<s>[INST] 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? 
 [/INST] We have the equation
$$
\frac{1}{2}(\sqrt{x^2-16} + \sqrt{x^2-9}) = 1
$$
Let's multiply it by $\sqrt{x^2-16} - \sqrt{x^2-9}$ to get
$$
-\frac{7}{2}=\sqrt{x^2-16} - \sqrt{x^2-9}
$$
Hence
$$
2\sqrt{x^2-16} = (\sqrt{x^2-16} + \sqrt{x^2-9}) + (\sqrt{x^2-16} - \sqrt{x^2-9})=2-\frac{7}{2}<0
$$
This is imossible so there is no real solution for this equation 
</s>