Solving For X: $(x-1)^2=50$

Hey guys! Today, we're diving into a classic algebra problem that’s super useful for nailing those tricky equations. We've got the equation $(x-1)^2=50$, and our mission, should we choose to accept it, is to find the exact values of $x$ that make this equation true. This isn't just about getting an answer; it's about understanding the process, which is key for tackling more complex problems down the line. So, grab your calculators (or just your brains!), and let's break this down step-by-step. We'll explore the most efficient methods to isolate $x$ and ensure we don't miss any solutions. Remember, in algebra, it's often the simple steps applied correctly that lead to the most elegant solutions. We'll look at the options provided and see which ones fit the bill. Get ready to flex those math muscles!

Understanding the Equation Structure

The equation $(x-1)^2=50$ is a quadratic equation, but it's presented in a way that makes it much easier to solve than a standard $ax^2+bx+c=0$ form. The key here is that the entire expression $(x-1)$ is being squared. This structure immediately suggests a method called taking the square root on both sides. Why is this so powerful? Because the square root is the inverse operation of squaring. When you take the square root of something that's been squared, you essentially undo the squaring operation, leaving you with the base expression. However, there’s a crucial detail we must remember: when you take the square root of a number, there are always two possible results – a positive one and a negative one. For example, the square root of 9 isn't just 3; it's also -3, because both $3^2$ and $(-3)^2$ equal 9. This is why quadratic equations often have two solutions. In our case, $(x-1)^2$ equals 50. So, when we take the square root of both sides, we need to consider both the positive and negative square roots of 50. This is a fundamental concept that prevents us from losing one of the two potential answers for $x$. So, before we even start manipulating the equation, recognizing this dual nature of square roots is vital. It’s the first big clue that there will likely be two distinct values for $x$ that satisfy the original equation. Keep this in mind as we move forward, because it's a common pitfall for students to forget the negative root!

The Square Root Method in Action

Alright, let's get our hands dirty with the math! Our equation is $(x-1)^2=50$. The first step, as we discussed, is to take the square root of both sides. This will help us get rid of that pesky square on the $(x-1)$ term. So, we have:

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

As we highlighted, the square root of $(x-1)^2$ is not just $(x-1)$, but it's $\pm(x-1)$. And the square root of 50? Well, 50 isn't a perfect square, meaning its square root isn't a whole number. We can simplify $\sqrt{50}$ by finding its largest perfect square factor. That would be 25, since $50 = 25 \times 2$. So, $\sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2}$.

Now, our equation looks like this:

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

This single equation actually represents two separate equations:

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

To find $x$, we just need to add 1 to both sides of each equation. This isolates $x$ and gives us our final solutions.

For the first equation: $x = 1 + 5\sqrt{2}$

And for the second equation: $x = 1 - 5\sqrt{2}$

So, the two values of $x$ that satisfy the original equation are $1 + 5\sqrt{2}$ and $1 - 5\sqrt{2}$. Pretty neat, right? This method is super efficient because it directly tackles the squared term. It's a go-to technique whenever you see an expression squared equal to a constant.

Evaluating the Options

Now that we've done the heavy lifting and found our solutions, let's look at the options provided and see which ones match our results. We found that $x = 1 + 5\sqrt{2}$ and $x = 1 - 5\sqrt{2}$. Let's check each option:

• A. $x=-49$: If we plug this back into the original equation, $(-49-1)^2 = (-50)^2 = 2500$. This is definitely not 50. So, option A is incorrect, guys.
• B. $x=51$: Let's test this one. $(51-1)^2 = (50)^2 = 2500$. Again, not 50. Option B is also out.
• C. $x=1+5 \sqrt{2}$: This is exactly one of the solutions we derived using the square root method. Bingo! This one is correct.
• D. $x=1-5 \sqrt{2}$: This is our other solution that we found. Double bingo! This one is also correct.

So, the values of $x$ that satisfy the equation $(x-1)^2=50$ are $x=1+5 \sqrt{2}$ and $x=1-5 \sqrt{2}$. When a question asks to select the values, and multiple options are correct, it's important to pick all the correct ones. In a multiple-choice test scenario, you'd select both C and D if allowed, or the question might be phrased to indicate selecting all that apply.

Why Other Methods Aren't Ideal Here

While we solved this problem efficiently using the square root method, it's worth thinking about why other common algebraic techniques aren't the best fit for this specific problem. For instance, you could, in theory, expand the $(x-1)^2$ term. This would give you $x^2 - 2x + 1 = 50$. Then, you'd rearrange it into the standard quadratic form: $x^2 - 2x - 49 = 0$. At this point, you could use the quadratic formula ($x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$) to solve for $x$. In this case, $a=1$, $b=-2$, and $c=-49$. Plugging these values in would eventually lead you to the same answers, $1+5\sqrt{2}$ and $1-5\sqrt{2}$. However, expanding the term and then using the quadratic formula involves significantly more steps and calculations. You risk making errors in expansion (like forgetting the $-2x$ term) or in the quadratic formula itself, which is more complex than taking a square root. The beauty of the square root method is its directness. It exploits the specific structure of the equation – an expression squared equals a constant – making it the most straightforward and least error-prone approach. Another method, factoring, isn't applicable here because $x^2 - 2x - 49 = 0$ doesn't factor easily into simple binomials with integer or simple radical coefficients. The number -49 makes simple factoring difficult. Therefore, for equations presented in this squared form, always look for the opportunity to use the square root method first. It's designed for exactly this type of problem and saves you time and potential headaches!

Conclusion: Mastering Quadratic Solutions

So there you have it, mathletes! We've successfully tackled the equation $(x-1)^2=50$ and pinpointed the exact values of $x$ that make it true. By applying the square root method, we found that $x = 1 + 5\sqrt{2}$ and $x = 1 - 5\sqrt{2}$. These are the correct solutions, corresponding to options C and D. Remember, the key takeaway here is the power of inverse operations. Recognizing that squaring and taking the square root are opposites allows us to simplify equations effectively. Always keep an eye out for perfect squares or expressions that can be easily squared, as they often signal an opportunity for this direct solving method. It's about working smarter, not harder, guys! Understanding why certain methods work best for specific equation structures is crucial for building a strong foundation in algebra. Keep practicing, keep exploring, and don't be afraid to try different approaches – but always be on the lookout for the most elegant and efficient path. We'll catch you in the next problem!