8.2
Radicals - Higher Roots
Objective: Simplify radicals with an index greater than two.
While square roots are the most common type of radical we work with, we can
take higher roots of numbers as well: cube roots, fourth roots, fifth roots, etc. Following is a definition of radicals.
√
m
a = b if bm = a
The small letter m inside the radical is called the index. It tells us which root we
are taking, or which power we are “un-doing”. For square roots the index is 2. As
this is the most common root, the two is not usually written.
World View Note: The word for root comes from the French mathematician
Franciscus Vieta in the late 16th century.
The following example includes several higher roots.
Example 1.
√
√
3
3
− 64 = − 4
125 = 5
√
√
4
7
81 = 3
− 128 = − 2
√
√
5
4
32 = 2
− 16 = undefined
We must be careful of a few things as we work with√higher roots.√First its important not to forget to check the index on the root. 81 = 9 but 4 81 = 3. This is
because 92 = 81 and 34 = 81. Another thing to watch out for is negatives under
roots. We can take an odd root of a negative number, because a negative number
raised to an odd power is still negative. However, we cannot take an even root of
a negative number, this we will say is undefined. In a later section we will discuss
how to work with roots of negative, but for now we will simply say they are undefined.
We can simplify higher roots in much the same way we simplified square roots,
using the product property of radicals.
√
√
√ m
m
Product Property of Radicals: ab = m a · b
Often we are not as familiar with higher powers as we are with squares. It is
important to remember what index we are working with as we try and work our
way to the solution.
Example 2.
√
3
54
23 = 8
33 = 27
We are working with a cubed root, want third powers
Test 2, 23 = 8, 54 is not divisible by 8.
Test 3, 33 = 27, 54 is divisible by 27!
1
√
3
√
3
27 · 2
√
27 · 3 2
√
33 2
Write as factors
Product rule, take cubed root of 27
Our Solution
Just as with square roots, if we have a coefficient, we multiply the new coefficients
together.
Example 3.
√
3 4 48
24 = 16
√
3 4 16 · 3
√
√
3 4 16 · 4 3
√
3·2 4 3
√
64 3
We are working with a fourth root, want fourth powers
Test 2, 24 = 16, 48 is divisible by 16!
Write as factors
Product rule, take fourth root of 16
Multiply coefficients
Our Solution
We can also take higher roots of variables. As we do, we will divide the exponent
on the variable by the index. Any whole answer is how many of that varible will
come out of the square root. Any remainder is how many are left behind inside
the square root. This is shown in the following examples.
Example 4.
p
5
x25y 17z 3
p
x5 y 3 5 y 2z 3
Divide each exponent by 5, whole number outside, remainder inside
Our Solution
25
In the previous example, for the x, we divided 5 = 5 R 0, so x5 came out, no x’s
17
remain inside. For the y, we divided 5 = 3 R 2, so y 3 came out, and y 2 remains
3
inside. For the z, when we divided 5 = 0R 3, all three or z 3 remained inside. The
following example includes integers in our problem.
Example 5.
√
3
2 40a4b8
√
3
2 8 · 5a4b8
√
3
2 · 2ab2 5ab2
√
3
4ab2 5ab2
Looking for cubes that divide into 40. The number 8 works!
Take cube root of 8, dividing exponents on variables by 3
Remainders are left in radical. Multiply coefficients
Our Solution
Beginning and Intermediate Algebra by Tyler Wallace is licensed under a Creative Commons
Attribution 3.0 Unported License. (http://creativecommons.org/licenses/by/3.0/)
2
8.2 Practice - Higher Roots
Simplify.
√
1) 3 625
√
3) 3 750
√
5) 3 875
√
7) − 44 96
√
9) 64 112
√
11) − 4 112
√
4
13) 648a2
√
5
15) 224n3
p
17) 5 224p5
√
19) − 37 896r
√
3
21) − 2 − 48v 7
√
3
23) − 7 320n6
p
25) 3 − 135x5 y 3
p
27) 3 − 32x4 y 4
p
29) 3 256x4 y 6
p
31) 73 − 81x3 y 7
√
3
33) 2 375u2v 8
√
3
35) − 3 192ab2
p
37) 63 − 54m8n3 p7
p
39) 64 648x5 y 7z 2
p
41) 74 128h6 j 8k 8
2)
√
3
375
4)
√
3
250
6)
√
3
24
√
8) − 84 48
√
10) 34 48
√
12) 54 243
√
4
14) 64n3
√
5
16) − 96x3
√
6
18) 256x6
√
7
20) − 8 384b8
√
3
22) 4 250a6
√
3
24) − 512n6
√
3
26) 64u5v 3
√
3
28) 1000a4b5
p
30) 3 189x3 y 6
p
32) − 43 56x2 y 8
√
34) 83 − 750xy
p
36) 33 135xy 3
p
38) − 64 80m4 p7 q 4
√
4
40) − 6 405a5b8c
p
42) − 64 324x7 yz 7
Beginning and Intermediate Algebra by Tyler Wallace is licensed under a Creative Commons
Attribution 3.0 Unported License. (http://creativecommons.org/licenses/by/3.0/)
3
8.2
√
3
1) 5 5
√
2) 53 3
√
3) 53 6
√
4) 53 2
√
5) 53 7
√
6) 23 3
√
7) − 84 6
√
8) − 164 3
√
9) 124 7
√
10) 64 3
√
11) − 24 7
√
12) 154 3
√
4
13) 3 8a2
√
4
14) 2 4n3
√
5
15) 2 7n3
Answers - Higher Roots
√
5
16) − 2 3x3
√
17) 2p5 7
√
18) 2x6 4
√
19) − 67 7r
√
20) − 16b7 3b
√
21) 4v 2 3 6v
√
22) 20a2 3 2
√
23) − 28n2 3 5
24) − 8n2
√
3
25) − 3xy 5x2
√
3
26) 4uv u2
√
27) − 2xy 3 4xy
√
3
28) 10ab ab2
√
29) 4xy 2 3 4x
√
30) 3xy 2 3 7
√
31) − 21xy 2 3 3y
p
32) − 8y 2 3 7x2 y 2
√
3
33) 10v 2 3u2v 2
√
34) − 403 6xy
√
3
35) − 12 3ab2
√
36) 9y 3 5x
p
37) − 18m2np2 3 2m2 p
p
38) − 12mpq 4 5p3
p
39) 18xy 4 8xy 3z 2
√
40) − 18ab2 4 5ac
√
4
41) 14hj 2k 2 8h2
p
42) − 18xz 4 4x3 yz 3
Beginning and Intermediate Algebra by Tyler Wallace is licensed under a Creative Commons
Attribution 3.0 Unported License. (http://creativecommons.org/licenses/by/3.0/)
4