27
SIMPLIFYING RADICALS
A SQUARE ROOT RADICAL is "simplified" when the radicand has no square factors.
Example 1. 33, for example, has no square factors. Its factors are 3· 11, neither of which is a square number. Therefore, 
is simplified, or, as we say, in its simplest form.
Example 2. 18 has the square factor 9. 18 = 9· 2. Therefore, 
is not in its simplest form. To put a radical in its simplest form, we make use of this theorem:
The square root of a product
is equal to the product of the square roots
of each factor.
(We will prove that when we come to rational exponents, Lesson 29.)
Therefore,
= 
= 
· 
= 3.
We have simplified .
Example 3. Simplify 
.
Solution. 48 = 4· 12. Therefore
on extracting 
= 2.
But the radicand 12 still has as square factor 4. We must continue:
We have now completely simplified .
Note that we could have achieved the result immediately, if we had realized that 48 = 16· 3.
Problem 1. Simplify the following. Inspect each radicand for a square factor: 4, 9, 16, 25, and so on.
To see the answer, pass your mouse over the colored area.
To cover the answer again, click "Refresh" ("Reload").
Do the problem yourself first!
a) 
=
b) 
=
= 
= 5
c) 
=
= 
= 3
d) 
=
= 7
e)  |
42 has no square factors. Therefore,  is in its simplest form. |
f) 
=
= 10
g) 
=
= 5
h) 
=
= 4
Problem 2. Reduce to lowest terms.
| a) |  2 |
= |  2 |
= |  2 |
= | |
| b) |  3 |
= |  3 |
= |  3 |
= | 2 |
| c) |  2 |
= | The radical is in its simplest form. The fraction cannot be reduced. |
For a method of finding square factors by means of prime factorization, see Lesson 31 of Arithmetic, Prime numbers.
Similar radicals
Similar radicals have the same radicand. We add them as like terms.
7 + 2
+ 5 + 6 −
= 7 + 8 + 4.
2 and 6 are similar, as are 5 and . We combine them by adding their coefficients.
As for 7, it does not "belong" to any radical.
Problem 3. Simplify each radical, then add the similar radicals.
a) + 
=
3 + 2 = 5
b) 4 − 2 +
| = |
4 − 2 + |
| = | 4· 5 − 2· 7 + |
|
| = | 20 − 14 + |
|
| = | 7 |
|
c) 3 +  − 2
| = |
3 +  − 2 |
| = | 3· 2 + 2 − 2· 4 |
|
| = | 6 + 2 − 8 |
|
| = | 2 − 2 |
|
d) 3 +  +  |
= |
3 +  +  |
| = | 3 + 2 + 3 |
|
| = | 3 + 5 |
|
e) 1 −  + |
= |
1 −  +  |
| = | 1 − 8 + 3 |
|
| = | 1 − 5 |
|
Problem 4. Simplify the following.
| a) |  2 |
= |  2 |
= | 2 − , |
on dividing each term in the numerator by 2. |
| b) |  5 |
= |  5 |
= | 2 + |
| c) |  6 |
= |  6 |
= |  3 |
Next Lesson: Multiplying and dividing radicals
Please make a donation to keep TheMathPage online.
Even $1 will help.
Copyright © 2001-2007 Lawrence Spector
Questions or comments?
E-mail: themathpage@nyc.rr.com