If $192x^2 -16 = 0$, find the value of $\frac{1}{x^4}$.
I notice that the equation has a common factor of 16, so I can simplify it by dividing both sides by 16. That gives me $12x^2 - 1 = 0$, which is a bit easier to work with. I want to solve for $x^2$, so I add 1 to both sides and get $12x^2 = 1$. Then I divide both sides by 12 and get $x^2 = \frac{1}{12}$. Now I have $x^2$, but I want $\frac{1}{x^4}$. I can rewrite $\frac{1}{x^4}$ as $(\frac{1}{x^2})^2$, so I need to find the reciprocal of $x^2$ and then square it. The reciprocal of $x^2$ is $\frac{1}{x^2}$, which is just $12$. Squaring $12$ gives me $144$, so $\frac{1}{x^4} = 144$. 