Solving Quadratic Equations A Step-by-Step Guide To 5x^2 + 6 = 131

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Hey everyone! Today, we're diving into the exciting world of quadratic equations and tackling a specific problem: 5x2+6=1315x^2 + 6 = 131. Don't worry if that looks intimidating at first glance. We'll break it down together, step by step, in a way that's super easy to understand. Think of this as a friendly chat about math, not a lecture! Quadratic equations are a fundamental concept in algebra, and mastering them opens doors to more advanced mathematical topics. Understanding how to solve them isn't just about getting the right answer; it's about developing a problem-solving mindset that's valuable in many areas of life. Before we jump into the nitty-gritty, let's make sure we're all on the same page. A quadratic equation is essentially a polynomial equation of the second degree. That fancy term "second degree" simply means that the highest power of the variable (in our case, x) is 2. The general form of a quadratic equation is ax2+bx+c=0ax^2 + bx + c = 0, where a, b, and c are constants, and a isn't zero (because if it were, we wouldn't have a quadratic equation anymore!). Our equation, 5x2+6=1315x^2 + 6 = 131, looks a bit different from the general form, but don't fret! We'll soon see how to rearrange it into the familiar ax2+bx+c=0ax^2 + bx + c = 0 format. The goal is always to isolate the variable, in this case x, to find the values that make the equation true. These values are called the roots or solutions of the equation. Solving quadratic equations might seem abstract, but they have real-world applications in various fields, including physics, engineering, and even finance. For example, they can be used to model projectile motion, calculate areas, and optimize financial investments. By understanding the underlying principles, you'll be able to tackle a wide range of problems, both inside and outside the classroom. So, grab your pencils, open your notebooks, and let's get started! We're about to embark on a mathematical adventure that will boost your confidence and expand your problem-solving skills. Remember, math is like learning a new language – the more you practice, the more fluent you become. And with a little patience and guidance, you'll be solving quadratic equations like a pro in no time! We will start by isolating the variable term x2x^2, then we'll take the square root to find the solutions for x. This method relies on the principle that if x2x^2 equals a certain value, then x must be the positive or negative square root of that value.

Step 1: Isolate the x2x^2 Term

The first thing we need to do is get the 5x25x^2 term by itself on one side of the equation. Right now, we have 5x2+6=1315x^2 + 6 = 131. To isolate the 5x25x^2 term, we need to get rid of the +6+6. How do we do that? Simple! We subtract 6 from both sides of the equation. Remember, whatever we do to one side of the equation, we must do to the other side to keep things balanced. Think of it like a seesaw – if you add weight to one side, you need to add the same weight to the other side to keep it level. This is a fundamental principle in algebra, and it's crucial for solving equations correctly. So, let's subtract 6 from both sides:

5x2+6βˆ’6=131βˆ’65x^2 + 6 - 6 = 131 - 6

This simplifies to:

5x2=1255x^2 = 125

Great! We've made progress. Now we have the 5x25x^2 term isolated on the left side. But we're not quite there yet. We want to isolate x2x^2 itself, not 5x25x^2. So, what's the next step? We need to get rid of that pesky 5 that's multiplying x2x^2. To do this, we'll divide both sides of the equation by 5. Again, we're applying the same principle of balance – whatever we do to one side, we do to the other. Dividing both sides by 5 gives us:

rac{5x^2}{5} = rac{125}{5}

This simplifies to:

x2=25x^2 = 25

Awesome! We've successfully isolated x2x^2. Now we're one step closer to finding the values of x that satisfy the equation. We've transformed the original equation, 5x2+6=1315x^2 + 6 = 131, into a much simpler form: x2=25x^2 = 25. This is a significant achievement because it allows us to see the relationship between x and its square more clearly. Isolating the variable term is a crucial step in solving many algebraic equations, not just quadratic ones. It allows us to manipulate the equation and work towards isolating the variable itself. It's like peeling away the layers of an onion – we're gradually simplifying the equation until we get to the core, which is the variable we're trying to solve for. This step often involves using inverse operations, such as subtraction to undo addition, or division to undo multiplication, as we've seen here. By mastering this technique, you'll be well-equipped to tackle more complex equations in the future. Remember, practice makes perfect! The more you work through these steps, the more comfortable and confident you'll become. So, keep practicing, and you'll be solving equations like a mathematical whiz in no time!

Step 2: Take the Square Root

Now that we have x2=25x^2 = 25, we need to figure out what values of x will make this equation true. This is where the square root comes in. Remember, the square root of a number is a value that, when multiplied by itself, gives you that number. For example, the square root of 9 is 3, because 3 * 3 = 9. But here's a crucial point: both positive and negative numbers can have the same square. For instance, both 3 and -3, when squared, give you 9. This is because (-3) * (-3) = 9 as well. This concept is essential when solving quadratic equations because it means we'll often have two solutions, not just one. So, when we take the square root of both sides of the equation x2=25x^2 = 25, we need to consider both the positive and negative square roots. This is represented mathematically using the plus-or-minus symbol (Β±). Applying this to our equation, we get:

x = ±ig(√25ig)

The square root of 25 is 5, so we have:

x=Β±5x = Β±5

This means that x can be either +5 or -5. Let's check if these solutions work by plugging them back into the original equation. First, let's try x = 5:

5(5)2+6=5(25)+6=125+6=1315(5)^2 + 6 = 5(25) + 6 = 125 + 6 = 131

That works! Now, let's try x = -5:

5(βˆ’5)2+6=5(25)+6=125+6=1315(-5)^2 + 6 = 5(25) + 6 = 125 + 6 = 131

That works too! Both 5 and -5 satisfy the original equation, so they are both valid solutions. Taking the square root is a powerful technique for solving equations where a variable is squared. It's the inverse operation of squaring, just like subtraction is the inverse operation of addition, and division is the inverse operation of multiplication. Understanding inverse operations is fundamental to solving algebraic equations effectively. When we take the square root, we're essentially "undoing" the squaring operation to isolate the variable. But remember the crucial detail about both positive and negative roots! Forgetting the negative root is a common mistake, so always be mindful of it when solving quadratic equations. This step highlights the beauty of mathematical operations – how we can use them to manipulate equations and uncover the hidden values of variables. It's like detective work, where we follow the clues and apply the rules to solve the mystery. By mastering the concept of square roots and their relationship to squaring, you'll significantly enhance your ability to solve a wide range of mathematical problems. Keep practicing, and you'll become a mathematical sleuth in no time!

Step 3: State the Solutions

We've done the math, and now it's time to state our final answer. We found that the solutions to the quadratic equation 5x2+6=1315x^2 + 6 = 131 are x=5x = 5 and x=βˆ’5x = -5. It's important to clearly state your solutions, so there's no ambiguity. You can write them separately, like we just did, or you can use set notation to express them as a set of values. Set notation is a concise way to represent a collection of items, and it's often used in mathematics. In this case, we can write the solutions as:

$x = ${-5, 5}

This notation simply means that the set of solutions for x consists of the numbers -5 and 5. Stating the solutions clearly is not just about being mathematically correct; it's also about communicating effectively. In mathematics, as in any field, clear communication is essential. When you present your work, you want to make sure that your audience (whether it's your teacher, your classmates, or yourself) can easily understand your reasoning and your results. So, always take the time to state your solutions in a clear and organized manner. This final step is also a good opportunity to reflect on the problem-solving process. We started with a quadratic equation that might have seemed intimidating at first, but we broke it down into manageable steps. We isolated the x2x^2 term, took the square root, and considered both positive and negative solutions. By following these steps, we were able to find the values of x that satisfy the equation. This process illustrates the power of a systematic approach to problem-solving. Often, complex problems can be solved by breaking them down into smaller, more manageable parts. This is a valuable skill that can be applied not only in mathematics but also in many other areas of life. So, congratulations! You've successfully solved a quadratic equation. You've mastered a new mathematical technique, and you've strengthened your problem-solving skills. Remember, the journey of learning mathematics is a continuous one. There are always new concepts to explore and new challenges to overcome. But with practice, perseverance, and a positive attitude, you can achieve anything you set your mind to. Keep exploring, keep learning, and keep solving!

In summary, the solutions to the quadratic equation 5x2+6=1315x^2 + 6 = 131 are x=5x = 5 and x=βˆ’5x = -5. We arrived at this answer by isolating the x2x^2 term, taking the square root of both sides, and considering both positive and negative solutions. This step-by-step approach demonstrates a powerful method for solving quadratic equations and other algebraic problems. Remember to always check your solutions by plugging them back into the original equation to ensure they are correct.