Solving For W In The Equation W - 1 = √(9w - 27)

This comprehensive guide delves into the process of solving for the real number w in the equation w - 1 = √(9w - 27). We will explore the necessary steps, potential pitfalls, and methods for verifying the solutions to ensure accuracy. This exploration includes a detailed breakdown of how to address equations involving square roots and isolating variables to determine the valid solutions. Mastering these techniques will significantly enhance your ability to solve algebraic equations involving radicals, which is a fundamental skill in mathematics. It will also give you a deeper understanding of the conditions under which solutions are valid. This guide will provide a robust approach to solving such problems. We will also delve into the reasons why certain solutions might emerge during the solving process that are not actual solutions to the original equation. Understanding these so-called extraneous solutions is crucial in advanced algebra and calculus, which often build upon the fundamentals discussed here. The guide is designed to be accessible to anyone with a basic understanding of algebra, but it also offers insights that will benefit more advanced learners. The goal is to equip you with the tools and understanding necessary to approach similar problems with confidence and precision. The structure of the guide will follow a step-by-step approach, beginning with simplifying the equation, then solving the resulting equation, and finally verifying the solutions to ensure their validity. Through this process, we will emphasize not just the mechanics of solving the equation but also the underlying mathematical principles that make the solution process work.

Step-by-Step Solution

1. Isolate the Square Root

In this particular equation, the square root term is already isolated on one side: w - 1 = √(9w - 27). This initial isolation is a critical first step in solving equations involving radicals. Isolating the radical allows us to eliminate it by raising both sides of the equation to the appropriate power, which in this case is squaring both sides. If the square root term were not isolated, our first step would be to perform algebraic manipulations to achieve this isolation. This might involve adding or subtracting terms from both sides of the equation or multiplying or dividing both sides by a constant. The goal is to ensure that the square root term is the only term on one side of the equation. This isolation step is crucial because it sets the stage for the next step, which is to eliminate the radical. By squaring both sides after isolating the square root, we can transform the equation into a more manageable form, typically a polynomial equation, which we can then solve using standard algebraic techniques. This process simplifies the equation and allows us to find potential solutions more easily. Without isolating the square root, squaring both sides could lead to a more complex expression that is more difficult to handle, potentially complicating the solution process.

2. Square Both Sides

To eliminate the square root, square both sides of the equation: (w - 1)² = (√(9w - 27))². Squaring both sides of an equation is a fundamental technique for solving equations involving square roots. The principle behind this operation is that if two quantities are equal, then their squares are also equal. However, it is important to note that squaring both sides can sometimes introduce extraneous solutions. These are solutions that satisfy the squared equation but not the original equation. Therefore, it is crucial to check any potential solutions in the original equation to verify their validity. When we square (w - 1), we are essentially multiplying the binomial by itself: (w - 1)(w - 1). The result of this multiplication, after applying the distributive property (or the FOIL method), is w² - 2w + 1. On the other side of the equation, squaring the square root eliminates the radical, leaving us with 9w - 27. This step transforms the original equation, which involved a square root, into a quadratic equation, which is a polynomial equation of degree two. Quadratic equations are generally easier to solve than equations involving square roots, and there are several well-established methods for solving them, such as factoring, completing the square, and using the quadratic formula. The resulting equation w² - 2w + 1 = 9w - 27 is a standard quadratic equation that we can now manipulate and solve.

3. Simplify and Rearrange

Expand and simplify the equation: w² - 2w + 1 = 9w - 27. To solve for w, we need to rearrange the equation into a standard quadratic form, which is aw² + bw + c = 0. This involves moving all terms to one side of the equation, leaving zero on the other side. In this case, we subtract 9w from both sides and add 27 to both sides to achieve this. Subtracting 9w from both sides gives us w² - 2w - 9w + 1 = -27. Adding 27 to both sides then results in w² - 2w - 9w + 1 + 27 = 0. Now we combine like terms to simplify the equation. The -2w and -9w terms combine to give -11w, and the 1 and 27 terms combine to give 28. Therefore, the simplified quadratic equation is w² - 11w + 28 = 0. This is a standard quadratic equation that we can solve using various methods, such as factoring, completing the square, or applying the quadratic formula. Rearranging the equation into this standard form is a crucial step in solving quadratic equations because it allows us to identify the coefficients a, b, and c, which are necessary for applying the quadratic formula or factoring techniques. Without this rearrangement, it would be more difficult to directly apply these solution methods.

4. Solve the Quadratic Equation

Rearrange the terms to get a quadratic equation: w² - 11w + 28 = 0. Now, we factor the quadratic equation: (w - 4)(w - 7) = 0. Factoring is a method of expressing a quadratic equation as the product of two binomials. The goal is to find two numbers that multiply to the constant term (28 in this case) and add up to the coefficient of the linear term (-11 in this case). The numbers -4 and -7 satisfy these conditions because (-4) * (-7) = 28 and (-4) + (-7) = -11. Therefore, we can factor the quadratic equation w² - 11w + 28 = 0 as (w - 4)(w - 7) = 0. Once the quadratic equation is factored, we can use 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 means that either w - 4 = 0 or w - 7 = 0. Solving these two linear equations gives us the potential solutions for w. If w - 4 = 0, then adding 4 to both sides gives us w = 4. If w - 7 = 0, then adding 7 to both sides gives us w = 7. So, the potential solutions are w = 4 and w = 7. However, it is crucial to remember that we squared both sides of the original equation, which means we might have introduced extraneous solutions. Therefore, we need to check these potential solutions in the original equation to ensure that they are valid solutions.

5. Check for Extraneous Solutions

Set each factor equal to zero: w - 4 = 0 or w - 7 = 0. Solving these gives potential solutions w = 4 and w = 7. However, because we squared both sides of the original equation, we must check these solutions in the original equation to ensure they are not extraneous. Extraneous solutions are solutions that satisfy the transformed equation (the squared equation) but do not satisfy the original equation. These solutions arise because squaring both sides can introduce additional roots that were not present in the original equation. To check the solutions, we substitute each potential solution back into the original equation and verify whether the equation holds true. For w = 4, we substitute it into the original equation w - 1 = √(9w - 27). This gives us 4 - 1 = √(9(4) - 27), which simplifies to 3 = √(36 - 27), further simplifying to 3 = √9, and finally 3 = 3. Since this is a true statement, w = 4 is a valid solution. For w = 7, we substitute it into the original equation, which gives us 7 - 1 = √(9(7) - 27). This simplifies to 6 = √(63 - 27), further simplifying to 6 = √36, and finally 6 = 6. Since this is also a true statement, w = 7 is also a valid solution. Therefore, both w = 4 and w = 7 are solutions to the original equation, and neither is an extraneous solution.

6. Final Solution

Check w = 4: 4 - 1 = √(9(4) - 27) → 3 = √9 → 3 = 3 (Valid). Check w = 7: 7 - 1 = √(9(7) - 27) → 6 = √36 → 6 = 6 (Valid).

Therefore, the solutions are w = 4 and w = 7.

Conclusion

The solutions to the equation w - 1 = √(9w - 27) are w = 4 and w = 7. This solution was found by isolating the square root, squaring both sides, simplifying the resulting quadratic equation, factoring, and checking for extraneous solutions. This process demonstrates a standard approach to solving equations involving square roots. The importance of checking for extraneous solutions cannot be overstated, as this step ensures the validity of the solutions in the context of the original equation. Extraneous solutions can arise whenever both sides of an equation are raised to an even power, such as squaring. Understanding and applying this method will allow you to solve a wide range of algebraic problems involving radicals, bolstering your mathematical skills and confidence in equation solving. The ability to manipulate and solve equations effectively is a crucial skill in mathematics and its applications, providing a foundation for more advanced topics and real-world problem-solving. The steps outlined in this guide provide a clear and systematic approach to tackling such equations, emphasizing both the mechanics of the solution process and the underlying mathematical principles.