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

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  • $\begingroup$ 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. $\endgroup$ Oct 25 at 7:53
  • $\begingroup$ 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. $\endgroup$ Oct 25 at 8:04
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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 $m=0$.

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  • $\begingroup$ Thanks for your reply. I have already edited the question and fixed my mistake. Can you look at it now? $\endgroup$ Oct 25 at 8:03
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    $\begingroup$ @BigProgrammer_67 I edited my answer with aditional info. $\endgroup$
    – 5xum
    Oct 25 at 8:11
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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 $x$.

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