Solving specific Quadratic Equations by Observation - II :-

In Vedic mathematics, one of the meanings of "Samuccchaya" is 'the product of independent terms'.
This concept is used to solve specific types of Quadratic Equations.



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Some more Examples :

Example # 3
Consider the following equation :
     1           -        1        =      1        -          1
x + a + b               x + c          x + a           x - b + c

Here, we first need to put the equation in the standard form, so that the sutra can be applied.
Thus, re-arranging the terms, we get :

      1          +         1           =        1        +       1
x + a + b          x - b + c            x + a           x + c

Here, D1 + D2 = (x + a + b) + (x - b + c) = 2x + a + c
and    D3 + D4 = (x + a) + (x + c) = 2x + a + c
Since, D1 + D2 = D3 + D4 = 2x + a + c
The solution is : 2x + a + c = 0    2x = -(a + c)    x = -(a + c) 2


Example # 4
Consider the following equation :
4x - 43     -     3x - 74     =     5x + 86     -     2x + 7
  x - 11              x - 25             x + 17              x + 3

To convert the equation in the standard form, add and subtract the coefficients of x, of each numerator from each term. Thus, we get :
4x - 43   -  4   +   4   -   3x - 74   -   3   +   3   =   5x + 86   -   5   +   5   -   2x + 7   -   2   +   2
  x - 11                            x - 25                            x + 17                              x + 3  

4x - 43 - 4x + 44   +   4   -   3x - 74 - 3x + 75   +   3   =   5x + 86 - 5x - 85   +   5   -   2x + 7 - 2x - 6   +   2
         x - 11                                   x - 25                                  x + 17                                  x + 3           

        1        -        1          +     7     =        1          -        1        +     7
        x - 11           x - 25                           x + 17            x + 3

       1         -        1         =        1         -       1
        x - 11           x - 25           x + 17           x + 3

Thus, rearranging the terms, we get :

        1        +        1        =       1          +      1
        x - 11           x + 3           x + 17           x - 25

The equation is now in the standard form, so the sutra can be applied now :
Here  D1 + D2 = (x - 11) + (x + 3) = 2x - 8
and    D3 + D4 = (x + 17) + (x - 25) = 2x - 8
Since, D1 + D2 = D3 + D4 = 2x - 8
The solution is : 2x - 8 = 0    2x = 8    x = 8 2 = 4


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