Absolute value equation with rational expression

Question

I am to solve the equation: $|\frac{2x}{x^2 - 3} | < 1$ And so: 1. I rewrote it as $|2x| < |(x - \sqrt 3) | |(x + \sqrt 3)|$ And I tried to divide it into a few intervals

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For $ x\in(-\infty;-\sqrt3)$
$-2x < (-x+\sqrt3)(-x-\sqrt3) \Rightarrow x\in(-\infty;-\sqrt3)$

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For $x\in[-\sqrt3;0]$
$-2x < (-x+\sqrt3)(x+\sqrt3) \Rightarrow x\in(-1;0)$

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For $x\in(0;\sqrt3]$
$2x < (-x+\sqrt 3)(x+\sqrt 3) => x\in(0;1)$

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For $x\in(\sqrt3;\infty)$
$2x < (x-\sqrt3)(x+\sqrt3) \Rightarrow x\in(3;\infty)$

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However, it looks as if some answers were missing. For exxample if I set $x=\sqrt2$, it still works, but I lack this answer. Where is my mistake?

Answer 1

$$\frac{4x^2}{(x^2-3)^2}\lt 1$$

$$4x^2\lt x^4-6x^2+9$$

$$x^4-10x^2+9>0$$

Put $y = x^2$

$$y^2-10y+9>0$$

$$(y-9)(y-1)>0$$

$$y>9 = x^2>9$$ $$=>x>3 \text{ and } x<-3$$

and $$y<1$$

$$=>x^2<1=>$$ $$ x<1 \text{ and }x>-1 $$

The solution to the problem is

$$x\epsilon {(-\infty,-3),(-1,1),(3,\infty)}$$

Answer 2

$|y|<a\iff -a<y<a$

$$\implies-1<\dfrac{2x}{x^2-3}<1$$

$$\dfrac{2x}{x^2-3}<1\iff0<1-\dfrac{2x}{x^2-3}=\dfrac{(x-3)(x+1)}{x^2-3}=f(x)$$

If $f(\sqrt3)\to-\infty$

Else we need $(x-3)(x+1)(x-\sqrt 3)(x+\sqrt3)>0$

$\implies$ even number of multiplicands must be $>0$

Please check the ranges $\{-\sqrt3,-1\};\{-1,\sqrt3\};\{\sqrt3,3\};\{3,\infty\}$

Can you check for $$-1<\dfrac{2x}{x^2-3}$$