Solving √x - √(x-5) = 1 Determining True Solutions
When dealing with equations involving radicals, such as the one presented here, √x - √(x-5) = 1, it's crucial to follow a systematic approach to ensure accurate solutions. Radical equations often require careful algebraic manipulation and, importantly, verification of solutions to avoid extraneous roots. This article will walk you through the process of solving the equation step-by-step, highlighting key concepts and potential pitfalls along the way. In this comprehensive guide, we will explore the method for solving equations with radicals, emphasizing the importance of identifying and eliminating extraneous solutions. Through a detailed, step-by-step approach, we'll dissect the equation √x - √(x-5) = 1, providing clarity and precision. Our aim is to equip you with the knowledge to confidently solve similar problems and understand the underlying mathematical principles. Solving equations involving square roots requires careful algebraic manipulation and a keen eye for detail. Our focus here is on the equation √x - √(x-5) = 1, which serves as an excellent example of the techniques involved. The core of our method is to isolate one radical term at a time, then square both sides of the equation to eliminate the square root. This process may need to be repeated if multiple radical terms are present. However, squaring both sides can introduce extraneous solutions, which are apparent solutions that do not satisfy the original equation. Therefore, it is essential to check every solution by substituting it back into the original equation. Let's begin by isolating one of the radical terms. We will then square both sides of the equation to eliminate the radical. This step is crucial in simplifying the equation and making it solvable. However, remember that squaring both sides can sometimes introduce solutions that don't actually work in the original equation, so we'll need to be careful later on to check our answers.
Isolating the Radical and Squaring Both Sides
The initial step in solving the equation √x - √(x-5) = 1 is to isolate one of the radical terms. This means rearranging the equation so that one of the square root terms is alone on one side. A common strategy is to add √(x-5) to both sides of the equation, resulting in the following:
√x = √(x-5) + 1
Now that we have isolated a radical term, we can eliminate it by squaring both sides of the equation. Squaring both sides is a fundamental technique in solving radical equations, as it removes the square root and allows us to work with a more manageable algebraic expression. When we square both sides, we get:
(√x)² = (√(x-5) + 1)²
On the left side, the square root and the square cancel each other out, leaving us with x. On the right side, we need to carefully expand the binomial (√(x-5) + 1)². Remember that squaring a binomial means multiplying it by itself:
(√(x-5) + 1)² = (√(x-5) + 1)(√(x-5) + 1)
Using the distributive property (also known as the FOIL method), we multiply each term in the first binomial by each term in the second binomial:
(√(x-5) + 1)(√(x-5) + 1) = (√(x-5))² + 2(√(x-5))(1) + 1²
Simplifying this expression, we get:
x - 5 + 2√(x-5) + 1
Combining like terms, the right side becomes:
x - 4 + 2√(x-5)
So, our equation now looks like this:
x = x - 4 + 2√(x-5)
Notice that we still have a square root term in the equation. This means we'll need to repeat the process of isolating the radical and squaring both sides. However, before we do that, let's simplify the equation further by subtracting x from both sides. Subtracting 'x' from both sides simplifies the equation by eliminating the 'x' term on both sides. This makes the equation easier to work with and prepares us for the next step of isolating the remaining radical term.
Simplifying and Isolating the Remaining Radical
Continuing from our previous step, we had the equation:
x = x - 4 + 2√(x-5)
To simplify this, we subtract x from both sides:
x - x = x - 4 + 2√(x-5) - x
This simplifies to:
0 = -4 + 2√(x-5)
Now, we want to isolate the remaining radical term, 2√(x-5). To do this, we add 4 to both sides of the equation:
0 + 4 = -4 + 2√(x-5) + 4
This gives us:
4 = 2√(x-5)
Next, we can divide both sides by 2 to further isolate the square root:
4 / 2 = (2√(x-5)) / 2
This simplifies to:
2 = √(x-5)
Now we have the radical term isolated. We are ready to square both sides again to eliminate the square root. This step will allow us to solve for x. Squaring both sides again will help us eliminate the square root and solve for 'x'. It's a critical step in our process, but we must remember the importance of checking our solutions later to avoid any extraneous results.
Squaring Again and Solving for x
Having isolated the radical term, our equation stands as:
2 = √(x-5)
To eliminate the square root, we square both sides of the equation:
2² = (√(x-5))²
This simplifies to:
4 = x - 5
Now, we have a simple linear equation. To solve for x, we add 5 to both sides:
4 + 5 = x - 5 + 5
This gives us:
9 = x
So, we have found a potential solution: x = 9. However, we must remember that squaring both sides of an equation can introduce extraneous solutions. Therefore, it is absolutely essential to check our solution in the original equation. This verification step is crucial to ensure the validity of our solution and to rule out any extraneous roots that may have been introduced during the algebraic manipulations.
Checking for Extraneous Solutions
We found a potential solution of x = 9 for the equation √x - √(x-5) = 1. Now, we need to verify whether this solution is true or extraneous. An extraneous solution is a value that we obtain through the solving process but does not satisfy the original equation. This often happens when we square both sides of an equation, as we did in this case. To check our solution, we substitute x = 9 back into the original equation:
√9 - √(9-5) = 1
Simplify the square roots:
3 - √4 = 1
3 - 2 = 1
1 = 1
The equation holds true. This confirms that x = 9 is a valid solution. If the equation did not hold true, we would classify x = 9 as an extraneous solution and discard it. Therefore, checking our solution is a critical step in solving radical equations. Based on our verification, we can now confidently state our conclusion about the nature of the solution x = 9.
Conclusion: Determining the Nature of the Solution
After solving the equation √x - √(x-5) = 1, we arrived at a potential solution of x = 9. We then meticulously checked this solution by substituting it back into the original equation. Our verification confirmed that:
√9 - √(9-5) = 1
3 - √4 = 1
3 - 2 = 1
1 = 1
Since the equation holds true, we can definitively conclude that x = 9 is a true solution to the equation √x - √(x-5) = 1. This process highlights the importance of not only solving equations but also verifying the solutions obtained, especially when dealing with radical equations. The check for extraneous solutions ensures the accuracy and validity of our results. In summary, the correct statement about the equation √x - √(x-5) = 1 is that x = 9 is a true solution. The detailed step-by-step approach we've taken not only solves the specific problem but also equips you with a methodology applicable to a wide range of radical equations. Understanding the concept of extraneous solutions and the necessity of verifying your results is crucial for success in algebra and beyond. This comprehensive analysis underscores the importance of a systematic approach to solving mathematical problems. By carefully isolating radicals, squaring both sides, and verifying potential solutions, we can confidently tackle even complex equations. Remember, accuracy and attention to detail are paramount in mathematics. With these skills, you'll be well-prepared to handle a variety of algebraic challenges. This step-by-step solution demonstrates a robust method for addressing radical equations, emphasizing the crucial step of verifying solutions to avoid extraneous results. The ability to apply these techniques is fundamental in algebra and will serve as a solid foundation for more advanced mathematical concepts. By mastering these skills, students can approach mathematical problem-solving with confidence and accuracy. This detailed explanation not only answers the specific question but also provides a broader understanding of the techniques and principles involved in solving radical equations.
The correct answer is C. x=9 is a true solution.
