Unit 3 Radical and Absolute Value Functions and Equations

1

Unit 3 Radical and Absolute Value

Functions and Equations

General Outcome:

• Develop algebraic reasoning and number sense.

• Develop algebraic and graphical reasoning through the study of relations.

Specific Outcomes:

3.1 Demonstrate an understanding on the absolute value of real numbers.

3.2 Solve problems that involve operations on radicals and radical expressions with numerical

and variable radicands.

3.3 Solve problems that involve radical equations (limited to square roots).

3.4 Graph and analyze absolute value functions (limited to linear and quadratic functions) to

solve problems.

Topics

• Working with Radicals (Outcome 3.2) Page 2

• Multiplying & Dividing Radicals (Outcome 3.2) Page 10

• Solving Radical Equations (Outcome 3.3) Page 20

• Absolute Value (Outcome 3.1) Page 31

• Absolute Value Functions & (Outcome 3.4) Page 37

Their Graphs

• Absolute Value Equations (Outcome 3.4) Page 45

2

Unit 3 Radical and Absolute Value

Functions and Equations

Working with Radicals:

Radicals include square roots, cube roots, 4th roots, etc.

square root→ −

3 cube root→ − index radicand

4 fourth root→ −

5 5th root→ −

Like Radicals:

Radicals with the same radicand and index are called like

radicals.

Ex) Like Radicals Unlike Radicals

5 7 and 7− 2 5 and 2 3

3 22 5

3x and 3 25x 4 5a and 3 5a

Like radicals can be combined together the same way like terms

are combined together.

Ex) 3 5 12 5 7 5 4 7− + + +

3

Reducing Radicals:

When we simplify radicals we make an entire radical into a

mixed radical.

Ex) Simplify the following radicals.

60 180 162 3 120

548y

94 c

4 23 320x y

Creating Entire Radical:

When going from a mixed radical to an entire radical you just do

the opposite.

Ex) Express the following as an entire radical

3 15 5 10 36 12

3 25b ab

4

Multiplying Radicals:

To multiply radicals with the same index together, simply

multiply the coefficients together then multiply the radicands

together. (Your final answer should be in reduced form.)

Ex) Multiply the following.

9 2 4 7 2 3 5 6 22 5 3 3 6x x

Ex) Simplify the following radicals.

50 3 2+ 27 3 5 80 2 12− + − −

4 4 9c c−

5

Working with Radicals Assignment:

1) Reduce the following radical expressions by expressing each as a mixed

radical.

a) 56 b) 3 75 c) 3 24

d) 3 2c d e) 4 23 250x y f) 43 8m

g) 53 24q h) 5 632 160s t− i)

72

33

815

8

xx y

y

6

2) Simplify each radical expression.

a) 5 9 5 4 5− + − b) 14 2 9 2 7+ −

c) 4 411 1 5 11 15− − + d) 9 5 1

6 10 10 62 2 3

− + − +

e) 3 75 27− f) 15 5 8 125+

7

g) 2 18 9 7 63+ − h) 8 2 48 112 48 7x x x−

i) 8 45 5.1 80 17.4− + − + j) 2 2 104 435 3 81w w w− +

k) 3

32 37581 4 99 5 11

3 4+ − + l) 3 36 2 6 54+

8

m) 3 32 6a a+ n) 3 2 3 8x x x+ −

o) 4334 625 40r r r− + p)

33

3 512 264 50 4 2

5 5 5

w ww w− + − −

3) The distance, d, in millions of kilometres, between a planet and the Sun is a

function of the length, n, in Earth-days, of the planet’s year. The formula is

23 25d n= . The length of 1 year on Mercury is 88 Earth-days, and the length

of 1 year on Mars is 704 Earth-days. Determine the difference between the

distances of Mercury and Mars from the Sun. Express your answer as an exact

value.

9

4) A square is inscribed in a circle. The area of the circle is 38 m2.

2

circle 38 mA =

a) Determine the exact length of the diagonal of the square.

b) Determine the exact perimeter of the square.

10

Multiplying Radicals (Continued):

Remember to reduce radicals when possible after multiplying.

Ex) Simplify the following.

a) ( )7 3 5 5 6 3− b) ( )( )8 2 5 9 5 6 10− +

c) ( )3 3 39 2 4 7 28w w +

Ex) The area of the square shown below is 32 cm2.

a) What is the exact perimeter of the

triangle?

b) Determine the height of the triangle.

c) What is the exact area of the triangle?

11

Dividing Radicals:

Division of radicals works the same as multiplication. To divide

radicals with the same index together, simply divide the

coefficients, then divide the radicands. (Your final answer

should be in reduced form.)

Ex) Simplify the following.

36 15

9 3 32

15028

Rationalizing Denominators Part I:

When dealing with fractions we do not leave radicals in the

denominator. (Radicals in the numerator are OK.)

Ex) 5

7

12

Ex) Rationalize the following.

a) 7

34 b)

113

5

c) x6

249 d) 38

5

x

Rationalizing Denominators Part II:

When two terms appear in the denominator that involve a radical

a different approach is needed.

Simplify the following.

( )( )543543 +− ( )( )327327 −+

13

Ex) 76

5

+

Ex) Rationalize the following denominators

a) 64

35

+ b)

75

11

−

c) 32

63

+

− d)

254

103

−

14

Multiplying & Dividing Radicals Assignment:

1) Simplify the following radical expressions.

a) ( )2 5 7 3 b) ( )32 7 2− c) ( )4 42 48 5

d) ( )24 19 2x x e) ( )7 43 354 6y y f) 26 34

tt t

g) ( )11 3 4 7− h) ( )7 1 2 6 5 6 8− − + +

15

i) ( )4 5 3 8 3 15 5a a+ − + j) ( )3 33 4 12 2 8k− +

k) ( )( )8 7 2 2 3+ − l) ( )( )3 2 15 3 15+ −

m) ( )2

36 2 4 13− n) ( )( )6 2 2 2 3 5 1− + − +

16

o) ( )( )15 2 2 6x x+ − p) ( )( )31 10 8 2 7 5x x− +

q) ( )2

9 2 4 6n n− r) ( )( )233 310 4 4 2 6 3 12r r r r r− +

2) Simplify the following. Rationalize each denominator.

a) 80

10 b)

2 12

4 3

−

17

c) 3 22

11 d)

5

3

3 135

21

a

a

e) 20

10 f)

21

7m

−

g) 2 5

3 12x

− h) 3

620

5

t

18

i) 5 5

4

9 432 7 27

33

p p

p

− j)

73

3

6 4

14

n

n

k) 5

2 3− l)

7 2

6 8+

m) 7

5 2 2

−

− n)

3 13

3 13

+

−

19

o) 4

6 9

r

r + p)

18 3

24

n

n

q) 8

4 6t− r)

5 3

10 2

y

+

3) Simplify 1 5 1 5

2 3 2 3

+ −

− − 4) Determine the restrictions on

2 14

3 5

x

x−

20

Solving Radical Equations:

A radical equation is one in which a variable appears inside a

radical.

Ex) 3 15x + = , 8 4 7x x− + = +

Equations involving 1 Radical:

Steps for Solving:

• Isolate (solve for) the radical.

• Square both sides of the equation.

• Solve.

• Check for Extraneous Roots.

Ex) Solve the following radical equations. Check for

extraneous roots.

a) 1 3 5x + + = b) 2 4x x= − +

21

Equations involving 2 Radicals:

Steps for Solving:

• Isolate (solve for) one of the radicals.

• Square both sides of the equation.

• Isolate (solve for) the other radical.

• Square both sides of the equation.

• Solve.

• Check for Extraneous Roots.

Ex) Solve the following radical equations. Check for

extraneous roots.

a) 4 5 2 1 2x x+ − − =

22

b) 3 4 11x x+ − = +

c) 1 5 0x x− − − =

23

Solving Radical Equations Assignment:

1) Solve the following radical equations. Check for and identify any extraneous

roots.

a) 2 3x = b) 8 4x− =

c) 7 5 2x= − d) 8 13a + =

24

e) 2 4y− = − f) 3 8 6x − = −

g) 5 2 6n− = − − h) 3 1 7 14n− − + = −

25

i) 7 4 2 1 17x− − − = j) 12 3 5 8 x= − + −

2) Solve the following radical equations. Check for and identify any extraneous

roots.

a) 2 3 5x − = b)

2 12 8x x+ =

26

c) 2

11 12

aa+ = − d) 22 2 7 14n n+ − =

e) 5 3 5x x+ − = f) 2 30 8x x+ =

27

g) 5 1d d+ = − h) 1

5 3 13

xx x

++ = −

3) Solve the following radical equations. Check for and identify any extraneous

roots.

a) 2 8x = b) 3 7n n− = −

28

c) 5 2002

a= d) 5 3y y+ =

e) 5 2 1x x+ = − f) 26 1 17y y− = − +

29

g) 5 9 3 4 2a a− − = + − h) 19 2 7x x+ + − =

4) Collision investigators can approximate the initial velocity, v, in kilometres per

hour, of a car based on the length, l, in metres, of the skid mark. The formula

12.6 8v l= + , 0l , models the relationship. What length of skid is expected

if a car is travelling 50 km/h when the brakes are applied? Express your answer

as an exact value.

30

5) Two more than then square root of a number, n, is equal to the number. Model

this situation using a radical equation and determine the value(s) of n

algebraically.

6) Solve for a in the given equation.

3 2x ax= +

31

Absolute Value:

The absolute value of a number can be thought of as the distance

from that number to zero on a number line. The absolute value is

a positive number.

Ex) a) Determine the absolute value of 6−

b) Determine the absolute value of 5

To symbolize absolute value when use vertical bars around a

number or expression

14− means the absolute value of 14−

In general the absolute value of a real number x is defined as

, 0

, 0

x xx

x x

=

−

32

Ex) Evaluate the following.

a) 3 b) 11− c) 28 6−

d) 4 6− − e) 5 3 2 7− − f) 22(5 7) 6− − +

Ex) On stock markets, individual stock and bond values

fluctuate a great deal. A particular stock opened the month

at $13.55 per share, dropped to $12.70, increased to

$14.05, and closed the month at $13.85. Determine the

total change in the value of this stock for the month. This

total shows how active the stock was that month.

33

Absolute Value Assignment:

1) Evaluate the following.

a) 9 b) 0 c) 7−

d) 4.728− e) 6.25 f) 1

52

−

g) 8 15− h) 3 8− − i) ( )7 3− −

j) 2 5(3)− k) 2 6 ( 11)− − − l) 9.5 12.3− −

34

m) 1 3

3 52 4

−+ n) 23( 2) 5( 2) 7− + − +

o) 4 13 6 ( 9) 8 17 2− + + − − − − + −

2) Order each of the following from least to greatest.

a) 3 1

0.8 , 1.1 , 2 , , 0.4, 1 , 0.85 4

− − − −

b) 7 1

2.4, 1.3 , , 1.9, 0.6 , 1 , 2.25 10

−− − −

35

3) Southern Alberta often experiences dry chinook winds in winter and spring

that can change temperatures by a large amount in a short time. On a particular

day in Warner, Alberta, the temperature was 11 C− in the morning. A chinook

wind raised the temperature to 7 C+ by afternoon. The temperature dropped to

9 C− during the night. Create an absolute value expression the represents the

total change in temperature for that day and determine the total change.

4) When Vanessa checks her bank account on-line, it shows the following

balances:

Date Balance

Oct. 4 $359.22

Oct. 12 $310.45

Oct. 17 $295.78

Oct. 30 $513.65

Nov. 5 $425.59

Create an absolute value expression that represents the total change in

Vanessa’s bank balance during this period.

36

5) As part of a scavenger hunt, Toby collects items along a specified trail.

Starting at the 2 km marker, he bicycles east to the 7 km marker, and then turns

around and bicycles west back to the 3 km marker. Finally, Toby turns back

east and bicycles until the total distance he has travelled is 15 km.

a) How many kilometres does Toby travel in the last interval?

b) At what kilometre marker is Toby at the end of the scavenger hunt?

37

Absolute Value Functions and Their Graphs:

• When graphing an absolute value function we can begin

my considering the graph of the function inside the

absolute value .

• To obtain the graph of the absolute value function simply

flip up any sections of the graph that are below the x-axis.

(Absolute value results in a positive value.)

Ex) Given the graph of ( )y f x= sketch the graph of ( )y f x=

a) b)

Ex) Sketch the graph of 2 3y x= − , then write 2 3y x= − as a

piecewise function.

38

Ex) Consider the absolute value function 2 2 8y x x= − + +

a) Determine the y-intercept and the x-intercept

b) Sketch the graph

b) State the domain and range.

c) Express as a piecewise function.

39

Absolute Value Functions & Their Graphs Assignment:

1) Create a table of values for ( )y f x= given the table of values for ( )y f x= .

a) x ( )y f x= x ( )y f x=

2− 3− 2−

1− 1− 1−

0 1 0

1 3 1

2 5 2

b) x ( )y f x= x ( )y f x=

2− 0 2−

1− 2− 1−

0 2− 0

1 0 1

2 4 2

2) The point ( )5, 8− − is on the graph of ( )y f x= . Identify the corresponding

pint on the graph of ( )y f x= .

3) The graph of ( )y f x= has an x-intercepts of 3 and 4− , and a y-intercept of 7− .

Identify the x-intercepts and y-intercept of the graph of ( )y f x= .

40

4) Given the graph of ( )y f x= , sketch the graph of ( )y f x= .

a) b)

c) d)

e) f)

41

5) Sketch the graph of each absolute value function. State the intercepts, domain

and range.

a) 3y x= − −

x-intercept(s):

y-intercept:

Domain:

Range:

b) 2 4y x= −

x-intercept(s):

y-intercept:

Domain:

Range:

c) 1

22

y x= −

x-intercept(s):

y-intercept:

Domain:

Range:

42

d) 2( 2) 1y x= − −

x-intercept(s):

y-intercept:

Domain:

Range:

c) 2( 3) 2y x= − − −

x-intercept(s):

y-intercept:

Domain:

Range:

6) Write a piecewise function that represents each graph given below.

a)

43

b)

c)

d)

e)

44

f)

7) Express the following as piecewise functions.

a) 4y x= − b) 3 5y x= +

c) 2 1y x= − + d) 2 6y x x= − −

45

Solving Absolute Value Equations:

Absolute value equations are equations that contain a variable

inside an absolute value sign.

Ex) 2 0x − = , 3 7 5x x− = −

Steps for Solving:

• Break the equation up into 2 equations. In one case

consider what is in the absolute value sign to be positive,

and in the other consider it to be negative.

• Solve each equation.

• Check each solution for Extraneous Roots

Ex) Solve the absolute value equations given below. Check for

extraneous roots.

a) 2 3 5x − =

b) 4 2 1x x− = +

46

c) 2 3 3x x− = −

d) 2 5 5 3x x− = −

d) 3 4 12 9x − + =

47

Ex) Kokanee salmon are sensitive to water temperature. If the

water is too cold, egg hatching is delayed, and if the water

is too warm, the eggs die. Biologists have found that the

spawning rate of the salmon is greatest when the water is

at an average temperature of 11.5 C with an absolute value

difference of 2.5 C . Write and solve the limits of the ideal

temperature range for the Kokanee salmon to spawn.

Ex) Determine an absolute value equation in the form

ax b c+ = given its solutions on the number line.

a)

b)

48

Solving Absolute Value Equations Assignment:

1) Solve the following absolute value equations. Be sure to check for and identify

any extraneous roots.

a) 7 12x + = b) 3 4 5 7x − + =

c) 2 6 12 4x + + = − d) 6 2 14 42x− − = −

49

e) 2 7 4a a+ = − f) 7 3 11x x+ = −

g) 1 2 2n n− = + h) 3 3 2 5x x+ = −

i) 3 2 7 3 12a a+ = + j) 2 3x x x= + −

50

k) 2 2 2 3 4x x x− + = − l) 2 29 9x x− = −

m) 2 1x x− = n) 2 2 16 8x x− − =

2) Bolts are manufactured at a certain factory to have a diameter of 18 mm and

are rejected if they differ from this by more than 0.5 mm. Create and then solve

an absolute value equation in the form d a b− = to describe the acceptable

limits for the diameter, d, in millimietres of these bolts, where a and b are real

numbers.

51

3) Create an absolute value equation whose solution is given by 7 4.8x =

4) The moon travels in an elliptical orbit around the Earth. The distance between

Earth and the moon changes as the moon travels in this orbit. The point where

the moon’s orbit is closest to Earth is called perigee and the point when it is

farthest from the Earth is called apogee. The equation 381550 25150d − =

can be used to find these distances, d, in kilometres. Solve this equation to find

the perigee and apogee of the moon’s orbit.

5) Explain, without solving, why the equation 3 1 2x + = − has no solution, while

the equation 3 1 4 2x + − = − does.

52

Answers

Working with Radicals Assignment:

1. a) 2 14 b) 15 3 c) 32 3 d) cd c e) 235 2x xy f) 26 2m

g) 232 3q q h) 2 234 20st s− i) 4

3153

2

xx

2. a) 4 5 b) 23 2 7− c) 44 11 14− + d) 2

6 2 103

−+ e) 12 3

f) 55 5 g) 6 2 6 7+ h) 416 7x− i) 28 5 12.5− +

j) 2 2426w w− k) 3133 7 11

4− l) 39 2 m) 8a a

n) 9 2x x− o) 318 5r r− p) 4

6 25

w w−

3. 312 3025

4. a) diagonal 2 38= b) perimeter 44 38=

Multiplying & Dividing Radicals Assignment:

1. a) 14 15 b) 56− c) 44 15 d) 4 38x x e) 3 233 12y y

f) 33

62

t g) 3 11 4 77− h) 9 6 1− + i) 15 33 5a +

j) 33 16 4k− k) 8 14 24 7 2 2 6− + − l) 3 5 27−

m) ( )3 336 4 48 13 2 208− + n) 4 3 3 30 6 4 2 6 5 2− + − + − +

o) 15 2 90 2 2 12x x x− + − p) 22 7 5 40 2 140 10x x x x+ − −

q) 258 144 3n n− r) 2 2 23 3320 6 30 12 16 3 24 6r r r r r r+ − −

2. a) 2 2 b) 1− c) 3 2 d) 9 35

7

a e) 2 10 f)

3m

m

−

g) 15

9

x

x

− h) 34 150t i)

87 11

11

p j)

2 36 98

7

n k) 10 5 3+

53

l) 7 3 28 2

29

− + m)

35 2 14

3

+ n)

11 39

5

− − o)

4 6 36

6 81

r r r

r

−

−

p) 9 2

2 q)

16 4 6

8 3

t

t

+

− r)

5 30 10 3

6

y y−

3. 28 16 3+

4. 3

5x

Solving Radical Equations Assignment:

1. a) 9

2x = b) 2x = − c) 22x = − d) 25a = e) 36y = f)

4

3x =

g) 49

6n

−= h) 50n = i) no solution j) 1x = −

2. a) 2 7x = b) 16x = − , 4x = c) 2 2 6a = d) 4n =

e) 10x = f) 32x = − , 2x = g) 4d = h) 2

3x

−=

3. a) 4x = b) 0n = c) 16a = d) 50 25 3

2y

+= e) 6x =

f) 8y = g) 5a = h) 6x =

4. 100

9l = metres

5. 2 n n+ = , 4n =

6. 3 4 3 4x x

ax

− +=

Absolute Value Assignment:

1. a) 9 b) 0 c) 7 d) 4.728 e) 6.25 f) 1

52

g) 7 h) 5−

i) 10 j) 13 k) 10 l) 2.8− m) 21

4 n) 9 o) 17

54

2. a) 3 1

0.8, 0.4, , 0.8 , 1.1 , 1 , 2 ,5 4

− − − −

b) 1 7

2.4, 1.9, 0.6 , 1 , 1.3 , , 2.210 5

−− − −

3. 11 7 9 27− + + − =

4. 359.22 310.45 310.45 295.78 295.78 513.65 513.65 425.59− + − + − + −

5. a) 6 km b) 9 km

Absolute Value Functions & Their Graphs Assignment:

1. a) x ( )y f x= b) x ( )y f x=

2− 3 2− 0

1− 1 1− 2

0 1 0 2

1 3 1 0

2 5 2 4

2. ( )5, 8−

3. x-intercepts: 3 and 4− , y-intercept: 7

4. a) b) c)

55

d) e) f)

5. a)

x-intercept(s): 3−

y-intercept: 3

Domain: x x R

Range: 0,y y y R

b)

x-intercept(s): 2− and 2

y-intercept: 4

Domain: x x R

Range: 0,y y y R

c)

x-intercept(s): 4

y-intercept: 2

Domain: x x R

Range: 0,y y y R

56

d)

x-intercept(s): 1 and 3

y-intercept: 3

Domain: x x R

Range: 0,y y y R

e)

x-intercept(s): none

y-intercept: 11

Domain: x x R

Range: 2,y y y R

6. a) 2 2, 12 2, 1

x xy

x x− +

=−

b) 3 6, 23 6, 2

x xy

x x− − −

=+ −

c)

11, 2

21

1, 22

x xy

x x

−+

= −

d)

2

2

2

2 2, 12 2, 1 1

2 2, 1

x xy x x

x x

− −

= − + − −

e)

2

2

2

( 1.5) 0.25, 1( 1.5) 0.25, 1 2

( 1.5) 0.25, 2

x xy x x

x x

− −

= − − + − −

f)

2

2

2

3( 2) 3, 13( 2) 3, 1 3

3( 2) 3, 3

x xy x x

x x

− −

= − − + − −

7. a) 4, 44, 4

x xy

x x− +

=−

b)

53 5,

35

3 5, 3

x xy

x x

−− −

= +

c)

2

2

2

1, 11, 1 1

1, 1

x xy x x

x x

− −

= − + − −

d)

2

2

2

6, 26, 2 3

6, 3

x x xy x x x

x x x

− − −

= − + + − − −

57

Solving Absolute Value Equations Assignment:

1. a) 19x = − , 5x = b) 2

3x = , 2x = c) no solution

d) 3.5x = , 10.5x = e) no solution f) 9x = − , 1x =

g) 1

3n

−= , 3n = h) no solution i)

11

3a

−= , 3a = −

j) 3x = − , 3x = k) 2x = , 3x = l) 3x − or 3x

m) 1 5

2x

+= ,

1 5

2x

− += n) 4x = − , 2x = − , 4x = , 6x =

2. 18 0.5d − = , 17.5d = , 18.5d =

3. 7 4.8x − =

4. perigee 356400= km apogee 406700= km