Question

Factorize using splitting the middle term method:
mx/n + n/m = 1 - 2x
Plzz answerr it fasstt....

Answer 1

given   m x² / n + n / m  = 1 - 2 x

         [ m*m x² + n² ] / ( n * m ) = 1 - 2 x
          m² x² + n² = n m - 2 n m x
          m² x² + 2 n m x + (n² - n m) = 0
          m² [ x² + 2 (n/m) x + (n²/m² - n/m) = 0

Let  n/m = k.
          m² [ x² + 2 k x + (k² - k) ] = 0

Take Product  1 * (k² - k) . We need to find its factors so that their sum or difference is equal to 2 k.

Factorizing:
              k² - k  = k ( k -1)   ,     Sum of factors is  2k - 1.
                        = (k + √k) (k - √k) .   Sum of factors is 2 k.

So we got it... NOw..

           m² [ x² + 2 k x + (k² - k) ] = 0   becomes
           m² [ x² + (k +√k) x + (k - √k) x + (k² - k) ] = 0
           m² [ x {x + (k + √k)} + (k - √k) { x + (k + √k) } ] = 0
=>           m²  {x + k + √k} * { x + k - √k}  = 0
           m²  { x + n/m +√(n/m) } * { x + n/m - √(n/m) } = 0
=>           [ m x + n + √(nm) ] * [ m x + n - √(nm) ] = 0

This is the answer.
There are two factors... (m x + n+√nm)  and (mx + n - √mn).