Chapter 5 Radical Expressions and Equations - Grade 11 Pre Calculus

[Pages:66]Chapter 5 Radical Expressions and Equations

Section 5.1 Working With Radicals

Section 5.1 Page 278 Question 1

Mixed Radical Form 47

50 = 25(2) = 5 2 -11 8

- 200 = - 100(2) = -10 2

Entire Radical Form 4 7 = 42 (7) = 112

50 -11 8 = - 112 (8) = - 968

- 200

Section 5.1 Page 278 Question 2

a) 56 = 4(14) = 2 14

b) 3 75 = 3 25(3) = 15 3

c) 3 24 = 3 8(3) = 3 23 (3) = 23 3

d) c3d 2 = c2 (c)(d 2 ) = cd c

Section 5.1 Page 278

Question 3

a) 3 8m4 = 3 4(2)(m2 )(m2 ) = 6m2 2, m R

b) 3 24q5 = 3 23 (3)(q3 )(q2 ) = 2q 3 3q2 , q R

c) -2 5 160s5t6 = -2 5 25 (5)(s5 )(t5 )(t) = -4st 5 5t , s,t R

MHR ? Pre-Calculus 11 Solutions Chapter 5

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Section 5.1 Page 279 Question 4

Mixed Radical Form 3n 5

3 -432 = 3 2(-6)3 = -6 3 2 1 3 7a 2a

3 128x4 = 3 43 (2)x3 (x) = 4x3 2x

Entire Radical Form 3n 5 = (3n)2 (5)

= 45n2 , n 0 or - 45n2 , n < 0 3 -432

1 2a

3

7a

=

3

1 2a

3

(7a)

= 3 1 (7a) 8a3

=

3

7 8a2

,a

0

3 128x4

Section 5.1 Page 279 Question 5

a) For 15 5 and 8 125 , express the second radical in terms of 5 .

15 5

8 125 = 8 25(5) = 40 5

b) For 8 112z8 and 48 7z4 , express both radicals in terms of 7 .

8 112z8 = 8 16(7)z8 = 32z4 7

48 7z4 = 48z2 7

c) For -354 w2 and 34 81w10 , express the second radical in terms of w2 .

-354 w2

34 81w10 = 34 34 (w8 )(w2 ) = 9w2 4 w2

d) For 63 2 and 6 3 54 , express the second radical in terms of 3 2 .

63 2

6 3 54 = 6 4 33 (2)

= 183 2

Section 5.1 Page 279 Question 6

MHR ? Pre-Calculus 11 Solutions Chapter 5

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a) 3 6 = 9(6) 10 = 100 7 2 = 49(2)

= 54

= 98

The numbers from least to greatest are 3 6 , 7 2 , and 10.

b) -2 3 = - 4(3) -4 = - 16 = - 12

-3 2 = - 9(2) = - 18

-2

7 =- 2

4

7 2

= - 14

The numbers from least to greatest are -3 2 , ?4, -2 7 , and -2 3 . 2

c) 3 21

33 2 = 3 27(2)

2.8 = 3 2.83

2 3 5 = 3 8(5)

= 3 54

= 3 21.952

= 3 40

The numbers from least to greatest are 3 21 , 2.8, 2 3 5 , and 33 2 .

Section 5.1 Page 279 Question 7

-2 3 = -3.464... ?4 -3 2 = -4.242... -2 7 = -3.741... 2

The numbers from least to greatest are -3 2 , ?4, -2 7 , and -2 3 . 2

Section 5.1 Page 279 Question 8

a) - 5 + 9 5 - 4 5 = 4 5

b) 1.4 2 + 9 2 - 7 = 10.4 2 - 7

c) 4 11 -1- 54 11 +15 = -4 4 11 +14

d) - 6 + 9 10 - 5 10 + 1 6 = - 2 6 + 2 10

2

2

3

3

Section 5.1 Page 279 Question 9

a) 3 75 - 27 = 3 25(3) - 9(3) = 15 3 - 3 3 = 12 3

MHR ? Pre-Calculus 11 Solutions Chapter 5

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b) 2 18 + 9 7 - 63 = 2 9(2) + 9 7 - 9(7) =6 2+9 7 -3 7 =6 2+6 7

c) -8 45 + 5.1- 80 +17.4 = -8 9(5) - 16(5) + 22.5 = -24 5 - 4 5 + 22.5 = -28 5 + 22.5

d) 2 3 81 + 3 375 - 4 99 + 5 11 = 2 3 27(3) + 3 125(3) - 4 9(11) + 5 11

3

4

3

4

= 2 3 3 + 53 3 -12 11 + 5 11 4

= 133 3 - 7 11 4

Section 5.1 Page 279 Question 10

a) 2 a3 + 6 a3 = 8 a3 = 8a a, a 0

b) 3 2x + 3 8x - x = 3 2x + 3 4(2)(x) - x = 3 2x + 6 2x - x = 9 2x - x, x 0

c) -4 3 625r + 3 40r4 = -4 3 125(5)(r) + 3 8(5)(r3)(r) = -20 3 5r + 2r 3 5r = (2r - 20) 3 5r = 2(r -10) 3 5r , r R

d) w 3 -64 + 3 512w3 - 2 50w - 4 2w = - 4w + 8w - 2 25(2)(w) - 4 2w

5

55

5 55

= 4w - 2 2w - 4 2w 5

= 4w - 6 2w, w 0 5

MHR ? Pre-Calculus 11 Solutions Chapter 5

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Section 5.1 Page 279 Question 11

w = 6.3 1013 - p w = 6.3 1013 - 965 w = 6.3 48 w = 25.2 3 The exact wind speed of a hurricane if the air pressure is 965 mbar is 25.2 3 m/s.

Section 5.1 Page 279 Question 12

c2 = 122 + 122 c2 = 144 + 144 c2 = 288 c = 288

c = 12 2 The length of the hypotenuse is 12 2 cm.

Section 5.1 Page 280 Question 13

Distance to Mars: d = 3 25n2

Distance to Mercury: d = 3 25n2

d = 3 25(704)2

d = 3 25(88)2

d = 3 25[64(11)]2

d = 3 25[8(11)]2

d = 16 3 25(11)2

d = 4 3 25(11)2

d = 16 3 3025

d = 4 3 3025

The difference between the distances of Mars and Mercury to the Sun is

16 3 3025 ? 4 3 3025 , or 12 3 3025 million kilometres.

Section 5.1 Page 280 Question 14

s = 10d s = 10(12) s = 120 s = 2 30 The speed of a tsunami with depth 12 m is 2 30 m/s, or 11 m/s to the nearest metre per second.

Section 5.1 Page 280 Question 15

MHR ? Pre-Calculus 11 Solutions Chapter 5

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a) Since the given area of the circle (r2) is 38 m2, the radius is 38 m. So, the diagonal of the square, which is also the diameter of the circle is 2 38 m.

b) First, find the side length of the square.

( ) s2 + s2 =

2

2

38

2s2 = 152 s2 = 76

s = 76

s = 2 19

( ) The perimeter of the square is 4 2 19 , or 8 19 m.

Section 5.1 Page 280 Question 16

First, find the half-perimeter for a triangle with side lengths 8 mm, 10 mm, and 12 mm. 0.5(8 + 10 + 12) = 15 Find the area of the triangle using Heron's formula. A = s(s - a)(s - b)(s - c)

A = 15(15 - 8)(15 -10)(15 -12)

A = 15(7)(5)(3)

A = 1575

A = 15 7 The area of the triangle is 1575 mm2 or 15 7 mm2.

Section 5.1 Page 280 Question 17

Using a diagram, the points lie on the same straight

y

line.

Use the Pythagorean theorem to find the distance

16

between the starting and ending points.

c2 = 72 + 142

12

c2 = 49 + 196

c2 = 245

8

18 - 4 = 14

c = 245

c= 7 5 The ant travels 7 5 units.

4

10 - 3 = 7

x

4

8

12

16

MHR ? Pre-Calculus 11 Solutions Chapter 5

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Section 5.1 Page 280 Question 18

Since the area of the entire square backyard is 98 m2, the side length is 98 m. Since the area of the green square is 8 m2, the side length is 8 m. Find the perimeter of one of the rectangular flowerbeds.

( ) P = 2 8 + 2 98 - 8

P = 2 98 P = 14 2 The perimeter is 14 2 m.

Section 5.1 Page 280 Question 19

98 m 8m

Brady is correct. Kristen's final radical, 5y 4 y3 , is not in simplest form. Express the radicand as a product of prime factors and combine identical pairs. 5 y 4 y3 = 5 y 2(2)( y)( y)( y)

= 10 y2 y

Section 5.1 Page 280 Question 20

2 216 = 2 36(6) 3 96 = 3 16(6)

4 58

= 12 6

= 12 6

The expression 4 58 is not equivalent to 12 6 .

6 24 = 6 4(6) = 12 6

Section 5.1 Page 281 Question 21

From the given information, ? square ABCD has perimeter 4 m, so CD = 1 m ? UCDE is an equilateral triangle, so CDE = 60? ? CA is a diagonal, so DAC = 45?

In UADF, draw the perpendicular from F to AD, meeting AD at point G. Let the height of FG be x. Then, in UFDG, FDG = 30? and DG = 1 ? x.

1 m

x

60?

45?

G 1?x

MHR ? Pre-Calculus 11 Solutions Chapter 5

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tan 30? = x 1- x

1= x 3 1- x 1- x = 3x

1 = x( 3 +1)

x= 1 3 +1

x = 3 -1 2

Next, in UAFG, FG = GA = x. AF2 = x2 + x2

AF2

=

3 -1 2

2

+

3 -12 2

=

2

4

-2 4

3

=2- 3

AF = 2 - 3

The exact length of AF is 2 - 3 m.

Section 5.1 Page 281 Question 22

Since AB = 18 cm, AC = 9 cm and CD = 4.5 cm. Consider right UADF, where DF = 4.5 cm and AD = 13.5 cm. Use the Pythagorean theorem. (AF)2 = (AD)2 ? (DF)2 (AF)2 = 13.52 ? 4.52 (AF)2 = 162

AF = 162

AF = 9 2

Consider similar triangles UADF and UABE.

MHR ? Pre-Calculus 11 Solutions Chapter 5

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