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Quadratic equations are polynomial equations of degree two, meaning they include a term with the variable raised to the power of two. These types of equations follow the general form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. The goal when solving quadratic equations is to find the values of \(x\) that make the equation true.

There are several methods to solve quadratic equations, including factoring, using the quadratic formula, completing the square, and in some simple cases, taking square roots directly. In this particular case, you rearrange the given equation, isolate the quadratic term, and use square roots because it’s already in a form conducive to this method.

The specific equation \(9y^2 - 4 = 0\) is straightforward because the middle term (\(bx\)) is absent. Hence, setting \(9y^2\) equal to \(4\) and taking square roots simplifies the process significantly.

Algebraic manipulation involves rearranging equations to isolate variables and make the expressions simpler or more convenient to work with. It’s a fundamental skill in algebra that allows you to solve various mathematical problems effectively.

In the provided example, algebraic manipulation is essential for rearranging the original equation. Here’s how it works step by step:

• Start by moving all terms involving the variable to one side of the equation. In our case, the equation \(9y^2 - 4 = 0\) rearranges to \(9y^2 = 4\).
• This step prepares the equation for isolating the square term and then taking the square root. Make sure you perform equivalent operations on both sides of the equation to maintain equality.

By systematically applying these manipulation techniques, you'll progress through the steps smoothly. Remember, clarity in each transformation step is crucial to avoid errors.

The concept of square roots is frequently used when solving quadratic equations, especially when the quadratic term can be isolated, as in our example. Taking the square root involves finding a number that, when multiplied by itself, results in the original number inside the radical sign.

In our problem, once you've isolated \(y^2\), apply the square root to both sides of the equation: \(y = \pm\sqrt{\frac{4}{9}}\). This results in two possible solutions for \(y\) because squaring either a positive or a negative number yields a positive result. Hence, both \(y = \frac{2}{3}\) and \(y = -\frac{2}{3}\) are valid solutions.

Square roots unlock the solutions to many quadratic equations, providing insight into how positive and negative inputs can fulfill the conditions of the equation.