Solving Radical Equations Step-by-Step Guide For √6y = √36 + 2y
In the realm of mathematics, solving equations involving radicals can be a challenging yet rewarding endeavor. This article aims to provide a comprehensive, step-by-step guide to solving the radical equation √6y = √36 + 2y. We will meticulously break down each step, offering clear justifications and insights along the way. Understanding the underlying principles behind each operation is crucial for mastering this type of problem. By delving into the process, we not only find the solution but also enhance our problem-solving skills and deepen our understanding of mathematical concepts. This detailed exploration will serve as a valuable resource for students, educators, and anyone seeking to expand their knowledge of algebra.
1. The Initial Equation: Setting the Stage
We begin with the given radical equation:
√6y = √36 + 2y
This equation presents us with square roots on both sides, a common scenario in algebra. To effectively solve this, we must systematically eliminate these radicals. The core strategy involves using inverse operations to isolate the variable, 'y' in this case. Before we dive into the steps, it’s important to understand the properties of equality and how they apply to radical equations. The squaring property of equality, which we'll soon employ, is a cornerstone in this process. It states that if two quantities are equal, then their squares are also equal. This property allows us to remove the square roots, transforming the equation into a more manageable form. As we progress, we’ll see how each step is carefully chosen to bring us closer to the solution, while adhering to fundamental algebraic principles. This initial assessment is crucial in setting the stage for the subsequent steps, where we will strategically apply mathematical operations to unravel the equation and find the value of 'y'.
2. Squaring Both Sides: Eliminating the Radicals
The first crucial step in solving the equation √6y = √36 + 2y involves eliminating the square roots. To achieve this, we employ the squaring property of equality. This property states that if two expressions are equal, then their squares are also equal. By squaring both sides of the equation, we effectively undo the square root operation. Let's delve into the mechanics of this step.
Squaring both sides of √6y = √36 + 2y gives us:
(√6y)² = (√36 + 2y)²
This transformation is a key move because it simplifies the equation by removing the radicals. On the left-hand side, (√6y)² simplifies directly to 6y. This is because squaring a square root cancels out the radical, leaving us with the radicand, which is 6y in this case. On the right-hand side, we have (√36 + 2y)². Squaring this expression also removes the square root, resulting in 36 + 2y. Thus, after squaring both sides, our equation transforms into a much simpler form:
6y = 36 + 2y
This equation is now linear, making it easier to solve for 'y'. The significance of this step cannot be overstated. By applying the squaring property of equality, we've converted a complex radical equation into a straightforward linear equation. This step showcases the power of using inverse operations to simplify mathematical problems. In the next step, we will focus on isolating the variable 'y' to find its value.
3. Isolating the Variable: Bringing 'y' to One Side
Having squared both sides of the original equation, we arrived at the simplified form: 6y = 36 + 2y. The next step in our problem-solving journey is to isolate the variable 'y'. This involves rearranging the equation so that all terms containing 'y' are on one side, and the constant terms are on the other. To achieve this, we need to perform algebraic manipulations that maintain the equality of the equation.
The first action we take is to subtract 2y from both sides of the equation. This move is strategically chosen because it eliminates the '2y' term from the right side, effectively moving all 'y' terms to the left side. Subtracting 2y from both sides gives us:
6y - 2y = 36 + 2y - 2y
Simplifying both sides, we get:
4y = 36
Now, we have a much simpler equation where the 'y' term is isolated on the left side. The equation 4y = 36 is a significant milestone in our solution process. It represents a clear relationship between 'y' and a constant, making it straightforward to determine the value of 'y'. This step demonstrates the importance of using inverse operations to isolate variables, a fundamental technique in algebra. In the subsequent step, we will complete the isolation of 'y' by dividing both sides of the equation by the coefficient of 'y'.
4. Solving for 'y': The Final Calculation
We've reached a pivotal point in solving our radical equation. After squaring both sides and isolating the variable, we have the equation 4y = 36. The final step is to solve for 'y' by isolating it completely. This involves dividing both sides of the equation by the coefficient of 'y', which is 4 in this case. Dividing both sides by 4 is a direct application of the division property of equality. This property ensures that if we perform the same operation on both sides of an equation, the equality is maintained.
Dividing both sides of 4y = 36 by 4, we get:
(4y) / 4 = 36 / 4
Simplifying both sides, we find:
y = 9
This is our potential solution for 'y'. The value y = 9 represents the culmination of our algebraic manipulations. However, in the context of radical equations, it's crucial to verify this solution. Radical equations can sometimes produce extraneous solutions, which are values that satisfy the transformed equation but not the original equation. Therefore, the next essential step is to check whether y = 9 indeed satisfies the initial equation. This verification process ensures the accuracy and validity of our solution. In the final section, we will perform this check, providing a conclusive answer to our problem.
5. Verification: Ensuring the Solution's Validity
Having found a potential solution, y = 9, it is imperative to verify its validity in the original equation. This step is particularly crucial when dealing with radical equations, as the process of squaring both sides can sometimes introduce extraneous solutions. Extraneous solutions are values that satisfy the transformed equation but not the original equation. To verify our solution, we substitute y = 9 back into the initial equation:
√6y = √36 + 2y
Substituting y = 9, we get:
√6(9) = √36 + 2(9)
Now, we simplify both sides of the equation:
√54 = √36 + 18
√54 = √54
The equation holds true. This verification confirms that y = 9 is indeed a valid solution for the original radical equation. The fact that both sides of the equation are equal when y = 9 provides us with confidence in our solution. This step underscores the importance of checking solutions in radical equations to avoid accepting extraneous values. In summary, by substituting our potential solution back into the original equation and verifying its accuracy, we complete the problem-solving process with assurance.
Conclusion: The Solution and the Process
In conclusion, we have successfully solved the radical equation √6y = √36 + 2y. Through a series of meticulous steps, we arrived at the solution y = 9. Our journey involved squaring both sides to eliminate the radicals, isolating the variable 'y' through algebraic manipulations, and finally, verifying our solution to ensure its validity. This process not only provides the answer but also reinforces fundamental algebraic principles. The importance of verification cannot be overstated, especially in radical equations where extraneous solutions may arise. By checking our solution, we confirm its accuracy and deepen our understanding of the equation's behavior. This step-by-step approach showcases the systematic nature of mathematical problem-solving. Each action is carefully chosen and justified, leading us closer to the final solution. Mastering such techniques is invaluable for anyone seeking to excel in mathematics. The ability to solve radical equations is a testament to one's algebraic proficiency, and the process outlined in this article serves as a robust guide for tackling similar challenges. Therefore, understanding both the solution and the method by which it was obtained is key to mathematical competence.
