7. [6] For any positive real numbers $a$ and $b$, define $a \circ b=a+b+2 \sqrt{a b}$. Find all positive real numbers $x$ such that $x^{2} \circ 9 x=121$.
Answer: $\square$
Since $a \circ b=(\sqrt{a}+\sqrt{b})^{2}$, we have $x^{2} \circ 9 x=(x+3 \sqrt{x})^{2}$. Moreover, since $x$ is positive, we have $x+3 \sqrt{x}=11$, and the only possible solution is that $\sqrt{x}=\frac{-3+\sqrt{53}}{2}$, so $x=\frac{31-3 \sqrt{53}}{2}$.