Question 2444
   Let us say the side of the square = x
   The area of the square = {{{x^2}}} Sq.mtrs.
  So after the increase, the sides of the rectangle is:
  length = (x+10);  width = (x+5);
  So the area of the new rectangle is =  (x+10)(x+5)
  =   {{{x^2+15x+50}}} = 3{{{x^2}}} Sq.mtrs. (Given.)
  =   15x + 50 = 2{{{x^2}}}  ...(By subtracting {{{x^2}}} from both the sides)
  =   2{{{x^2}}}-15x-50 = 0   (By bringing all the terms to one side to make it 
                               a quadratic equation.)
By factorising the above expression, we get,
  =   2{{{x^2}}}-20x+5x-50 = 0
  =  2x(x-10)+5(x-10) = 0
  =  (x-10)*(2x+5) = 0
So either (x-10) or (2x+5) has to be equal to 0.
If (x-10) = 0, then x = 10;
If (2x+5) = 0, then x = -2.5;  (because 2x should be = -5)
But the length cannot be negative. 
Hence, the side of the square was equal to 10mtrs.


gsmani iyer