What is the value of $\sqrt{49-x^2}+\sqrt{25-x^2}, x \in \mathbb{R}$ if $\sqrt{49-x^2}-\sqrt{25-x^2}=3$ Suppose that real number $x$ satisfies $$\sqrt{49-x^2}-\sqrt{25-x^2}=3$$What is the value of $\sqrt{49-x^2}+\sqrt{25-x^2}$?

This is what I did:
I try to multiply by the conjugate. Its value I believe is technically the solution. $(\sqrt {49-x^2} + \sqrt {25-x^2})(\sqrt {49-x^2} - \sqrt {25-x^2})=24$. Given that $(\sqrt {49-x^2} - \sqrt {25-x^2}) = 3$, $(\sqrt {49-x^2} + \sqrt {25-x^2})(\sqrt {49-x^2} - \sqrt {25-x^2})= 3(\sqrt {49-x^2} + \sqrt {25-x^2}) =24\implies \sqrt {49-x^2} + \sqrt {25-x^2}=8$ 

My question is that will this method work for similar problems, and is there a faster method?
Thanks!

Another way:
$$1) \ \sqrt{49-x^2}=\sqrt{25-x^2}+3 \Rightarrow \\ 
49-x^2=25-x^2+9+6\sqrt{25-x^2} \Rightarrow \\
2.5=\sqrt{25-x^2};\\
2) \ \sqrt{25-x^2}=\sqrt{49-x^2}-3 \Rightarrow \\
25-x^2=49-x^2+9-6\sqrt{49-x^2} \Rightarrow \\
\sqrt{49-x^2}=5.5.$$
Hence:
$$\sqrt{49-x^2}+\sqrt{25-x^2}=5.5+2.5=8.$$
