Solving (w+2)^2 - 63 = 0 A Step-by-Step Guide

In this article, we will delve into the process of solving the quadratic equation $(w+2)^2 - 63 = 0$. This type of equation is a fundamental concept in algebra, and mastering its solution is crucial for various mathematical applications. We will break down the solution into manageable steps, ensuring a clear understanding of the underlying principles. The goal is to find the real number values of w that satisfy the given equation. Quadratic equations, in general, can have two, one, or no real solutions, and our approach will systematically uncover these solutions, if they exist.

To begin, the given equation $(w+2)^2 - 63 = 0$ is a quadratic equation in disguise. While it's not in the standard form of $ax^2 + bx + c = 0$ initially, we can manipulate it to reveal its quadratic nature. The term $(w+2)^2$ suggests a squared expression, and the constant term -63 indicates a shift in the parabola represented by the equation. Our first step is to isolate the squared term, which will allow us to apply the square root property and ultimately solve for w. This involves adding 63 to both sides of the equation, a fundamental algebraic manipulation that preserves the equation's balance. Isolating the squared term simplifies the equation and brings us closer to a solution. By understanding the initial form of the equation and the algebraic steps required to transform it, we lay the foundation for a clear and efficient solution process. The subsequent steps will involve taking the square root, solving the resulting linear equations, and verifying the solutions. Understanding the nature of quadratic equations and the properties of square roots is essential for this process.

Step 1: Isolate the Squared Term

The initial equation we are working with is $(w+2)^2 - 63 = 0$. Our primary objective in this first step is to isolate the squared term, which is $(w+2)^2$. To achieve this, we need to eliminate the constant term, -63, from the left side of the equation. The fundamental principle we employ here is the addition property of equality, which states that adding the same value to both sides of an equation maintains the equality. In our case, we will add 63 to both sides of the equation. This will effectively cancel out the -63 on the left side and move it to the right side, isolating the squared term. Performing this operation, we get:

$(w+2)^2 - 63 + 63 = 0 + 63$

This simplifies to:

$(w+2)^2 = 63$

Now, we have successfully isolated the squared term. This is a crucial step because it allows us to proceed with the next phase of the solution process, which involves taking the square root of both sides. Isolating the squared term makes the equation more amenable to solving by revealing the direct relationship between the variable expression and a constant. This step demonstrates the power of algebraic manipulation in simplifying complex equations. The result, $(w+2)^2 = 63$, sets the stage for applying the square root property, which is a key technique in solving quadratic equations.

Step 2: Apply the Square Root Property

Having isolated the squared term in the previous step, we now have the equation $(w+2)^2 = 63$. To proceed, we will utilize the square root property, a fundamental concept in algebra that states if $x^2 = a$, then $x = ±√a$. This property allows us to undo the squaring operation and find the possible values of the expression inside the parentheses, which in our case is $(w+2)$. Applying the square root property to both sides of the equation $(w+2)^2 = 63$, we get:

$√(w+2)^2 = ±√63$

The square root of $(w+2)^2$ is simply $(w+2)$, but it's crucial to remember that when taking the square root of a constant, we must consider both the positive and negative roots. This is because both a positive and a negative number, when squared, will result in a positive number. Therefore, we have:

$w+2 = ±√63$

Now, we need to simplify the square root of 63. We look for perfect square factors of 63. The largest perfect square that divides 63 is 9, since $63 = 9 × 7$. Thus, we can rewrite √63 as $√(9 × 7)$. Using the property of square roots that states $√(ab) = √a × √b$, we have:

$√63 = √9 × √7 = 3√7$

Substituting this back into our equation, we get:

$w+2 = ±3√7$

This step is critical as it transforms the equation into two simpler linear equations, one for the positive root and one for the negative root. These linear equations can then be easily solved for w. The proper application of the square root property, including the consideration of both positive and negative roots, ensures that we capture all possible solutions to the original quadratic equation.

Step 3: Solve for w

From the previous step, we have the equation $w+2 = ±3√7$. This equation represents two distinct linear equations, one for the positive square root and one for the negative square root. To find the values of w, we need to solve each of these equations separately. The operation we will use here is subtraction, as we need to isolate w on one side of the equation. We will subtract 2 from both sides of each equation. Let's consider the two cases:

Case 1: Positive Square Root

$w+2 = 3√7$

Subtracting 2 from both sides, we get:

$w+2 - 2 = 3√7 - 2$

$w = 3√7 - 2$

Case 2: Negative Square Root

$w+2 = -3√7$

Subtracting 2 from both sides, we get:

$w+2 - 2 = -3√7 - 2$

$w = -3√7 - 2$

Thus, we have found two possible solutions for w: $w = 3√7 - 2$ and $w = -3√7 - 2$. These are the exact solutions to the equation. It's important to express the solutions in their simplified radical form, unless a decimal approximation is specifically requested. These solutions represent the values of w that, when substituted back into the original equation, will make the equation true. This step highlights the importance of considering both positive and negative roots when solving quadratic equations. The two solutions demonstrate the symmetry inherent in quadratic equations due to the squared term.

Final Answer

Having completed the steps of isolating the squared term, applying the square root property, and solving for w, we have arrived at the solutions to the equation $(w+2)^2 - 63 = 0$. The two solutions we found are:

$w = 3√7 - 2$

and

$w = -3√7 - 2$

These are the exact solutions expressed in their simplified radical form. We can write the final answer as a set of two distinct values, separated by a comma:

$w = 3√7 - 2, -3√7 - 2$

This final answer represents the complete solution set to the given quadratic equation. Each value of w, when substituted back into the original equation, will satisfy the equation. It's important to present the solutions clearly and accurately, adhering to the conventions of mathematical notation. This process demonstrates a systematic approach to solving quadratic equations, involving algebraic manipulation, the square root property, and careful consideration of both positive and negative roots. By following these steps, we can confidently solve a wide range of quadratic equations and gain a deeper understanding of their properties. In summary, the solutions $w = 3√7 - 2$ and $w = -3√7 - 2$ are the real number values that make the equation $(w+2)^2 - 63 = 0$ true.