To solve the equation \(x^2 + \frac{2x}{r^2} + \frac{1}{r^4} = 0\) for \(x^2 + \frac{2x}{r^2} + \frac{1}{r^4}\), follow these steps:

First, let's multiply the entire equation by \(r^4\) to eliminate the fractions:

\(r^4 \times \left(x^2 + \frac{2x}{r^2} + \frac{1}{r^4}\right) = r^4 \times 0\).

This simplifies to:

\(x^2 r^4 + 2x r^2 + 1 = 0\).

Now, let \(y = xr^2\), then the equation becomes:

\(y^2 + 2y + 1 = 0\).

This is a quadratic equation that can be factored as:

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\((y + 1)^2 = 0\).

So, \(y = -1\), which means:

\(xr^2 = -1\).

Now solve for \(x\):

\(x = -\frac{1}{r^2}\).

Therefore, the solution for \(x^2 + \frac{2x}{r^2} + \frac{1}{r^4} = 0\) is \(x = -\frac{1}{r^2}\).