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? 
 A: 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.
A: 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 
A: You can solve it algebraically by isolating one of the square roots, squaring both sides, solving for the other square root, and squaring both sides again. This will give you a quadratic equation in $x^2$.
But you can also argue more cleverly directly from the function. First, notice that the LHS is undefined for $|x|<4$. For $|x|\geq 4$, 
$$\frac{\sqrt{x^2-16}+\sqrt{x^2-9}}{2} \geq \frac{\sqrt{x^2-9}}{2}\geq \frac{\sqrt{7}}{2} > 1,$$
so your equation has no (real) solutions.
