91Ó°ÊÓ

Solving for a variable means finding the value of that variable in terms of the other variables and constants in the equation. In our exercise, we are working with the formula:
$$ Z = \sqrt{\frac{L}{C}} $$
The goal is to express the variable \( L \) in terms of \( Z \) and \( C \).
To do this, we utilize algebraic techniques to manipulate the equation until \( L \) is by itself on one side of the equation.
Let's break this down into clearer steps in the following sections.

The initial step in solving an equation involving a square root is to isolate the square root on one side. Our given formula is:
$$ Z = \sqrt{\frac{L}{C}} $$
First, we need to remove the square root to deal directly with the variables and constants. We do this by squaring both sides of the equation. This operation is done because the square and the square root are inverse operations; they cancel each other out.
So, if we square both sides of the equation, we get:
$$ Z^2 = \frac{L}{C} $$
Now, the square root is removed, and we have a linear relationship between \( Z^2 \) and \( \frac{L}{C} \).

Now that we've dealt with the square root, the task is to solve for \( L \) by isolating it. We take the equation:
$$ Z^2 = \frac{L}{C} $$
We need to get \( L \) by itself. Currently, \( L \) is being divided by \( C \). To isolate \( L \), we perform the inverse operation of division, which is multiplication. Thus, we multiply both sides of the equation by \( C \):
$$ Z^2 \cdot C = L $$
By doing this, we've isolated \( L \) and expressed it in terms of \( Z \) and \( C \). The final expression for \( L \) is:
$$ L = Z^2 \cdot C $$
These steps show the power of algebraic manipulation. Breaking down complex equations into simpler parts and carefully applying inverse operations allow us to solve for any variable effectively.