The sum of a number and its positive square root is 6/25.

Solution:

Let first numbers be x 

Then according to question

$x+\sqrt{x}=\frac{6}{25}$

Let $x=y^{2}$ then

$y^{2}+y=\frac{6}{25}$

$25 y^{2}+25 y=6$

$25 y^{2}+25 y-6=0$

$25 y^{2}+30 y-5 y-6=0$

$5 y(5 y+6)-1(5 y+6)=0$

$(5 y+6)(5 y-1)=0$

$(5 y+6)=0$

$y=\frac{-6}{5}$

Or

$(5 y-1)=0$

$x=\frac{1}{5}$

Since, being a positive number, so y cannot be negative.

Therefore,

$x=y^{2}$

$=\left(\frac{1}{5}\right)^{2}$

$=\left(\frac{1}{25}\right)$

Thus, the required number be  $\left(\frac{1}{25}\right)$