11.2. How much greater is one of two positive numbers than the other if their arithmetic mean is $2 \sqrt{3}$ and the geometric mean is $\sqrt{3}$?

(Hint: the geometric mean of two numbers $m$ and $n$ is the number $p=\sqrt{\mathrm{mn}}$).
# Solution.

Let the unknown numbers be denoted by $x$ and $y$. Then, from the problem statement, we get:

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ \frac { x + y } { 2 } = 2 \sqrt { 3 } , } \\
{ \sqrt { x y } = \sqrt { 3 } , }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
x+y=4 \sqrt{3}, \\
x y=3 .
\end{array}\right.\right. \\
& \text { Then }|x-y|=\sqrt{(x-y)^{2}}=\sqrt{(x+y)^{2}-4 x y}=\sqrt{4^{2} \cdot 3-4 \cdot 3}=6 .
\end{aligned}
$$

Other methods of solution are possible.

Answer: 6 .