Solving $w-1=\sqrt{9 W-27}$ A Step-by-Step Guide

Introduction

In this article, we will explore the process of solving for the real number $w$ in the equation $w - 1 = \sqrt{9w - 27}$. This equation involves a square root, which means we need to be particularly careful about checking for extraneous solutions. Extraneous solutions are values that we obtain algebraically but do not satisfy the original equation. Our primary goal is to find the value or values of $w$ that make the equation true. This involves algebraic manipulation, including squaring both sides of the equation and solving the resulting quadratic equation. Finally, we must verify each potential solution to ensure it is not extraneous. The combination of algebraic techniques and careful verification is key to solving such equations accurately. The process requires a systematic approach to avoid errors and ensure we arrive at the correct solution(s). We will start by isolating the square root term, then square both sides to eliminate the radical. From there, we will simplify the equation and solve for $w$. Finally, we will substitute our solutions back into the original equation to check their validity. By following these steps, we can confidently determine the real solutions for $w$. Remember, the step-by-step method not only helps in finding the correct answer but also in understanding the underlying mathematical principles. Understanding the properties of square roots and quadratic equations is essential in this process. Extraneous solutions arise because squaring both sides of an equation can introduce solutions that do not satisfy the original equation. Therefore, the verification step is an integral part of the solution process. Through this detailed explanation, we aim to make the solution clear and understandable, enhancing your problem-solving skills.

Step-by-Step Solution

1. Start with the Given Equation

We begin with the equation:

$w - 1 = \sqrt{9w - 27}$

This equation presents a common challenge in algebra: solving for a variable when it's inside a square root. The key to tackling such equations is to systematically remove the square root. Our first step involves understanding the conditions under which the equation is valid. The expression inside the square root, $9w - 27$, must be non-negative because the square root of a negative number is not a real number. Similarly, the left side of the equation, $w - 1$, must also be non-negative since the square root will always yield a non-negative result. By addressing these conditions early, we ensure that our solutions are valid within the domain of real numbers. Now, let's delve into the methodical steps needed to solve for $w$, starting with isolating the square root. The concept of isolating the square root is crucial because it sets the stage for squaring both sides, which is the next logical step. Without this initial isolation, the squaring process becomes more complex and prone to errors. Therefore, it's a best practice to always make sure the square root term is alone on one side of the equation before proceeding further. This careful approach not only simplifies the algebraic manipulation but also enhances the accuracy of the solution. By understanding these preliminary considerations, we lay a strong foundation for solving the equation efficiently and correctly.

2. Square Both Sides

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

$(w - 1)^2 = (\sqrt{9w - 27})^2$

Squaring both sides of an equation is a fundamental technique in algebra, especially when dealing with square roots. This operation effectively cancels out the square root on the right side, transforming the equation into a more manageable form. However, this step is not without its caveats. Squaring both sides can introduce extraneous solutions, which are solutions that satisfy the transformed equation but not the original one. This is because the squaring operation can mask the sign of the expressions involved. For example, both 3 and -3, when squared, give 9, but they are distinct numbers with different signs. Therefore, after solving the resulting equation, it is imperative to check each solution in the original equation to ensure its validity. This verification step is not merely a formality; it is a critical part of the solution process. Without it, we risk including incorrect answers. The technique of squaring both sides is powerful, but it demands careful attention to detail and a thorough understanding of its implications. By being mindful of the potential for extraneous solutions, we can use this method effectively while maintaining the accuracy of our results. This proactive approach to problem-solving ensures that we arrive at the correct answer while reinforcing our grasp of algebraic principles.

3. Expand and Simplify

Expanding the left side gives us:

$w^2 - 2w + 1 = 9w - 27$

Now, we move all terms to one side to form a quadratic equation:

$w^2 - 2w + 1 - 9w + 27 = 0$

Simplifying, we get:

$w^2 - 11w + 28 = 0$

This step involves several crucial algebraic manipulations. First, expanding $(w - 1)^2$ correctly to $w^2 - 2w + 1$ is fundamental. This expansion uses the binomial square formula, which is a cornerstone of algebraic identities. Any error in this expansion will propagate through the rest of the solution, leading to an incorrect answer. The next key step is rearranging the equation by moving all terms to one side. This is done to set up the equation for solving as a quadratic. A quadratic equation in the standard form $ax^2 + bx + c = 0$ can be solved using a variety of methods, such as factoring, completing the square, or the quadratic formula. By bringing all terms to one side, we make the equation amenable to these solution techniques. The simplification process, which involves combining like terms, is equally important. This ensures that the quadratic equation is in its simplest form, making it easier to solve. A clear and methodical approach to these algebraic manipulations is essential for accuracy and efficiency. Understanding the principles behind each step, from expanding the square to simplifying the equation, not only helps in solving this particular problem but also enhances overall algebraic proficiency.

4. Solve the Quadratic Equation

We can factor the quadratic equation:

$(w - 4)(w - 7) = 0$

This gives us two potential solutions:

$w = 4$ or $w = 7$

Solving a quadratic equation is a pivotal step in many algebraic problems, and factoring is one of the most efficient methods when applicable. The ability to factor a quadratic equation hinges on recognizing the relationship between the coefficients and the roots. In this case, we look for two numbers that multiply to 28 (the constant term) and add up to -11 (the coefficient of the $w$ term). The numbers -4 and -7 satisfy these conditions, allowing us to factor the equation as $(w - 4)(w - 7) = 0$. Once the equation is factored, we apply the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. This leads to the two potential solutions: $w - 4 = 0$ or $w - 7 = 0$, which give us $w = 4$ and $w = 7$. While factoring is a powerful technique, it's not always straightforward. If a quadratic equation is not easily factorable, other methods such as the quadratic formula or completing the square may be necessary. Understanding the strengths and limitations of each method is crucial for effective problem-solving. Moreover, it’s important to remember that the solutions obtained at this stage are potential solutions. They must be checked in the original equation to rule out any extraneous solutions, a step that underscores the importance of thoroughness in algebra.

5. Check for Extraneous Solutions

We must check both solutions in the original equation $w - 1 = \sqrt{9w - 27}$:

For $w = 4$:

$4 - 1 = \sqrt{9(4) - 27}$

$3 = \sqrt{36 - 27}$

$3 = \sqrt{9}$

$3 = 3$ (This is true)

For $w = 7$:

$7 - 1 = \sqrt{9(7) - 27}$

$6 = \sqrt{63 - 27}$

$6 = \sqrt{36}$

$6 = 6$ (This is true)

The crucial step of checking for extraneous solutions highlights a fundamental aspect of solving equations involving radicals. Squaring both sides of an equation, while a necessary technique to eliminate the square root, can introduce solutions that do not satisfy the original equation. These are extraneous solutions. To ensure the validity of our solutions, we must substitute each potential solution back into the original equation and verify that the equation holds true. In this case, we have two potential solutions, $w = 4$ and $w = 7$. For $w = 4$, substituting into the original equation gives us $4 - 1 = \sqrt{9(4) - 27}$, which simplifies to $3 = 3$, a true statement. Similarly, for $w = 7$, the substitution yields $7 - 1 = \sqrt{9(7) - 27}$, which simplifies to $6 = 6$, also a true statement. This verification process confirms that both $w = 4$ and $w = 7$ are valid solutions to the original equation. If either of these substitutions had resulted in a false statement, we would have identified that value as an extraneous solution and discarded it. This methodical checking process is not just a formality; it is an integral part of the solution, ensuring the accuracy and correctness of our answer. By understanding the potential for extraneous solutions and diligently checking our results, we demonstrate a comprehensive approach to problem-solving in algebra.

Final Answer

Both solutions are valid, so the solutions are $w = 4, 7$.

In summary, we have successfully solved the equation $w - 1 = \sqrt{9w - 27}$ for the real number $w$. The process involved several key steps, each requiring careful attention to detail. We started by squaring both sides of the equation to eliminate the square root, which transformed the equation into a quadratic form. We then simplified and solved the quadratic equation by factoring, obtaining two potential solutions: $w = 4$ and $w = 7$. However, we emphasized the critical importance of checking for extraneous solutions, a step often overlooked but crucial for accuracy. By substituting each potential solution back into the original equation, we verified that both $w = 4$ and $w = 7$ indeed satisfy the equation. This comprehensive approach highlights the significance of not only algebraic manipulation but also the validation of solutions. The problem-solving strategy we employed serves as a valuable template for tackling similar equations involving radicals. It underscores the need for a methodical approach, combining algebraic techniques with thorough verification. Understanding the underlying principles and potential pitfalls, such as extraneous solutions, is key to mastering algebraic problem-solving. By following these steps and insights, students can confidently approach and solve a wide range of equations involving radicals and quadratic forms. This article aimed to provide not just the solution, but also a deeper understanding of the process and the mathematical concepts involved.