28
MULTIPLYING AND DIVIDING
RADICALS
HERE IS THE RULE for multiplying radicals:
It is the symmetrical version of the rule for simplifying radicals.
Problem 1. Multiply.
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Do the problem yourself first!
| a) |  ·  =  |
b) 2 · 3 = 6 |
||
| c) |  ·  =  |
= 6 | d) (2)² = 4· 5 = 20 |
|
| e) |  =  |
The difference of two squares |
Example 1. Multiply ( + )( − ).
Solution. The student should recognize the form those factors will produce:
| ( + )( − ) |
= | ()² − ()² |
| = | 6 − 2 | |
| = | 4 | |
Problem 2. Multiply.
a) ( + )( − ) =
5 − 3 = 2
b) (2 + )(2 − ) =
4· 3 − 6 = 12 − 6 = 6
c) (1 + 
)(1 − ) =
1 − (x + 1) =
1 − x − 1 =
−x
d) (
+ 
)( − ) =
a − b
Problem 3. (x − 1 − )(x − 1 + )
a) What form does that produce?
The difference of two squares. x − 1 is "a." 
is "b."
b) Multiply out.
| (x − 1 − )(x − 1 + ) |
= | (x − 1)² − 2 | ||
| = | x² − 2x + 1 − 2, | on squaring the binomial, | ||
| = | x² − 2x − 1 | |||
Problem 4. Multiply out.
| (x + 3 + )(x + 3 − ) |
= | (x + 3)² − 3 |
| = | x² + 6x + 9 − 3 | |
| = | x² + 6x + 6 | |
Dividing radicals
For example,
 |
= |  |
= |  |
Problem 5. Simplify the following.
| a) |  |
= | |
b) |  8 |
= | 3 4 |
 |
c) |  |
= | a |
= | a· a | = | a² |
Conjugate pairs
The conjugate of a + is a − . They are a conjugate pair.
Example 2. Multiply 6 − with its conjugate.
Solution. The product of a conjugate pair --
(6 − )(6 + )
-- is the difference of two squares. Therefore,
(6 − )(6 + ) = 36 − 2 = 34
When we multiply a conjugate pair, the radical vanishes and we obtain a rational number.
Problem 6. Multiply each number with its conjugate.
a) x + 
= x² − y
b) 2 −
(2 − 
)(2 + ) = 4 − 3 = 1
| c) + |
You should be able to write the product immediately: 6 − 2 = 4. |
d) 4 − 16 − 5 = 11
Example 3. Rationalize the denominator:
1  |
Solution. Multiply both the denominator and the numerator by the conjugate of the denominator; that is, multiply them by 3 − .
| 1 |
= |  9 − 2 |
= | 7 |
The numerator becomes 3 − . The denominator becomes the difference of the two squares.
| Example 4. |  |
= |  3 − 4 |
= |  −1 |
| = | −(3 − 2) |
||||
| = | 2 − 3 |
||||
Problem 7. Write out the steps that show the following.
| a) | 1  |
= ½( ) |
| 1 |
= |  5 − 3 |
= | 2 |
= | ½( −) |
|
| The definition of division | |||||||
| b) | 2 3 + |
= ½(3 − ) |
| 2 3 + |
= |  9 − 5 |
= | 4 |
= | ½(3 − ) |
| c) | _7_ 3 + |
= |  6 |
| _7_ 3 + |
= |  9· 5 − 3 |
= | 42 |
= |  6 |
| d) |  − 1 |
= | 3 + 2 |
| − 1 |
= |  2 − 1 |
= | 2 + 2 + 1, |
Perfect square trinomial | ||
| = | 3 + 2 |
||||||
| e) |  1 + |
= |  x |
1 + |
= |  1 − (x + 1) |
||||
| = |  1 − x − 1 |
, | Perfect square trinomial | |||
| = |  −x |
|||||
| = |  x |
on changing all the signs. | ||||
| Example 5. Simplify |  |
| Solution. | |
= |  |
on adding those fractions, | |
| = |  |
on taking the reciprocal, | |||
| = |  6 − 5 |
on multiplying by the conjugate, | |||
| = | 6 − 5 |
on multiplying out. | |||
| Problem 8. Simplify |  |
|
= |  |
on adding those fractions, | ||
| = |  |
on taking the reciprocal, | |||
| = |  3 − 2 |
on multiplying by the conjugate, | |||
| = | 3 + 2 |
on multiplying out. | |||
Problem 9. Here is a problem that Calculus students have to do. Write out the steps that show:
 |
= − | ____1____ x  + (x + h) |
In this case, you will have to rationalize the numerator.
|
= | 1 h |
· |  |
| = | 1 h |
· | _____x − (x + h)_____
| |
| = | 1 h |
· | ____x − x − h_____ x  + (x + h)
| |
| = | 1 h |
· | _______−h_______ x + (x + h)
| |
| = | − | _______ 1_______ x + (x + h)
| ||
Next Lesson: Rational exponents
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