This is a contest problem. It asks for all the real solutions of
\begin{equation*} \sqrt{x^2-p}+2\sqrt{x^2-1}=x, \end{equation*}
for arbitrary real $p$.
From the eq. it's clear that $p$ cannot be negative. By squaring once, splitting the radicals from the other terms and squaring again, I found a formula for $x^2$:
\begin{equation*} x^2 = 1+\frac{p^2}{16(1-p/2)}. \end{equation*}
From here we have $p<2$ for the second radical to be defined, and by comparing this expression with $p$ we get that $p<2$ makes the first radical be defined as well. Yet solutions of the original equation can only be found when $p\leq 4/3$, and $4/3$ is the (double) root of $9p^2-24p+16=0$ (obtained from comparing $1+\frac{p^2}{16(1-p/2)}$ with $p$).
I don't understand why this happens. Perhaps I've done something wrong. If I haven't, I'd appreciate any help in proving that there are no solutions if $p$ is greater than $4/3$.
– 2010-10-12