Mathematics 20
Module 1
Lesson 3
Mathematics 20
Operations with Radicals
99
Lesson 3
Mathematics 20
100
Lesson 3
Operations with Radicals
Introduction
This lesson continues with the study of square root radicals. In particular, the algebraic
operations and equations involving radicals are studied and then applied to related
practical problems.
Whole Number Math
•
How many squares are contained in a 6 by 6 square grid?
•
Try with smaller size square grids to see if you can find a pattern.
•
See the summary at the end of the lesson for the solution.
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101
Lesson 3
Mathematics 20
102
Lesson 3
Objectives
After completing this lesson, you will be able to
•
add, subtract, multiply, and divide square root radicals.
•
solve and verify radical equations containing one radicand.
•
solve problems that involve equations which contain radicals.
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Lesson 3
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Lesson 3
3.1 Addition and Subtraction of Radicals
In the study of polynomials, two terms are called "like" terms if the variables and the
exponents of the variables are the same.
For example:
•
the terms 2 x 2 and 16 x 2 are "like" terms
•
the terms 2 x and 16 x 3 are not "like" terms
The situation with square root radicals is the same.
Two square root radicals are called "like" radicals if the radicands are the same.
Therefore:
•
the radicals 2 3 and  6 3 are "like" radicals
•
the radicals
3 are not "like" radicals
6 and
Just as with polynomials, "like" radicals can be combined by addition and subtraction.
Part of simplifying radical expressions is combining all the "like" terms.
Example 1
Simplify 4 3 + 7 5  3 + 3 5 .
Solution:
Write the original expression.
4 3+ 7 5 3+ 3 5
Combine like radicals.
= 4 3 3+7 5+ 3 5
Simplify.
= 3 3 + 10 5
Since
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3 and
5 are not "like" radicals, no further simplification can occur.
105
Lesson 3
The distributive property is also used when simplifying radical expressions. The same
rules apply as when working with polynomials or rational expressions.
The Distributive Property
The product of a and (b  c) is given by:
a(b  c)  ab  ac or (b  c)a  ba  ca .
The product of a and ( b  c ) is given by:
a(b  c)  ab  ac or (b  c)a  ba  ca .
Example 2
Simplify the radical expression 2( 5  3 )  7(3 3  2 5 ) .
Solution:
Write the original expression.
2( 5  3 )  7(3 3  2 5 )
Apply the distributive property.
= 2 5  2 3  21 3 + 14 5
Group the "like" terms.
= 2 5 + 14 5  2 3  21 3
Combine "like" terms.
= 16 5  23 3
Radicals such as 75 and 6 3 do not appear to be "like" radicals. In this case, it is
necessary to change 75 to a mixed radical.
•
75 =
25  3 =
25 
3= 5 3
6 3 and 5 3 now become "like" radicals.
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Lesson 3
Example 3
Simplify the expression 18 + 6 2  50  3 8 by first changing all of the
radicals to the simplest possible mixed radical form.
Solution:
Change each radical to the simplest possible mixed radical form.
•
18 = 9  2 = 3 2
•
•
•
6 2 is in its simplest form
25  2 = 5 2
50 =
3 8 = 3 4  2 = (3  2) 2 = 6 2
Write the original expression.
18 + 6 2  50  3 8
Express in simplest mixed radical form.
= 3 2 + 6 2 5 2 6 2
Combine like terms.
= (3 + 6  5  6 ) 2
Simplify.
= 2 2
Example 4
Simplify
6 x + 5 x  x x and then evaluate for x = 16 .
Solution:
Write the original expression.
6x + 5 x  x x
Change each radical to a mixed radical.
=
Combine the like terms.
= ( 6 + 5  x) x
Substitute x = 16 .
6 x + 5 x x x
= ( 6 + 5  16 ) 16
Simplify.
= ( 6  11 )4
Apply the distributive property.
= 4 6  44
The
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6 and 44 are not "like" and cannot be further combined.
107
Lesson 3
Exercise 3.1
1.
Simplify.
a.
b.
c.
d.
e.
2.
d.
e.
6  2 3 + 4 3 1 + 6 3
3 + 7 4 + 6 3  3 25
2 5 + 7 6 + 9 6 5 + 3 6 3
1
3
3 8 2 +
3 + 10 2
3
4
7 2 + 3 + 5 4 2 3 3  5
Simplify.
a.
b.
c.
d.
4.
3 2 3 + 3
7  10 7 + 6 7
6 3 6
1
3+
3
3
1
3+
3
5
Simplify.
a.
b.
c.
3.
6
4
9
5
2
5
4

1
1
1

4
3
5 +
6 5 2 3
4
2
2

 3( 2 + 8 3 )  (3 2  24 3 )

2( 9  6 3 )  3( 16  12 3 )
2 25 +
100  5(2 2  36 )
Simplify.
a.
b.
c.
d.
e.
f.
g.
h.
i.
8  32 + 2
2 18 + 50
27 + 48  2 3
 50 + 75  5 3
80  20  45
4 8 + 2 50 + 3 18
54  8 24 + 7 6
2 125  3 80 + 2 20
5 99  7 11 + 44
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Lesson 3
5.
Prove that x + y = x + y is false by finding just one set of values for x and y
for which the equation does not hold.
6.
Simplify. Assume that all variables are non negative.
a.
b.
c.
d.
2 a 6 a + a
5x b  3x b 7x b
100 x + 9 x  144 x
2
2
x +x+ 3 x
e.
x x+
f.
x x 2 y 2 + x 2 y + yx
g.
2
x +
2
3
x x 4 x
x
2
9 x2
3.2 Multiplication of Radicals
In the previous lesson the multiplication of two square root radicals was defined.
Multiplication of Radicals
a  b = ab , for a  0, and b  0
The radicals do not have to be "like" radicals to be multiplied. The usual rules of algebra
such as the commutative and distributive laws apply for radicals.
When multiplying radicals, you can group the radicals and group the non radical
coefficients. Each of these are then multiplied separately and then combined.
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Lesson 3
Example 1
Simplify the radical expression 3 5  7 2 .
Solution:
Write the original expression.
3 5  7 2
Group the radicals and group the
non radical coefficients.
= (3  7 )( 5 
Multiply.
= 21 5  2
2)
= 21 10
Example 2
Use the distributive law to find the product and simplify 3 3 (7 2 + 8 6 ) .
Solution:
Write the original expression.
3 3 (7 2 + 8 6 )
Distribute 3 3 .
= (3 3  7 2 ) + (3 3  8 6 )
Multiply coefficients and radicals.
= (3  7)( 3 2 ) + (3  8)( 3 6 )
Simplify.
= 21 6 + 24 18
= 21 6  72 2
Since the radicals are not “like” radicals, no further simplification is possible.
It is useful to simplify each radical to simplest form first so that the radicands do not
become too large.
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110
Lesson 3
Example 3
Simplify  2 2 (3 8  5 18 ) .
Solution:
Change each radical to the simplest possible mixed radical form.
•
2 2 is in its simplest form
•
3 8  3 4  2  (3  2) 2  6 2
•
5 18  5 9  2  (5  3) 2  15 2
Write the original expression.
 2 2 (3 8  5 18 )
Express in simplest mixed radical form.
=  2 2 (6 2  15 2 )
Combine radicals by subtraction.
=  2 2 ( 9 2 )

Multiply coefficients and radicals.
= ( 2 )( 9 ) 2 2
Simplify.
= 18 (2)

= 36
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Lesson 3
Activity 3.2
Sometimes patterns are very helpful in mathematical problem solving. The
following table has been started for you. Complete the table by determining
the pattern that can be used when multiplying two identical radicals together.
Like Radicals
Multiply
Solution
1 
1
1 1
1
2 
2
2  2
2
3 
3
3  3
3
5 
5
5  5
7
7
7  7
10 
10
25 
25
93 
93
What pattern can you see?
Multiplication of Like Radicals
x x x
Radical expressions can also be in the form ( a + b )( c + d ) . These two factors are
multiplied in the same way that two binomials are multiplied.
( a + b )( c + d ) = ( a + b )c + ( a + b )d
= ac + bc + ad + bd
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(a + b ) is distributed.
c and d are distributed.
Lesson 3
Another method to use when multiplying a binomial by a binomial is the FOIL method.
Multiply the:




O
I
L
F
First terms of the binomials.
Outside terms of the binomials.
Inside terms of the binomials.
Last terms of the binomials.
Example 4
Simplify ( 5  2 )( 2 5 + 3 2 ) .
Solution:
Use the foil method.
Multiply the:
•
First terms
( 5)(2 5 ) = 2 5  5 = 2(5) = 10
•
Outside terms
( 5)(3 2 ) = 3 5  2 = 3 10
•
•
Inside terms
Last terms
(  2 )( 2 5 ) =  2 2  5 =  2 10
(  2)(3 2 ) =  3 2  2 =  3(2) =  6
Write the original expression.
Multiply using FOIL.
Simplify.
= ( 5  2 ) (2 5  3 2 )
= 10 + 3 10  2 10  6
= 4 + 10
Example 5
Simplify (3 5  1) (3 5  1) .
Solution:
Use the foil method.
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Lesson 3
Multiply the:
•
First terms
(3 5 ) (3 5 ) = 9 5  5 = 9(5) = 45
•
Outside terms
(3 5 ) ( 1) =  3 5
•
•
Inside terms
Last terms
(3 5)(  1) =  3 5
(1) (1) =  1
Write the original expression.
Multiply using FOIL.
Simplify.
= (3 5  1) (3 5  1)
= 45  3 5  3 5  1
= 45  1
= 44
You should have noticed that this question was in the form of a difference of
squares where the answer is the “square of the first term minus the square of the
last term.”
Exercise 3.2
1.
Simplify each product of two mixed radicals.
a.
6 7
b.
7 7
6(5 3 )
c.
(4 3 )( 2 6 )
d.
(2 9 )( 2 16 )
e.
f.
g.
h.
i.
j.
2.
(2 6 )( 3 8 )
(3 8 )( 2 18 )
( 12 ) 2
(2 6 ) 2
( ab )( cd )
Simplify.
( 3  5)( 3 + 5)
a.
(2 3 + 1)( 2 3 + 1)
b.
( 2 + 3 )( 2 + 3 )
c.
(1  2 5 )(1 + 2 5 )
d.
(2 3  3 2 )( 2 3 + 3 2 )
e.
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Lesson 3
3.
(2 5  2 ) 2
f.
g.
(2 7  49 )( 2 7 +
h.
(2  x )( 4 + 2 x ), x > 0
49 )
Can you see a pattern?
a.
b.
c.
What must 3 be multiplied by to get 3?
What must 6 5 be multiplied by to get 30?
What must c b , b> 0 , be multiplied by to get cb?
4. Can you see a pattern?
a.
Which expression must 5  2 be multiplied by to get 5  2 ?
i) 5  2 , or ii) 5  2
b.
Which expression must 7  5 be multiplied by to get 7  5 or 2?
i) 7  5 , or ii) 7  5
3.3 Division of Radicals
The rule for division of radicals:
Division of Radicals
a
b
=
a
, for a  0 and b > 0
b
Just as for multiplication the radicals do not have to be "like" radicals to divide.
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Lesson 3
Example 1
Simplify
8 
2.
Solution:
Write the original expression.
Simplify using the rule for division.
=
=
=
=
8 
8
2
8
2
4
2
2
Another way to simplify division is to change each radical to simplest mixed radical form
first and then eliminate common factors.
The same example will be used to illustrate this.
Example 2
Simplify
8 
2.
Solution:
Change each radical to the simplest possible mixed radical form.
•
•
8= 4  2= 2 2
2 is in its simplest form
Write the original expression.
=
Express in simplest mixed radical form.
=
2
2 2
2
Simplify by eliminating common factors.
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8 
8
2
= 2
116
Lesson 3
In division, radicals often occur in the denominator. Such expressions are not in simplest
form. Changing an expression with a radical in the denominator to an equal form with no
radical in the denominator is called rationalizing the denominator.
The following steps in the first example will illustrate how this is done.
Example 3
Simplify
3
.
5
•
•
In the expression
3
the denominator can be rationalized by multiplying both the
5
numerator and the denominator by 5 .
5
This does not change the expression because
= 1 and multiplication by 1
5
changes nothing.
Solution:
3
Write the original expression.
Rationalize the denominator.
=
Simplify.
=
5
3 5
5 5
15
5
Notice that now the denominator does not have a radical in it and is therefore
in simplest form.
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117
Lesson 3
Example 4
Simplify
7 6
by rationalizing the denominator.
3 5
Solution:
7 6
Write the original expression.
Multiply numerator and denominator by
5.
=
3 5
7 6
3 5
7 30
=
3(5)
Simplify.
=

5
5
7 30
15
Example 5
Simplify
3+ 2 5
by rationalizing the denominator.
3
Solution:
3+ 2 5
Write the original expression.
Multiply numerator and denominator by
3.
=

3
3+ 2 5
3
Apply the distributive property.
=

3
3
3 + 2 15
3
Exercise 3.3
1.
Simplify by rationalizing the denominator.
a.
b.
4 2
7
2 3
3 7
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118
Lesson 3
1
c.
3
2 3
d.
27
5
e.
2 2
2
3
f.
2.
Rationalize the denominator in each case.
7+
a.
5
2
2 3+ 6 5
b.
3
c.
d.
10  6 5
2 5
1 7 3
e.
 3
2 3
2 5
3.
Simplify. Assume all variables are positive.
a.
bx
b.
4 x2 y
2 xy
c.
8 x3
d.
a
b

b
a
3
1

8
2
e.
Mathematics 20
x
2x
Hint: Simplify and then find a common denominator.
119
Lesson 3
3.4 Radical Equations and Problem Solving
A radical equation is an equation in which the variable is part of the radicand.
Some examples are:
•
x= 5
•
2 + x 5 = 7
•
3 x 3 = x + 1 x
As usual, the method of solving an equation is to isolate the variable. This is also true in a
radical equation. Once the radical has been isolated, the next step is to remove the radical
sign.
Squaring the radical removes the radical sign.
Squaring a Radical
•
To remove a radical sign, square the radical.
Example:
 2
2
2 
=
2=
2  2=
4= 2
In general:
 n
2
=n
Example 1
Simplify
a.
 3
2
b.
3 5 
2
c.


2
x+ 1 .
Solution:
a.
b.
c.
 3 = 3
3 5  = 3  5  = 9
 x + 1 = x + 1
2
2
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2
2
 5 = 45
2
120
Lesson 3
To solve a radical equation, the radical must be isolated first and then both sides of the
equation are squared.
Recall that whatever is done to one side of the equation must be done to the
other side.
It is also very important to include a check when solving equations and even
more important when solving radical equations. A root that does not work when
substituted back into the original equation is called an extraneous root.
Example 2
Solve the radical equation
x  11  3 .
Solution:
Write the original equation.
x  11  3
Isolate the radical.
x  3  11
x 8
 x   (8 )
Square both sides.
2
Simplify.
2
x  64
Check the solution by substituting the root into the original equation.
x  11 =
64  11 =
8  11 =
3 =
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121
3
3
3
 3 ()
Lesson 3
Example 3
Solve  x =  5 .
Solution:
Write the original equation.
 x  5
Multiply both sides of the equation by  1 .
x 5
 x  5
Square both sides.
2
Simplify.
2
x  25
Check the solution by substituting the root into the original equation.
 x  5
 25  5
 5  5 ()
Example 4
Solve
x  5 .
Solution:
x  5
This has no solution because
equal a negative.
x is a principal root and is positive. A positive cannot
Example 5
Solve
x 8  0 .
Solution:
Write the original equation.
Isolate the radical.
Square both sides.
Simplify.
x 8  0
x  8
 x
  8 
x  64
2
2
Check the solution by substituting the root into the original equation.
64  8  0
8 8 0
16  0
Since the root does not check, it is called extraneous and this equation has no solution.
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122
Lesson 3
Example 6
Solve
x + 7 = 12  2 x .
Solution:
Write the original equation.
x + 7 = 12  2 x
Isolate the variable by collecting similar terms.
x + 2 x = 12  7
Simplify.
3 x 5
3 x   (5)
Square both sides.
2
2
9 x = 25
25
x=
9
Check the solution by substituting the root into the original equation.
x + 7 = 12  2 x
25
+ 7=
9
5
+ 7=
3
5
21
+
=
3
3
26
=
3
25
9
5
12  2  
3
36 10

3
3
26
()
3
12  2
Example 7
Solve
3x  5 =
3x + 7 .
Solution:
Write the original equation.
Square both sides.
3x  5 =

3x  5
Simplify.

2
=
3x + 7

3x + 7

2
3x  5 = 3x + 7
5 = 7
Since this is a false conclusion, the equation has no solution.
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123
Lesson 3
Exercise 3.4
1.
Determine which is a solution to x  2 =
x 1 + 1
x = 5 or x = 2 .
2.
3.
Solve for x and check your solution.
a.
x=7
b.
x+ 2= 3
c.
x = 3
d.
2 x + 6 x = 30  2 x
e.
2 x  1 = 9 x + 15
f.
x 7 = 2 2
g.
3 x  5 = 13
h.
5x =
i.
2x + 1 =
j.
3x  5 =
Explain why
Mathematics 20
2x 7
5x + 7
3x + 5
x =  2 has no solution.
124
Lesson 3
Conclusion
In the introduction you were given a Whole Number Math problem. The
solution to the problem is:
•
Square
Number
6 by 6
1
5 by 5
4
4 by 4
9
3 by 3
16
2 by 2
25
1 by 1
36
Total
91
There are 91 squares contained in a 6 by 6 square grid.
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125
Lesson 3
Summary
The following is a list of concepts that you have learned in this lesson.
•
Radical expressions are added or subtracted by combining "like" radicals. Like
radicals are radicals with the same radicand.
•
Radicals are multiplied by multiplying the radicands together in the radical sign.
•
•
•
a  b  ab , for a  0, and b  0
For a quotient of radicals a monomial denominator is rationalized by multiplying
the numerator and denominator by the radical part of the denominator.
a
b
=
a
, for a  0 and b > 0
b
•
To remove a radical sign, square the radical.
•
Equations which contain radicals can be solved by first isolating the radical and,
then, squaring both sides.
•
The solution for radical equations must always be checked to eliminate extraneous
roots.
•
An equation such as x =  2 has no solution since the left side is a principal root.
Radicals in which the radicand is negative have no solution in the real number
system since the product of two equal real numbers is always non negative.
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126
Lesson 3
Answers to Exercises
Exercise 3.1 1.
a.
b.
c.
d.
e.
2.
a.
b.
3.
6 1  2 3  4 3  6 3
5 + 8 3
3  14  6 3  15
 3  6 3  14  15
 7 3 1
c.
d.
 4 5 + 10 6
e.
3 2 2 3
a.
2 3  5 3 5  3
 2 3  3  5 3 5
 3+ 2 5
b.
 3 2  24 3  3 2  24 3
 6 2
c.
2(3  6 3 )  3(4  12 3 )
 6  12 3  12  36 3
 24 3  6
d.
Mathematics 20
5 3
0 7 0
6 6
15
2
5
3
3 2
3
6
6
6
9
1
3
20
4
9
3
3  8 2  10 2
12
12
1
1
3+ 2 2
12
2(5)  10  10 2 )  5(6 )
 10  10  10 2  30
 50  10 2
127
Lesson 3
4.
a.
4  2  16  2  2
 2 2 4 2  2
 2
b.
c.
11 2
d.
e.
5 2
f.
4 4  2  2 25  2  3 9  2
 8 2  10 2  9 2
 27 2
g.
h.
6 6
i.
5.
9  3  16  3  2 3
3 3 4 3 2 3
5 3
16  5  4  5  9  5
 4 5 2 5 3 5
 5
2 25  5  3 16  5  2 4  5
 10 5  12 5  4 5
2 5
10 11
Let x = 9 and y = 16 .
?
9  16  9  16
?
3  4  25
75
There are many other values for which this is false.
Mathematics 20
128
Lesson 3
a.
b.
c.
d.
e.
3 a
 5x b
10 x  3 x  12 x  x
5x
f.
x  x  y  x2 y  y  x  x
 x2 y  x2 y  x2 y
 3 x2 y
g.
4x
Exercise 3.2 1.
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
42
49  7
30 3
 8 18  24 2
 48
(2 6 )( 6 2 )  12 12  24 3
(6 2 )( 6 2 )  36 4  72
12
24
abcd
2.
a.
b.
c.
d.
e.
9  5 3  5 3  25  22
4 9  2 3  2 3  1  13 + 4 3
5+ 2 6
1  4 25  19
4 9 6 6 6 6 9 4
 12  18
 6
f.
g.
22  4 10
 21
h.
8  4 x  4 x  2 x2  8  2x
6.
3.
Mathematics 20
a.
b.
c.
x x  x x  2 x2 x
 x x  x x  2x x
0
3
5
b
129
Lesson 3
4.
a.
b.
Exercise 3.3 1.
a.
b.
c.
d.
e.
f.
2.
4 14
7
2 3  7 2 21

21
3 7 7
3
3
2 3 2 3 3 2 9 2



3 3 3 3 3 3 9 3
5 2
4
6
3
14 +
2
a.
b.
c.
d.
e.
3.
5+ 2
7 5
a.
b.
c.
d.
e.
(2 3  6 5 )  3
 3
3
2 9  6 15

3
6  6 15

3
 2  2 15
10  6 5  5 10 5  30

 5 3
10
2 5  5
3  21 21  3

3
3
10  15
10
b
2 x
4 x2  2x
1
3
2 2
Mathematics 20
10

1
2

3
2 2

2
2 2
130

1
2 2

2
4
Lesson 3
Exercise 3.4 1.
2.
By substitution, the solution is 5.
a.
b.
c.
d.
3.
Mathematics 20
?
( x ) 2  (7) 2
x  49
check :
( x ) 2  (1) 2
x= 1
49  7
7  7 (3 )
?
check :
1  2 3
?
1  2 3
3  3 (3 )
no solution
10 x
(10 x ) 2
100 x
x
?
check : 2 9  6 9  30  2 9
 30
 (30 ) 2
 900
9
e.
no solution
f.
x = 15
g.
( 3 x  5 ) 2  13 2
3 x  5  169
3 x  174
x  58
?
6  18  30  6
24  24 (3 )
?
check :
3(58 )  5  13
?
169  13
13  13 (3 )
h.
x  4
i.
no solution (The radicand cannot be negative.)
j.
no solution
x is the principal root which is positive and cannot equal a negative.
131
Lesson 3
Mathematics 20
132
Lesson 3
Mathematics 20
Module 1
Assignment 3
Mathematics 20
133
Lesson 3
Mathematics 20
134
Lesson 3
Optional insert: Assignment #3 frontal sheet here.
Mathematics 20
135
Lesson 3
Mathematics 20
136
Lesson 3
Assignment 3
Values
(40)
A.
Multiple Choice: Select the best answer for each of the following and place a
() beside it.
1.
The simplified form of
a.
b.
c.
d.
2.
3.
4
2
2
2
3
9
3
81
If 5 x = 25 , then x is equal to ***.
a.
b.
c.
d.
Mathematics 20
(2) 2 is ***.
If 3 x 2  27 , then x is equal to ***.
a.
b.
c.
d.
4.
4
2
2
4
The simplified form of
a.
b.
c.
d.
2
2 is ***.
 5
5
 25
25
137
Lesson 3
5.
If
2 x  1 = 7 , then x is equal to ***.
a.
b.
c.
d.
6.
The simplified form of 2 2 + 5 2  7 2 is equal to ***.
a.
b.
c.
d.
7.
0
2
 2
14 2
The simplified form of 3  2  5  2 3  4  5 is ***.
a.
b.
c.
d.
8.
49
4
25
5
3
10
23
27
To rationalize the denominator in the expression
2 3
, it is best to
7 5
multiply the numerator and denominator by ***.
a.
b.
c.
d.
9.
The simplified form of 7 8  3 12  13 2 + 5 3 is ***.
a.
b.
c.
d.
Mathematics 20
5
7 5
3
7 5
2 3
15 2  7 3
2+ 3
4 4 8 5
138
Lesson 3
10.
The rationalized form of
2
is ***.
5
a.
b.
c.
d.
11.
12 cm 2
3 cm 2
3 3 cm 2
12 3 cm 2
The surface area of a sphere with radius r is S = 4 r 2 . The equation
with the radius isolated is ***.
a.
b.
c.
d.
Mathematics 20
10
2 5
5
The formula for the area of an equilateral triangle with sides of length
2
s 3
s is A =
. If each side is of length 4 3 cm , the area is ***.
4
a.
b.
c.
d.
12.
5
2
2
5
2 5
S
4
2
r= S 2
16 
1 S
r=
2 
r=
r=
4
S
139
Lesson 3
13.
The distance between two points (a, b ) , (c, d ) on a plane is
( c  a ) 2  ( d  b) 2 . The distance between (0 , 0 ) , and ( 3 , 7 ) is ***.
a.
b.
c.
d.
14.
The solution to 2 3 x =  18 is ***.
a.
b.
c.
d.
15.
27
0
 27
non existent
The solution to
a.
b.
c.
d.
16.
4
2 10
2 29
58
b.
c.
d.
1 + 2 x is ***.
non existent
0
1
3
The solution to
a.
2x + 5 =
1
= 8 is ***.
x
1
64
64
2 2
1
2 2
17.
The irrational number 2.236068... is between the rational numbers ***.
a.
b.
c.
d.
Mathematics 20
2.235 and 2.236
2.236 and 2.237
2.23 and 2.235
 2 .236 and 2.236
140
Lesson 3
18.
The number
a.
b.
c.
d.
19.
c.
d.
2
2
x + y is ***.
2
x +
x+y
y
2
( x  y) 2
none of the above
The simplified form of
a.
b.
c.
d.
Mathematics 20
rational
an integer
a negative irrational
a real number
Another form of
a.
b.
20.
3 is ***.
3
x , x > 0 is ***.
x x
x x
x+x+x
3 x
141
Lesson 3
Mathematics 20
142
Lesson 3
Part B can be answered in the space provided. You also have the option to do the
remaining questions in this assignment on separate lined paper. If you choose this
option, please complete all of the questions on the separate paper. Evaluation of
your solution to each problem will be based on the following.
•
A correct mathematical method for solving the problem is shown.
•
The final answer is accurate and a check of the answer is shown where asked
for by the question.
•
The solution is written in a style that is clear, logical, well organized, uses
proper terms, and states a conclusion.
(40) B.
(4 each)
Mathematics 20
1.
Simplify 3 5  4 3  3 5 + 6 3 .
2.
Simplify  3 50  32 + 5 200 .
143
Lesson 3
Mathematics 20



3.
Simplify 2 3  5 3 3  2 5 .
4.
Simplify 1  3 5 .
5.
Simplify 7 b 2  5 4 b 2 + 4 b, b > 0 .
6.
Simplify


2

3 5 2 2 3

2


2
 2 14 2 .
144
Lesson 3
7.
Rationalize the denominator in the expression
4
.
15
8.
Rationalize the denominator in the expression
3+ 7 2
.
2 2
9.
10.
Mathematics 20
Find all the solutions to x 2  5 = 95 .
Find the solution to the equation 2 3  5 x = 34 and provide a check.
145
Lesson 3
C.
(8)
1.
The community of Melfort is planning to build
a circular wading pool in the park. The pool
will cover an area of 1000 m2. The building
committee has decided to put a 5 m cement pad
around it. How much additional area will the
cement pad take up? (Remember A   r 2 .)
Use all of the steps for problem solving.
(12)
(3 each)
2.
Mathematics 20
Provide a table of values and draw the graphs of each of the following
equations, each on a separate grid supplied on the following pages.
a.
y
b.
y x
c.
2
y x
d.
y
x
x 1
146
Lesson 3
C.
2.
a.
y
x
x
b.
y x
x
Mathematics 20
y
y
147
Lesson 3
.
2.
c.
y2  x
x
d.
y
x
y
x 1
y
100
Mathematics 20
148
Lesson 3