Solving $(x-5)^2=28$: A Step-by-Step Guide

Hey everyone! Today, we're diving into a super common type of algebra problem that pops up all the time: solving quadratic equations. Specifically, we're going to tackle this beast: $(x-5)^2=28$. You might see a few different ways to approach this, but I promise, once you get the hang of it, it's pretty straightforward. We'll break down exactly how to find those elusive values of x that make this equation true, and we'll even look at why the other options aren't quite right. So, grab your thinking caps, and let's get this math party started!

Understanding the Equation: Isolating the Squared Term

Alright, guys, the first thing you gotta notice about our equation, $(x-5)^2=28$, is that the entire left side is squared. That's a massive clue! It means we can use a technique called the square root property. This property basically says that if you have something squared and it equals a number, you can take the square root of both sides to undo that squaring. It's like hitting a reset button on the exponent. So, the very first step in solving this equation is to make sure our squared term is all by itself. In this case, $(x-5)^2$ is already isolated on the left side, and it's equal to 28 on the right. This is perfect! We don't need to do any addition, subtraction, multiplication, or division to get it there. It's already set up for us. This is why recognizing the structure of the equation is so darn important. If the equation looked like, say, $2(x-5)^2 = 56$, our first move would be to divide both sides by 2 to get $(x-5)^2 = 28$. But since our problem is already in the ideal form, we can skip that initial cleanup and jump straight to the good stuff: taking the square root. So, to recap, the equation is already simplified for us to apply the square root property. Isn't that neat?

Applying the Square Root Property: Unveiling the Solutions

Now that we've got our squared term isolated, it's time to bring in the big guns: the square root property. Remember, this property tells us that if $a^2 = b$, then $a = pm{\pm} \sqrt{b}$. In our equation, $a$ is represented by $(x-5)$, and $b$ is 28. So, when we take the square root of both sides of $(x-5)^2=28$, we get:

$\sqrt{(x-5)^2} = \pm\sqrt{28}$

This simplifies to:

$x-5 = \pm\sqrt{28}$

Here's a crucial point, folks: when you take the square root of both sides of an equation like this, you must include both the positive and negative roots. That's why we have that $\pm$ symbol. It's accounting for two possible scenarios: one where $x-5$ equals the positive square root of 28, and another where $x-5$ equals the negative square root of 28. Without that $\pm$, you'd miss one of your solutions! Now, let's simplify that $\sqrt{28}$. We can break 28 down into its prime factors: $28 = 4 \times 7$. Since 4 is a perfect square, we can simplify $\sqrt{28}$ like this:

$\sqrt{28} = \sqrt{4 \times 7} = \sqrt{4} \times \sqrt{7} = 2\sqrt{7}$

So, our equation now becomes:

$x-5 = \pm 2\sqrt{7}$

This is a critical step because it shows us the two distinct paths our solutions will take. We're so close to finding x now, just one more simple step to go. This process of simplifying radicals is super important in algebra, and it often trips people up, so take a moment to make sure you're comfortable with how we turned $\sqrt{28}$ into $2\sqrt{7}$. It's all about finding those perfect square factors hiding inside the number. We found that 4 was a perfect square factor of 28, and its square root is 2, leaving the 7 behind the radical.

Isolating x and Finding the Final Solutions

We're in the home stretch, people! We have the equation $x-5 = \pm 2\sqrt{7}$. Our goal is to get x all by itself on one side of the equation. To do that, we simply need to get rid of that '-5'. How do we undo subtracting 5? You guessed it – we add 5 to both sides!

$x - 5 + 5 = 5 \pm 2\sqrt{7}$

This simplifies to:

$x = 5 \pm 2\sqrt{7}$

And there you have it! This single statement actually represents two distinct solutions. Let's break them down explicitly:

1. The positive case: $x = 5 + 2\sqrt{7}$
2. The negative case: $x = 5 - 2\sqrt{7}$

So, the two solutions to the equation $(x-5)^2=28$ are $x = 5 - 2\sqrt{7}$ and $x = 5 + 2\sqrt{7}$. This matches option B from the multiple-choice questions. Make sure you write it out like this when you're solving, explicitly showing both the plus and minus possibilities. It's really easy to accidentally just pick one or forget the $\pm$ symbol, but remembering that is key to getting the full picture. This process of isolating the variable after dealing with the square root is usually the final step in these types of problems. We've successfully unwrapped the variable x from its squared binomial and found the exact values that satisfy the original equation. It's pretty cool when you think about it – we started with a squared term and ended up with two specific numbers (expressed with a radical, of course!) that are the only ones that work.

Why Other Options Are Incorrect

Let's quickly touch on why the other options (A, C, and D) aren't the correct solutions for $(x-5)^2=28$. Understanding why wrong answers are wrong is just as important as knowing the right one, guys!

• Option A: $x=-\sqrt{33}$ or $x=\sqrt{33}$ This option looks like it might come from trying to solve an equation where you just have $x^2 = 33$. If the original equation was just $x^2=33$, then $x = \pm\sqrt{33}$ would be correct. But our equation has that $(x-5)$ term inside the square. This option completely ignores the '-5' part of the equation, so it's definitely not it.

• Option C: $x=9$ or $x=19$ Where could these numbers come from? Let's test them. If $x=9$, then $(x-5)^2 = (9-5)^2 = 4^2 = 16$. This is not 28. If $x=19$, then $(x-5)^2 = (19-5)^2 = 14^2 = 196$. This is also not 28. These numbers don't satisfy the equation at all. It's possible someone might get confused and try to add 5 to 28 (giving 33, which isn't squared) or do some other arithmetic error, but these are just incorrect.

• Option D: $x=-5-2\sqrt{7}$ or $x=-5+2\sqrt{7}$ This option looks very similar to our correct answer, $x = 5 \pm 2\sqrt{7}$. The only difference is the sign in front of the 5. This error usually happens when you're isolating x. Instead of adding 5 to both sides of $x-5 = \pm 2\sqrt{7}$ to get $x = 5 \pm 2\sqrt{7}$, someone might mistakenly subtract 5 from both sides, or perhaps they tried to move the '-5' incorrectly. Remember, to isolate x, we need to add 5 to both sides to cancel out the '-5'. Getting that sign wrong here leads you straight to option D, which is tantalizingly close but ultimately incorrect.

Conclusion: Mastering the Square Root Property

So there you have it, folks! We successfully solved the equation $(x-5)^2=28$ using the square root property. The key steps were:

1. Isolate the squared term: Ensure $(x-5)^2$ is by itself.
2. Take the square root of both sides: Remember the $\pm$ symbol!
3. Simplify the radical: Break down $\sqrt{28}$ into $2\sqrt{7}$.
4. Isolate x: Add 5 to both sides.

This process yielded the correct solutions: $x = 5 - 2\sqrt{7}$ and $x = 5 + 2\sqrt{7}$. This method is super powerful for solving quadratic equations when you can easily isolate the squared binomial. Keep practicing, and you'll be a pro at this in no time! Math on!