Page 1 of 22
Review of Radical Expressions and Equations
Skills involving radicals can be divided into the following groups:
Evaluate square roots or higher order roots.
Simplify radical expressions. Rationalize denominators.
Find domain of a radical function.
Evaluate expressions involving rational exponents.
Solve a radical equation.
Evaluate square roots or cube roots.
9
Example: Evaluate
Identification/Analysis
This is a square root. The number under the radical is 9. (An asterisk * means multiply.)
Solution
Example: Evaluate
9 = 32 = 3
3
8
Identification/Analysis
Solution
Find a number such that the number squared is
9. This number is 3 because 3 2 = 3 ∗ 3 = 9
This is a cube root. The number under the radical is 8.
3
8 = 3 23 = 2
Find a number such that the number cubed is
8. This number is 2 because 2 3 = 2 ∗ 2 ∗ 2 = 8 .
Page 2 of 22
− 400
Example: Evaluate
Identification/Analysis
This is a square root. The number under the radical is –400.
− 400 does not have a real number solution.
Solution
Find a number such that the number squared is
–400. Note that 20 ∗ 20 = 400 and
− 20 ∗ −20 = 400 (not –400). Any real number
squared is a positive number or zero.
Example: Evaluate − 400
Identification/Analysis
This is a square root. The number under the radical is 400.
Solution
− 400 = − 202
.
= −20
Find a number such that the number squared is
400.
( 20 ∗ 20 = 400 )
Thus
202 = 20
Don’t forget the negative sign in front of
Example: Evaluate
3
− 27
Identification/Analysis
Solution
This is a cube root. The number under the radical is –27.
3
−27 =
3
( −3 )
= −3
3
Find a number such that the number cubed
will give you –27. The number is –3 because
(−3) 3 = (−3) ∗ (−3) ∗ (−3) = −27 .
.
Page 3 of 22
Simplify radical expressions and Rationalize denominators
Usually to simplify means to rewrite the expression in such a way that it has as few radicals as possible, and that the
expression under each radical does not contain perfect powers. Some rules for radicals are illustrated below.
a.
3 2 = 3 or ( 3 ) 2 = 3
d.
5 * 7 = 5 * 7 = 35
b.
3 2 *10 = 3 10
e.
5 +6 5 =7 5
c.
5 ÷ 7 = 5÷7 =
5
7
Also, (7 2 ) 5 = 7 2*5 = 710 and 6 2 * 6 5 = 6 ( 2+5) = 6 7
Keep in mind that:
5 + 7 does not equal
Example: Simplify
28 .
Identification/Analysis
Solution
5 + 7 , and
5
5
does not equal .
7
7
The only thing which can be “simplified” is 28 under the square root. Can it be factored in
such a way that one of the multiples is a perfect square?
28 = 4*7
= 22 *7
= 22 * 7
=2 7
Factor 28 so one factor is a perfect square.
The radical of a product equals the product of
the radicals.
Simplify the perfect square.
Page 4 of 22
Example: Simplify 7 x 2 . Assume x ≥ 0 .
Identification/Analysis The expression under the square root has a perfect square.
Solution
7 x2 = 7 x2
= 7 x
Example: Simplify
k 5 . Assume k ≥ 0 .
Identification/Analysis
Solution
The radical is a square root. The expression under the square root is not a perfect square.
k5 = k4 ∗ k
= k4 ∗ k .
= k2 k
Example: Simplify
Factor so there are perfect squares.
The radical of a product equals the product of
the radicals.
Simplify the perfect square.
k remains under the radical since the
exponent on k is 1 which is smaller than the
index (2) of the radical
320k 7 q 8 . Assume k ≥ 0 and q ≥ 0 .
Identification/Analysis
Solution
The radical of a product equals the product of
the radicals.
Simplify the perfect square.
The radical is a square root. The expression under the square root is not a perfect square.
320k 7 q8 = 320 ∗ k 7 ∗ q8
= 5 ∗ 64 ∗ k ∗ k 6 ∗ q8
= 5 ∗ 64 ∗ k ∗ k 6 ∗ q8
= 5 ∗ 8 ∗ k ∗ k 3 ∗ q4
= 8k 3 q 4 5k
Factor so there are perfect squares.
The radical of a product equals the product of
the radicals.
Simplify the perfect squares.
The radical is simplified because there are no
perfect squares left under the radical and the
remaining variables and numbers have
exponents of one.
Page 5 of 22
Example: Simplify
3
27 k 15 . Assume k ≥ 0 .
Identification/Analysis
Solution
This is a cube root. Each factor is a perfect cube. 27 = 33 and k 15 = (k 5 ) .
3
3
27k 15 = 3 27 ∗ 3 k 15
= 3 33 ∗ 3 (k 5 )
3
= 3k 5
Example: Simplify
3
The radical of a product equals the product of
the radicals.
Write each factor as a perfect cube.
Simplify the cube roots.
64 x 4 y 5 . Assume x ≥ 0 and y ≥ 0 .
Identification/Analysis
The radical is a cube root. The expression under the cube root is not a perfect cube.
Solution
3
64 x 4 y 5 = 3 64 ∗ 3 x 4 ∗ 3 y 5
= 3 43 ∗ 3 x 3 ∗ x ∗ 3 y 3 ∗ y 2
= 3 43 ∗ 3 x 3 ∗ 3 x ∗ 3 y 3 ∗ 3 y 2
= 3∗ x ∗ 3 x ∗ y ∗3 y2
= 3 xy
3
xy 2
The radical of a product is the product of the
radicals.
Factor so there are perfect cubes.
Simplify the perfect cubes.
Rearrange the factors.
Generally the radical factor is written last
The radical is simplified because there are no
perfect cubes left in the radical.
Page 6 of 22
Example: Simplify
3t 8t . Assume t ≥ 0 .
Identification/Analysis
The expression is the product of two square roots.
3t 8t = 3t ∗ 8t
Solution
= 24t 2
= 24 ∗ t 2
= 22 ∗ 6 ∗ t 2
= 2t 6
Example: Simplify
y −3
3
Identification/Analysis
Solution
The product of two radicals equals the radical
of their product.
Multiply.
The radical of the product equals the product
of the radicals.
Factor so there are perfect squares.
Simplify the square roots and rearrange.
4
. Assume y ≥ 3 .
y+5
The expression is the product of two square roots.
y −3
4
∗
=
3
y+5
y −3
4
∗
3
y+5
=
( y − 3) ∗ 4
3 ∗ ( y + 5)
=
2 2 ( y − 3)
3 y + 15
=2
y −3
3 y + 15
The product of two radicals equals the radical
of the product.
Multiply.
Factor the perfect square in the numerator.
Distribute 3 in the denominator.
Simplify the perfect square.
Page 7 of 22
3
Example: Simplify
162
.
6
3
Identification/Analysis
Solution
The expression is the quotient of two cube roots.
3
162
3
6
=3
The quotient of two radicals equals the
radical of their quotient.
Divide.
Simplify the perfect cube. 27 = 33
162
6
= 3 27
=3
Example: Simplify 4 5 + 8 5 .
Identification/Analysis
The expression is the sum of two radicals. The radicals are like radicals. Both contain
Solution
4 5 + 8 5 = (4 + 8) 5
5.
Add the like radicals.
= 12 5
Example: Simplify − 4 15 + 3 13 + 9 15 + 4 13 .
Identification/Analysis
Solution
The expression contains the sum of four radical terms. The like radicals can be added.
− 4 15 + 3 13 + 9 15 + 4 13 = −4 15 + 9 15 + 3 13 + 4 13
= (−4 + 9) 15 + (3 + 2) 13
= 5 15 + 7 13
Rearrange so like radicals are
together.
Combine the like radicals.
Page 8 of 22
Example: Simplify − 5 7 + 9 63 .
Identification/Analysis
The expression contains the sum of two square roots.
Solution
− 5 7 + 9 63 = −5 7 + 9 ∗ 9 ∗ 7
= −5 7 + 9 ∗ 3 ∗ 7
= −5 7 + 27 7
Simplify the radicals to determine if there are
like radicals.
Combine the like radicals.
= 22 7
Example: Simplify
x − 8 + 3 x − 8 . Assume x ≥ 8 .
Identification/Analysis
x −8 +3 x −8 = 4 x −8
Solution
Example: Simplify
The radicals in both terms are like radicals
( x − 8 ), so the terms can be added.
x + 2 + 9 x + 18 . Assume x ≥ −2 .
Identification/Analysis
Solution
The expression contains the sum of two terms containing radicals.
The expression contains the sum of two terms.
x + 2 + 9 x + 18 = x + 2 + 9( x + 2)
Simplify the radicals.
= x + 2 + 3 2 ∗ ( x + 2)
= x+2 +3 x+2
=4 x+2
Combine the like radicals.
Page 9 of 22
Example: Rationalize the denominator
3
5
.
Identification/Analysis
Solution
Rationalize the denominator means to find an equivalent fraction whose denominator does not
contain a radical.
3
5
=
=
3∗ 5
5∗ 5
3 5
( 5)2
3 5
=
5
Multiply the numerator and the denominator
by 5 . (Note: You can’t just square the
numerator and the denominator. It
will change the value of the fraction)
Simplify the denominator. Page 10 of 22
Example: Rationalize denominator 5− 2
5+ 2
.
Identification/Analysis
The denominator of this expression is irrational because it includes an irrational number ( 2 ).
Rationalize denominator means to find an equivalent expression, but with a rational
denominator.
Solution
5 − 2 (5 − 2)(5 − 2)
=
5 + 2 (5 + 2)(5 − 2)
52 − 2 ∗ 5 2 + ( 2) 2
25 − 2
25 − 10 2 + 2
=
23
27 − 10 2
=
23
=
(
There is a binomial in the denominator.
Multiply the numerator and denominator by
5 − 2 . (The conjugate of the denominator.)
)
Find the product of the numerator and
denominator.
Combine like radicals.
(
)
Note: Choosing 5 − 2 uses the formula
(a + b)(a − b) = a 2 − b 2 . In this example, a = 5
and b = 2 . Squaring The fraction would
change the value of the fraction.
Page 11 of 22
Find the domain of a function
Domain
The domain is a list or set of all possible inputs that yield a real number output.
There are three operations we “can’t do” with real numbers in algebra. Each of these restrict the domain.
• Can’t divide by zero.
• Can’t take the square root (or any even-index radical) of a negative number.
• Can’t take the logarithm of zero or a negative number.
Two common notations to write the domain are set-builder and interval notation.
1. Set-builder notation: Sets are typically written in braces
{ }.
The notation is
{independent variable some property or restriction about independent variable} where the vertical line is read “such that.”
Example: “All real numbers, x , less than 2.”
{ x x < 2}
Example: “All real numbers, n , greater than or equal to −4 and less than 6.”
{n − 4 ≤ n < 6}
2. Interval notation: Parenthesis indicate the starting or ending value is not included and a square bracket indicates the
starting or ending value is included. Within the parentheses or square bracket, we indicate the smallest value of x followed by
a comma and then the largest value of x . The examples above are shown using interval notation.
Example: “All real numbers, x , less than 2.”
( − ∞, 2 )
Example: “All real numbers, n , greater than or equal to −4 and less than 6.”
[ − 4, 6 )
Page 12 of 22
Find the domain of a radical function
Example: Find the domain of y = 2 x − 12 .
Identification/Analysis
Solution
The function contains a square root. The expression under the square root, 2 x − 12 , must be
greater or equal to zero.
2 x − 12 ≥ 0 2 x ≥ 12 x ≥ 6 In interval notation the answer is [ 6, ∞ )
Isolate the term with a variable .
Divide both sides by 2. (2 is positive, so
don’t change the inequality sign) Example: Find the domain of y = 5 − 3t .
Identification/Analysis
Solution
The function contains a square root. The expression under the square root, 5 − 3t , must be
greater or equal to zero.
5 − 3t ≥ 0 −3t ≥ −5 t≤
5
3
5⎤
⎛
In interval notation the answer is ⎜ −∞, ⎥ .
3⎦
⎝
Isolate the term with a variable.
Divide both sides by −3 . (Remember when
you multiply or divide an inequality by a
negative number, the inequality sign
changes direction.) Page 13 of 22
Example: Find the domain of y = 3 x + 14 .
Identification/Analysis
The function contains cube root. The expression under the cube root can be any real number.
Solution
x + 14
( x + 14 ) can be any real number
( − ∞, ∞ )
Write the expression under the radical.
(The expression is called the radicand.)
Since the radical is a cube root the expression
can be any real number.
Write the domain using interval notation.
Evaluate expressions involving rational exponents
For the problems in this group, an expression containing rational exponents should be written using radical notation, and
an expression containing radical notation should be written using rational exponents.
1
The definition of rational exponent is
n
x = xn
1
.
xn
All rules for exponents apply to rational and negative exponents. Often used rules are listed below. Assume: a ≠ 0
The definition of a negative exponent is x − n =
Product Rule: a m a n = a m + n
Example: Write
Power Rule:
n
Quotient Rule:
am
= a m−n
n
a
7 3 using rational exponents.
Identification/Analysis
Solution
( a m ) = a mn
The expression contains a cube root, which could be rewritten using rational exponents.
7 = (7
3
=7
3
3* 12
3
= 72
)
1
2
Rewrite the radical using a rational exponent.
Use the power rule.
Page 14 of 22
2
Example: Evaluate 8 3 .
Identification/Analysis
Solution
The exponent is
2
1 ∗2
8 3 = 83
2
, which is a rational number. Use the definition for rational exponents.
3
Use the power rule.
1
= (8 3 ) 2
= ( 3 8)
= (2) 2
=4
The definition of a rational exponent is used.
2
Simplify the cube root.
Note: If you use a calculator, remember to use
parentheses. Enter 8^(2/3), not 8^2/3.
Page 15 of 22
Example: Evaluate 9
−
3
2
Identification/Analysis
Solution
The exponent is
9
−
3
2
=
=
−
3,
2
which is a negative rational number.
Use the definition of a negative exponent.
1
3
2
9
1
Use the power rule.
1 ∗3
92
=
=
=
1
(9 )
1
2
3
Write the power
1
( 9)
1
( 3)
1
=
27
1
as square root.
2
3
Simplify the square root and raise the result
to the third power.
3
If you use a calculator, remember to use
parentheses. Enter 9^(-3/2), not 9^-3/2
Page 16 of 22
Solve radical equations
A radical equation is an equation containing one or more radical terms. For example, x = 2 x is a radical equation.
To solve means to determine all the real values which, when substituted in the equation for x , will make the statement true. All such
real values should be included in the answer. Note that 0 = 2 ∗ 0 is true. Hence, x = 0 is a solution of the equation x = 2 x . It turns
⎛
1
1
1⎞
= 2 ∗ ⎟⎟ . This second solution is often ignored.
out that x = is also a solution of x = 2 x ⎜⎜ verify that
4
4
4⎠
⎝
To find all the solutions, follow the steps for solving radical equations given below.
For equations containing one radical, the steps are:
1. Isolate the radical.
2. Square both sides of the equation
3. Solve the resulting equation which no longer contains radicals. This equation is often linear, quadratic, or rational.
4. Check the answers. (It is possible that some x values may be in the solution set of the resulting equation, but will not make
the original equation true.)
For equations with two radicals:
1. Isolate one of the radicals.
2. Square both sides of the equation.
3. Combine like terms.
4. Now the equation either has no radicals, or just one radical term.
a. If there are no radicals follow steps 3 and 4 under equations containing one radical.
b. If there is one radical follow steps 1 through 4 under equations containing one radical.
Page 17 of 22
Example: Solve the equation
Identification/Analysis
x+2 =6
This is an equation (contains an equal sign). Use the steps for equations containing one radical.
x+2 =6
Solution
The radical is isolated.
Square both sides.
( x + 2) 2 = 62
Check
x + 2 = 36
x + 2 − 2 = 36 − 2
x = 34
Solve the resulting linear equation.
34 + 2 = 6
Substitute x = 34 into the original equation
and simplify the results.
The statement is true so the solution is
x = 34 .
36 = 6
6=6
Example: Solve the equation
Identification/Analysis
x + 2 = −6
This is an equation (contains an equal sign). Use the steps for equations containing one radical.
x + 2 = −6
Solution
(
x+2
)
2
= ( −6 )
x + 2 = 36
x = 34
Check
34 + 2 = −6
36 = −6
6 = −6
2
The radical is isolated.
Square both sides.
Solve the resulting linear equation.
Substitute x = 34 into the original equation
and simplify the results.
The statement is false, so x = 34 is not a
solution.
Note: The square root of any real number can’t be
negative. Hence, x + 2 = −6 can’t be true.
Page 18 of 22
Example: Solve the equation
Identification/Analysis
Solution
2 x − 1 + 4 2 x − 1 = 40 .
This is an equation (contains an equal sign). It contains two radical terms.
2 x − 1 + 4 2 x − 1 = 40
5 2 x − 1 = 40
2x −1 = 8
(
2x −1
)
2
= 82
2 x − 1 = 64
x=
Check
Add like radicals. Follow the steps for
equations containing one radical.
Divide both sides by 5.
The radical is isolated.
Square both sides.
Solve the resulting linear equation.
65
= 32.5
2
2 ( 652 ) − 1 + 4 2 ( 652 ) − 1 = 40
65 − 1 + 4 65 − 1 = 40
64 + 4 64 = 40
8 + 4 ( 8 ) = 40
8 + 32 = 40
40 = 40
Substitute x = 652 into the original equation
and simplify the results.
The statement is true so x =
to the equation.
65
2
is a solution
Page 19 of 22
Example: Solve the equation
Identification/Analysis
Solution
x − 2x = 0
This is an equation (contains an equal sign). It contains one radical term.
x − 2x = 0
Isolate the radical.
x = 2x
( x)
2
= ( 2x)
Square both sides; remember to square each
factor.
2
x = 4x2
This is a quadratic equation that can be
solved by factoring.
x − 4 x2 = 0
x(1 − 4 x) = 0
Set each factor equal to zero.
x = 0 or (1 − 4 x) = 0
x = 0 or
−4 x = −1
1
x = 0 or
x=
4
Check
Check x = 0 :
Check x =
1
:
4
Solve the two linear equations.
There are two possible answers.
0 − 2 ( 0) = 0
0=0
1
⎛1⎞
− 2⎜ ⎟ = 0
4
⎝4⎠
1 2
− =0
4 4
1 1
− =0
2 2
0=0
Substitute each answer into the original
equation.
Simplify the results to be sure each yields a
true statement.
Both answers check.
Page 20 of 22
Example: Solve the equation
Identification/Analysis
Solution
x+3 = x−3
This is an equation (contains an equal sign). It contains one radical term.
x+3 = x−3
2
x + 3 = ( x − 3) 2
x + 3 = x 2 − 6x + 9
x2 − 7x + 6 = 0
( x − 1)( x − 6 ) = 0
x − 1 = 0 or x − 6 = 0
x = 1 or
x=6
Check
Check x = 1 :
1+ 3 = 1− 3
4 = −2
2 = −2
Check x = 6 :
6+3 = 6−3
9 =3
3=3
The radical is isolated.
Square both sides. Note: ( x − 3) 2 ≠ x 2 − 3 2 .
This is a quadratic equation that can be
solved by factoring.
Set each factor equal to zero.
Solve the two linear equations.
There are two possible solutions.
Substitute each answer into the original
equation.
Simplify the results to be sure each yields a
true statement.
Notice x = 1 does not yield a true statement,
while x = 6 yields a true statement.
The solution is x = 6 .
Page 21 of 22
Example: Solve the equation
Identification/Analysis
Solution
3x + 1 − x − 4 = 3
This is an equation (contains an equal sign). It contains two radical terms. Use the steps for
equations that contain two radicals.
3x + 1 − x − 4 = 3
Isolate one radical by adding
Square both sides.
3x + 1 = 3 + x − 4
2
(
3x + 1 = 3 + x − 4
)
2
3x + 1 = 9 + 6 x − 4 + x − 4
2x − 4 = 6 x − 4
2( x − 2) = 6 x − 4
( x − 2) = 3 x − 4
(
( x − 2) 2 = 3 x − 4
x 2 − 4 x + 4 = 9( x − 4)
x 2 − 13 x + 40 = 0
( x − 5)( x − 8) = 0
( x − 5) = 0 or ( x − 8) = 0
x = 5 or
x=8
)
2
Note:
(3 +
x−4
)
2
x − 4 to both sides.
≠ 32 + x − 4
2
Now there is one radical term left in the equation.
Combine like terms and isolate the radical.
Factor two out of the terms on the right side.
Divide both sides of the equation by a common
factor of 2.
Squared both sides.
This is a quadratic equation that can be solved by
factoring.
Set each factor equal to zero.
Solve the two linear equations.
There are two possible solutions.
Page 22 of 22
Check
Check x = 5 :
3 (5) + 1 − 5 − 4 = 3
15 + 1 − 1 = 3
Substitute each answer into the original equation.
Simplify the results to be sure each yields a true
statement.
16 − 1 = 3
4 −1 = 3
3=3
Check x = 8 :
3 (8) + 1 − 8 − 4 = 3
24 + 1 − 4 = 3
25 − 4 = 3
5−2 = 3
3=3
Both x = 5 and x = 8 check, so they are both
solutions.