TITLE: When does $\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=2mx+4$ have $2$ real solutions?
QUESTION [5 upvotes]: $$\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=2mx+4$$
My solution:
I consider only real numbers, so $|x|\geq1$. After adding the fractions I get the following quadratic equation:
$$2x^2-mx-3=0,$$ and it has 2 real solutions if $D>0$, so $m^2+24>0$.
It comes down to every real number, but the answer in the book states that there is no real m value for which the main equation has 2 real solutions, and no detailed explanation has been included. I kindly ask for explanation.

REPLY [3 votes]: Another point of view.
After getting $2x^2-mx-3=0$ (I skipped the derivation because you can do it by yourself and it has also been explained by the other existing answer), we can isolate $m$ as follows.
$$m=\frac{2x^2-3}{x}=2x-\frac{3}{x}$$
where $x\not = 0$.

$m'(x)=2+3/x^2>0$ for all $x\in \mathbb{R}$ suggests that $m(x)$ has no extremum. It spans from $-\infty$ to $\infty$.
$m(-x)=-m(x)$ suggests that $m(x)$ is an odd function.
Every horizontal line $y=k$ will cut $m(x)$ at exactly two points.

Thus for any $m$ there are exactly two real $x$.