|
A radical
or the
principal nth root of
k:
k, the radicand, is a real
number.
n, the index, is a positive integer greater than
one.
|
 |
|
|
Simplifying Radicals:
Radicals that are simplified
have:
- no fractions left under the radical.
- no perfect power factors in the
radicand, k.
- no exponents in the radicand, k, greater than
the index, n.
- no radicals appearing in the denominator
of a fractional answer. |
|
Examples:
(The following examples demonstrate various
solution methods.)
|
Simplify:
Factor the radicand to isolate
the perfect power factor(s),
which will allow them to be removed from under the radical.
You will need to remember your rules for working with
exponents in order to isolate the perfect powers. |
perfect squares
4, 9, 16, 25, 36, ...
x2,
x4, x6,
x8, ...
x2y2,
x2y4,
16x6y8,
...
powers are "even" |
perfect cubes
8, 27, 64, 125, ...
x3,
x6, x9,
x12, ...
x3y3,
x3y6,
27x6y9,
...
powers are "multiples of 3" |
|
1.
(notice the perfect cube of 8 being isolated)
2.
(in this problem, several perfect squares
were isolated, namely 9, a2 and b4)
3.
(the -8 and the y3 are perfect
cubes)
4.
(x5 is a perfect power of 5)
|
To add radicals,
simplify first if possible,
and add "like" radicals.
1.
2.
(After
simplifying the radicals, it became
(Be sure to combine ONLY like radicals. Also
apparent which radicals could be added.)
be sure to continue writing
the index of 3. Failing
to do so, creates an incorrect answer using square
root, instead of the correct cube root.) |
To multiply radicals apply
and/or distributive multiplication (FOIL).
1.
(multiply outside, multiply inside, and simplify)
2.
(notice the perfect powers isolated)
3.
(distribute across the parentheses)
4.
(this problem requires multiplication of two
binomials by distributive multiplication or FOIL.
Notice how
none of the terms can be combined in
the final answer.)
5.
(again binomial multiplication is needed.
This time, however, terms can be
combined for the final answer.)
6.
(again binomial multiplication is needed.
The two terms being multiplied in this problem are
conjugates.
Notice how the middle terms of the
answer drop out.) |
|
