How to solve equation $ \frac{1}{2} (\sqrt{x^2-16} + \sqrt{x^2-9}) = 1$? $$ \dfrac{1}{2} (\sqrt{x^2-16} + \sqrt{x^2-9}) = 1$$
How can I solve this equation in the easiest way? 

Multiply by $2$ to obtain
$$\tag1\sqrt{x^2-16}+\sqrt{x^2-9}=2$$
and multiply by the conjugate $\sqrt{x^2-16}-\sqrt{x^2-9}$ to obtain
$$\tag2 -\frac72=\frac12((x^2-16)-(x^2-9))=\sqrt{x^2-16}-\sqrt{x^2-9}.$$
Add $(1)$ and $(2)$ and divide by $2$ to obtain
$$\sqrt {x^2-16}=-\frac34$$
Which has no real solution.
