Chapter 2
Radicals, Rational Exponents and Complex Numbers
Square Roots, Cube Roots and Higher Roots
Recall
when raising a number to a power n,
where n
is an integer greater than 1, we multiply the number by itself n
times. For example, 4^{3
}=
4 ∙ 4 ∙ 4. Now suppose we want to know what number multiplied by
itself 2 times equals 169. Problems of this kind can be represented
using radicals.
A
radical
symbol √ is
used to show the square
root,
or
principal square root
of a number or expression that appears under the radical symbol.
Recall that the square root is defined as a number or expression
multiplied by itself twice to equal the number or expression under
the radical symbol, known as the
radicand.
For
example, if we want to know what number multiplied by itself 2 times
equals 169, we can set this up with the radical symbol as follows:
√169,
read as “square root of 169”. The answer to this is 13.

√169, read as “negative square root of 169”. The answer to
this is 13.
√0.09
= 0.3 and 0.3 since (0.3)^{2}
and (0.3)^{2}
equals 0.09. The principal square root is 0.3. Another way to
simplify this is to change √0.09 to √(9/100) and simplify to
3/10.
√(25/49)
= 5/7 and 5/7 since (5/7)^{2}
and (5/7)^{2}.
The principal square root is 5/7.
• Note
that a square root also has a negative value since a negative times a
negative equals a positive, but we will deal with only the principal
square root unless otherwise noted.
• Note
that you can also simplify the square root of a fraction by taking
the square root of the numerator and then the square root of the
denominator instead of the square root of the fraction as a whole.
In the previous example, you can take the square root of 25 first,
then the square root of 49.
• Note
that the square root of many positive integers are not whole numbers
or rational numbers. For example, √19 can be found on a calculator
or by leaving the answer as √19.
An
easy way to solve many square root problems is to know the perfect
squares from 1 to 25. They are as follows:
1^{2} = 1 10^{2} = 100 19^{2} = 361
2^{2}
= 4 11^{2}
= 121 20^{2}
= 400
3^{2}
= 9 12^{2}
= 144 21^{2}
= 441
4^{2}
= 16 13^{2}
= 169 22^{2}
= 484
5^{2}
= 25 14^{2}
= 196 23^{2}
= 529
6^{2}
= 36 15^{2}
= 225 24^{2}
= 576
7^{2}
= 49 16^{2}
= 256 25^{2}
= 625.
8^{2}
= 64 17^{2}
= 289
9^{2}
= 81 18^{2}
= 324
Sometimes
we have to find the square root of a number that is not a perfect
square. In these cases, we break down the radicand into factors, one
of which is a perfect square.
Examples:
Find each square root.
1.
√68
First,
find factors of 68.
Since
68 is even, we can divide it by 2. Therefore, 68 = 2 ∙ 34. Notice
34 is also divisible by 2, therefore 34 = 2 ∙ 17.
So
68 is factored into 2 ∙ 2 ∙ 17. Notice that 17 is prime and
cannot be factored further and 2 ∙ 2 = 4, which is a perfect
square. Therefore √68 = √4 ∙ √17 = 2√17.
2.
√108
First,
find factors of 108.
Since
108 is even, we can divide it by 2. Therefore 108 = 2 ∙ 54. Notice
54 is also divisible by 2, therefore 54 = 2 ∙ 27. Next, we know
that 27 = 3 ∙ 3 ∙ 3.
The
factors of 108 are 2 ∙ 2 ∙ 3 ∙ 3 ∙ 3. Notice 2 ∙ 2 = 4,
which is a perfect square and 3 ∙ 3 = 9, which is also a perfect
square. Therefore, 108 = 4 ∙ 9 ∙ 3 and √108 = √4 ∙ √9 ∙
√3 = 2 ∙ 3 ∙ √3 = 6√3.
• Note
that 108 = 36 ∙ 3 and 36 is a perfect square. But if you can't see
right away that 3 is a factor of 108, you can break down by dividing
108 by 2 first and then simplify further at that point. It's easy to
determine that 108 is divisible by 3. If the sum of the digits of a
number are divisible by 3, the number is divisible by 3.
Recall,
for a variable x, where x ≠ 0, √x^{2}
= x and x. If we are dealing with just the principal
square root, then the answer is x. Recall that x
is the absolute value of x.
Examples:
Find the square root of the
following.
a.
√(25x^{2})
Simplify
the numerical part and the variable part separately.
√25
= 5 since 5^{2}
= 25 and √x^{2}
= x
since (x)^{2}
= x^{2}.
Therefore,
√(25x^{2})
= 5x.
b.
√(64x^{4}y^{6})
√64
= 8 since 8^{2}
= 64, √x^{4}
= x^{2}
since (x^{2})^{2}
= x^{4}
and √y^{6}
= y^{3}
since (y^{3})^{2}
= y^{6}.
Therefore,
multiply the terms in bold to get √(64x^{4}y^{6})
= 8x^{2}y^{3}.
c.
√(x
+ 4)^{2
}
= (x
+
4) because (x
+ 4)(x
+
4) = (x
+ 4)^{2}
• Note
that the square root of any number or expression squared is just that
number. For example, notice in the previous examples that √(x
+ 4)^{2 } = (x + 4) and √25 = √(5)^{2} =
5.
d.
√(x^{2}
+ 12x
+ 36) = √(x
+ 6)(x
+ 6)
Notice
that we factored (x^{2}
+ 12x
+ 36) to get (x
+ 6)(x
+ 6). That is a perfect square which enables us to simplify the
radical expression.
√(x^{2}
+ 12x
+ 36) = √(x
+ 6)(x
+ 6) = √(x
+
6)^{2}
= (x
+ 6)
• Note
that you can check by squaring your answer. After squaring, you
should get the expression under the √.
In
the previous examples, we simplified radicals that were perfect
squares. Many times we can simplify radicals that are not perfect
squares. The idea is to break down the number or variable into
factors, one of which is a perfect square. By the
multiplication
property of radicals ^{n}√(ab)
= ^{n}√a
∙ ^{n}√b,
where ^{n}√a
and ^{n}√b
are real numbers.
• Note
that in a square root n is 2 but is not written. When solving
any root higher than 2, it is noted in the upper left of the radical.
For example, the cube root (n = 3) of a is noted as
^{3}√a.
The
cube
root of
a number n
is the number when multiplied by itself 3 times equals n.
For example, the cube root of 125 is 5 because (5) ∙ (5) ∙
(5 ) = 125, or (5)^{3}
= 125. By
definition, the cube root of a is defined as ^{3}√a
= b,
if b^{3}
= a.
Examples:
^{3}√27
= 3 since 3^{3}
= 27
^{3}√(27)
= 3 since (3)^{3}
= 27
^{3}√125
= 5 since 5^{3}
= 125
^{3}√(125)
= 5 since (5)^{3}
= 125
• Note
that 27 and 125 have 1 real number cube root and 2 real number square
roots. Any negative number will have a cube root that is negative
and any positive number will have a cube root that is positive.
Examples:
Simplify the following.
a.
^{3}√8
To
solve this, think of what number multiplied by itself 3 times gives
you 8. Notice that (2) ∙ (2) ∙ (2) = 8. Therefore, ^{3}√8
= 2.
b.
^{3}√343
= 7 since 7 ∙ 7 ∙ 7 = 343.
c.
^{3}√729
= 9
• Note
that if the sum of the digits of a number is divisible by 9, the
number is divisible by 9.
The
sum of the digits of 729 is 18, which is divisible by 9. Therefore,
729 is divisible by 9 and 729 is divisible by 9.
729/9 = 81 and we know the square root of 81 is 9.
We
reviewed how to take the cube root of positive and negative numbers.
Now we will review how to take the cube root involving variables.
Recall that by definition, if x
is a variable representing a real number, ^{3}√x^{3}
= x.
Examples:
Simplify the following.
a.
^{3}√(27x^{3})
= ^{3}√(27)
∙ ^{3}√(x^{3})
= 3 ∙ x
= 3x
Check
by taking the cube of 3x.
b.
^{3}√(5x
+ 2y)^{3}
= (5x
+ 2y)
because of the rule that ^{3}√x^{3}
= x.
c.
^{3}√(216x^{6}y^{12})
^{3}√(216)
= 6, ^{3}√(x^{6})
= x^{2 }^{
}since (x^{2})^{3}
= x^{6
}and ^{3}√(y^{12})
= y^{4}
since (y^{4})^{3
}= y^{12}.
^{ }Multiply
the terms in bold to get 6x^{2}y^{4}.
The
previous examples were perfect cubes, but oftentimes we can find the
cube root of expressions that are not perfect cubes. The idea is to
break down the number or variable into factors, one of which is a
perfect cube. For example, if you solve ^{3}√81,
break it down into ^{3}√27
∙ ^{3}√3
because ^{3}√27
= 3. If you solve ^{3}√(y^{5}),
break it down into ^{3}√(y^{3})
∙ ^{3}√(y^{2})
because ^{3}√(y^{3})
= y.
Remember from the multiplication property of radicals that ^{3}√ab
= ^{3}√a
∙ ^{3}√b.
Example:
Simplify the following.
^{3}√(16x^{4})
^{3}√16
= ^{3}√8
∙ ^{3}√2
= 2
∙ ^{3}√2
(^{3}√8
= 2 because 2^{3}
= 8)
^{3}√(x^{4})
= ^{3}√(x^{3})
∙ ^{3}√x
= x
∙ ^{3}√x
(^{3}√(x^{3})
= x by
the definition of cube root (x)^{3}
= x^{3})
Multiply
like terms to get ^{3}√(16x^{4}) =
2x∙^{3}√(2x)
• Note
that you can check by cubing the answer, [2x ∙ ^{3}√(2x)]^{3
}= 8x^{3} ∙ 2x = 16x^{4}.
Example: Simply the following.
^{3}√(192x^{4}y^{5})
^{3}√(192)
= ^{3}√(64)
∙ ^{3}√(3)
= 4∙
^{3}√(3)
(note that 64 ∙ 3 = 192 and (4)^{3}
= 64)
^{3}√(x^{4})
= ^{3}√(x^{3})
∙ ^{3}√(x)
= x
∙ ^{3}√(x)
(note that x^{3}∙
x
= x^{4}
and (x)^{3}
= x^{3})
^{3}√(y^{5})
= ^{3}√(y^{3})
∙ ^{3}√(y^{2})
= y
∙ ^{3}√(y^{2})
(note that y^{3}
∙ y^{2}
= y^{5}
and (y)^{3}
= y^{3})
Now
multiply the like terms to get 4xy∙^{3}√(3xy^{2}).
In
some cases cube roots will be the form of a fraction of two numbers,
a number and a expression or two expressions. In these cases, try to
simplify the ratio first. By the division property of radicals
^{3}√(x/y) = ^{3}√x / ^{3}√y.
Example:
Simplify the following.
^{3}√(8x^{3}/64y^{6})
By
the division property of radicals ^{3}√(8x^{3}/64y^{6})
= ^{3}√(8x^{3})
/ ^{3}√(64y^{6})
^{3}√(8x^{3})
= 2x
because (2x)^{3}
= 8x^{3}
^{3}√(64y^{6})
= 4y^{2}
because (4y^{2})^{3}
= 64y^{6}
Therefore
^{3}√(8x^{3}/64y^{6})
= (2x)/(4y^{2})
= x/(2y^{2}).
Example:
Simplify the following.
^{3}√(216y^{5}/8y^{8})
By
the division property of radicals ^{3}√(216y^{5}/8y^{8})
= ^{3}√(27/y^{3})
because 216/8 = 27 and (y^{5})/(y^{8})
= 1/(y^{3}).
^{3}√(27)
= 3 and ^{3}√(1/y^{3})
= 1/y.
Therefore
^{3}√(216y^{5}/8y^{8})
= 3/y.
• Note
that it's easier to solve cube roots if you know the first several
perfect cubes. The first ten perfect cubes are 1, 8, 27, 64, 125,
216, 343, 512, 729 and 1000.
We
reviewed how to take the square root and cube root of numbers and
expressions. There are also fourth roots, fifth roots, sixth roots,
seventh roots and so on. The procedure for solving these roots is the
same as for square roots and cube roots. For example, a fourth root
of a natural number x
is a number multiplied by itself four times to equal x,
and so on for higher roots. If n
is an odd natural number greater than 1 (n
> 1), then ^{n}√(x)
is an odd root. When
n
is an even natural number greater than 1 (n
> 1) and x
> 0, then ^{n}√(x)
is an even root.
Examples:
Simplify the following
radical expressions.
a.
^{4}√10000
^{4}√10000
= 10 since 10^{4}
= 10 ∙ 10 ∙ 10 ∙ 10 = 10000
b.
^{5}√(32x^{15})
^{5}√32
= 2 because 2^{5}
= 32
^{5}√(x^{15})
= x^{3}
because (x^{3})^{5}
= x^{15}
Therefore
^{5}√(32x^{15})
= 2x^{3}
c.
^{6}√(729x^{6}y^{18})
^{6}√729
= 3 because 3^{6} = 729
^{6}√(x^{6})
= x because (x)^{6} = x^{6}
^{6}√(y^{18})
= y^{3} because (y^{3})^{6} =
y^{18}
Therefore
^{6}√(729x^{6}y^{18})
= 3xy^{3}
d.
^{7}√(x + 10)^{14}
^{7}√(x
+ 10)^{14 }= (x + 10)^{2} because [(x +
10)^{2}]^{7} = (x + 10)^{14}
Review Problems: Set 1
Simplify
each of the following square roots.
1.
√121 2. √(4/25) 3. √0.49
4.
√60 5. √125 6. √252
Simplify
each of the following radical expressions.
7.
√(9y^{2}) 8. √(64x^{4})
9. √(x^{2} + 14x + 49)
10.
√(y^{2} – 8y + 16) 11. √(81x^{8}y^{6})
12. √(144x^{4}y^{2}z^{6})
13. √(80x^{5}) 14. √(75xy^{2}z^{3}) 15. √(98z^{3})
Simplify
each of the following cube roots.
16.
^{3}√64 17. ^{3}√(27/125)
18. ^{3}√216
19.
^{3}√40 20. ^{3}√320
21. ^{3}√54
Simplify
each of the following radical expressions.
22.
^{3}√(8x^{4}y^{5}z^{6})
23. ^{3}√(108z^{4}) 24. ^{4}√256
25.
^{5}√(243a^{5}) 26. ^{6}√(32x^{12})
27. ^{7}√(10^{7})
Rational Exponents
Just
as we can raise a number or expression to exponents that are
integers, we can raise them to fractional exponents as well. If n
is a positive integer greater then 1, then x^{1/n
}
is the nth
root of x
and
1/n
is a rational
exponent.
In other words x^{1/n}
= ^{n}√x.
To
make this more clear, consider √3 and 3^{1/2}.
By squaring both of these, we can determine that they are equal.
(√3)^{2}
= (√3)(√3) = √9 = 3
(3^{1/2})^{2}
= 3^{(1/2
∙ 2)}
= 3^{1}
= 3
Notice
that squaring both gives a result of 3. Therefore √3 = 3^{1/2}.
We
can rewrite exponential expressions in radical form and radical
expressions in exponential form. The radicand of a radical
expression becomes the base of an exponential expression and one over
the index of the radical expression becomes the exponent. We'll
illustrate this in the following
^{3}√6
= 6
^{1/}^{3}.
Notice the index in red becoming the denominator of the exponent and
the radicand in blue becomes the base.
Examples:
Evaluate
the following.
1.
16^{1/2
}=
√16 = 4. Recall that the index of a square root is 2, although the
2 is not written.
2.
(27)^{1/3
}=
^{3}√(27)
= 3 because (3)^{3
}=
27.
3.
(16/81)^{1/4}
= ^{4}√(16/81)
= 2/3 because (2/3)^{4}
= 16/81.
4.
32^{1/5
}=
2 because (2)^{5}
= 32.
Examples:
Simplify
the following rational expressions.
1.
(25x^{4})^{1/2
}
^{ }We
will break this down into the number part and the variable part.
25^{1/2
}=
5
and (x^{4})^{1/2
}=
√(x^{4})
= x^{2}
since (x^{2})^{2}
= x^{4}.
Therefore
(25x^{4})^{1/2
}=
5x^{2}.
• Note
that (x^{4})^{1/2
}can also be simplified using the rules for
raising an exponent to an exponent. Therefore, the exponents 4 and ½
are multiplied together to get 2, which is the new exponent and
(x^{4})^{1/2
}= x^{2}.
2.
(125x^{3})^{1/3
}
^{ }(125)^{1/3
}=
^{3}√(125)
= 5
since (5)^{3}
= 125 (x^{3})^{1/3
}=
^{3}√(x^{3})
= x
Therefore,
(125x^{3})^{1/3
}
= 5x.
3.
[(2a
+ 7)^{4}]^{1/4
}
= ^{4}√(2a
+ 7)^{4}
= 2a
+ 7
Notice
the answer is 2a
+ 7
and not 2a
+ 7. 2a
+ 7^{4}
= (2a
+ 7)^{4
}and
a
can be any number, therefore 2a
+ 7 could be a negative number. But we can't have a real number
solution if 2a
+ 7 is negative because no real number raised to the 4^{th}
power is negative. Therefore we must have absolute value to ensure a
positive real number solution.
4.
[(4b
– 3)^{5}]^{1/5
}=
^{5}√(4b
– 3)^{5}
= 4b
– 3.
In
this case, since n is
odd, there is no need for the absolute value symbols in the solution.
The value of 4b 3
could still be negative but a negative quantity raised to an odd
power will be negative.
In
the previous examples all of the exponents were in the form 1/n.
Now we will learn how to simplify exponential expressions where the
exponent is in the form p/n
where p and n
are positive integers and n ≠ 1.
To illustrate how to simplify , consider the following example.
27^{2/3}.
We can rewrite this in the form x^{1/}^{n}
raised to the 2nd power, (x^{1/}^{n})^{2}
= x^{2/}^{n}.
Therefore,
27^{2/3 }= (27^{1/3})^{2
}
^{
}=^{
} (^{3}√27)^{2}
= 3^{2} = 9
We
can rewrite the exponential expression in radical form. The
numerator of the rational exponent becomes the power of the radical
expression. The denominator of the rational exponent becomes the
index of the radical expression and the base of the exponential
expression becomes the radicand. In the previous example,
27^{2}^{/3}
= (^{3}√27)^{2}
We
would solve this as follows:
Take
the cube root of 27: ^{3}√27
= 3
Square
3: 3^{2} = 9
Or we can think of 27^{2/3 }as
27^{2}^{/3}
= ^{3}√(27)^{2}
We
would solve this as follows:
Square
27 first : 27^{2} =
729
Take
the cube root of 729 : ^{3}√729
= 9 since 9^{3} =
729.
• Note
that we know that 729 is divisible by 9 because the sum of the digits
in 729 is divisible by 9. The sum of the digits of 729 is 18, which
is divisible by 9.
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