Chapter 4
Radicals
4.1 – Mixed and
Entire Radicals
Chapter 4: Radicals
Review:
Squares and Square Roots
Remember: Squares and Square Roots are opposite operations.
x2 = x × x
x2 = x
( x )2 = x
The principal square root is the positive
square root of a number.
The secondary square root is the
negative square root of a number.
Unless specified, typically a square root
symbol is only looking for the principal
square root.
Example: Simplifying
a)
3600 = 2 × 2 × 2 × 2 × 3× 3× 5 × 5
= 2 ×3 ×5
4
2
2
= 22 × 3 × 5
When you’re simplifying radicals, the
first step is to factor the number into
its prime factors.
= 4 × 3× 5
= 60
b)
288 = 2 × 2 × 2 × 2 × 2 × 3× 3
= 2 5 × 32
= 2 ×2 ×3
4
1
2
c)
3
432 = 3 2 × 2 × 2 × 2 × 3× 3× 3
= 3 2 4 × 33
= 3 2 3 × 21 × 33
= 2 × 3× 2
= 2 × 3× 3 2
= 12 2
= 63 2
2
Example: Expanding
a) 4 5 = 4 2 × 5
= 42 × 5
= 16 × 5
= 80
b) -5 2 = -1× 5 2
c) -2 3 5 = -1× 2 3 5
= -1× 5 2 × 2
= -1× 3 2 3 × 3 5
= -1× 25 × 2
= -1× 3 2 3 × 5
= -1× 50
= - 50
= -1× 3 8 × 5
= - 3 40
When turning a mixed radical into an entire radical, you
must first pull the outside coefficient in underneath the
radical sign. Remember, roots and exponents are
opposite operations!
pg. 182-183, #1, 2, 3, 5, 6,
7, 11, 12, 13, 16, 18, 20
Independent practice
4.2 – Adding
and Subtracting
Radicals
Chapter 4: Radicals
Example
Check on your
calculator! What is
the answer
different?
L= 8+ 8+ 2+ 2
L= 4× 2+ 4× 2+ 2+ 2
ÞL=2 2 +2 2 + 2 + 2
ÞL=6 2
ÞL
ÞL
Determine the length of fascia
needed.
6 ×1.5
9m
Karen’s uncle needs 9
metres of fascia.
Example
Determine the difference in length between each pair of sides.
a) PS and SR
b) RQ and PQ
a) Let D represent the difference between PS and SR
D = PS - SR
D = 50 - 10 8
D = 5 2 - 20 2
D = -15 2
The difference between PS and SR is
15 2 cm.
Why is our answer positive?
b) Let E represent the difference between RQ and PQ
E = RQ - PQ
E = 29 5 - 10 45
E = 5 2 - 30 5
E=- 5
The difference between RQ and PQ
is 5 cm.
Example
a) 2 27 - 4 3 - 12
=2 9× 3-4 3- 4× 3
= 2×3 3 - 4 3 - 2 3
=6 3-4 3-2 3
=0 3
=0
b) 2 24 - 3 96 + 432
= (2 22 × 2 × 3) - (3 2 4 × 2 × 3) + ( 2 4 × 32 × 3)
= 2 × 2 6 - 3× 22 6 + 22 × 3 3
= 4 6 - 12 6 - 12 3
= -8 6 + 12 3
Your turn…
Create a negative mixed radical using one
addition sign, one subtraction sign, and the
radicals 32, 3 8 and 18 . Challenge the
person sitting next to you to simplify it!
The Rules
• You can take a number out from under a root sign by square-rooting
(or cube-rooting, if it’s a cubed root) the number.
• Ex:
18 = 9 × 2 = 9 × 2 = 3 2
• You can bring a number inside the a root sign by squaring (or
cubing, if it’s a cubed root) the number.
• Ex: 6 5 = 6 2 × 5 = 36 × 5 = 36 × 5 = 180
• You can add or subtract like radicals.
• Ex: 7 3 - 4 3 + 2 3 = (7 - 4 + 2) 3 = 5 3
• You cannot add or subtract unlike radicals.
pg. 188-190, #1, 2, 3, 5, 6,
8, 9, 11, 13, 15, 16, 18, 19
Independent practice
4.3 – Multiplying
and Dividing
Radicals
Chapter 4: Radicals
Rules for multiplication and
division
The product of two square roots is equal to the square root
of the product.
i.e.
a × b = ab
The quotient of two square roots is equal to the square root
of the quotient.
i.e.
a
a
=
b
b
Prove:
a × b = ab
• What is a square root?
• Is there a way to represent a
square root with exponents?
1
2
a × b = a ×b
1
2
Recall that exponent laws
dictate that ax  bx=(ab)x
= (ab)
= ab
1
2
Now, try to prove
the division rule on
your own:
a
a
=
b
b
Example
Simplify:
(5 3 + 2 6)2 = (5 3 + 2 6)(5 3 + 2 6)
= 5 3×5 3 + 5 3×2 6 + 2 6 ×5 3 + 2 6 ×2 6
= 25 9 + 10 18 + 10 18 + 4 36
= 25 × 3 + 20 18 + 4 × 6
= 75 + 20 × 3 2 + 24
= 99 + 60 2
Try this!
Express the following expression in its simplest form:
2 3( 12 - 7)
Method 2 is called rationalizing the
denominator, which is done by
multiplying both the numerator
and the denominator by the
radical in the denominator.
Example
Simplify:
Method 2: (Rationalizing the denominator)
Method 1:
4 12 - 10 6 4 12 10 6
=
2 3
2 3
2 3
4 12
10 6
= ( )(
) - ( )( )
2
2
3
3
= 2×
12
6
- 5×
3
3
=2 4 -5 2
= 4-5 2
4 12 - 10 6 2 12 - 5 6
=
2 3
3
2 12 - 5 6 3
=
×
3
3
=
3(2 12 - 5 6 )
3× 3
2 12 × 3 - 5 6 × 3
3× 3
2 12 × 3 - 5 6 × 3
=
9
2 36 - 5 18
=
3
12 - 15 2 = 4 - 5
=
3
=
2
Example continued…
Simplify:
Method 3: (Finding a common factor)
4 12 - 10 6 2 × 2 3× 4 - 2 × 5 3× 2
=
2 3
2 3
=
2 3(2 4 - 5 2)
2 3
=2 4 -5 2
= 4-5 2
• Which method seemed easiest
to you?
• What might be some
advantages to each method?
• What might be some
disadvantages?
• Will all the methods work all of
the time?
pg. 198-200, #1-5, 8-16,
19-21
Independent practice
Handout
Today we will be working on an project about
Pythagorean fractal trees. Make sure to answer all the
questions to your fullest ability, as this is a summative
assessment. You may need a ruler for parts of the
assignment.
4.4 – Simplifying
Algebraic Expressions
Involving Radicals
Chapter 4: Radicals
If an algebraic expression involves a
radical like x , can x be any real number?
We sometimes have to apply restrictions to variables.
Restrictions are the values of the variable in an
expression that ensure the expression in defined.
Ex. For 1/x, x cannot be equal to zero, or
symbolically we can write x ≠ 0.
What about:
•
2
x
•
•
x3
x4
Example:
Identify the restrictions
a)
x is defined
when x ³ 0,
where
x ÎR
Remember you
can’t take the
square root of a
negative number
c)
x 3 is defined
when x ³ 0,
where x ÎR

b)
x is defined
2
when

x ÎR
x2 = x
Square roots and
squares are
opposites, so they
cancel to leave
the absolute value
of x.
x3 = x2 × x
x3 = x × x
Simplify the expression,
and remember that
you can’t take the
square root of a
negative.
Can we make any general rules about the
restrictions on radicals in algebraic expressions?
Consider x n .
Example
a)
x + 5 x is defined
when x ³ 0,
where
x ÎR
x +5 x =6 x
You must always
identify your restrictions
when simplifying
algebraic expressions
involving radicals.
b)
2 4x 4 - 8x 4
when x ÎR
is defined
2 4x 4 - 8x 4 = 2 2 2 × x 4 - 2 2 × 2 × x 4
= 2 × 2 × x 2 - 2x 2 2
= 4x 2 - 2x 2 2
Example
Simplify:
-7y 2 8y 5
-7y 2 8y 5 is defined when y ≥ 0, where y ÎR.
Þ -7y2 8y5 = -7y2 22 × 2 × y 4 × y
= -7y2 × 2 × y2 × 2y
= -14y 4 2y
Example
Simplify: -3 x(2 2 - 3x)
-3 x(2 2 - 3x) is defined when x ≥ 0, where x ÎR.
-3 x(2 2 - 3x) = -3 x × 2 2 - (-3 x × 3x)
= -3× 2 × 2 × x - (-9 × x × x )
= -6 2x + 9x x
Remember, while multiplying radical expressions, you multiply the
coefficients by the coefficients, and combine the radicals.
Example
6 5 - 2 24x 3
Simplify:
2 x
6 5 - 2 24x 3
2 x
is defined for x > 0, where x ÎR.
What about an expression like
What are the restrictions?
Just remember that the
number underneath
the radical needs to
stay positive.
x - 2?
pg. 211-213, #1, 3, 6, 8, 9,
10, 11, 13, 14, 15, 17
Independent practice
4.5 – Exploring
Radical
Equations
Chapter 4: Radicals
Solving Radical Equations
What is the opposite of a square root?
Ex.
7 x = 42
Þ x = 42 / 7 = 6
Þ ( x )2 = 6 2
Þ x = 36
To get rid of a cube root, you need to cube both sides.
Make sure that you simplify the equations as much as possible
before solving. Remember your orders of operations.
Try it!
Solve for x: 3 =
x
8
3=
x
x
=
4 ×2 2 2
Þ 3× 2 2 = x
Þ (6 2 )2 = ( x )2
Þ 36 × 2 = x
Þ x = 72
pg. 215, #1-5
Independent practice
4.6 – Solving
Radical
Equations
Chapter 4: Radicals
Solving Radical Equations
If the solution does not work when substituted back into the equation,
then it is an extraneous root, and the equation does not have a solution.
Ex.
x + 2 = -3 is defined for x ≥ -2, where x Î R.
because x + 2 ³ 0
Þ x ³ -2
x + 2 = -3
Þ ( x + 2 )2 = (-3)2
Þx+2=9
You always must
check your solution
when solving radical
equations.
Þx=7
Check by substitution:
7+2 = 9 = 3
Since 3 ≠ -3, this is an
extraneous root.
Example
The forward and backward motion of a swing can be modeled
using the formula
L
T = 2p
9.8
where T represents the time in seconds for a swing to return to its original
position, and L represents the length of the chain supporting the swing, in
metres. When Cara was swinging, it took 2.5 s for the swing to return to its
original position. Determine the length of the chain supporting her swing to
the nearest centimetre.
Solution:
T = 2p
L
9.8
T
L
=
2p
9.8
T
L
Þ ( )2 =
2p
9.8
Þ
T2
Þ 9.8 2 = L
4p
Substitute:
2.5 2
Þ L = 9.8 2 = 1.551...
4p
The length of the
chain is 1.55 metres, or
155 cm.
Alternate Solution
Another way of solving these types of problems is to first substitute in
your values, and then manipulate the equation.
Solution #2:
T = 2p
L
9.8
2.5
L
=
2p
9.8
2.5
L
Þ ( )2 =
2p
9.8
2
2.5
Þ 9.8 2 = L
4p
2.5 2
Þ L = 9.8 2 = 1.551...
4p
Þ
The solution is the same!
• Which method do you prefer?
• What are some of the advantages
and disadvantages of each?
• What if you had to solve multiple
problems using the same
equation?
pg. 222-224, #1-15
Independent practice