Find a sufficient condition on $p$, $q$ and $r$ for $p^2\cosh x+q^2 \sinh x=r^2$ to have at least one solution.
I know $\sinh x=\frac{e^x-e^{-x}}{2}$
and $\cosh x=\frac{e^x+e^{-x}}{2}$
Also $\ln(x)$ only has a solution when $x>0$.
Perhaps the discriminant will come into this as well, I'm not sure.
The answer is $r^4 \ge p^4 -q^4$, however, I am struggling to get to it.
Help would be much appreciated. Please don't use (too much) calculus if any in your solution.
On
You are practically there with your question. Substitute for $ \cosh $ and $\sinh$ \begin{eqnarray*} (p^2+q^2)(e^x)^2-2r^2e^x +(p^2-q^2)=0 \end{eqnarray*} This is a quadratic in $e^x$ and we require that the discriminant is non negative. So \begin{eqnarray*} r^4 \geq (p^2+q^2)(p^2-q^2). \end{eqnarray*}
You're on the right track
\begin{eqnarray} p^2\cosh x + q^2\sinh x &=& r^2 \\ \frac{p^2}{2}(e^{x} + e^{-x}) + \frac{q^2}{2}(e^{x} - e^{-x}) &=& 2r^2 \\ (p^2 + q^2)e^x + (p^2 - q^2)e^{-x} &=& 2r^2 \\ (p^2 + q^2)e^{2x} + (p^2 - q^2) &=& 2r^2e^x \end{eqnarray}
where in the last step I multiplied by $e^{x}$. Now call $y = e^x$, the equation above can be written as
$$ (p^2 + q^2)\color{blue}{y^2} - 2r^2 \color{blue}{y} + (p^2-q^2) = 0 $$
Which is a quadratic equation on $\color{blue}{y}$. Can you take it from here?