I have the following inequality that I need to solve:

$|\frac{3}{x^2}| < 1$

My approach:

$-1 < \frac{3}{x^2} < 1$

I separated it into two inequalities: $\frac{3}{x^2} > -1$ and $\frac{3}{x^2} < 1$

  • Solving for the $\frac{3}{x^2} < 1$, for which I got $a < \sqrt{3}$.

  • Solving for the $\frac{3}{x^2} > -1$ is where the trouble starts for me. I multiplied both sides by $x^2$ and got $3 > -x^2$. I then proceeded to take the square root of both sides $\sqrt{3} > -x$. Now dividing by $-1$ I changed the symbol to its opposite, as per the rule: $\sqrt{3} < x$. I know that this is not the correct answer, so where am I going wrong?


$\frac 3 {x^2}\,$ is always positive so you can rewrite it as follows: $$\frac 3 {x^2} \lt 1$$ that is equivalent to $$\frac {x^2} 3 \gt 1$$ $$x^2 \gt 3$$ that is satisfied by $$x \gt \sqrt 3 $$ $$x \lt -\sqrt 3 $$


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