# Solving an Absolute Value Inequality Clarification

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 $$x \gt \sqrt 3$$ $$x \lt -\sqrt 3$$