1. (CZS) Determine all real solutions of the equation $\sqrt{x^{2}-p}+2 \sqrt{x^{2}-1}=$ $x$, where $p$ is a real number.
1. Obviously, $x \geq 0$; hence squaring the given equation yields an equivalent equation $5 x^{2}-p-4+4 \sqrt{\left(x^{2}-1\right)\left(x^{2}-p\right)}=x^{2}$, i.e., $4 \sqrt{\left(x^{2}-1\right)\left(x^{2}-p\right)}=$ $(p+4)-4 x^{2}$. If $4 x^{2} \leq(p+4)$, we may square the equation once again to get $-16(p+1) x^{2}+16 p=-8(p+4) x^{2}+(p+4)^{2}$, which is equivalent to $x^{2}=(4-p)^{2} /(4(4-2 p)]$, i.e., $x=(4-p) /(2 \sqrt{4-2 p})$. For this to be a solution we must have $p \leq 2$ and $(4-p)^{2} /(4-2 p)=4 x^{2} \leq(p+4)$. Hence $4 / 3<p<2$. Otherwise there is no solution.