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28

MULTIPLYING AND DIVIDING
RADICALS

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.

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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

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 − (
 
  = 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

Dividing radicals

For example,


 
= =

Problem 6.   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 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()
        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 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
·   ____xxh_____
x + (x + h)
 
    =   1
h
·  _______−h_______
x + (x + h)
 
    =   −  _______ 1_______
x + (x + h)

 

Next Lesson:  Rational exponents


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