Solving √3x-4 = 2√x-5 A Step-by-Step Guide
In this article, we will delve into the step-by-step process of solving the equation √3x-4 = 2√x-5. This equation falls under the category of radical equations, which involve variables within radical expressions (usually square roots). Solving these types of equations requires careful manipulation to eliminate the radicals and isolate the variable. We will explore the necessary steps, including squaring both sides, simplifying the resulting equation, and checking for extraneous solutions. Understanding these concepts is crucial for anyone studying algebra or related fields in mathematics. It will enhance your problem-solving skills and give you a solid foundation for tackling more complex equations.
Understanding Radical Equations
To start, let's define what radical equations are and why they require a unique approach. Radical equations are equations that contain a variable within a radical, typically a square root. The challenge in solving these equations lies in removing the radical to isolate the variable. This is often achieved by squaring both sides of the equation. However, this process can sometimes introduce extraneous solutions, which are solutions that satisfy the transformed equation but not the original one. Therefore, it is vital to check all potential solutions in the original equation.
Radical equations appear frequently in various mathematical contexts, from basic algebra to more advanced calculus. Mastering the technique to solve them is not just academically valuable but also practically useful. In real-world applications, radical equations can model various phenomena, such as calculating the speed of an object based on its kinetic energy or determining the dimensions of a shape given its area.
Understanding the theory behind radical equations can greatly improve your approach to solving them. Knowing the purpose of each step, such as squaring both sides or checking for extraneous solutions, makes the process more intuitive and less of a mechanical exercise. This deeper understanding allows you to adapt the techniques to various types of radical equations, even those that might initially seem complicated.
Step-by-Step Solution
Now, let's proceed with solving the equation √3x-4 = 2√x-5 step by step. This methodical approach will help you understand the process clearly and ensure accuracy.
1. The Original Equation
Our starting point is the given equation:
√3x-4 = 2√x-5
This equation contains square roots on both sides, which we need to eliminate to solve for x. The initial step in solving this equation involves isolating the radical terms. In our case, the radical terms are already isolated, which simplifies our task. It's important to make sure the radicals are correctly identified because this will dictate our next steps. If the radicals weren't isolated, we would need to rearrange the terms to achieve this.
The act of isolating radicals is crucial because it sets the stage for squaring both sides of the equation, which is the primary method for removing the square roots. When radicals are mixed with other terms, squaring the entire equation becomes much more complex and can lead to unnecessary errors. Properly isolating radicals simplifies the process and keeps the equation manageable.
2. Squaring Both Sides
To eliminate the square roots, we square both sides of the equation. This is a fundamental step in solving radical equations.
(√3x-4)² = (2√x-5)²
When we square a square root, we effectively undo the radical, leaving us with the expression inside the square root. However, it's crucial to apply the squaring operation to the entire side of the equation, especially when there are coefficients or multiple terms. On the left side, squaring √3x-4 results in 3x-4. On the right side, we must square both the coefficient 2 and the radical √x-5.
The correct application of the squaring operation is crucial because an error here can invalidate the entire solution process. Remember that (2√x-5)² is equivalent to 2² * (√x-5)², which simplifies to 4 * (x-5). This is a common area for mistakes, so it's important to pay close attention to the order of operations and ensure that each term is correctly handled.
3. Simplifying the Equation
After squaring both sides, we simplify the equation to remove parentheses and combine like terms.
3x - 4 = 4(x - 5)
Expanding the right side of the equation gives:
3x - 4 = 4x - 20
Now, we rearrange the equation to isolate the variable x. This involves moving all terms containing x to one side and constant terms to the other side. We do this by adding or subtracting terms from both sides of the equation. Subtracting 3x from both sides and adding 20 to both sides will help us achieve this.
Simplifying equations efficiently is a skill that develops with practice. The goal is to make the equation as straightforward as possible without losing accuracy. This involves careful application of algebraic principles, such as the distributive property and the properties of equality. Being methodical in this step can save time and prevent errors in the later stages of the solution.
4. Solving for x
Next, we solve for x by isolating it on one side of the equation.
3x - 4 = 4x - 20
-4 + 20 = 4x - 3x
16 = x
So, we find that a potential solution is x = 16. However, this is not the end of the process. We must remember that squaring both sides of an equation can introduce extraneous solutions. Extraneous solutions are values that satisfy the transformed equation but do not satisfy the original equation. Therefore, it is essential to check our solution in the original equation.
Isolating the variable correctly is a key skill in algebra. It involves understanding the properties of equality and applying them systematically. In this step, we used the addition and subtraction properties of equality to move terms around and ultimately isolate x. The result, x = 16, is a critical point in our solution, but it remains provisional until we verify it in the original equation.
5. Checking for Extraneous Solutions
To ensure our solution is valid, we substitute x = 16 back into the original equation:
√3x-4 = 2√x-5
√3(16)-4 = 2√16-5
√48-4 = 2√16-5
√44 = 2√11
√(4 * 11) = 2√11
2√11 = 2√11
Since the left side equals the right side, x = 16 is a valid solution. It is crucial to perform this check because, in some cases, squaring both sides of an equation can introduce extraneous solutions that do not satisfy the original equation.
Checking for extraneous solutions is often the most overlooked step in solving radical equations, but it is arguably the most important. This step ensures that the solution we have found is not an artifact of the algebraic manipulation but is indeed a true solution to the original problem. By substituting the solution back into the original equation, we verify its validity and complete the problem-solving process.
Conclusion
In conclusion, the solution to the equation √3x-4 = 2√x-5 is x = 16. We arrived at this solution by carefully squaring both sides of the equation, simplifying, isolating the variable, and most importantly, checking for extraneous solutions. This step-by-step approach ensures that we have a correct and valid solution.
Solving radical equations involves a combination of algebraic skills and careful attention to detail. Understanding the theory behind each step and the potential pitfalls, such as extraneous solutions, is key to mastering this type of problem. By following a structured approach, you can confidently solve a wide range of radical equations.
Whether you are a student learning algebra or someone looking to refresh your math skills, the ability to solve radical equations is a valuable asset. It not only improves your mathematical proficiency but also enhances your problem-solving abilities in general. Practice and a thorough understanding of the principles involved will lead to mastery and confidence in this area.
Solving radical equations involves a systematic approach. Each step is crucial to ensure an accurate result. Below, we will delve deeper into the key steps, expanding on why each is necessary and how to execute it effectively. These steps include isolating the radical, squaring both sides, simplifying the equation, solving for the variable, and checking for extraneous solutions.
Isolating the Radical
The first step in solving a radical equation is to isolate the radical on one side of the equation. This means getting the radical term by itself, without any other terms added or subtracted. Isolating the radical is crucial because it sets the stage for the next step, which is squaring both sides of the equation. When the radical is isolated, squaring both sides directly eliminates the square root, simplifying the equation. If other terms are present, squaring the entire side can lead to a more complex equation that is harder to solve.
Consider the equation √(2x + 3) + 1 = 5. To isolate the radical, we need to subtract 1 from both sides, resulting in √(2x + 3) = 4. Now, the radical is isolated, and we can proceed with squaring both sides. Failing to isolate the radical first can lead to a significantly more complicated equation, often involving more terms and higher powers, making it harder to solve.
Isolating the radical isn't just a matter of algebra; it's a strategic move that simplifies the problem. It reflects a thoughtful approach to problem-solving, where we aim to make the equation as manageable as possible before proceeding with the next steps. This foundational step is critical for success in solving radical equations.
Squaring Both Sides
Once the radical is isolated, the next step is to square both sides of the equation. Squaring both sides eliminates the square root, which is the main obstacle in solving radical equations. When we square a square root, we are essentially undoing the radical operation, leaving us with the expression inside the square root.
In our example, √(2x + 3) = 4, squaring both sides gives us (√(2x + 3))² = 4², which simplifies to 2x + 3 = 16. This step transforms the radical equation into a more familiar algebraic equation that is easier to solve. However, it's important to remember that squaring both sides can sometimes introduce extraneous solutions, which we'll address later.
Squaring both sides is a powerful technique, but it must be applied carefully. The key is to ensure that we square the entire side of the equation, not just individual terms. This is particularly important when there are multiple terms on one or both sides of the equation. A common mistake is to square terms individually rather than the entire expression, which can lead to incorrect results. Correctly applying this step is crucial for moving forward in the solution process.
Simplifying the Equation
After squaring both sides, the equation often needs to be simplified. Simplification involves expanding any expressions, combining like terms, and rearranging the equation into a standard form. The goal is to make the equation as simple and manageable as possible before proceeding to solve for the variable.
Continuing from our example, 2x + 3 = 16, we can simplify this equation by subtracting 3 from both sides, resulting in 2x = 13. This step reduces the equation to a simple linear equation that is easily solved. In other cases, simplification might involve distributing terms, combining like terms, or even factoring, depending on the complexity of the equation.
Simplification is a critical step because it makes the equation more transparent and easier to work with. By reducing the number of terms and putting the equation into a standard form, we can more easily identify the next steps required to solve for the variable. Effective simplification is a hallmark of good algebraic technique and helps prevent errors in the solution process.
Solving for the Variable
Once the equation is simplified, the next step is to solve for the variable. This involves using algebraic techniques to isolate the variable on one side of the equation. The specific techniques used will depend on the type of equation, but they typically involve adding, subtracting, multiplying, or dividing both sides of the equation by the same value.
In our example, 2x = 13, we solve for x by dividing both sides by 2, which gives us x = 13/2 or x = 6.5. This step isolates x, giving us a potential solution to the equation. However, as mentioned earlier, we must still check this solution to ensure it is not extraneous.
Solving for the variable is the heart of the equation-solving process. It requires a solid understanding of algebraic principles and the ability to apply them systematically. The goal is to isolate the variable without changing the equality of the equation. This step provides a candidate solution that we then must verify in the original equation.
Checking for Extraneous Solutions
The final and perhaps most crucial step in solving radical equations is checking for extraneous solutions. Extraneous solutions are values that satisfy the transformed equation (after squaring both sides) but do not satisfy the original equation. These solutions arise because squaring both sides can introduce solutions that were not present in the original equation.
To check for extraneous solutions, we substitute our potential solution back into the original equation and see if it holds true. In our example, we found x = 6.5, so we substitute this value back into the original equation, √(2x + 3) + 1 = 5:
√(2(6.5) + 3) + 1 = 5
√(13 + 3) + 1 = 5
√16 + 1 = 5
4 + 1 = 5
5 = 5
Since the equation holds true, x = 6.5 is a valid solution. If the substitution had resulted in a false statement, we would have concluded that the solution was extraneous and discarded it.
Checking for extraneous solutions is an essential part of the solution process for radical equations. It ensures that we only accept solutions that truly satisfy the original problem. Overlooking this step can lead to incorrect answers, so it should always be performed meticulously.
Solving radical equations can be tricky, and there are several common mistakes that students often make. Understanding these mistakes and how to avoid them is crucial for mastering the technique. Below, we will discuss some of the most common pitfalls and provide strategies to ensure accuracy in your solutions.
Forgetting to Check for Extraneous Solutions
One of the most common mistakes is forgetting to check for extraneous solutions. As mentioned earlier, squaring both sides of an equation can introduce solutions that do not satisfy the original equation. Failing to check for these extraneous solutions can lead to incorrect answers.
To avoid this mistake, always make it a habit to substitute your potential solutions back into the original equation. This step is not just a formality; it's a critical part of the solution process. If the solution does not satisfy the original equation, it is extraneous and should be discarded.
Remember, the goal is not just to solve the transformed equation but to find solutions that work for the original problem. Checking for extraneous solutions is the final safeguard against accepting incorrect answers.
Incorrectly Squaring Both Sides
Another frequent mistake is incorrectly squaring both sides of the equation. This can happen when there are multiple terms on one side, and students forget to apply the distributive property correctly. For example, if the equation is √(x + 1) + 1 = x, squaring both sides incorrectly might lead to x + 1 + 1 = x², which is wrong. The correct way to square the left side is (√(x + 1) + 1)² = (√(x + 1))² + 2√(x + 1) + 1.
To avoid this, always remember to square the entire side of the equation, not just individual terms. If there are multiple terms, use the distributive property or the formula for squaring a binomial, (a + b)² = a² + 2ab + b². Careful attention to detail in this step is essential for avoiding errors.
Squaring both sides correctly requires a solid understanding of algebraic principles. It's not just about removing the square root; it's about ensuring that the equality of the equation is maintained throughout the process.
Not Isolating the Radical First
Failing to isolate the radical before squaring both sides is another common error. Squaring both sides without isolating the radical can lead to a more complicated equation that is difficult to solve. The purpose of isolating the radical is to simplify the equation after squaring.
To avoid this mistake, always make sure that the radical term is by itself on one side of the equation before squaring. This typically involves adding or subtracting terms to move them to the other side. Isolating the radical makes the squaring process more straightforward and reduces the chances of making errors.
Isolating the radical is a strategic move that simplifies the problem. It reflects a thoughtful approach to equation-solving, where we aim to make each step as manageable as possible.
Making Arithmetic Errors
Simple arithmetic errors can also lead to incorrect solutions. These errors can occur during any step of the process, from simplifying expressions to solving for the variable. Even a small mistake can throw off the entire solution.
To minimize arithmetic errors, it's helpful to work carefully and double-check each step. Use a calculator if necessary, and be particularly attentive to signs (positive and negative) when adding, subtracting, multiplying, or dividing. Practicing with a variety of problems can also help improve your arithmetic skills and reduce the likelihood of errors.
Arithmetic accuracy is a foundational skill in algebra. Even the most advanced techniques will fail if basic arithmetic is flawed. Paying attention to detail and double-checking your work are essential habits for success.
Not Following the Order of Operations
Finally, not following the order of operations (PEMDAS/BODMAS) can lead to mistakes in simplifying and solving equations. The order of operations dictates the sequence in which mathematical operations should be performed: Parentheses/Brackets, Exponents/Orders, Multiplication and Division, and Addition and Subtraction.
To avoid this mistake, always follow the order of operations carefully. Simplify expressions within parentheses or brackets first, then handle exponents, followed by multiplication and division, and finally, addition and subtraction. Adhering to the order of operations ensures that expressions are simplified correctly and equations are solved accurately.
The order of operations is a fundamental principle in mathematics. Mastering it is essential for accurate algebraic manipulation and problem-solving. A consistent application of the order of operations will help you avoid errors and achieve correct solutions.
By understanding these common mistakes and taking steps to avoid them, you can significantly improve your ability to solve radical equations accurately and efficiently. Remember, practice and careful attention to detail are key to success in algebra.
