Let's determine all real roots of the equation

$$
\sqrt{x^{2}-p}+2 \sqrt{x^{2}-1}=x
$$

where \( p \) is a real parameter.
By squaring twice, we can write down a consequence of the equation in which no root sign appears:

$$
\begin{gathered}
4 \cdot \sqrt{\left(x^{2}-p\right)\left(x^{2}-1\right)}=p+4-4 x^{2} \\
16 p-16(p+1) x^{2}=p^{2}+8 p+16-8(p+4) x^{2}
\end{gathered}
$$

from which

$$
\begin{gathered}
8(2-p) x^{2}=p^{2}-8 p+16=(p-4)^{2} \\
x^{2}=\frac{(p-4)^{2}}{8(2-p)}
\end{gathered}
$$

Accordingly, as potential roots, only the numbers

$$
x_{1}=\frac{p-4}{\sqrt{8(2-p)}}, \quad x_{2}=\frac{4-p}{\sqrt{8(2-p)}}
$$

come into question if $p \neq 2$. In the case $p=2$, (2) expresses a contradiction, in which case (1) has no root.

$x_{1}$ and $x_{2}$ are real if $p<2$, and $p>1$, so the mentioned interval is (1, $2 / \sqrt{3})$.