<s>[INST] $\sqrt{x^2-16} - (x-4) = \sqrt{x^2-5x+4}$ How do I solve this equation I found in my textbook:

$\sqrt{x^2-16} - (x-4) = \sqrt{x^2-5x+4}$ 

This is what I tried:
$\sqrt{x^2-16} - (x-4) = \sqrt{x^2-5x+4}$
$\mapsto \sqrt{(x+4)(x-4)} - (x-4) = \sqrt{(x-1)(x-4)}$

Dividing both sides by $\sqrt{x-4}$

$\mapsto \sqrt{x+4} - \sqrt{x-4} = \sqrt{x-1} $

Squaring both Sides

$\mapsto x - 2 \sqrt{x^2 - 16}= -1$
$\mapsto \frac{x+1}{2} = \sqrt{x^2 - 16}$

Squaring both sides

$ \mapsto \frac{x^2 + 2x + 1}{4} = x^2 - 16$
$\mapsto 3x^2 -2x - 65 = 0$

Solving the quadratic equation

$$x = 5 OR \frac{-13}{3}$$
Checking in the initial equation we can see that $5$ is a valid root. But the second value that is given in the key to the book is $4$. How do I obtain $4$?
 [/INST] I think it should be like this when the equation gets divided by $\sqrt{x-4}$:
$(\sqrt{x-4})(\sqrt{x+4} - \sqrt{x-4}) - \sqrt{(x-4)(x-1)} = 0$
$\mapsto (\sqrt{x-4})(\sqrt{x+4} - \sqrt{x-4} - \sqrt{x-1}) = 0$

$\therefore$ Either $\sqrt{x-4} = 0$ OR $(\sqrt{x+4} - \sqrt{x-4} - \sqrt{x-1}) = 0$

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