Solve $(x-1)^2=50$: Find The Values Of X

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Hey guys! Let's dive into this math problem where we're given the equation $(x-1)^2 = 50$ and our mission is to find the values of $x$ that make this equation true. This is a classic algebra problem that involves understanding how to deal with squared terms and square roots. Don't worry, we'll break it down step by step so it's super easy to follow. We'll cover everything from the initial equation to the final solutions, making sure you understand the logic behind each step. So, grab your pencils, and let's get started!

Understanding the Equation $(x-1)^2=50$

Okay, first things first, let’s really get what this equation, $(x-1)^2 = 50$, is telling us. At its heart, this equation is a quadratic in disguise. Quadratic equations often come in the form $ax^2 + bx + c = 0$, but sometimes they need a little algebraic massaging to reveal their true form. In our case, the equation presents a squared term, $(x-1)^2$, which means we're dealing with a situation where some quantity (in this case, $x-1$) multiplied by itself equals 50. This is super important because it tells us we’re not just looking for one value of $x$, but potentially two, because both a positive and a negative number, when squared, can give us a positive result. Think about it: both $5^2$ and $(-5)^2$ equal 25. So, when we see a squared term, we've got to keep in mind that there might be two paths to explore.

To solve this, we need to undo the square. The mathematical tool for undoing a square is the square root. When we take the square root of both sides of an equation, we maintain the equality, but we also introduce a crucial consideration: the plus or minus. Why? Because, as we discussed, both the positive and negative square roots of a number, when squared, yield the same positive number. For instance, the square root of 50 could be either $\sqrt{50}$ or $-\sqrt{50}$. This is the key to unlocking both possible values of $x$. So, before we jump into the algebra, remember this foundational concept: squaring a number makes its sign disappear, and taking the square root brings back the duality of positive and negative possibilities. Understanding this principle is not just about solving this particular problem; it’s a fundamental skill in algebra that will help you tackle a wide range of equations. We're setting the stage for a journey into the heart of quadratic solutions, and this initial understanding is our compass.

Step-by-Step Solution

Alright, let's get into the nitty-gritty of solving the equation $(x-1)^2 = 50$. We're going to break this down into simple, manageable steps. This way, you can see exactly how we get from the problem to the solution. No magic tricks here, just clear, logical steps!

1. Taking the Square Root of Both Sides

The first move is to undo that square on the left side. As we talked about earlier, to do this, we take the square root of both sides of the equation. This is a crucial step, and it’s where we need to be super careful about remembering both the positive and negative roots. So, when we apply the square root, we get:

$\sqrt{(x-1)^2} = \pm\sqrt{50}$

Notice that little “$\pm$” symbol? That’s our way of saying “plus or minus.” It’s a compact way of telling us that $\sqrt{50}$ and $-\sqrt{50}$ are both valid possibilities. This is because both $(\sqrt{50})^2$ and $(-\sqrt{50})^2$ equal 50. Now, the square root undoes the square on the left side, leaving us with:

$x - 1 = \pm\sqrt{50}$

We're one step closer! We've peeled away the square, but we still need to isolate $x$. The “$\pm$” sign is our constant reminder that we’re actually dealing with two separate equations here, one for the positive square root and one for the negative square root. This is a common theme in algebra, where a single equation can branch out into multiple possibilities. So, we’re not just solving for one value of $x$; we’re potentially solving for two. This step is all about setting the stage for finding those two values, and we’ve done the hard part by remembering to include both the positive and negative roots.

2. Simplifying the Square Root

Now that we've taken the square root, let's make things a bit cleaner by simplifying $\sqrt{50}$. Simplifying radicals is a handy skill in algebra, and it makes our lives easier down the road. Think of it like tidying up before you start the next phase of a project. So, how do we simplify $\sqrt{50}$? We're looking for perfect square factors within 50. A perfect square is a number that can be expressed as the square of an integer (like 4, 9, 16, etc.). Can you spot one in 50?

Yep, 25 is a perfect square, and it's a factor of 50! In fact, $50 = 25 \times 2$. This is our key to simplifying. We can rewrite $\sqrt{50}$ as $\sqrt{25 \times 2}$. Now, here’s a neat trick: the square root of a product is the product of the square roots. In math speak:

$\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}$

So, we can rewrite $\sqrt{25 \times 2}$ as $\sqrt{25} \times \sqrt{2}$. And what’s $\sqrt{25}$? It’s 5! So, we've simplified $\sqrt{50}$ to $5\sqrt{2}$. This means our equation now looks like this:

$x - 1 = \pm 5\sqrt{2}$

See how much cleaner that is? Simplifying radicals is not just about aesthetics; it also makes the numbers easier to work with in the following steps. We've taken a potentially messy term, $\sqrt{50}$, and turned it into something much more manageable, $5\sqrt{2}$. This is a classic example of how a little simplification can go a long way in algebra. By breaking down the number under the square root into its factors, we were able to extract the perfect square and reduce the radical to its simplest form. This skill will serve you well in all sorts of algebraic adventures!

3. Isolating x

We're getting closer to the finish line! We've simplified the square root, and now we just need to isolate $x$. Remember, isolating a variable means getting it all by itself on one side of the equation. In our case, we have $x - 1 = \pm 5\sqrt{2}$. What's standing in the way of $x$ being alone? That “- 1” is the culprit. So, how do we get rid of it? Easy peasy – we add 1 to both sides of the equation. This keeps the equation balanced and moves us closer to our goal.

Adding 1 to both sides, we get:

$x - 1 + 1 = 1 \pm 5\sqrt{2}$

The “- 1” and “+ 1” on the left side cancel each other out, leaving us with:

$x = 1 \pm 5\sqrt{2}$

Boom! We've isolated $x$. But hold on – what does that “$\pm$” mean for our answer? Remember, it means we actually have two solutions hiding in this one equation. We have a solution where we add $5\sqrt{2}$ to 1, and we have another solution where we subtract $5\sqrt{2}$ from 1. This is why dealing with that “$\pm$” is so important – it’s the key to finding all possible values of $x$ that satisfy the original equation.

We've gone from a somewhat intimidating equation with a square and a square root to a simple expression for $x$. This step, isolating $x$, is a fundamental technique in algebra. It's all about using inverse operations (like adding 1 to undo subtracting 1) to peel away the layers surrounding our variable until it stands alone. We're not just solving for $x$ here; we're mastering a core algebraic skill that you'll use again and again. Now, let’s unveil those two solutions hiding in our equation!

4. Finding the Two Values of x

Okay, the moment of truth! We've got $x = 1 \pm 5\sqrt{2}$, and we know that “$\pm$” means we have two separate solutions to find. This is where we split our equation into two different paths, one for the plus and one for the minus. Think of it like a fork in the road – we're going to explore both directions.

Path 1: The Plus Sign

Let’s start with the “+” side. We'll replace the “$\pm$” with a “+” and solve for $x$:

$x = 1 + 5\sqrt{2}$

This is one of our solutions! We can leave it just like that. It's an exact solution, meaning it's not rounded off or approximated. Exact solutions are often preferred in math because they give us the most precise answer. So, we've found our first value of $x$: $1 + 5\sqrt{2}$.

Path 2: The Minus Sign

Now, let’s head down the other path, the one with the “-”. We'll replace the “$\pm$” with a “-” and solve for $x$:

$x = 1 - 5\sqrt{2}$

And there’s our second solution! Just like the first one, this is an exact solution, and it represents another value of $x$ that makes our original equation true. So, we've found our second value of $x$: $1 - 5\sqrt{2}$.

We've successfully navigated both paths and found the two values of $x$ that satisfy the equation $(x-1)^2 = 50$. This step is all about understanding the implications of that “$\pm$” sign and how it leads us to multiple solutions. We didn't just find a solution; we found all the solutions. This is a crucial distinction in algebra, especially when dealing with quadratic equations and other equations with multiple possible answers. By splitting the equation and solving each part separately, we ensured that we didn't miss any solutions. High five! We’ve conquered this equation and learned a valuable lesson about the dual nature of solutions in algebra.

Checking the Solutions

Before we celebrate our victory, there’s one more crucial step: let's double-check our solutions! In math, checking your work is like putting a lock on the treasure chest – it ensures that your hard-earned answers are the real deal. We've found two potential values for $x$: $1 + 5\sqrt{2}$ and $1 - 5\sqrt{2}$. To check them, we're going to plug each one back into the original equation, $(x-1)^2 = 50$, and see if it holds true.

Checking $x = 1 + 5\sqrt{2}$

Let's substitute $1 + 5\sqrt{2}$ for $x$ in the original equation:

$((1 + 5\sqrt{2}) - 1)^2 = 50$

First, we simplify inside the parentheses:

$(5\sqrt{2})^2 = 50$

Now, we square $5\sqrt{2}$. Remember, squaring a product means squaring each factor:

$5^2 \times (\sqrt{2})^2 = 50$

$25 \times 2 = 50$

$50 = 50$

It checks out! When we plug in $x = 1 + 5\sqrt{2}$, the equation holds true. This gives us confidence that this is indeed a valid solution.

Checking $x = 1 - 5\sqrt{2}$

Now, let's do the same for our other solution, $x = 1 - 5\sqrt{2}$:

$((1 - 5\sqrt{2}) - 1)^2 = 50$

Again, we simplify inside the parentheses:

$(-5\sqrt{2})^2 = 50$

We square $-5\sqrt{2}$. Remember, a negative number squared becomes positive:

$(-5)^2 \times (\sqrt{2})^2 = 50$

$25 \times 2 = 50$

$50 = 50$

It checks out again! When we plug in $x = 1 - 5\sqrt{2}$, the equation also holds true. This confirms that both of our solutions are correct.

Checking our solutions is not just a formality; it's a crucial part of the problem-solving process. It's like the final brushstroke on a masterpiece, ensuring that everything is just right. By plugging our solutions back into the original equation, we've verified that they satisfy the equation and are indeed the correct values for $x$. This step also helps catch any potential errors we might have made along the way. So, always remember to check your work – it's the key to mathematical confidence!

Final Answer

Alright, guys! We've reached the end of our journey through the equation $(x-1)^2 = 50$. We've unpacked the problem, solved for $x$, and even double-checked our answers to make sure they're spot on. So, what's the grand finale? Let's state our final answer clearly and confidently.

After all our hard work, we've discovered that there are two values of $x$ that satisfy the equation. These values are:

$x = 1 + 5\sqrt{2}$

and

$x = 1 - 5\sqrt{2}$

These are the exact solutions to the equation. Remember, exact solutions are the most precise way to express our answers, especially when dealing with square roots and other irrational numbers. We could also approximate these values using a calculator, but leaving them in this form maintains their mathematical purity.

We've not only found the solutions but also understood the steps involved in getting there. We tackled the square, navigated the square root, simplified radicals, and isolated $x$. We even remembered to consider both the positive and negative possibilities, which is crucial when dealing with squared terms. And, most importantly, we checked our answers to ensure accuracy.

This problem is a fantastic example of how algebra can be a puzzle to solve, with each step building upon the previous one. By breaking down the problem into smaller, manageable steps, we were able to conquer it with confidence. So, the next time you encounter a similar equation, remember the techniques we've learned here, and you'll be well on your way to success! You got this!