To find the perimeter of the square, we first need to determine the side length of the square. The area of a square is given by $A = s^2$, where $s$ is the side length.

Here, the area $A$ is given by the quadratic expression $9x^2 - 6x + 1$. Therefore, we have:

$s^2 = 9x^2 - 6x + 1$

To find $s$, we take the square root of both sides:

$s = \sqrt{9x^2 - 6x + 1}$

Next, we need to express the perimeter of the square. The perimeter $P$ of a square is given by:

$P = 4s$

So, substituting the expression for $s$:

$P = 4 \times \sqrt{9x^2 - 6x + 1}$

Thus, the perimeter of the square is $P = 4 \sqrt{9x^2 - 6x + 1}$.

Would you like further details on how to simplify or evaluate this expression, or any other questions?

Here are 5 related questions:

1. How can we simplify $\sqrt{9x^2 - 6x + 1}$?
2. What is the value of $s$ when $x = 1$?
3. Can the quadratic expression $9x^2 - 6x + 1$ be factored further?
4. What happens to the perimeter if $x$ is doubled?
5. How does the area change as $x$ increases?

Tip: When dealing with square roots of quadratic expressions, it's helpful to check if the quadratic can be factored into a perfect square, which simplifies calculations.