HERE IS THE RULE for multiplying radicals:
It is the same version of the rule for simplifying radicals. It is valid for a and b greater than or equal to 0.
Problem 1. Multiply.
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| a) |  ·  =  | b) 2 · 3 = 6 | ||
| c) |  ·  =  | = 6 | d) (2)² = 4· 5 = 20 | |
| e) |  =  | How to squares differ. |
Problem 2. Multiply, then simplify:
Example 1. Multiply ( + )( − ).
+ )( − ).
Solution.
How 2 squares differ.
| ( + )( − ) | = | ()² − ()² |
| = | 6 − 2 | |
| = | 4. | |
Problem 3. Multiply.
a) ( + )( − ) = 5 − 3 = 2
+ )( − ) = 5 − 3 = 2
b) (2 + )(2 − ) = 4· 3 − 6 = 12 − 6 = 6
+ )(2 − ) = 4· 3 − 6 = 12 − 6 = 6
c) (1 + 
)(1 − ) = 1 − (x + 1) = 1 − x − 1 = −x
)(1 − ) = 1 − (x + 1) = 1 − x − 1 = −x
d) (
+ 
)( − ) = a − b
+ )( − ) = a − b
Problem 4. (x − 1 − )(x − 1 + )
)(x − 1 + )
a) What form does that produce?
The difference of two squares. x − 1 is "a." 
is "b."
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 5. Multiply out.
| (x + 3 + )(x + 3 − ) | = | (x + 3)² − 3 |
| = | x² + 6x + 9 − 3 | |
| = | x² + 6x + 6 | |
For example,
 | = |  | = |  |
Problem 6. Simplify the following.
| a) |  | = | | b) |  8 | = | 3 4 |  | c) |  | = | a | = | a· a | = | a² |
The conjugate of a + is a − . They are a conjugate pair.
is a − . They are a conjugate pair.
Example 2. Multiply 6 − with its conjugate.
with its conjugate.
Solution. The product of a conjugate pair --
(6 − )(6 + )
)(6 + )
-- is the how 2 squares differ. Therefore,
(6 − )(6 + ) = 36 − 2 = 34.
)(6 + ) = 36 − 2 = 34.
When we multiply a conjugate pair, the radical disappears and we obtain a rational number.
Problem 7. Multiply each number with its conjugate.
a) x + 
= x² − y
= x² − y
b) 2 − (2 − 
)(2 + ) = 4 − 3 = 1
(2 − )(2 + ) = 4 − 3 = 1| c) + | You should be able to write the product immediately: 6 − 2 = 4. |
d) 4 − 16 − 5 = 11
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.
. The denominator becomes the difference of the two squares.| Example 4. |  | = |  3 − 4 | = |  −1 |
| = | −(3 − 2) | ||||
| = | 2 − 3. | ||||
Problem 8. Write out the steps that show the following.
| a) | 1  | = ½( ) |
| 1 | = |  5 − 3 | = | 2 | = | ½( −) | |
| The definition of trinomials | |||||||
| 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 9. Simplify |  |
| = |  | on adding those fractions, | ||
| = |  | on taking the reciprocal, | |||
| = |  3 − 2 | on multiplying by the conjugate, | |||
| = | 3 + 2 | on multiplying out. | |||
Problem 10. Here is a problem that comes up in Calculus. 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) Youtube video. http://www.youtube.com/watch?v=cx1TAJ9cP0o | ||
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