1963 IMO Problems/Problem 1

Problem

Find all real roots of the equation

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

where $p$ is a real parameter.

Solution

Assuming $x \geq 0$, square the equation, obtaining $4\sqrt {(x^2 - p)(x^2 - 1)} = p + 4 - 4x^2$. If we have $p + 4 \geq 4x^2$, we can square again, obtaining $x^2 = \frac {(p - 4)^2}{4(4 - 2p)}$, or $x = \pm\frac {p - 4}{2\sqrt {4 - 2p}}$. We must have $4 - 2p > 0 \iff p < 2$, so we have $x = \frac {4 - p}{2\sqrt {4 - 2p}}$. However, this is only a solution when $p + 4 \geq 4x^2 = \frac {(p - 4)^2}{4 - 2p} \iff (p + 4)(4 - 2p)\leq(p - 4)^2 \iff 0\leq p(3p - 4)$, so we have $p\leq 0$ or $p \geq \frac {4}{3}$. But if $p < 0$, then $\sqrt {x^2 - p} > x$, contradiction. So we have $x = \frac {4 - p}{2\sqrt {4 - 2p}}$ for $p = 0, \frac {4}{3}\leq p < 2$.