Solving $(x+3)^2 - 3 = 49$: A Step-by-Step Guide

by Andrew McMorgan 49 views

Hey math enthusiasts! Today, we're diving into solving the equation (x+3)2βˆ’3=49(x+3)^2 - 3 = 49. If you've ever felt a little intimidated by algebraic equations, don't worry! We're going to break it down step by step, making it super easy to understand. Whether you're a student tackling homework or just someone who loves a good math challenge, this guide is for you. So, grab your pencils, and let's get started!

Understanding the Basics

Before we jump into the solution, let's make sure we're all on the same page with the basic principles involved. This equation is a quadratic equation in disguise, and to solve it effectively, we need to understand a few key concepts. First, let's talk about order of operations, often remembered by the acronym PEMDAS (Parentheses, Exponents, Multiplication and Division, Addition and Subtraction). This tells us the sequence in which we should perform operations when simplifying an expression.

Next, we need to understand how to isolate the variable, which in this case is x. This involves performing inverse operations to get x by itself on one side of the equation. For example, if we have x + 3 = 5, we subtract 3 from both sides to isolate x. Lastly, we need to be comfortable with the concept of square roots, as we'll be using them to undo the square in our equation. Remember, when we take the square root of a number, we get both a positive and a negative solution. This is super important because it means our equation might have two possible answers.

Now, let's dive a bit deeper into why these basics are so crucial. Understanding PEMDAS ensures we tackle the equation in the correct order, preventing errors that can arise from misinterpreting the expression. Isolating the variable is the heart of solving any equation; it's about strategically manipulating the equation to reveal the value of x. And grasping square roots is essential for undoing the squared term, allowing us to find all possible solutions. Without a solid understanding of these fundamentals, solving more complex equations becomes a real challenge. So, before we move on, take a moment to review these concepts if you need to. Trust me, mastering these basics will make the entire process smoother and more enjoyable!

Step 1: Isolate the Squared Term

Alright, let's dive into the first step of solving our equation: isolating the squared term. This is a crucial step because we want to get extit{that} (x+3)2(x+3)^2 part all by itself on one side of the equation. Think of it like peeling away the layers of an onion – we need to get to the core, which in this case is the squared expression. So, how do we do it? Well, remember our original equation: (x+3)2βˆ’3=49(x+3)^2 - 3 = 49. We need to get rid of that β€œ- 3” first. The golden rule of algebra is that whatever we do to one side of the equation, we must do to the other. This keeps the equation balanced, like a perfectly balanced scale.

So, to eliminate the β€œ- 3”, we’re going to do the opposite operation: we're going to add 3 to both sides of the equation. When we add 3 to the left side, the β€œ- 3” and β€œ+ 3” cancel each other out, leaving us with just (x+3)2(x+3)^2. And on the right side, we have 49 + 3, which equals 52. So, after this first step, our equation looks like this: (x+3)2=52(x+3)^2 = 52. See? We've made some real progress! We’ve successfully isolated the squared term. This is a big win because it sets us up perfectly for the next step, which involves undoing that square. But before we move on, let’s just pause for a moment and appreciate how far we’ve come. extit{Isolating the squared term} is often the trickiest part of these types of problems, and you’ve nailed it. You’re one step closer to cracking the code and finding those solutions for x. Keep up the awesome work, guys! You've got this.

Step 2: Take the Square Root of Both Sides

Okay, now that we've isolated the squared term, it's time for the next exciting step: taking the square root of both sides of the equation. Remember, our equation currently looks like this: (x+3)2=52(x+3)^2 = 52. The whole point of this step is to get rid of that pesky square, so we can start figuring out what x actually is. extit{The square root} is like the opposite of squaring something, so it's the perfect tool for the job. But here's a super important thing to keep in mind: whenever we take the square root of a number, we always have to consider both the positive and the negative possibilities.

Think about it this way: both 7 squared (7 * 7) and -7 squared (-7 * -7) give us 49. So, when we take the square root of 49, we get both 7 and -7 as possible answers. This is crucial because it means our equation might have two different solutions for x. So, when we take the square root of both sides of (x+3)2=52(x+3)^2 = 52, we get x+3=Β±extit√52x+3 = Β± extit{√}52. That little β€œΒ±β€ symbol means β€œplus or minus,” reminding us to consider both the positive and negative square roots of 52. Now, let’s talk about extit√52}. It’s not a perfect square, meaning it doesn’t simplify to a nice whole number. However, we can simplify it a bit by noticing that 52 is 4 times 13. Since the square root of 4 is 2, we can rewrite extit{√}52 as 2 extit{√}13. This makes our equation look a little cleaner $x+3 = Β±2 extit{√13$.

We're getting closer to the finish line! By taking the square root of both sides and remembering to include both positive and negative roots, we've opened up the path to finding both possible solutions for x. The next step is all about isolating x completely, and we're well-equipped to tackle it. So, keep that positive and negative mindset, and let's move on to the next stage of the game! You're doing fantastic, guys!

Step 3: Isolate x

Alright, we're on the home stretch now! We've taken the square root of both sides, and our equation looks like this: x+3=Β±2extit√13x + 3 = Β±2 extit{√}13. Now, the final boss in this level is isolating x completely. This means we need to get x all by itself on one side of the equation. Remember, the key to isolating a variable is to use inverse operations. In this case, we have β€œx + 3”, so to get x alone, we need to do the opposite of adding 3, which is subtracting 3.

Just like before, whatever we do to one side of the equation, we must do to the other to keep things balanced. So, we're going to subtract 3 from both sides. This gives us: x=βˆ’3Β±2extit√13x = -3 Β± 2 extit{√}13. And there you have it! We've successfully isolated x. But wait, there's more! Because of that β€œΒ±β€ symbol, this actually represents two different solutions for x. We have one solution where we add 2extit√132 extit{√}13 to -3, and another solution where we subtract 2extit√132 extit{√}13 from -3. This is why taking both the positive and negative square roots was so important back in Step 2. If we had forgotten about the negative root, we would have missed one of our solutions!

So, our two solutions are:

  • x=βˆ’3+2extit√13x = -3 + 2 extit{√}13
  • x=βˆ’3βˆ’2extit√13x = -3 - 2 extit{√}13

These are the exact solutions to the equation. If we wanted to get approximate decimal values, we could plug 2extit√132 extit{√}13 into a calculator, but for now, we'll leave them in this exact form. Guys, you've done it! You've successfully navigated all the steps and found both solutions to this equation. Give yourselves a huge pat on the back! You've shown that you can handle algebraic challenges with confidence and skill. Now, let's recap what we've learned and solidify our understanding.

Step 4: Verify the Solutions

Okay, mathletes, we've arrived at a super crucial step in our problem-solving journey: verifying our solutions! Think of this as the ultimate check – we want to make absolutely sure that the values we found for x actually work when we plug them back into the original equation. It's like double-checking your work on a masterpiece to ensure every brushstroke is perfect. So, why is this step so important? Well, imagine going through all the effort of solving an equation, only to find out your answer is incorrect. Verifying extit{our solutions} helps us catch any potential errors we might have made along the way. It gives us the peace of mind of knowing that our hard work has paid off and that we've truly cracked the code.

Now, let's get down to business. Remember our original equation? It was (x+3)2βˆ’3=49(x+3)^2 - 3 = 49. And we found two solutions for x: x=βˆ’3+2extit√13x = -3 + 2 extit{√}13 and x=βˆ’3βˆ’2extit√13x = -3 - 2 extit{√}13. To verify these solutions, we're going to take each one and substitute it back into the original equation, and then simplify to see if we get a true statement. Let's start with the first solution: x=βˆ’3+2extit√13x = -3 + 2 extit{√}13. We'll plug this into our equation wherever we see an x: ((βˆ’3+2extit√13)+3)2βˆ’3=49((-3 + 2 extit{√}13) + 3)^2 - 3 = 49. Now, let's simplify. Inside the parentheses, the β€œ-3” and β€œ+3” cancel each other out, leaving us with (2extit√13)2βˆ’3=49(2 extit{√}13)^2 - 3 = 49.

Squaring 2extit√132 extit{√}13 means squaring both the 2 and the extit√}13. So, (2extit√13)2(2 extit{√}13)^2 becomes 4βˆ—134 * 13, which equals 52. Now our equation looks like this 52βˆ’3=4952 - 3 = 49. And guess what? 52 - 3 does indeed equal 49! This means our first solution checks out. Woo-hoo! Now, let's tackle the second solution: $x = -3 - 2 extit{√13$. We'll plug this into our equation: ((βˆ’3βˆ’2extit√13)+3)2βˆ’3=49((-3 - 2 extit{√}13) + 3)^2 - 3 = 49. Again, inside the parentheses, the β€œ-3” and β€œ+3” cancel, leaving us with (βˆ’2extit√13)2βˆ’3=49(-2 extit{√}13)^2 - 3 = 49.

Squaring βˆ’2extit√13-2 extit{√}13 also gives us 4βˆ—134 * 13, which is 52 (remember, a negative times a negative is a positive). So, our equation becomes 52βˆ’3=4952 - 3 = 49, which, as we know, is a true statement. This means our second solution also checks out! We've verified both solutions, and we can confidently say that we've solved the equation correctly. High fives all around, guys! You've not only found the solutions but also proven that they work. That's the mark of a true math whiz. Now, let's wrap things up with a quick recap of our journey.

Conclusion

Alright, math aficionados, let's take a moment to celebrate our victory and recap the awesome journey we've been on. We set out to solve the equation (x+3)2βˆ’3=49(x+3)^2 - 3 = 49, and we absolutely crushed it! We didn't just stumble upon the answer; we systematically broke down the problem, conquered each step, and emerged victorious. Solving equations like this is more than just finding the right numbers; it's about developing a problem-solving mindset that you can apply to all sorts of challenges, both inside and outside the math classroom.

So, let's quickly review the steps we took: First, we isolated the squared term by adding 3 to both sides of the equation. This got us one step closer to unraveling the mystery of x. Then, we took the square root of both sides, remembering the crucial detail that we need to consider both the positive and negative roots. This led us to two possible paths for our solutions. Next, we isolated x by subtracting 3 from both sides, revealing our two solutions: x=βˆ’3+2extit√13x = -3 + 2 extit{√}13 and x=βˆ’3βˆ’2extit√13x = -3 - 2 extit{√}13. And finally, we verified our solutions by plugging them back into the original equation, confirming that they both work perfectly.

By following these steps, we transformed a seemingly complex equation into a manageable problem. We used the principles of algebra, such as inverse operations and the properties of square roots, to navigate our way to the solutions. More importantly, we learned the value of checking our work and ensuring the accuracy of our answers. Solving equations is like building a strong foundation – each step builds upon the previous one, and a solid understanding of the basics is key to success. So, keep practicing, keep exploring, and keep challenging yourselves with new and exciting math problems. You've proven that you have what it takes to excel in the world of algebra and beyond. Keep that mathematical fire burning bright, guys! You're all amazing problem-solvers!