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SURDS (RADICALS)

Surds are roots of arithmetic numbers that cannot be determined exactly e.g  are

surds.  While  are not surds since

 

RULES OF SURDS

If a, b, c, d are integers then

 

Example

Simplify the following, leaving your answers in surd form

Solution

 

BASIC FORM OF SURD

When a rational number under square root contain a factor which is a square of a number, the surd can be reduced to a simple form or its basic form.

Example

Reduce the following to its basic form

1

Solution              

1.

2.  

3. 

    The transformation  which reduced a surd to its form. can also be reversed to form a single surd

Example

Write the following basic form in their single surd form

 

Solution

So then we can say that  s are similar, so also

 

Example

       Simplify the following

           

Solution

           

 

CONJUGATE SURDS

Two surds are said to be conjugate of each other, if their product gives a difference of their square.  That is

             

 

 

 

Example

Solution

Example

Simplify each of the following products

Solution

 

 

RATIONALISING DENOMINATORS

(i)         To rationalize the denominator of the fraction

           

ii.          To rationalize the denominator of the fraction

 

           

(iii)       To rationalize the denominator of the

fraction

               

Example

By rationalizing the denominator, simplify each of the following

   

               

                

 

Example

Express as equivalent fraction with rational denominator

Solution

              

Example

If x = 5 − , find the value of x2 +

Solution

           

           

Example

            If x =   

Find the value of 3x2 – 5xy + 3y2

 

Solution

Example

            Simplify

 

Solution              

 

Example   

Simplify    

Solution

              

     =       [Difference of two squares]

   =  

  =   =      =

 

 

 

  Example

     Simplify   

                                                             =

            =      = −1

 

SQUARE ROOT OF SURD

Given a surd x +  then the square root of the given surd is

                 Let    

                  

                      x

             Comparing the equation

             x = a + b …………{i}

           

            y  =  4ab ………….{ii}

Solving equation {i}and{ii}simultaneously will yield the value of a and b

             Example

Find the square root of the following

1.    12+ 2

2.     31−4

Solution  

 1.  12 + 2

Let the square root of 12+2  be

           

                          12 +2

                         12 +     

            From equality of surds

               a + b = 12………{i}

               2

            ab =  35……..{ii}

From{i} a = 12 − b

             Substitute  a = 12 −b into {11}

               b = 35

             12bb2 = 35

              b2 −12b −35 =0                   

             b =7 or b = 5

Substituting the values of b

  a = 12−7 =or a =12−5 =7

  Hence the Square root is  twice

 

(ii)  31 −4

Let the square root 31−4   

        

Square both sides

            31− 4

            31 − 4   

 From the equality of surd

       a + b = 31………..(i)

         −2

            ab = 84………(ii)

From (i) a = 31−b

Substituting a = 31– b into (11)

         b = 84

           b2 – 31b + 84 =0

            = 0

           b= 28 or b = 3

 Substituting the values of b

     a = 31−28 =3 or a = 31 -3 = 28 

Hence the square root of

 

                                          EQUATION INVOLVING SURDS

               Example

               Solve for x in

                 Solution

             

            Squaring both sides

                  6x – 3 = 49

                 6x = 52

                    x =

 

   EXAMPLE

 Solve for x in

With these types of surds, it is easier to transpose one to the opposite side

Squaring both sides

Squaring both sides

64(3x-5) = 1+10x + 25x2

192x – 320 = 1+10x+25x2

25x2 – 182x + 321 = 0

(25x – 107) (x−3) = 0

x = 3 or x =

 

By substituting these two x-values in the original equation we find that x=  is not a solution.  Thus, the given equation has 3 as its solution

 

Example

     Solve for x in the equation  

           

Solution

Squaring both sides

Squaring both sides

      9x2 + 54x + 81 = 12x2 + 40x + 32

      3x2 – 14x – 49 = 0

      (3x+7) (x-7) = 0

      x = 7 or x =  

By substituting these two x-values in the original equation, we see that x =  does not satisfy the given equation. Thus, the equation has one solution x = 7

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