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? $$\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=2mx+4$$
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.


I consider only real numbers, so $|x|\geqslant1.$ After adding the fractions I get the following quadratic equation: $$4x^2-mx+1=0$$

Can you show us the steps of how you got to that equation? Because that is not the equation I got. In general, when calculating $$\frac{a+b}{a-b}+\frac{a-b}{a+b},$$
you have $$\begin{align}
\frac{a+b}{a-b}+\frac{a-b}{a+b} &= \frac{(a+b)(a+b)}{(a-b)(a+b)}+\frac{(a-b)(a-b)}{(a-b)(a+b)} \\&= \frac{a^2+2ab+b^2+a^2-2ab+b^2}{a^2-b^2} \\&= \frac{2(a^2+b^2)}{a^2-b^2}\end{align}$$
and this expression, in your case, when $a=x$ and $b=\sqrt{x^2-1}$, simplifies to something rather simple.

Edit:
Yes, the equation is indeed simplified down to $2x^2-2mx-3=0$, your solution looks correct to me. Also, the book's solution is clearly wrong, because for $m=0$, the equation clearly does have two solutions, and they are $x_1=\sqrt{\frac32},x_2=-\sqrt{\frac32}$.
You can see for yourself (or use Wolfram alpha, see here and here) that when $x$ is any of those two values, then the left side of your expression becomes $4$, and thus, it solves your equation if $m=0$.
