 8 that we changed the form of the original equation, 2x ϩ 6 ϭ x, to

2x ϭ x Ϫ 6 before we squared both sides. Squaring both sides of 2x ϩ 6 ϭ x produces

x ϩ 122x ϩ 36 ϭ x 2, a more complex equation that still contains a radical. So again, it pays

to think ahead a few steps before carrying out the details of the problem.

Another Look at Applications

In Section 9.1 we used the formula S ϭ 230Df to approximate how fast a car was traveling,

based on the length of skid marks. (Remember that S represents the speed of the car in miles

per hour, D the length of skid marks measured in feet, and f a coefficient of friction.) This

same formula can be used to estimate the lengths of skid marks that are produced by cars traveling at different rates on various types of road surfaces. To use the formula for this purpose,

we change the form of the equation by solving for D.

230Df ϭ S

30Df ϭ S 2

S2

30f

The result of squaring both sides of the original equation

D, S, and f are positive numbers, so this final equation

and the original one are equivalent

Classroom Example

Suppose that for a parttcular road

surface the coefficient of friction is

0.45. How far will a car traveling at

75 miles per hour skid when the

brakes are applied?

407

EXAMPLE 9

Suppose that for a particular road surface the coefficient of friction is 0.35. How far will a car

traveling at 60 miles per hour skid when the brakes are applied?

Solution

We substitute 0.35 for f and 60 for S in the formula D ϭ

602

ϭ 343

30(0.35)

S2

.

30f

to the nearest whole number

The car will skid approximately 343 feet.

Remark: Pause for a moment and think about the result in Example 9. The coefficient of fric-

tion of 0.35 applies to a wet concrete road surface. Note that a car traveling at 60 miles per

hour will skid farther than the length of a football field.

Concept Quiz 9.5

For Problems 1–10, answer true or false.

1. To solve a radical equation, we can raise each side of the equation to a positive integer

power.

2. Solving the equation that results from squaring each side of an original equation may

not give all the solutions of the original equation.

3

3. The equation 2x Ϫ 1 ϭ - 2 has a solution.

4. Potential solutions that do not satisfy the original equation are called extraneous

solutions.

5. The equation 2x ϩ 1 ϭ - 2 has no solutions.

6. The solution set for 2x ϩ 2 ϭ x is {1, 4}.

7. The solution set for 2x ϩ 1 ϩ 2x Ϫ 2 ϭ -3 is the null set.

3

8. The solution set for 2x ϩ 2 ϭ - 2 is the null set.

9. The solution set for 2x ϭ x is {0}.

3

10. The solution set for 2x ϭ x is {Ϫ1, 0, 1}.

Problem Set 9.5

For Problems 1– 40, solve each equation. Be sure to check

all potential solutions in the original equation. (Objective 1)

1. 2x ϭ 7

2. 2x ϭ 12

3. 22x ϭ 6

4. 23x ϭ 9

5. 23x ϭ -6

6. 22x ϭ -8

7. 24x ϭ 3

8. 25x ϭ 4

9. 32x ϭ 2

10. 42x ϭ 3

11. 22n Ϫ 3 ϭ 5

12. 23n ϩ 1 ϭ 7

13. 25y ϩ 2 ϭ - 1

14. 24n Ϫ 3 Ϫ 4 ϭ 0

15. 26x Ϫ 5 Ϫ 3 ϭ 0

16. 25x ϩ 3 ϭ -4

17. 52x ϭ 30

18. 62x ϭ 42

19. 23a Ϫ 2 ϭ 22a ϩ 4

20. 24a ϩ 3 ϭ 25a Ϫ 4

21. 27x Ϫ 3 ϭ 24x ϩ 3

408

Chapter 9 • Roots and Radicals

Solve Problems 41– 43 by using formulas that are radical

equations. (Objective 3)

22. 28x Ϫ 6 ϭ 24x ϩ 11

23. 22y ϩ 1 ϭ 5

24. 32y Ϫ 2 ϭ 4

25. 2x ϩ 3 ϭ x ϩ 3

26. 2x ϩ 7 ϭ x ϩ 7

27. 2-2x ϩ 28 ϭ x Ϫ 2

28. 2-2x ϭ x ϩ 4

29. 23n Ϫ 4 ϭ 2n

30. 25n Ϫ 1 ϭ 22n

31. 23x ϭ x Ϫ 6

32. 22x ϭ x Ϫ 3

33. 42x ϩ 5 ϭ x

34. 2-x Ϫ 6 ϭ x

35. 2x2 ϩ 27 ϭ x ϩ 3

41. Go to the formula given in the solution to Example 9;

use a coefficient of friction of 0.95. How far will a car

skid at 40 miles per hour? At 55 miles per hour? At 65

miles per hour? Express the answers to the nearest foot.

L

for L. (Remember that

B 32

in this formula, which was introduced in Section 9.1,

T represents the period of a pendulum expressed in

seconds, and L represents the length of the pendulum in

feet.)

42. Solve the formula T ϭ 2p

36. 2x2 Ϫ 35 ϭ x Ϫ 5

37. 2x2 ϩ 2x ϩ 3 ϭ x ϩ 2

43. In Problem 42 you should have obtained the equation

8T 2

L ϭ 2 . What is the length of a pendulum that has

p

a period of 2 seconds? Of 2.5 seconds? Of 3 seconds?

Use 3.14 as an approximation for p, and express the

answers to the nearest tenth of a foot.

38. 2x2 ϩ x ϩ 4 ϭ x ϩ 3

39. 28x Ϫ 2 ϭ x

40. 22x Ϫ 4 ϭ x Ϫ 6

Thoughts Into Words

44. Explain in your own words why possible solutions for

as follows:

(3 ϩ 22x)2 ϭ x 2

45. Your friend attempts to solve the equation

9 ϩ 122x ϩ 4x ϭ x 2

3 ϩ 22x ϭ x

At this step, she stops and doesn’t know how to proceed.

What help can you give her?

Further Investigations

To solve an equation with cube roots, we raise each side of

the equation to the third power. Consider the following

example:

For Problems 46 – 49, solve each of the equations.

46. 2x ϩ 1 ϭ 5 Ϫ 2x Ϫ 4

47. 2x Ϫ 2 ϭ 2x ϩ 7 Ϫ 1

3

2

xϪ4ϭ5

48. 22n Ϫ 1 ϩ 2n Ϫ 5 ϭ 3

49. 22n ϩ 1 Ϫ 2n Ϫ 3 ϭ 2

50. Suppose that we solve the equation x 2 ϭ a2 for x as

follows:

3

(2

x Ϫ 4) 3 ϭ 53

x Ϫ 4 ϭ 125

x ϭ 129

For Problems 51–55, solve each equation.

x2 Ϫ a2 ϭ 0

1x ϩ a21x Ϫ a2 ϭ 0

x ϩ a ϭ 0

or   x Ϫ a ϭ 0

x ϭ -a   or

xϭa

3

52. 2x Ϫ 12 ϭ -2

3

3

54. 24x Ϫ 5 ϭ 3

51. 2x ϩ 7 ϭ 4

3

53. 22x Ϫ 4 ϭ 6

3

55. 24x Ϫ 8 ϭ 10

What is the significance of this result?

1. True

2. False

3. True

4. True

5. True

6. False

7. True

8. False

9. False

10. True

### Tài liệu bạn tìm kiếm đã sẵn sàng tải về

4: Products and Quotients Involving Radicals

Tải bản đầy đủ ngay(0 tr)

×