91Ó°ÊÓ

Quadratic equations are mathematical expressions of the second degree, typically in the form of \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a \eq 0\). They are called 'quadratic' because \(quadra\) refers to a square, and the highest degree of the variable is 2, which corresponds to a square when graphed. The quadratic equation has great significance in various fields like physics, engineering, and finance, where it's used to model many real-life situations.

The general quadratic equation \(ax^2 + bx + c = 0\) can be solved by a variety of methods, including factoring, completing the square, using the quadratic formula, or graphing. Each method has its advantages in different scenarios. Some quadratic equations, like the one in the exercise, can be arranged in a perfect square form \( (x - h)^2 = k \), where the solution can be obtained using the Square Root Property. This property states that if \( (x - h)^2 = k \), then \(x - h = \pm\sqrt{k}\) or \(x = h \pm \sqrt{k}\).

When it comes to solving quadratic equations, there are various methods available to students. The approach used largely depends on the structure of the equation. For example, some quadratics can be easily factored, such as \(x^2 - 5x + 6 = 0\), which factors to \( (x-2)(x-3) = 0 \) and is solved when \(x = 2\) or \(x = 3\).

However, in scenarios like in our example, where the equation is already in a perfect square, we'd apply the Square Root Property. It's efficient and straightforward. Once the Square Root Property is applied, and we obtain \(x - 3 = \pm \sqrt{24}\), the next step is to solve for \(x\) by isolating the variable, which simply means undoing whatever has been done to \(x\). In our specific case, we add 3 to both sides, keeping the \pm\, which accounts for both the positive and negative roots, yielding two potential solutions. This is one of the characteristics of quadratic equations; they can have up to two real solutions.

Simplifying square roots is a crucial step in solving quadratic equations, especially when applying the Square Root Property. The goal is to express the square root of a number in its simplest radical form, which often involves identifying and pulling out perfect squares that are factors of the number under the root.

To simplify \(\sqrt{24}\), we factor 24 into its prime factors \(24 = 2^3 \cdot 3\). We then group the factors into pairs of similar numbers, seeing \(2^3\) as a pair of 2s and a single 2, we can take a pair of 2s out of the square root, leaving us with \(2\sqrt{6}\). It's important to remember that we're looking for pairs because the square root of a number is a value that, when multiplied by itself, gives the original number, and a pair of 2s (\(2 \cdot 2\)) gives us the perfect square of 4.
In the context of the exercise, simplifying \(\sqrt{24}\) to \(2\sqrt{6}\) allows us to write more concise and precise solutions: \(x = 3 + 2\sqrt{6}\) and \(x = 3 - 2\sqrt{6}\). Remember, simplifying square roots not only helps in achieving an exact form of the solutions but also in recognizing their approximate numerical values if necessary.