Question 98787


        solve the quadratic equation by completing the square 
x^2 - 6x - 3=0
 
   x^2-6x-3 = 0  reduce it to the std form i.e a^2+2ab+b^2 = (a+b)^2

  x^2-2.3x = 3   compare with the std form we get 2ab = 2.3x  so a= x & b = 3

   add (b)^2 = (3)^2 on both sides

  x^2- 2.3x+(3)^2 = 3+(3)^2

   (x-3)^2 = 3+9

  (x-3)^2 = 12

  taking sq.rt on both sides   

  (x-3) = +-sq.rt 12

      x = +-2 sq.rt3 +3

   required  solution is x = 3+-2 (sq.rt 3)