jmc

We are given that $x+y=25$ and $xy=126$ for some numbers $x$ and $y$. We note that $$\begin{array}{rl}(x-y{)}^2& =x^2-2xy+y^2\\ & =x^2+2xy+y^2-4xy\\ & =(x+y{)}^2-4xy\\ & =(25{)}^2-4\cdot 126\\ & =121.\end{array}$$

Thus, we have $(x-y{)}^2=121$. Taking the square root of both sides, we have $\sqrt{(x-y{)}^2}=|x-y|=\boxed{11}$.