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When it comes to solving quadratic equations, the square root property is a reliable method you can turn to. It's straightforward to use when the equation is arranged in the form of \( x^2 = k \), where \( k \) is a constant. The square root property states that if \( x^2 = k \), then \( x = \pm\sqrt{k} \).

The \( \pm \) symbol tells us that there are typically two solutions to a quadratic equation: a positive and a negative root. Why is this? Well, when you square a negative number, just like a positive number, the result is positive due to the nature of squaring. It's a 'mirror effect' because both \( (\sqrt{k})^2 \) and \( (-\sqrt{k})^2 \) return the same \( k \). To ensure we're thorough, always consider both the positive and negative square roots when using the square root property.

In our exercise, we applied this property to solve for \( y \) in the equation \( y^2 = 12.25 \), which yielded two solutions, \( y = 3.5 \) and \( y = -3.5 \). It's like unwrapping a present: the square root undoes the square, revealing the number inside.

Simplifying radicals might sound complex, but it's an essential step in cleaning up solutions in algebra. A radical is simply an expression that includes a square root, cube root, or any higher root. To simplify a radical, our goal is to make the number inside the root as small as possible while keeping the number outside the root as a whole number.

This can often be done by finding perfect squares (like 4, 9, 16) within the radicand (the number under the root) and 'pulling them out' of the root as their square root. For example, \( \sqrt{12} \) can be written as \( \sqrt{4 \times 3} \), which simplifies to \( 2\sqrt{3} \) because \( \sqrt{4} = 2 \).

Remember, though, the objective is simplification, so if the radicand is already a perfect square—like the 12.25 in the exercise which is \( 3.5^2 \)—you can directly take the square root without needing further simplification. The square root of 12.25 is easily understood to be 3.5. Always aim to leave no factor inside the radical that could be eliminated by simplification.

The process of rationalizing denominators is concerned with expressions that contain a radical in the denominator. The goal here is sleekness and clarity—eliminating the radical from the bottom of a fraction makes it easier to handle, especially when combining with other fractions or doing further operations. Basically, we're giving the denominator a makeover.

To rationalize a denominator, multiply the fraction by a form of 1 that will clear the radical. This form of 1 is typically the radical itself or the radical with a necessary adjustment to make the radicand a perfect square. For instance, if we have \( \frac{1}{\sqrt{3}} \), we can rationalize it by multiplying by \( \frac{\sqrt{3}}{\sqrt{3}} \) to get \( \frac{\sqrt{3}}{3} \).

Our textbook exercise doesn't require this step since our denominator, after simplification, isn't irrational. However, in many cases where simplification leaves a square root in the denominator, rationalization becomes a crucial technique for achieving a proper final answer.

Isolating the variable is a foundational technique across all algebra problems. It's about keeping the variable we're solving for on one side and moving everything else to the opposite side of the equation. Think of it as organizing a room: you want all the toys in the toy box (the variable by itself) and everything else neatly stored away elsewhere (constants and other terms on the other side).

Starting with the given quadratic equation in the form \( ax^2 = c \), the first step is to divide by the coefficient of \( x^2 \) to isolate \( x^2 \) on one side, which usually prepares us to apply the square root property. In our exercise, we detached \( y^2 \) by dividing the entire equation \( 4y^2 = 49 \) by 4, leaving \( y^2 \) alone on one side.

Disentangling variables from other terms in an equation is not just a step, it's part of an ongoing strategy in problem-solving. It's a clean and effective way of guiding us closer to finding the solution we're searching for.