Solving √{x-5} + 7 = 11: Find X And Check Solutions
Let's dive into solving the equation √{x-5} + 7 = 11. This problem involves finding the value of 'x' that satisfies the equation and also checking if the solution is extraneous. Extraneous solutions can sometimes pop up when dealing with square roots, so it's a crucial step to verify our answer. We'll break down each step in a clear, easy-to-follow manner.
Understanding the Equation
At its core, the equation √{x-5} + 7 = 11 is an algebraic expression where we need to isolate 'x'. The presence of the square root means we'll have to use some specific techniques to get 'x' by itself. The key here is to understand the order of operations and how to reverse them to solve for the unknown. We want to peel away the layers, so to speak, until we have 'x' on one side of the equation. This involves using inverse operations, like subtracting to undo addition and squaring to undo a square root. Keep in mind that whatever we do to one side of the equation, we must also do to the other side to maintain the balance and ensure the equality remains true. This fundamental principle of algebraic manipulation is what allows us to systematically solve for 'x'. So, let’s embark on this step-by-step journey, keeping our focus sharp and our algebraic toolkit ready. Remember, math isn't just about getting the right answer; it's about understanding the process and the underlying principles. As we work through this, we’ll not only find the value of 'x', but also reinforce our understanding of how equations work and how to solve them effectively.
Step-by-Step Solution
1. Isolate the Square Root
First, we need to isolate the square root term. This means getting √{x-5} alone on one side of the equation. To do this, we'll subtract 7 from both sides:
√{x-5} + 7 - 7 = 11 - 7
√{x-5} = 4
This step is crucial because it sets us up to eliminate the square root in the next stage. By isolating the square root, we're essentially preparing it for the inverse operation, which is squaring. Think of it like clearing the path before building a bridge; we need to remove the obstacles before we can proceed with the main task. The beauty of this approach lies in its simplicity and directness. We're not just randomly moving numbers around; we're strategically manipulating the equation to bring us closer to our goal. The principle of maintaining balance in an equation is paramount here, and subtracting 7 from both sides ensures that the equality remains intact. It's a fundamental technique that applies to a wide range of algebraic problems, and mastering it is key to success in equation solving. With the square root now isolated, we're perfectly positioned to take the next step towards finding the value of 'x'.
2. Eliminate the Square Root
To get rid of the square root, we square both sides of the equation:
(√{x-5})^2 = 4^2
x - 5 = 16
Squaring both sides effectively cancels out the square root, simplifying the equation and bringing us closer to solving for 'x'. This step is a classic example of using an inverse operation to undo a mathematical process. Just as subtraction undoes addition, squaring undoes the square root. The act of squaring both sides is a powerful technique, but it's also one that requires caution. It's essential to remember that squaring can sometimes introduce extraneous solutions, which we'll need to check later. However, for now, our focus is on simplifying the equation. By squaring √{x-5}, we've transformed it into a much more manageable form: x - 5. This is a linear expression, which is significantly easier to deal with than a square root. The right side of the equation, 4^2, becomes 16, a simple constant. With the equation now in this simplified state, we're just one step away from isolating 'x' and finding its value. This process highlights the elegance of algebra, where complex problems can be broken down into simpler steps through strategic manipulation.
3. Solve for x
Now, we isolate 'x' by adding 5 to both sides:
x - 5 + 5 = 16 + 5
x = 21
This step is a straightforward application of the addition property of equality. By adding 5 to both sides, we effectively cancel out the -5 on the left side, leaving 'x' all by itself. This is the moment of truth, where we arrive at a potential solution for 'x'. However, it's crucial to remember that we're not quite done yet. Because we squared both sides of the equation earlier, we need to check if our solution is valid or if it's an extraneous solution. Extraneous solutions can arise when squaring introduces roots that don't actually satisfy the original equation. So, while we've found a value for 'x', we need to put it to the test. This is a critical part of the process, ensuring that our answer is not only mathematically correct but also logically consistent with the original problem. With x = 21 in hand, we're ready to take the final step: verifying our solution.
Checking for Extraneous Solutions
4. Verify the Solution
We substitute x = 21 back into the original equation:
√{21 - 5} + 7 = 11
√{16} + 7 = 11
4 + 7 = 11
11 = 11
Since the equation holds true, x = 21 is a valid solution.
Substituting our potential solution back into the original equation is the ultimate test of its validity. This step is not just a formality; it's a crucial safeguard against extraneous solutions. By replacing 'x' with 21 in the original equation, we're essentially asking: does this value make the equation true? The process unfolds step by step, revealing whether our solution holds up under scrutiny. We first simplify inside the square root, then evaluate the square root itself, and finally perform the addition. The result, 11 = 11, is a resounding confirmation that our solution is indeed valid. This means that x = 21 is not just a mathematical answer; it's a solution that fits the original problem perfectly. The satisfaction of verifying a solution cannot be overstated. It's the final piece of the puzzle, the moment when the entire process comes together and we can confidently say that we've solved the equation. This step reinforces the importance of thoroughness in mathematics, where checking our work is just as vital as the solving itself.
Conclusion
The solution to the equation √{x-5} + 7 = 11 is x = 21, and it is not an extraneous solution. Therefore, option B is the correct answer.
In summary, we've successfully navigated the process of solving an equation involving a square root, paying careful attention to the possibility of extraneous solutions. The journey began with isolating the square root, a strategic move that paved the way for simplification. We then eliminated the square root by squaring both sides, a technique that demands caution but ultimately leads us closer to our goal. Solving for 'x' was a straightforward application of algebraic principles, but the real test came with verifying our solution. Substituting x = 21 back into the original equation confirmed its validity, solidifying our confidence in the answer. This entire process underscores the importance of a systematic approach to problem-solving in mathematics. Each step builds upon the previous one, and attention to detail is paramount. The concept of extraneous solutions is a crucial takeaway, reminding us that algebraic manipulations can sometimes lead to answers that don't fit the original problem. By mastering these techniques and principles, we not only solve equations but also deepen our understanding of mathematical reasoning. For further exploration of algebraic equations and problem-solving strategies, consider visiting resources like Khan Academy's algebra section.
