Question 931131
Let x = the number

A number is increased by the square root of this result is less than the square root of half the number is 3. 

      √x-√(x/2)=3
            √ x=√(x/2)+3
         (√x)^2=(√(x/2)+3)^2
              x=x/2+6√(x/2)+9
              0=-x+x/2+6√(x/2)+9
              0=-x/2+(6√(x/2))(√(2/2)+9
              0=-x/2+6√(2x/2)/2+18/2
              0=-x+6√2x+18
       (x-18)^2=(6√2x)^2
    x^2-36x+324=72x
x^2-36x+324-72x=0
   x^2-108x+324=0

Use Quadratic formula to solve for x
{{{x= (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{x= (-(-108) +- sqrt( (-108)^2-4*(1)*(324) ))/(2*(1)) }}} 
{{{x= (108 +- sqrt( 11664-1296 ))/(2) }}} 
{{{x= (108 +- sqrt( 10368 ))/(2) }}} 

x = 104.91168824543142175686079407155 or 105
or
x = 3.08831175456857824313920592845 or 3

Substitute x for 105
√(105)-√((105)/2)= 3
       10 - 7  = 3
            3  = 3

Substitute x for 3
√(3)-√((3)/2)= 3
      2 -  1 = 3
           1 = 3

The number is 105.