Download Exponents and Radicals are the topic for this months provincial

Exponents and Radicals are the topic for this months provincial exam practice questions.
These topics are probably familiar to most of you but a little refresher never hurts. We will start
out by reviewing a few helpful rules.
Radicals: are any number that can be expressed in the form
radicals you will experience are listed below.
√
n
x. The most common types of
√
Square Roots - x means that we are seeking
√ to find a whole number with the property that
multiplied by itself it yields x. Example: 81 = 9 since 9 × 9 = 81.
√
Cube Roots - 3 x means that we are seeking to find
a whole number with the property that
√
3
multiplied by itself twice it yields x. Example: 125 = 5 since 5 × 5 × 5 = 125.
√
√ √
Another Helpful Rule - ab = a b - This rule is often used in simplifying radicals.
Exponent Rules: Student should be familiar with the following rules and be able to apply them
in problems.
• (am )(an ) = am+n
• (am ) ÷ (an ) = am−n , a 6= 0
• (am )n = amn
• (ab)m = am bm
• ( ab )m =
an
bn ,
b 6= 0
Exponents and Radicals: Students should also know the following relationships between exponents and radicals.
• a−n =
1
an ,
a 6= 0
√
m
• a n = ( n a)m , n > 0
Problems similar to the ones explored below will appear on the provincial exam.
√
1. Simplify 116
In this problem you are asked to simplify a radical that is not a perfect square. In other
words there is no whole number x such that x × x = 116. When we are asked to simplify
such a radical we are really being asked to examine the factors of 116, find the largest perfect
square factor and remove it from the radical.
I will start out by guessing that 4 is a factor of 116. On examining it I find that 116 = 4 × 29,
now since 29 is prime (only factors are 1 and 29) my work is finished. 4 is a perfect square
so I can rewrite my radical in the following form:
1
√
√
116 = 4 × 29
√
√ √
116 = 4 29
√
√
116 = 2 29
√
√
2 29 is the simplified form of 116 and the answer to this problem.
√
2. Simplify 3 −216
This problem is similar to problem 1 except instead of looking for perfect square factors of
216, we are looking for perfect cube factors. The negative
sign may seem a bit daunting at
√
3
first but remember that −1 × −1 × −1 = −1 thus −1 = −1. Unlike square roots, cube
roots are defined for negative numbers.
To start off lets look for factors of 216 that are perfect cubes. First I’ll list some perfect cubes
(remember these are just whole numbers multiplied by themselves 2 times).
1, 8, 27, 64, 125, 216, ....
In my list above I notice that 216 = 6 × 6 × 6 and therefore −216 = −6 × −6 × −6 . Thus
√
3
−216 = −6
3. Which pattern could be used to predict 5−3 .
53
52
51
A. 50
5−1
5−2
5−3
125
25
5
1
53
52
51
C. 50
5−1
5−2
5−3
125
25
5
1
-5
-25
-125
1
5
1
25
1
125
53
52
51
B. 50
5−1
5−2
5−3
53
52
51
D. 50
5−1
5−2
5−3
Knowing the property that a−n =
1
know that 5−3 = 513 = 125
.
1
an ,
15
10
5
0
− 15
1
− 10
1
− 15
15
10
5
0
-5
-25
-125
a 6= 0 can certainly simplify this problem. Thus we
Upon examining the options above it is easy to conclude that B and D are unlikely to be
correct. They do not even compute the values correctly. In deciding between option A and
C we see that in A the negative exponent is dealt with according the the definition, therefore
that is the pattern that can be used to predict 5−3 .
2
3
4. Evaluate 32 5
√
m
For this problem refer to the definition a n = ( n a)m , n > 0
√
3
32 5 = ( 5 32)2
√
We now need to consider 5 32 - Can we find a whole number that multiplied by itself 4 times
is 32? In other words is there and
√ x such that x × x × x × x × x = 32? I happen to know that
2 is such a number. Therefore: 5 32 = 2
√
3
32 5 = ( 5 32)2 = 22 = 4
3
5. Evaluate 9− 2
In this problem we will use the same definition to first note that
√
3
9− 2 = ( 9)−3
Since the square root of 9 is easy to compute we now have:
√
3
9− 2 = ( 9)−3 = (2)−3
Since a−n =
1
an
we have:
√
3
9− 2 = ( 9)−3 = (2)−3 =
3
1
23
=
1
8