Hey everyone! Let's dive into this math problem together and figure out what's going on. We've got an equation that looks a bit tricky at first glance, but don't worry, we'll break it down step by step. Our main goal here is to complete the work that's been started and see if we can find a possible solution. So, grab your thinking caps, and let's get started!
Understanding the Equation
First, let’s take a good look at the equation we're dealing with:
(x-5)^(1/2) + 5 = 2
This equation involves a square root, which might seem intimidating, but it's totally manageable. Remember that (x-5)^(1/2) is just another way of writing the square root of (x-5). Our main keyword here is finding a possible solution, so we need to isolate x and see what value(s) make the equation true. The initial steps provided give us a head start, which is super helpful. We'll follow along and add our insights to make sure we nail this.
Following the Steps
The problem already gives us the first few steps, which is awesome. Let's walk through them to make sure we understand what’s happening and why. The first step is:
(x-5)^(1/2) + 5 = 2
This is our starting point. To isolate the square root term, we need to get rid of the + 5 on the left side. How do we do that? We subtract 5 from both sides of the equation. This keeps the equation balanced, which is super important in algebra. So, we get:
(x-5)^(1/2) = 2 - 5
Which simplifies to:
(x-5)^(1/2) = -3
Cool, we’re making progress! Now, let’s look at the next step provided:
[(x-5)^(1/2)]^2 = (-3)^2
Here, we're squaring both sides of the equation. Why? Because squaring a square root cancels it out, leaving us with just the term inside the square root. This is a common technique for solving equations with radicals. On the right side, we’re squaring -3, which is straightforward. Let's continue this process.
Completing the Solution
So, after squaring both sides, we have:
[(x-5)^(1/2)]^2 = (-3)^2
This simplifies to:
x - 5 = 9
Now we're getting somewhere! We have a simple linear equation. To solve for x, we need to isolate it by adding 5 to both sides:
x - 5 + 5 = 9 + 5
Which gives us:
x = 14
Okay, so we've found a possible solution: x = 14. But hold on a second! We're not quite done yet. In math, especially when dealing with square roots, it's super important to check our answers. Why? Because sometimes we can get what are called extraneous solutions – answers that we get algebraically but don't actually work in the original equation. These usually pop up when we square both sides of an equation, so we always need to be careful.
Checking for Extraneous Solutions
To check if x = 14 is a valid solution, we need to plug it back into our original equation:
(x-5)^(1/2) + 5 = 2
Substitute x = 14:
(14-5)^(1/2) + 5 = 2
Simplify inside the parentheses:
(9)^(1/2) + 5 = 2
The square root of 9 is 3:
3 + 5 = 2
This simplifies to:
8 = 2
Wait a minute! That’s definitely not true. 8 does not equal 2. This means that x = 14 is an extraneous solution. It's a solution we got algebraically, but it doesn't actually satisfy the original equation. So, what does this tell us?
The Real Solution (or Lack Thereof)
Since our possible solution x = 14 turned out to be extraneous, we need to conclude that the original equation has no real solution. Why is this? Let’s go back to the step where we had:
(x-5)^(1/2) = -3
Here's the key: the square root of a real number can never be negative. When we take the principal (positive) square root, we're only looking for non-negative results. So, (x-5)^(1/2) will always be 0 or a positive number for any real x. It can never be -3. This is why we ended up with an extraneous solution and, ultimately, no real solution to the equation.
Conclusion: No Real Solution
So, to wrap it all up, we started with a radical equation, followed the steps to isolate x, and found a possible solution. But, by checking our work, we discovered that this possible solution was extraneous. Therefore, the final answer is that there is no real solution to the equation:
(x-5)^(1/2) + 5 = 2
Remember, guys, always check your solutions, especially when dealing with square roots! It’s a crucial step in making sure you've got the right answer. Keep practicing, and you’ll become a pro at solving these types of equations.
The Importance of Checking Solutions
Checking solutions is not just a formality; it’s a critical part of the problem-solving process, especially when dealing with equations involving radicals or even rational expressions. The act of squaring both sides of an equation, while a legitimate algebraic manipulation, can introduce solutions that don't actually satisfy the original equation. These are what we call extraneous solutions. Think of it like this: you're expanding the possible solution space, but not all of those possibilities are valid.
For instance, consider a simpler example:
√(x) = -2
If we square both sides, we get:
x = 4
But if we plug x = 4 back into the original equation:
√(4) = -2
2 = -2
This is clearly false. The square root of 4 is 2, not -2. So, x = 4 is an extraneous solution, and the original equation has no solution in the real numbers.
The reason extraneous solutions arise is that squaring both sides can mask the original conditions of the equation. In the case of square roots, the principal square root is always non-negative. So, if the original equation sets a square root equal to a negative number, like in our example, there can be no real solution. Squaring both sides can
