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28

Conjugate pairs

HERE IS THE RULE for multiplying radicals:

It is the symmetrical version of the rule for simplifying radicals. It is valid for a and b greater than or equal to 0.

Problem 1.   Multiply.

Do the problem yourself first!

 a) ·  = b)  2· 3 = 6 c) ·  = = 6 d)  (2)2 = 4· 5 = 20
 e) =

The difference of two squares

Problem 2.   Multiply, then simplify:

Example 1.   Multiply  ( + )().

Solution.   The student should recognize the form those factors will produce:

The difference of two squares

 ( + )( − ) = ()2 − ()2 = 6 − 2 = 4.

Problem 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)   ( + )()  =  ab

Problem 4.   (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 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 = 34 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 7.   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 8.   Write out the steps that show the following.

 a) 1 =  ½()
 = 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
 = 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.

 = 1h · = 1h · _____x − (x + h)_____ = 1h · ____x − x − h_____x + (x + h) = 1h · _______−h_______x + (x + h) = − _______ 1_______x + (x + h)

Next Lesson:  Rational exponents

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