Solving For C: A Step-by-Step Guide

Hey Plastik Magazine readers! Ever stumbled upon a math problem that looks like a tangled mess of variables and roots? Don't sweat it! We're here to break down one such equation and show you how to solve it step-by-step. Today, we're tackling the equation 10b = 5(√c + 2) and figuring out the correct expression for c. So, grab your thinking caps, and let's dive in!

Understanding the Equation

Before we jump into the solution, let's make sure we understand what the equation is telling us. We have three variables here: b and c, and we're trying to isolate c on one side of the equation. This means we want to rewrite the equation so that it looks like c = something. The equation involves a square root (√c), which adds a little twist, but nothing we can't handle! Remember, the goal is to get c by itself, and we'll do that by carefully unwrapping the operations around it. The key is to perform operations on both sides of the equation to maintain balance. Think of it like a seesaw – whatever you do on one side, you have to do on the other to keep it level. This principle will guide us as we solve for c. We'll be using algebraic manipulations to achieve this, such as distribution, subtraction, division, and squaring. Each step brings us closer to isolating c and revealing its relationship to b. So, let's get started and see how it's done!

Step 1: Distribute the 5

The first thing we're going to do is simplify the right side of the equation. Notice that we have 5 multiplied by the expression (√c + 2). To get rid of those parentheses, we need to distribute the 5. This means we multiply 5 by both terms inside the parentheses: 10b = 5 * √c + 5 * 2. This simplifies to 10b = 5√c + 10. Distributing the constant is a fundamental algebraic technique that helps us to unravel expressions and make them easier to work with. By multiplying 5 with both √c and 2, we've effectively removed the parentheses and created separate terms. This is a crucial step because it allows us to isolate the term containing the square root of c (5√c). Remember, our ultimate goal is to get c by itself, and by removing the parentheses, we've made progress in that direction. This sets the stage for the next steps, where we'll isolate the square root term and eventually square both sides to eliminate the root. So, keep this distribution in mind, as it's a common and powerful tool in solving algebraic equations!

Step 2: Isolate the Square Root Term

Now that we've distributed the 5, we have the equation 10b = 5√c + 10. Our next goal is to isolate the term with the square root, which is 5√c. To do this, we need to get rid of the + 10 on the right side. The opposite of adding 10 is subtracting 10, so we'll subtract 10 from both sides of the equation. This gives us 10b - 10 = 5√c + 10 - 10, which simplifies to 10b - 10 = 5√c. Subtracting 10 from both sides maintains the equality, ensuring that the equation remains balanced. This step is crucial because it brings us closer to isolating the square root term, which is essential for ultimately solving for c. By getting the square root term alone on one side of the equation, we're setting ourselves up for the next operation: dividing by the coefficient of the square root. This process of isolating terms is a fundamental technique in algebra and is used extensively in solving various types of equations. So, keep in mind the importance of performing the same operation on both sides to maintain balance, and you'll be well on your way to mastering algebraic manipulations!

Step 3: Divide to Simplify

We're making great progress! Our equation now looks like 10b - 10 = 5√c. The square root term (√c) is almost isolated, but it's still being multiplied by 5. To get rid of the 5, we need to do the opposite operation, which is division. We'll divide both sides of the equation by 5: (10b - 10) / 5 = (5√c) / 5. On the right side, the 5s cancel out, leaving us with just √c. On the left side, we can simplify the expression. Dividing both terms in the numerator by 5, we get (10b / 5) - (10 / 5) = √c, which simplifies to 2b - 2 = √c. Dividing both sides of the equation by 5 is a crucial step in isolating the square root of c. This operation effectively removes the coefficient multiplying the square root, bringing us closer to solving for c itself. Remember, whatever operation you perform on one side of the equation, you must perform on the other to maintain equality. By dividing both terms on the left side by 5, we've simplified the expression and made it easier to work with in the next step. This process of simplification is essential in algebra, as it allows us to manipulate equations into more manageable forms. So, keep in mind the importance of division as a tool for isolating variables and simplifying expressions!

Step 4: Square Both Sides

Alright, guys, we're on the home stretch! We've got 2b - 2 = √c. Now we need to get rid of that pesky square root. The opposite of taking the square root is squaring, so we'll square both sides of the equation. This means we raise each side to the power of 2: (2b - 2)² = (√c)². Squaring a square root cancels it out, so on the right side, we're left with just c. On the left side, we have (2b - 2)², which means (2b - 2) multiplied by itself. This is a bit more work to expand, but we can handle it! Squaring both sides of the equation is a pivotal step in solving for c because it eliminates the square root symbol, allowing us to isolate c completely. By squaring the entire expression on the left side, we maintain the balance of the equation while effectively undoing the square root operation on the right side. This step highlights the importance of inverse operations in algebra, where we use the opposite operation to undo an existing one. In this case, squaring is the inverse operation of taking the square root. This step sets the stage for the final expansion and simplification of the equation, leading us to the correct expression for c.

Step 5: Expand and Simplify

Okay, let's expand that left side. We have (2b - 2)², which is the same as (2b - 2)(2b - 2). We need to use the FOIL method (First, Outer, Inner, Last) to multiply these two binomials:

• First: 2b * 2b = 4b²
• Outer: 2b * -2 = -4b
• Inner: -2 * 2b = -4b
• Last: -2 * -2 = 4

So, (2b - 2)(2b - 2) = 4b² - 4b - 4b + 4. Combining the like terms (-4b and -4b), we get 4b² - 8b + 4. Therefore, our equation is now c = 4b² - 8b + 4. Expanding and simplifying the squared term is a crucial step in expressing c in its most simplified form. By using the FOIL method (First, Outer, Inner, Last) to multiply the binomials (2b - 2) and (2b - 2), we ensure that each term is correctly multiplied and combined. This process reveals the quadratic relationship between c and b, showing how the value of c changes as b varies. Simplifying the expression by combining like terms (-4b and -4b) results in a more concise equation that is easier to interpret and use. This step highlights the importance of algebraic manipulation in transforming equations into their most understandable and usable forms. With the equation now simplified, we have successfully solved for c in terms of b.

Step 6: Factoring (Optional, but Recommended)

While c = 4b² - 8b + 4 is a correct answer, we can simplify it further by factoring. Notice that all the coefficients (4, -8, and 4) have a common factor of 4. We can factor out a 4 from the expression: c = 4(b² - 2b + 1). Now, look at the expression inside the parentheses: b² - 2b + 1. This looks like a perfect square trinomial! It can be factored as (b - 1)². So, our final simplified expression for c is: c = 4(b - 1)². Factoring the expression is an important step in simplifying the equation and expressing c in its most concise form. By recognizing that all coefficients (4, -8, and 4) have a common factor of 4, we can factor out the 4 and reduce the complexity of the equation. Additionally, noticing that the expression inside the parentheses (b² - 2b + 1) is a perfect square trinomial allows us to factor it further into (b - 1)². This step demonstrates the power of algebraic factorization in revealing the underlying structure of an equation and making it easier to analyze and use. The final factored expression, c = 4(b - 1)², provides a clear and compact representation of the relationship between c and b.

Matching the Answer Choices

Now, let's see which of the answer choices matches our solution, c = 4(b - 1)². If we look at the original options, we need to manipulate our answer a little to see the match. Let's rewrite 4(b - 1)² as (2(b - 1))². Distributing the 2 inside the parentheses, we get (2b - 2)². Now, let's multiply and divide by 25: ((2b - 2)² * 25) / 25. We can rewrite this as (5(2b - 2))² / 25. Distributing the 5, we have (10b - 10)² / 25, which can also be written as (10(b-1))^2 / 25. However, we need to compare with the given choices, one of which is c = (10b - 10)² / 25. To determine if these are equivalent, let's factor the numerator of our derived expression: (10b - 10)² = (10(b - 1))² = 100(b - 1)². Now we have 100(b - 1)² / 25, which simplifies to 4(b - 1)². This confirms that c = (10b - 10)² / 25 is indeed a correct solution. The given choices sometimes require a bit of manipulation and simplification to match the derived answer. By understanding how to rewrite and simplify algebraic expressions, you can confidently compare your solution with the provided options and select the correct one. This process involves using various algebraic techniques, such as factoring, distributing, and simplifying fractions, to ensure that the derived expression is in the same form as one of the choices.

Therefore, the correct expression for c is c = (10b - 10)² / 25.

So there you have it! We've successfully solved for c in the equation 10b = 5(√c + 2). Remember, guys, the key to solving these types of problems is to break them down into smaller, manageable steps. Don't be intimidated by the square root – just follow the order of operations and you'll get there. Keep practicing, and you'll become a math whiz in no time! Stay tuned for more math adventures in Plastik Magazine! By breaking down the problem into manageable steps, we've shown how to solve for c in a clear and understandable manner. Remember, guys, math can be fun and rewarding when you approach it with the right mindset and techniques. Keep practicing, and you'll become more confident in your problem-solving abilities. And that's a wrap for today's math adventure! Stay tuned to Plastik Magazine for more exciting topics and helpful guides.