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·
Explain
the notion of rationalization.
·
Identify
a rationalizing factor for a given expression.
Suggested ways of teaching this topic: Explanation by
the Teacher and Practice of Students
Starter Activities
The teacher may start with revising the rules for multiplying
and dividing radicals using examples:
The rules are:
2.
Let students do the problems
themselves first then the teacher might summarize.
Example 1: Multiply.
a) |
· = |
|
b)
2· 3 = 6 |
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c) |
· = |
= 6 |
d)
(2)² = 4· 5
= 20 |
e) |
= |
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The difference of two squares |
Example 2: Multiply, and then simplify:
a)
b)
( + )( − ).
Solution:
1.
2.
The student should recognize the form
those factors will produce.
The difference of two squares
(
+ )(
− ) |
= |
()²
− ()² |
|
= |
6 − 2 |
|
= |
4. |
Practice and Consolidation is needed here! Thus, the teacher
should guide students to recognize the forms of “difference of two squares”
letting them practice on products of the following type.
Lesson
Notes
Example 3: 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
Example 4: Given (x − 1 − )
(x − 1 + )
a) What form does that produce?
Expected Answer:
The
difference of two squares of the form a2- b2, where (x–
1) is "a" and is
"b"
b) Find the final product.
(x − 1 − )(x
− 1 + ) |
= |
(x − 1)² − 2 |
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|
|
= |
x² − 2x + 1 − 2, |
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On squaring the binomial, |
|
= |
x² − 2x − 1 |
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Example 5: Multiply.
(x + 3 + )(x
+ 3 − ) |
= |
(x + 3)² − 3 |
|
= |
x² + 6x + 9 − 3 |
|
= |
x² + 6x + 6 |
Example 6: Simplify the following.
a) |
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= |
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b) |
|
= |
3 |
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c) |
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= |
a |
= |
a· a |
= |
a² |
< TR> |
What are Conjugate pairs?
The conjugate of a + is a
−.
They are a conjugate pair.
Example 7: Multiply 6 − with
its conjugate.
Solution: The product of a conjugate pair is the difference
of two squares
(6 − )(6
+ )
= 36 − 2 = 34
CONCLUSION
When we multiply a conjugate pair, the radical vanishes and
we obtain a rational number.This process is called Rationalization.
Example 8: Multiply each number with its conjugate.
This time, students should be able
to write the product immediately: 6 − 2 = 4
Example 9: Rationalize the denominator of
Solution:
Multiply both the denominator and the numerator by the
conjugate of the denominator; that is, multiply them by 3−.
1
|
= |
|
= |
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The numerator becomes 3−.
The denominator becomes the difference of the two squares.
Example 10:
Rationalize the denominator of
Solution: |
|
= |
|
= |
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|
= |
−(3 − 2) |
|||
|
= |
2
− 3 |
Example
11: Write out the
steps that show each of the following equivalences.
a)
|
1 |
= ½() |
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1 |
= |
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= |
|
= |
½(
−) |
b)
|
2 |
= ½(3 − ) |
|
2 |
= |
|
= |
|
= |
½(3
− ) |
c)
|
_7_ |
= |
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_7_ |
= |
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= |
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= |
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d) |
|
= |
3 + 2 |
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|
= |
|
= |
2 + 2
+ 1, |
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Example 12: Simplify |
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At every step of the solution the
teacher is expected to ask the students to give reasons.
Solution: |
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= |
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Why? |
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= |
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Why? |
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= |
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Why? |
||
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= |
6
− 5 |
Why? |
. |
Example 13: Simplify |
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= |
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Why? |
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= |
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Why? |
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|
= |
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Why? |
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|
= |
3
+ 2 |
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Why? |
Concluding
Activities
Make sure
that the students summarize that rationalization of the denominator can be done
by multiplying both the denominator and the numerator by the conjugate of the
denominator. Check if every student has developed the skill of rationalizing
radical expressions by giving them practice exercises.
Practice
Exercises
1.
Rationalize
the following by multiplying with an appropriate conjugate and simplify
a)
b)
2.
Rationalize
the denominator and simplify
a) |
b) |
c) |
d) |
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e) |
f) |
g) |
h) |
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