Make your own free website on Tripod.com

Saamya Samucchaya And  Shoonyam Anyat

 

In this section we will solve the quadratic equations based on two sutras already known to us i.e. Saamya Samucchaya (under simple equations) and Shoonyam Anyat (under simple simultaneous equations).

Consider the following two examples :


1)       3    +    4     =      6    +    1   
       x + 3     x + 4      x + 6     x + 1
  

Using Shoonyam Anyat, we find that the sum of the ratios of the numerators to the independent factors in each term, on either side are equal. i.e.

     3/3  +  4/4   =  6/6  +  1/1

     Therefore according to the sutra  we have  x = 0

Now the above problem can be written as


       1 -    x    + 1 -    x    =  1 -    x    + 1 -    x    
            x + 3         x + 4          x + 6         x + 1
            
   ==>     x     +      x     =       x     +     x  
          x + 3       x + 4        x + 6      x + 1
         
   By taking x as the common factor from each term (or applying
   the first meaning of the samucchaye sutra) we get  x = 0
   
   ==>     1    +    1     =      1    +    1   
         x + 3     x + 4      x + 6    x + 1
         
     Therefore by Samucchaye formula, we have    2x + 7 = 0  
                                        ==>   x = -7/2

                                        
2)        1    +    1    =      2     +     1                                        
       2x + 1   3x + 1    3x + 2    6x + 1  
       
       Now     1     can be written as   1 -   2x    and so on
               2x + 1                                  2x + 1
            
   Therefore the problem can be rewritten as
   
    1 -   2x    + 1 -   3x     =   1 -   3x    + 1 -   6x  
        2x + 1        3x + 1            3x + 2        6x + 1
        
      ==>    2x    +    3x    =    3x    +    6x  
           2x + 1     3x + 1     3x + 2     6x + 1
           
      Taking x as the common factor we have  x = 0
      
      ==>    2    +    3    =      3    +     6  
          2x + 1    3x + 1    3x + 2    6x + 1
          
By the method discussed in the 'Disguised Problems' under 
'Simple Equations' we get
        
           6     +     6    =      6     +    6    
        6x + 3    6x + 2    6x + 4    6x + 1
        
        Hence we have  12x + 5 = 0
                       ==> x = -5/12
        

Back to Quadratic Equations