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 $4/3$.

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When solving equations, by doing algebra like squaring etc, one could potentially include more solutions than the original equation.

For instance

$x = 1$ implies $x^2 = 1$, by squaring.

The second equation also has $x = -1$ as a solution, which is not a solution of the first one.

In your case, when you square etc, you might lose constraints like $x \ge 1$ and $x^2 \ge p$.

I suppose when you squared, you had the term $x - 2\sqrt{x^2 -1}$.

The constraint that $x \ge 2 \sqrt{x^2-1}$ (i.e. $x^2 \le 4/3$) gives us the constraint that $p \le 4/3$.

As you noticed,

$1 + \frac{p^2}{16(1-p/2)} -p \ge 0$ for all $p$.

Thus

$1 + \frac{p^2}{16(1-p/2)} \le 4/3$ implies that $0 \le 1 + \frac{p^2}{16(1-p/2)} -p \le 4/3 - p$ which implies $4/3 \ge p$.

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thank you. I'm still confused, since 1+... >= p, 0<p<2 is satisfied for all p in [0,2). wolframalpha.com/input/?i=Plot[{1+%2B+p^2/(16*(1+-+(p/2))),+p},+{p,+0‌​,+3}] Anyway I'm probably doing something very wrong. Thank you in advance for any further clarification. – Weltschmerz Oct 12 '10 at 0:35
@Welt: I have updated the answer. I had a mistake in algebra earlier. – Aryabhata Oct 12 '10 at 0:51
very well, thank you. This problem struck me as somewhat odd. – Weltschmerz Oct 12 '10 at 0:57
you stole my rep! And you have more than you know what to do with. I had my answer typed up and was 15 sec from posting when yours hit. Turned out you had covered everything I had to say, and more clearly. Good answer. – Ross Millikan Oct 12 '10 at 13:33
@Ross: Thanks! Sorry about the wasted effort. – Aryabhata Oct 12 '10 at 13:56