27
SIMPLIFYING RADICALS
WE SAY THAT A SQUARE ROOT RADICAL is simplified, or in its simplest form, when the radicand has no square factors.
A radical is also in simplest form when the radicand is not a fraction.
Example 1. 33, for example, has no square factors. Its factors are 3· 11, neither of which is a square number. Therefore, 
is 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.
| Here is a simple illustration: |  |
) |
Therefore,
= 
= 
· 
= 3.
We have simplified .
Example 3. Simplify 
.
Solution. We have to factor 42 and see if it has any square factors. We can begin the factoring in any way. For example,
42 = 6· 7
We can continue to factor 6 as 2· 3, but we cannot continue to factor 7, because 7 is a prime number (Lesson 31 of Arithmetic). Therefore,
42 = 2· 3· 7
We now see that 42 has no square factors -- because no factor is repeated. Compare Example 1 and Problem 2 of the previous Lesson.
therefore is in its simplest form.
Example 4. Simplify 
.
Solution.
180 = 2· 90 = 2· 2· 45 = 2· 2· 9· 5 = 2· 2· 3· 3· 5
Therefore,
= 2· 3
= 6.
Problem 1. Simplify the following. Do that by inspecting 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) 
=
= 4
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. |
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. (Lesson 20) |
To see that 2 is a factor of the radical, we first have to simplify the radical. See Problem 2.
| b) |  5 |
= |  5 |
= | 2 + |
| c) |  6 |
= |  6 |
= |  3 |
on dividing each term by 2. |
Next Lesson: Multiplying and dividing radicals
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