The equation provided is:

$\frac{1}{x^2} - 4 \cdot \frac{1}{x^4} = 0$

To solve this equation, let's follow these steps:

1. Rewrite $\frac{1}{x^2}$ as a variable to simplify. Let $y = \frac{1}{x^2}$, then $\frac{1}{x^4} = y^2$.

   Substituting these, the equation becomes: $y - 4y^2 = 0$

2. Factor out $y$: $y(1 - 4y) = 0$

3. Set each factor equal to zero:

   • $y = 0$
   • $1 - 4y = 0$

   Solving these:

   • For $y = 0$: $y = \frac{1}{x^2} = 0$, which has no solution since $x$ cannot be infinite.
   • For $1 - 4y = 0$: $y = \frac{1}{4}$

4. Substitute back $y = \frac{1}{x^2}$ and set it equal to $\frac{1}{4}$: $\frac{1}{x^2} = \frac{1}{4}$

5. Solve for $x$: $x^2 = 4$ $x = \pm 2$

Solution:

The solutions to the equation are: $x = 2 \quad \text{and} \quad x = -2$

Would you like more details on this solution, or have any questions?

Here are five related questions that may deepen understanding:

1. What happens if there is an additional constant term in the equation?
2. How would the approach change if the equation had $\frac{1}{x^3}$ instead of $\frac{1}{x^4}$?
3. Can similar techniques solve equations involving higher powers of $x$ in the denominator?
4. How can substitution methods be generalized for more complex equations?
5. What role does factoring play in solving equations like this?

Tip: In equations with fractions, substitution can simplify terms and make factoring easier.