Solving $6+\sqrt{w+9}=2$ A Step-by-Step Guide

In mathematics, solving for a variable is a fundamental skill. This article delves into solving an equation involving a square root, specifically, $6 + \sqrt{w + 9} = 2$, where $w$ is a real number. We will break down the steps, explain the underlying concepts, and provide a clear solution. This process not only helps in finding the value of $w$ but also reinforces the understanding of algebraic manipulations and the properties of square roots. Solving equations like these is crucial in various fields, including physics, engineering, and computer science, making it an essential topic for students and professionals alike.

The equation $6 + \sqrt{w + 9} = 2$ presents a unique challenge due to the presence of the square root. To isolate $w$, we need to first isolate the square root term and then eliminate it by squaring both sides of the equation. This process introduces the possibility of extraneous solutions, which are solutions that satisfy the transformed equation but not the original equation. Therefore, it's crucial to check our final solution by substituting it back into the original equation. This step ensures that we have a valid solution that satisfies the initial conditions of the problem. Understanding these nuances is key to mastering algebraic equations and applying them effectively in different contexts.

Square root equations involve variables under a radical sign, which necessitates a specific approach to solve them. The primary strategy is to isolate the square root term and then square both sides of the equation to eliminate the radical. However, this process can introduce extraneous solutions, which are solutions that arise from the squaring operation but do not satisfy the original equation. For instance, consider the equation $\sqrt{x} = -2$. Squaring both sides gives $x = 4$, but substituting $x = 4$ back into the original equation yields $\sqrt{4} = 2$, not -2, making it an extraneous solution. Therefore, it is imperative to check every solution obtained by substituting it back into the original equation. This step ensures that the solution is valid and satisfies the initial conditions.

When dealing with square root equations, it's also crucial to understand the domain of the square root function. The expression inside the square root must be non-negative, meaning that $w + 9$ in our equation must be greater than or equal to zero. This constraint helps us identify potential solutions that may not be valid. For example, if we find a solution for $w$ that makes $w + 9$ negative, we can immediately discard it as an extraneous solution. Moreover, understanding the domain helps in interpreting the solutions within the context of the problem. In real-world applications, the domain might represent physical constraints, such as length or time, which cannot be negative. Therefore, a comprehensive understanding of square root equations includes not only the algebraic manipulations but also the domain considerations and the potential for extraneous solutions.

To solve the equation $6 + \sqrt{w + 9} = 2$, we will follow a step-by-step approach that ensures we arrive at the correct solution while also understanding the underlying principles. Here's a detailed breakdown of each step:

Step 1: Isolate the Square Root Term

The first step is to isolate the square root term. This involves moving all other terms to the other side of the equation. In our case, we subtract 6 from both sides:

$6 + \sqrt{w + 9} - 6 = 2 - 6$

This simplifies to:

$\sqrt{w + 9} = -4$

Step 2: Square Both Sides of the Equation

To eliminate the square root, we square both sides of the equation:

$(\sqrt{w + 9})^2 = (-4)^2$

This gives us:

$w + 9 = 16$

Step 3: Solve for $w$

Now, we solve for $w$ by subtracting 9 from both sides:

$w + 9 - 9 = 16 - 9$

This simplifies to:

$w = 7$

Step 4: Check for Extraneous Solutions

This is a crucial step when dealing with square root equations. We need to check if $w = 7$ satisfies the original equation. Substitute $w = 7$ back into the original equation:

$6 + \sqrt{7 + 9} = 2$

$6 + \sqrt{16} = 2$

$6 + 4 = 2$

$10 = 2$

This is clearly false. Therefore, $w = 7$ is an extraneous solution.

Step 5: Determine the Final Solution

Since the only solution we found is extraneous, there is no real number $w$ that satisfies the original equation.

Final Answer: No solution

Extraneous solutions are a common pitfall when solving equations involving radicals, and understanding why they occur is crucial for mathematical accuracy. Extraneous solutions arise due to the process of squaring both sides of an equation, which can introduce solutions that do not satisfy the original equation. To further understand this concept, let's delve into why extraneous solutions occur and how to identify them effectively.

Why Extraneous Solutions Occur

The primary reason for the emergence of extraneous solutions is that squaring both sides of an equation can turn a false statement into a true one. Consider the example from our solution: $\sqrt{w + 9} = -4$. The square root of any real number cannot be negative, so this statement is inherently false. However, when we square both sides, we get $w + 9 = 16$, which leads to $w = 7$. Substituting $w = 7$ into the original equation gives $6 + \sqrt{16} = 2$, which simplifies to $10 = 2$, a false statement. The squaring operation has masked the initial contradiction, leading to an extraneous solution.

In essence, the squaring operation eliminates the sign difference. If we have an equation $a = b$, squaring both sides gives $a^2 = b^2$. This holds true if $a = b$ or $a = -b$. The squaring process doesn't distinguish between these two cases, which means that solutions to $a^2 = b^2$ may not necessarily be solutions to $a = b$. This is why checking solutions by substituting them back into the original equation is essential.

Identifying Extraneous Solutions

The most reliable method for identifying extraneous solutions is to substitute each potential solution back into the original equation. If the substitution leads to a true statement, the solution is valid. If it leads to a false statement, the solution is extraneous and must be discarded.

In the context of our problem, we found $w = 7$ as a potential solution. Substituting it into the original equation, $6 + \sqrt{w + 9} = 2$, yielded $6 + \sqrt{7 + 9} = 2$, which simplifies to $10 = 2$, a false statement. This clearly indicates that $w = 7$ is an extraneous solution.

Another way to identify potential extraneous solutions is to consider the domain of the square root function. The expression inside the square root must be non-negative. In our equation, $w + 9$ must be greater than or equal to zero. If we find a value for $w$ that makes $w + 9$ negative, we know immediately that it cannot be a valid solution. This domain restriction serves as an additional check that can help in identifying and discarding extraneous solutions before even substituting into the original equation.

In this article, we have thoroughly explored the process of solving the equation $6 + \sqrt{w + 9} = 2$. We have demonstrated the step-by-step method, which involves isolating the square root term, squaring both sides, solving for $w$, and crucially, checking for extraneous solutions. Our analysis revealed that the potential solution $w = 7$ is extraneous, leading us to the conclusion that there is no real solution for the given equation.

The concept of extraneous solutions is a significant aspect of solving radical equations. These solutions arise due to the squaring operation, which can mask contradictions present in the original equation. The only way to ensure the validity of a solution is to substitute it back into the original equation and verify that it holds true. This process is not just a final step but a critical component of solving radical equations, reinforcing the importance of mathematical rigor and precision.

Understanding the nuances of extraneous solutions and the domain restrictions of square root functions equips students and professionals with a more comprehensive grasp of algebraic manipulations. The skills and insights gained from this exercise are invaluable in tackling more complex problems in mathematics, physics, engineering, and various other fields where equations involving radicals are common. Therefore, the ability to confidently solve and verify solutions for radical equations is a cornerstone of mathematical proficiency.