Olympic College Topic 16 Radicals
Topic 16 - Radicals
1.
Definition:
Definition of a Radicals
In Mathematics an efficient way of representing repeated multiplication is by using
the notation an , such terms are called exponentials where a is called the base and n
the exponent. In general we say that an is ”a to the power of n” but in particular
when the exponent is 2 we often use the word squared and when the exponent is 3
we use the word cubed.
For example,
n terms
Here are some examples of exponentials with whole numbers as powers.
Definition:
53
=
= 125
(called 5 cubed)
102
=
= 100
(called 10 squared)
25
=
(called 2 to the power of 5)
= 32
A radical is a special version of an exponential; a radical is an exponential with a
fractional power such as
power.
For example
While
,
,
,
,
,
that cannot be simplified to a exponential with a integer
are all examples of radicals.
are not truly radicals since they can be simplified to
,
It can also be shown that the radical
so “a to the power of “ is the same as
the nth root of a. Also the more generalized version of this property is that
There are three basic operations involving all exponentials including radicals – they are as follows.
Multiplication Rule
=
Division Rule
Powers Rule
=
=
Page | 1
Olympic College Topic 16 Radicals
Example 1:
Solution:
Which of the following are radicals?
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
(a)
This has a fractional power that can’t be simplified and so is a radical.
(b)
This has a fractional power that can’t be simplified and so is a radical.
(c)
=
(d)
(e)
=5
This has a fractional power that can’t be simplified and so is a radical.
=
(f)
This has a fractional power that can’t be simplified and so is a radical.
This has a fractional power that can’t be simplified and so is a radical.
(g)
This is not a radical.
(h)
This has a fractional power that can’t be simplified and so is a radical.
(i)
=2
(j)
=
This is not a radical.
This has a fractional power that can’t be simplified and so is a radical.
(k)
(l)
Exercise 1:
This is not a radical.
This is not a radical.
=
This is not a radical.
Which of the following are radicals?
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
Page | 2
Olympic College Topic 16 Radicals
2.
Simplifying Numerical Radicals
A numerical radical is a radical that has a number as its base. For example
numerical radicals since they have bases of 8, 32 and 162 = 256 respectively while
numerical radicals.
Definition:
are all
are not
A radical is in its simplest form when it is written as a multiple of a radical with the
smallest base.
The method we will use to simplify numerical radicals involves splitting the number into a product of
two whole numbers where one of the numbers is the largest perfect square and then simplifying the
resulting terms by using the property that
=
for any numbers x and y
In order to use this method we need to know the perfect squares and cubes, etc….
Perfect Squares
Perfect Cubes
1
1
4
8
8
27
16
64
25
125
36
216
49
343
64
256
……….
……….
For example the 6th perfect cube is 63 = 216 while the 8th perfect square is 82 = 64
Example 1:
Simplify the radical
Solution:
The
can be written as
and as
we choose the form
the largest perfect square we can use. We then simplify the result as follows.
=
=
= 6
We can now say that the simplified form of
( since
as 36 is
)
is
The process of simplifying a numerical radical that is a cube root is to split the number into a product of
two whole numbers where one of the numbers is the largest perfect cube and then simplifying the
resulting terms by using the property that
=
for any numbers x and y
Example 2:
Simplify the radical
Solution:
The
can be written as
and as
we choose the form
is the largest perfect cube we can use. We then simplify the result as follows.
=
=
= 4
We can now say that the simplified form of
( since
as 64
)
is
Page | 3
Olympic College Topic 16 Radicals
Example 3:
Simplify the radical
Solution:
=
=
= 2
( since
We can now say that the simplified form of
Example 4:
is
Simplify the radical
Solution:
=
=
= 5
( since
We can now say that the simplified form of
Example 5:
is
=
=
= 4
( since
We can now say that the simplified form of
is
=
=
= 10
( since
We can now say that the simplified form of
)
is
Simplify the radical
Solution:
=
=
= 3
( since
We can now say that the simplified form of
Exercise 2:
)
Simplify the radical
Solution:
Example 7:
)
Simplify the radical
Solution:
Example 6:
)
)
is
Simplify the following numerical radicals.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
Page | 4
Olympic College Topic 16 Radicals
3.
Simplifying Radicals Expressions
The method used to simplify radicals expressions that contain variables involves expressing the radical
as an exponential with a rational exponent we then express the rational exponent as a mixed fraction
and simplify the result using the property that xa+b = xa xb where a will be a whole number and b will
be a fraction.
Example 1:
Solution:
Simplify the radical
=
=
=
=
Example 2:
Solution:
Solution:
Solution:
Using the property that x1/2 =
=
Express
=3
3 2/3
c c
c3
Using the property that x2/3 =
Simplify the radical
=
=
=
=
Example 4:
x x
x3
Simplify the radical
=
=
=
Example 3:
Express = 3
3 1/2
Express
2
b b
b2
=2
3/4
Using the property that b3/4 =
Simplify the radical
=
Using the property that
=
=
Expressing = 1 , = 3 and = 3
=
=
Using the property that x1/2 =
=
=
Rearranging the terms
Using the property that
Page | 5
Olympic College Topic 16 Radicals
Example 5:
Simplify the radical
Solution:
=
Using the property that
=
Expressing
=
Expressing
= 3 and = 2
=
Example 6:
=
Using the property that x1/2 =
=
=
Rearranging the terms
Using the property that
Simplify the radical
Solution:
=
Using the property that
=
Expressing
=
Expressing
=3
Using the property that xm/n =
Rearranging the terms
Using the property that
=
=
=
Exercise 3
= 2 and
Simplify the following radical expressions.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
(m)
(n)
(o)
(p)
(q)
(r)
Page | 6
Olympic College Topic 16 Radicals
4.
Adding and Subtracting Radicals
To add or subtract radicals we simplify the radicals using the mehod from secction 3 first and then we
collect like terms as in algebra we then finally add or subtract the like terms.
Example 1:
Add and subtract the following radical expressions.
(a)
(b)
–
(c)
(d)
(e)
+
Solution (a):
=
=
=
=
Solution (b):
Solution (c):
Solution (d):
Solution (e):
+
+
+
Simplify the radicals
Add like terms
=
=
=
=
=
–
=
=
=
=
+
–
–
–
Simplify the radicals
Subtract like terms
–
=
=
=
=
=
+
Simplify the radicals
Subtract like terms
2
+5
)+
)+
+
=
=
=
=
=
=
+
+
+
+
Simplify the radicals
Collect like terms
Page | 7
Olympic College Topic 16 Radicals
Example 2:
Solution (a):
Add and subtract the following radical expressions.
(a)
(b)
(c)
(d)
=
=
=
=
=
=
=
Solution (b):
=
=
=
=
=
=
Solution (c):
Simplify the radicals
Add like terms
– 14x
=
=
=
=
=
=
Solution (d):
Simplify the radicals
Subract like terms
Simplify the radicals
Add like terms
=
=
=
=
=
=
Simplify the radicals
Add like terms
Page | 8
Olympic College Topic 16 Radicals
Exercise 4
1. Add and subtract the following radicals.
(a)
(b)
(c)
(d)
(e)
+
+
+
(f)
(g)
(h)
(i)
Page | 9
Olympic College Topic 16 Radicals
5. Multiplying Radicals
To multiply radicals we do the following steps.
Step 1: Multiply the expression using the distributive law or FOIL and the property
=
Step 2: Simplify each of the radical terms.
Step 3: Collect like terms (if necessary)
Step 4: Add or subtract like terms (if necessary)
Example 1:
Solution (a):
Solution (b):
Solution (c):
Multiply the following radicals.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
=
=
=
=
Using he distributive law
Using the property
=
16 – 4
12
=
=
=
=
Using he distributive law
Using the property
Simplifying the radicals
=
=
Using the property
=
=
=
=
=
=
Simplify the radicals
Page | 10
Olympic College Topic 16 Radicals
Solution (d):
=
Using the property
=
=
=
=
=
=
=
Solution (e):
Solution (f):
Solution (g):
Solution (h):
Simplify the radicals
=
=
=
=
=
=
=
=
=
=
=
=
=
Use F.O.I.L.
Simplify the radicals
Add and subtract like terms
Use F.O.I.L.
6b
6b
Simplify the radicals
Add and subtract like terms
Simplify the radicals
Add and subtract like terms
=
=
=
=
=
Simplify the radicals
=
Page | 11
Olympic College Topic 16 Radicals
Exercise 5
1. Multiply the following radicals.
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
Page | 12
Olympic College Topic 16 Radicals
6. Dividing Radicals
To divide radicals we do the following steps.
Step 1: Divide the two radicals using the property
Step 2: Simplify each of the radical terms.
Example 1:
Divide the following radicals.
(b)
(a)
Solution (a):
Solution (b):
Solution (c):
(c)
=
Using the property
=
=
=
Simplifying the radical
=
Using the property
=
=
=
Simplifying the radical
=
Using the property
(d)
=
=
=
=
=
Solution (d):
Simplifying the radical
=
Using the property
=
=
=
=
=
Simplifying the radical
Page | 13
Olympic College Topic 16 Radicals
There is a special case of division of radicals it is called “Rationalizing the denominator” it normally
involves a fraction where the denominator does not simplify into the numerator. In these si8tuations
we use a technique of multiplying the numerator and the denominator by the denominator of thye
fraction.
Example 2:
Solution:
Rationalize the denominator of the radical
=
Multiply the fraction by the special version of 1 =
=
Simplify the fraction
=
=
Example 3:
Solution:
Rationalize the denominator of the radical
=
Multiply the fraction by the special version of 1 =
=
Simplify the fraction
=
=
Example 4:
Solution:
Rationalize the denominator of the radical
=
Multiply the fraction by the special version of 1 =
=
=
Simplify the fraction
=
Example 5:
Solution:
Rationalize the denominator of the radical
=
.
Multiply the fraction by the special version of 1 =
=
=
=
Page | 14
Olympic College Topic 16 Radicals
Exercise 6
1. Divide the following radicals.
(a)
(b)
(c)
(c)
2. Rationalize the denominators of the following radicals.
(a)
(b)
(c)
(d)
Page | 15
Olympic College Topic 16 Radicals
7.
Definition:
Solving Radical Equations
A radical equation is one that contains a radical typically a
or a
These are examples of radical equations.
x
2x  3
2x  17
=
=
=
=
=
4
3
5
11
x+1
When you solve a radical equation you are trying to find usable value(s) of x that satisfy
both sides of the equation.
The main method used in solving a radical equation is to remove the radical by squaring
or cubing both sides of the equation. The new equation formed can then be solved in
the usual manner. It is important to check that your solutions work as it is possible to
get unusable solutions . For example we can try to solve the radical equation
=
we would discover that x = – 4 is a possible solution.
However if you check this solution by substituting it into the original radivcal equation
you will discover that it can’t be used since
=
=
So x = – 4 is not a usable solution as there is no real number that is equal to
When these situations occur we cannot use x = – 4 as a solution and this may mean that
there is no real solutions to the radical equation. The x = – 4 is sometimes called a
“phantom solution” as it does not really exist.
Example 1:
Solution:
Solve the radical equation
x
x
=
=
x
=
4
16
Check: Substitute x = 16 into
4
Squaring both sides
x
=
=
4
4
It works
Page | 16
Olympic College Topic 16 Radicals
Example 2:
Solve the radical equation
Solution:
2x + 1
2x
x
=
=
=
=
=
3
27
26
13
Cubing both sides
Subtract 1 from both sides
Divide both sides by 2
Check: Substitute x = 13 into
Example 3:
Solve the radical equation
Solution:
3x + 4
3x
x
=
=
=
=
=
5
25
21
7
Solution:
Solve the radical equation
2x  3
2x – 3
2x
x
=
=
=
=
2x  3 =
11
121
124
62
Check: Substitute x = 62 into
=
=
3
3
=
3
It works
5
Squaring both sides
Subtract 4 from both sides
Divide both sides by 3
Check: Substitute x = 7 into
Example 4:
3
=
=
5
5
=
5
It works
11
Squaring both sides
Adding 3 to both sides
Divide both sides by 2
2x  3
2(62)  3
=
=
11
11
=
11
It works
Page | 17
Olympic College Topic 16 Radicals
Example 5:
Solve the radical equation
2x  17 =
x+1
2x  17
=
x+1
=
(x +1)2
2x + 17
=
(x + 1)(x + 1)
2x + 17
=
x2 + 2x + 1
2x + 17
=
x2 + 1
17
=
x2
16
Solutions are x = 4 and x = – 4
2x  17
Check: Substitute x = 4 into
2(4)  17
Solution:
Check: Substitute x = – 4 into
This value of x does not work since
=
=
x+1
4+1
=
5
=
2x  17
2(4)  17 =
=
It works
x+1
– 4+1
–3
is positive 3 and not – 3
This means that this radical equation has only one real solution x = 4
Exercise 7
1. Solve the following radical equations.
(a)
x
=
7
(b)
=
4
(c)
=
5
(d) 2 x  5
=
9
(e)
=
x+1
(f) 5x  5
=
x+1
(g)
=
2x + 7
(h)
=
Page | 18
Olympic College Topic 16 Radicals
8.
Example 1:
Solution:
Example 2:
Solution:
Applications Using Radicals
Pythagoras Theorem. What is the length
of the hypotenuse of the right angled
triangle shown opposite.
x
10
Pythogaras theorem states a2 + b2 = c2
where c is the hypotenuse.
This can be rearranged to give
c =
=
=
Pythagoras Theorem. What is the length
of the hypotenuse of the right angled
triangle shown opposite.
24
=
2x + 6
x+4
Pythogaras theorem states a2 + b2 = c2
where c is the hypotenuse.
This can be rearranged to give
c
=
2x + 6 =
2x + 6 =
2x + 6 =
=
(2x + 6)2
4x2 + 24x + 36=
0
=
0
=
Solutions x = 2 and x =
Check x = 2 The right angled triangle formed
when x = 2 is triangle shown opposite.
This triangle exists and so x = 2 is
a usable solution with a missing hypotenuse
of 10 units.
Check x =
when x =
= 26
4x
(13 +10)(x – 2)
10
6
8
The right angled triangle formed
is triangle shown opposite.
This triangle does not exist and so x =
Not a usable solution.
is
Page | 19
Olympic College Topic 16 Radicals
So the solution to this problem is that x = 2 and in this situation the hypotenuse = 10.
Example 3:
The weight of an object in pounds is given by the formula W =
length of the object in inches.
where L is the
(a) What is the weight of an object if its length is 10 inches?
(b) If the weight of the object is 10 pounds what is its length?
Solution (a): W
W
W
W
=
=
=
=
5
The weight of the object is 5 pounds.
Solution (b): W
10
100
95
=
=
=
=
=
2L + 5
2L
L
The length of the object is 47.5 in
Example 4:
Solution:
The period of a pendulum T in seconds is given by the formula T = 6.28
Where L is the length of the pendulum chain in feet.
If the period of a pendulum is 10 seconds how long is L the length of the pendulum
chain?
T
=
6.28
10
=
6.28
=
Divide both sides by 6.28
1.592 =
(1.592)2
2.534464
=
=
(2.534464)9.8 =
24.8 =
Square both sides
L
L
Multiply both sides by 9.8
The pendulum chain will be 24.8 feet long!!!
Page | 20
Olympic College Topic 16 Radicals
Exercise 8
1.
What is the length of the hypotenuse of the right angled
triangle shown opposite?
x
7
10
2.
What is the length of the hypotenuse of the right angled
triangle shown opposite?
2x + 3
x
2x+2
3.
The weight of an object in pounds is given by the formula W =
the object in inches.
where L is the length of
(a) What is the weight of an object if its length is 7 inches?
(b) If the weight of the object is 4 pounds what is its length?
4.
The period of a pendulum T in seconds is given by the formula T = 6.28
Where L is the length of the pendulum chain in feet.
(a)
If the length of the pendulum is 19.6 feet what will be the period of the pendulum?
(b)
If the period of a pendulum is 12.56 seconds how long is L the length of the pendulum
chain?
Page | 21
Olympic College Topic 16 Radicals
Solutions
Exercise 1:
(a) not a radical
(e) not a radical
(i) radical
(b) radical
(f) radical
(j) radical
(c) radical
(g) not a radical
(k) radical
(d) radical
(h) radical
(l) not a radical
Exercise 2:
(a)
(e)
(i)
(b)
(f)
(j) 3
(c) 2
(g) 5
(k) 2
(d)
(h) 2
(l)
Exercise 3
(a)
(d) )
(g)
(j)
(m)
(p)
(b)
(e)
(h)
(k)
(n)
(q)
(c)
(f)
(i)
(l)
(o)
(r)
Exercise 4:
(a)
(d)
(g)
(b)
(e)
(h)
(c)
(f)
(i)
Exercise 5:
(a)
(b)
(c)
(d)
(e)
(f)
(g) x – y
(h) 6x – 2y
Exercise 6:
1.(a)
2.(a)
(b) 2x
(b)
(c)
(c)
(d)
(d)
Exercise 7:
(a) x = 49
(e) x = 2 only
(b) x = 3
(f) x = 4 and x = 1
(c) x = 3
(g) x = – 2
(d) x = 43
(h) x = 1
Exercise 8:
1. Hypotenuse = x =
2. x = 5 Hypotenuse = 2x + 3 = 13
3.(a) Weight =
3.(b) length = 5.5 inches
4.(a) T = Period of pendulum 8.88 sec
4.(b) L = length of pendulum =39.2 feet
=
Page | 22