Solving For X: A Step-by-Step Guide

Hey math enthusiasts! Today, we're diving into a classic algebra problem: solving for x. Specifically, we'll tackle the equation x - 12oxed{\sqrt{x}} + 35 = 0. This might look a little intimidating at first glance with that square root lurking around, but trust me, we'll break it down step-by-step and make it totally manageable. This article is your ultimate guide. By the end, you'll be able to confidently solve this type of equation and understand the underlying concepts. So, grab your pencils, and let's get started!

Unveiling the Strategy: The Power of Substitution

Alright, guys, the key to solving this equation lies in a clever trick: substitution. See that pesky square root of x? We're going to make things easier by temporarily replacing it with another variable. This simplifies the equation and allows us to use familiar algebraic techniques. Think of it like a secret code – we're just changing the symbols to make the problem more friendly. Don't worry, we'll decode it back at the end to find the actual value of x. The approach is to identify a part of the equation that can be replaced with a single variable, transforming the expression into a more manageable form. Let's walk through the exact steps. First, we identify the square root of x, which is $\sqrt{x}$. Next, we let a new variable, say y, equal $\sqrt{x}$. This means $y = \sqrt{x}$. Now, we substitute y into the original equation, $x - 12\sqrt{x} + 35 = 0$. Remember that since $y = \sqrt{x}$, then $y^2 = x$. Replacing x with $y^2$ and $\sqrt{x}$ with y, we get $y^2 - 12y + 35 = 0$. This is a standard quadratic equation, and we know how to solve those! Isn't that a relief? This substitution transforms the original equation into a more familiar form, making it easier to solve using techniques like factoring or the quadratic formula. By strategically introducing a new variable, we can often simplify complex equations into simpler forms that we already know how to handle. This technique is really useful in algebra and calculus.

The Benefits of Strategic Substitution

Substitution isn't just a clever trick; it's a powerful problem-solving tool. It helps us break down complex equations into more manageable parts. By isolating and replacing specific terms, like the square root of x in our example, we reduce the complexity of the original equation. This reduction allows us to use simpler algebraic techniques, like factoring or the quadratic formula. This process reduces the likelihood of making mistakes, as each step involves simpler operations. Plus, understanding and using substitution builds a stronger foundation in algebra and opens doors to tackling even more challenging math problems. By mastering substitution, you're not just solving one specific equation; you're building a versatile skill that will serve you well in various mathematical contexts. This method is used in many mathematical problems, especially when the equation has a complex expression. You can apply it to a wide range of problems, from calculus to differential equations.

Factoring the Quadratic: Unraveling the Solution

Now that we have the quadratic equation $y^2 - 12y + 35 = 0$, our next step is to factor it. Factoring is the process of breaking down a quadratic expression into two simpler expressions, usually in the form of $(y - a)(y - b) = 0$. If you're a little rusty on factoring, don't sweat it. We'll go through it step by step. We're looking for two numbers that multiply to give us 35 (the constant term) and add up to -12 (the coefficient of the y term). Think about the factors of 35. We have 1 and 35, and 5 and 7. The pair -5 and -7 satisfy both conditions: (-5) * (-7) = 35 and (-5) + (-7) = -12. That means we can factor the quadratic equation as $(y - 5)(y - 7) = 0$. This factored form is super useful because it tells us that either $(y - 5) = 0$ or $(y - 7) = 0$. This is called the zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. This simplifies the solving process greatly. Factoring the quadratic breaks down the complex expression into two linear expressions, making the process of finding the solutions much easier. The goal of factoring is to transform the complex equation into a simpler form.

Why Factoring Works Wonders

Factoring simplifies the process of solving a quadratic equation significantly. When we factor a quadratic expression, we essentially rewrite it as a product of two linear expressions. Once we have the expression in factored form, we can easily find the solutions by setting each factor equal to zero and solving for the variable. This allows us to quickly identify the values that make the original equation true. Factoring also helps us understand the structure of the quadratic equation. Factoring allows us to identify the roots, or solutions, of the equation directly. This understanding provides valuable insights into the behavior of the quadratic function. The more you practice, the faster and more intuitive factoring becomes. With enough practice, you can factor the expressions in your head quickly. There are many methods for factoring, but the basic process is the same. The process helps you quickly identify the roots of the equation.

Solving for y: Unveiling the Intermediate Solutions

Now, let's solve for y. We have $(y - 5) = 0$ and $(y - 7) = 0$. Solving for y in the first equation, we get $y = 5$. Solving for y in the second equation, we get $y = 7$. Great! We've found the values of y that satisfy the factored equation. But remember, our original equation was in terms of x, not y. These are intermediate solutions, and we're not done yet. We used substitution, where $y = \sqrt{x}$. Therefore, $y$ is simply a stand-in for the square root of x. The values of y give us the values of $\sqrt{x}$ that satisfy the original equation. Solving for y gives us the intermediate solutions, but these are essential steps towards finding the actual solution for x. The two y values give us a glimpse of the solution to the original problem. We used substitution to simplify the equation, and now we must use the result to calculate the original variable.

The Importance of Intermediate Steps

Each step in solving a mathematical problem is critical. In this case, solving for y might seem like an intermediate step, but it plays a crucial role in arriving at the final answer. These intermediate steps help you build a solid understanding of the problem and the process of finding its solution. They provide a structured approach that guides you through the complex parts of the equation. Understanding the y values will allow us to easily find the solutions to the main variable x. Skipping steps can lead to errors and a lack of understanding. It's crucial to write down each step, making sure that your math is accurate. Taking these intermediate steps allows us to verify our solution and prevent mistakes. These steps are a part of the solving process.

Back to x: Finding the Final Solutions

Okay, guys, time to go back to our original variable, x. Remember that $y = \sqrt{x}$. We found that $y = 5$ and $y = 7$. Now, substitute these values back to find the values of x. When $y = 5$, we have $5 = \sqrt{x}$. To find x, we square both sides of the equation: $5^2 = (\sqrt{x})^2$, which simplifies to $25 = x$. So, one solution is $x = 25$. Similarly, when $y = 7$, we have $7 = \sqrt{x}$. Squaring both sides, we get $7^2 = (\sqrt{x})^2$, which simplifies to $49 = x$. Therefore, the other solution is $x = 49$. Great! We've found two possible values for x: 25 and 49. But before we get too excited, we need to do one last, very important step: checking our answers.

The Significance of Checking Your Answers

Always check your answers. This is a non-negotiable step in solving any equation, especially those involving square roots. Why? Because sometimes, when we manipulate equations (like squaring both sides), we can introduce extraneous solutions. Extraneous solutions are solutions that appear to satisfy the equation during the solving process but don't actually work in the original equation. That's why we need to verify if our solutions are correct. In this example, squaring both sides could potentially create extraneous solutions. By plugging the solutions back into the original equation, we ensure that they are actually valid. Let's start with x = 25. Substitute 25 into the original equation: $25 - 12\sqrt{25} + 35 = 0$. Simplifying, we get $25 - 12(5) + 35 = 0$, which gives us $25 - 60 + 35 = 0$. This simplifies to $0 = 0$, which is correct. Therefore, $x = 25$ is a valid solution. Now, let's check $x = 49$. Substituting 49 into the original equation: $49 - 12\sqrt{49} + 35 = 0$. Simplifying, we get $49 - 12(7) + 35 = 0$, which gives us $49 - 84 + 35 = 0$. This also simplifies to $0 = 0$, which is also correct. Thus, $x = 49$ is a valid solution as well. Checking your answers ensures the accuracy of your solution and gives you confidence. This step will help you avoid careless mistakes. Always check your work to make sure your solution is valid.

The Final Answer: Solutions Unveiled

So, after all that work, what's our final answer? The solutions to the equation $x - 12\sqrt{x} + 35 = 0$ are $x = 25$ and $x = 49$. We found these solutions by using substitution, factoring the quadratic, and carefully checking our answers. You've now successfully solved an equation involving a square root! Congratulations! This problem showcases the power of strategic thinking and breaking down a complex problem into manageable steps. Remember the key takeaways: Use substitution to simplify the equation, factor the quadratic expression, and always check your answers. Keep practicing these skills, and you'll become a pro at solving these types of equations. You are now equipped to tackle similar problems. The journey of solving this problem will help you solve more complex math problems. Keep practicing and keep learning, and you'll be well on your way to mastering algebra. Keep going and practicing math, and you will become better at solving this kind of problem. You have now acquired the knowledge needed to solve this problem.