PRE-CALCULUS 11
Unit 4 – Day 1: WORKING WITH RADICALS
RADICALS
n
Radicals are expressions in the form
a , which is the "nth root of a".
is the radical symbol; n is the index; and a is the radicand.
n
= a number
such that
=
because
8 =
because
81 =
because
a
25
3
4
(the number)n = a
The index, n, must be a positive integer.
Restrictions on Radicands
−25 = a number
only if
(the number)2 = −25
There is no real number that will satisfy this condition, ∴
− 25 is not a real number.
• When the index of a radical is even, the radicand must be greater than or equal to
zero and the radical is positive.
3
− 8 = a number
3
− 8 is the negative number −2.
only if
(the number)3 = −8
• When the index of a radical is odd, the radicand could be any real number:
o If the radicand is positive, then the radical is positive,
o If the radicand is negative, then the radical is negative.
exercise: For what values of x is
7 − 3 x a real value?
[Answer: x ≤ 7 ]
exercise: For what values of x is
x2 a real number?
[Answer:
x ∈R]
[Answer:
x ≥ 0]
exercise: Does
x2 = x for all values of x ?
3
Unit 4: Day 1 notes - Working with Radicals
Page 2 of 2
SIMPLIFYING SQUARE ROOTS
There are rules that can be used to work with radicals.
n a
a
n ab = ( n a ) ( n b )
n
=
n b
b
n
a
m
=
m
an
The first condition for simplest radical form with square roots is:
the radicand cannot have a factor that is perfect square larger than 1.
example: Simplify
Just simplify
72
72
(
Answer:
An entire radical is an expression that is only a radical;
(36)(2)
36 ) ( 2 )
6 2
72 .
A mixed radical is an expression that is a product of a non-radical and a radical; 6 2 .
exercise: For a ≥ 0 and b ≥ 0 , simplify
exercise: Simplify
3
48a6b9
[Answer: 4a3b4
40
3b ]
[Answer: 2 3 5 ]
ADDING AND SUBTRACTING RADICALS
Adding/subtracting monomial terms: only like terms can be combined.
7x + 5y + 9x
simplifies to
16x + 5y
Adding/subtracting radical terms: only terms with the same radical can be combined.
7 2 + 5 3 + 9 2
simplifies to
16 2 + 5 3
exercise: Simplify
11 − 8 13 − 5 11 − 8 13
exercise: Simplify
5 +
exercise: Simplify
20 −
45
8 6 − 5 12 + 2 27
[Answer: 0]
[Answer: 8
6 −4 3]
PRE-CALCULUS 11
Unit 4 – Day 2: MULTIPLYING RADICAL EXPRESSIONS
MULTIPLYING RADICALS
There are rules that can be used to work with radicals.
n
ab =
(
n
a
)(
n
b)
n
a
=
b
n
n
a
b
n
a
m
m
an
=
The first condition for simplest radical form with square roots is:
the radicand cannot have a factor that is perfect square larger than 1.
examples: Simplify
a)
(7
5 ) (4 6 )
7×
5 ×4×
7×4×
28 ×
5 ×
b)
(5
14
6
5×
6
5×2×
5× 6
28 30
) (2
21 )
14 × 2 ×
14 ×
21
21
10 ×
2 ×
7 ×
3 ×
7
10 ×
7 ×
7 ×
2 ×
3
10 × 7 ×
2×3
70 6
exercises: Simplify
a)
(
7 ) ( 11 )
c)
(
42 ) ( 30 )
6 35
b)
(5
17 ) ( 2 17 )
d)
(11 2 ) ( 3
6)
66 3
Unit 4: Day 2 notes - Multiplying Radical Expressions
Page 2 of 2
MULTIPLYING RADICAL EXPRESSIONS
Multiplying a monomial to a polynomial: use the Distributive Property.
3x (x + 5y + 2)
simplifies to
3x2 + 15xy + 6x
Multiplying a radical term to a sum of terms: use the Distributive Property.
example: Simplify
3 2
(
2 + 5 3 + 2)
o Use the Distributive Property
( 3 2 )(
o Multiply
3(2)
Answer:
exercise: Simplify
2 ) + ( 3 2 )( 5 3 ) + ( 3 2 ) (2)
+
(15
6)
+
(6
2)
6 + 15 6 + 6 2
5 15
(
3 −3 6
)
[Answer: 15
5 − 45 10 ]
Multiplying a binomial to a binomial: use the Distributive Property.
(a + b)(c + d) becomes a(c + d) + b(c + d) which simplifies to ac + ad + bc + bd
Multiplying a radical binomial to a radical binomial: use the Distributive Property.
example: Simplify
(7
o Distribute
3 + 2 5 )( 3 − 4 5 )
(7
3 )( 3 ) − ( 7 3 )( 4 5 ) + ( 2 5 )( 3 ) − ( 2 5 )( 4 5 )
o Multiply
exercise: Simplify
exercise: Simplify
(28
15 ) + ( 2 15 )
−
40
−19 − 26 15
Answer:
exercise: Simplify
−
21
(
(
(
11 − 7 )( 5 11 + 9 )
5 − 6)
[Answer: −26
2
10 +
3 )( 10 −
11 − 8 ]
[Answer: 11 − 2
3)
30 ]
[Answer: 7]
PRE-CALCULUS 11
Unit 4 – Day 3: DIVIDING RADICAL EXPRESSIONS
n
SIMPLIFYING RADICALS
ab =
(n a ) (n b )
a
=
b
n
n
n
a
b
The three conditions for simplest radical form are:
the radicand cannot have a factor that is perfect square larger than 1.
the radicand cannot be a fraction or decimal,
the denominator cannot contain a radical.
examples: Simplify
a)
30
6
=
30
6
=
5
b)
18
75
=
6
25
=
6
25
=
6
5
or
1 6
5
When the denominator of a fraction is an irrational number, the denominator is a nonterminating and non-repeating number. It is impossible to divide by such a decimal
number. The process or rewriting a fraction so that the denominator is not irrational is
called rationalizing the denominator.
examples: Simplify
a)
5
7

= 

=
=
5 

7 
(5)(7)
(7)(7)
35
7
7 

7 
=
b)
15
20
=
15
2 5
 15  
= 

2 5 
 15 5
= 
 2 (5)(5)
1 35
7
15 5
10
3 5
=
2
5

5



=
exercises: Simplify
5 6
a)
2 11
5 66
22
b)
5 7
2 90
70
12
c)
5
3 4
=
3 5
2
53 2
2
Unit 4: Day 3 notes - Dividing Radical Expressions
Page 2 of 2
CONJUGATES
The conjugate of a binomial is another binomial with the same first term, but the
second term has the opposite coefficient (opposite sign); a + b and a − b .
The product of conjugates (a + b)(a −b) is a2 − b2 .
example: Multiply ( 7 3 − 2 5 ) to its conjugate.
(7
3 − 2 5 ) (7 3 + 2 5 )
(7
3)
2
(2
−
−
127
147
Answer:
5)
2
20
Note that the product of binomial radical conjugates is a rational number.
DIVIDING RADICAL EXPRESSIONS
Consider the division 6 ÷ (1 +
5 ) ; this division results in the quotient
6
1+
5
This fraction is not in simplest form because there is a radical in the denominator.
example: Simplify
6
1+
6
1+
5
6 (1 −
o Use the conjugate to simplify
(1 +
5
5)
5 )(1 −
6 (1 −
5)
5)
(1)2 − ( 5 )
2
6 (1 − 5 )
−4
− 3 (1 − 5 )
2
Answer:
exercise: Simplify
−3+3 5
2
2 3 −3 2
2 3 +3 2
=
−
3 3
+
5
2 2
[Answer: −5 + 2
6
]
PRE-CALCULUS 11
Unit 4 – Day 4: RADICAL EQUATIONS (Part 1)
RADICAL EQUATIONS
A radical equation is an equation having radicals with the variable in the radicand.
The solution of any equation are all values of the variable that satisfies the equation.
9−
x + 5 = 2 is an example of a radical equation. Since there is a square root of a
variable expression, there may be restrictions from the radicand: x + 5 ≥ 0 ∴ x ≥ −5
This equation can be solved graphically:
• Rewrite the equation so that one side is 0; graph y = (the non-zero side); the roots
are the x-intercepts.
• Graph y = (the left side) and y = (the right side); the roots are the x-coordinate from
the points of intersection.
SOLVING RADICAL EQUATIONS ALGEBRAICALLY
Radical equations will be easier to solve without the radical.
x+5 = 2
9 −
o Isolate the square root.
−
example: Solve
9 −
x+5 = 2
x + 5 = −7
x+5 = 7
(
o Square both sides of the equation.
o Simplify.
o Solve for x .
o Check this root.
Answer:
exercise: Solve
9 −
2
x + 5 ) = (7)
x + 5 = 49
x = 44
2
(44) + 5 = 2
x = 44
19 − 2 1 − 3 x = 11
[Answer: −5]
Unit 4: Day 4 notes - Radical Equations (Part 1)
Page 2 of 2
x − x2 + x − 1 = 0
exercise: Solve
[Answer:
1 or 1]
2
EXTRANEOUS ROOTS
An extraneous root is a value the algebraic steps may produce as roots; however it does
not satisfy the original equation and so cannot be a root.
example: Solve
x+5 − x = 3
x+5 − x = 3
x+5 = x + 3
o Isolate the square root.
o Square both sides of the equation.
(
o Simplify.
2
x + 5 ) = (x + 3)
x + 5 = x2 + 6x + 9
o Solve for x .
0
0
x+4 = 0
x = −4
Check the roots.
Answer:
−4 + 5 − (−4) ≠ 3
or
or
= x2 + 5x + 4
= (x + 4)(x + 1)
(x + 1) = 0
x = −1
−1 + 5 − (−1) = 3
x = −1
5 − 2 x = −7
[Answer: −2]
1 − x2 − x + 1 = 0
[Answer: 1]
exercise: Solve 2x −
exercise: Solve
2
PRE-CALCULUS 11
Unit 4 – Day 5: RADICAL EQUATIONS (Part 2)
RADICAL EQUATIONS WITH TWO RADICAL TERMS
x −1 −
5 − x = 0 is an example of a radical equation with two radical terms.
With two radical terms, there are two restrictions to consider: x − 1 ≥ 0 and 5 − x ≥ 0
which is x ≥ 1 and x ≤ 5 , ∴ 1 ≤ x ≤ 5 .
Radical equations will be easier to solve without the radicals; isolate a radical and
square both sides of the equation.
example: Solve
x −1 −
5− x = 0
x −1 −
x −1 =
o Isolate the first square root.
o Square both sides of the equation.
(
x −1 ) =
2
x − 1
2x
x
o Simplify.
o Solve for x .
(3) − 1 −
o Check this root.
Answer:
5− x = 0
5− x
(
5− x
)2
= 5 − x
= 6
= 3
5 − (3) = 0
x=3
exercise: Solve
x+6 −
3 − 2x = 0
[Answer: −1]
exercise: Solve
x +1 −
x+7 = 0
[Answer: φ]
Unit 4: Day 5 notes - Radical Equations (Part 2)
Page 2 of 2
Isolating one of the radicals and squaring both sides of the equation may not eliminate
both radical symbols. If that happens work on one radical at a time.
4x + 5 −
example: Solve
2x − 1 = 2
4x + 5 −
2x − 1 = 2
4x + 5 = 2 +
o Isolate the first square root.
(
o Square both sides.
o Simplify.
o Isolate the second square root.
2x − 1
4 x + 5 ) = (2 + 2 x − 1 )
2
2
4x + 5
= 4 + 4 2 x − 1 + 2x − 1
4x + 5
= 2x + 3 + 4 2 x − 1
2x + 2 = 4 2 x − 1
o Simplify
x + 1 = 2 2x − 1
o Square both sides.
(x + 1)2 = ( 2 2 x − 1 )
2
x2 + 2x + 1 = 4(2x − 1)
x 2 + 2x + 1 = 8 x − 4
x2 − 6x + 5 = 0
o Simplify
(x − 1)(x − 5) = 0
x−1 = 0
or
x−5 = 0
x = 1
or
x = 5
o Solve for x .
Check the roots.
Answer:
x=1
or
exercise: Solve
x−5 +
exercise: Solve
x−5 +3 =
4(1) + 5 −
2(1) − 1 = 2
4(5) + 5 −
2(5) − 1 = 2
x=5
x+4 = 9
2x + 7
[Answer: 21]
[Answer: 9 or 21]