2.154. What is the value of $\sqrt{25-x^{2}}+\sqrt{15-x^{2}}$, given that the difference $\sqrt{25-x^{2}}-\sqrt{15-x^{2}}=2$ (the value of $x$ does not need to be found)?
Solution.

Domain of definition: $\left\{\begin{array}{l}25-x^{2} \geq 0, \\ 15-x^{2} \geq 0\end{array} \Leftrightarrow-\sqrt{15} \leq x \leq \sqrt{15}\right.$.

Multiplying both sides of the equation by $\sqrt{25-x^{2}}+\sqrt{15-x^{2}}$, we have

$$
\begin{aligned}
& \left(\sqrt{25-x^{2}}-\sqrt{15-x^{2}}\right)\left(\sqrt{25-x^{2}}+\sqrt{15-x^{2}}\right)= \\
& =2\left(\sqrt{25-x^{2}}+\sqrt{15-x^{2}}\right) \Leftrightarrow \\
& \Leftrightarrow 25-x^{2}-15+x^{2}=2\left(\sqrt{25-x^{2}}+\sqrt{15-x^{2}}\right)
\end{aligned}
$$

from which $\sqrt{25-x^{2}}+\sqrt{15-x^{2}}=5$.

Answer: 5.