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.

• You cannot add the fractions as written because they do not share a common denominator. You can rewrite them so that they do have a common denominator, though. Oct 25 '21 at 7:53
• I skipped the step of making a common denominator because its trivial, and the quadratic equation of $2x^2-mx-3=0$ is the result of my calculations. Oct 25 '21 at 8:04

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 $$m=0$$.

• Thanks for your reply. I have already edited the question and fixed my mistake. Can you look at it now? Oct 25 '21 at 8:03
– 5xum
Oct 25 '21 at 8:11

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 $$x$$.