Solving (4x - 16)^(1/2) = 36 A Step-by-Step Guide

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In this article, we will delve into the process of finding the solution to the equation (4xβˆ’16)12=36(4x - 16)^{\frac{1}{2}} = 36. This equation involves a square root, which might seem daunting at first, but with a systematic approach, it can be solved effectively. Our goal is to isolate the variable x and determine its value that satisfies the given equation. We will walk through each step in detail, ensuring a clear understanding of the method used. By the end of this guide, you'll not only know the solution but also the underlying principles involved in solving such equations. So, let’s begin our mathematical journey and unravel the mystery behind this equation. Understanding how to solve equations like this is fundamental in algebra, and it forms the basis for more complex mathematical problems. Before diving into the solution, it's crucial to grasp the core concepts of algebraic manipulation and equation solving. This includes understanding the properties of equality, which allow us to perform the same operations on both sides of the equation without changing its balance. Also, we need to be comfortable with the concept of inverse operations, such as squaring to undo a square root. With these principles in mind, we can approach the problem with confidence and break it down into manageable steps. Remember, mathematics is like a puzzle, and each piece fits together to form the complete picture. Let's put the pieces together and solve for x.

Step-by-Step Solution

1. Isolating the Square Root

The initial equation we are given is (4xβˆ’16)12=36(4x - 16)^{\frac{1}{2}} = 36. The left side of the equation contains a square root, and our first step is to eliminate this radical. To achieve this, we need to perform the inverse operation of taking a square root, which is squaring. We will square both sides of the equation to maintain the equality. This crucial step allows us to simplify the equation and progress towards isolating x. When squaring both sides, we apply the operation to the entire expression on each side, not just individual terms. This ensures that the balance of the equation remains intact. Understanding this fundamental principle of equation manipulation is key to solving various algebraic problems. So, let's proceed with squaring both sides and observe how the equation transforms, bringing us closer to the solution. Remember, each step we take is a deliberate move towards our final goal, and it's important to understand the reasoning behind each action. By carefully executing each step, we can confidently arrive at the correct answer.

2. Squaring Both Sides

To eliminate the square root, we square both sides of the equation:

  • ((4xβˆ’16)12)2=362((4x - 16)^{\frac{1}{2}})^2 = 36^2

  • This simplifies to 4xβˆ’16=12964x - 16 = 1296.

Squaring both sides is a critical step in solving equations involving radicals. It's essential to understand why this works. The square root of a number, when squared, returns the original number. In our case, squaring (4xβˆ’16)12(4x - 16)^{\frac{1}{2}} effectively cancels out the square root, leaving us with the expression inside the radical, which is 4xβˆ’164x - 16. On the other side of the equation, we have 36236^2, which equals 1296. By performing this operation, we've transformed the equation from one involving a square root to a simple linear equation. This transformation makes the equation much easier to solve. Now, we can proceed with standard algebraic techniques to isolate x. It's important to remember that squaring both sides of an equation can sometimes introduce extraneous solutions, which are solutions that satisfy the transformed equation but not the original one. Therefore, it's always a good practice to check our final answer in the original equation to ensure it's a valid solution.

3. Isolating x

Now we have the equation 4xβˆ’16=12964x - 16 = 1296. Our next step is to isolate the term containing x. To do this, we will add 16 to both sides of the equation. This is based on the principle of maintaining equality – whatever operation we perform on one side, we must perform on the other. Adding 16 to both sides will cancel out the -16 on the left side, leaving us with just the term containing x. This is a crucial step in solving for x as it brings us closer to having x by itself. After adding 16 to both sides, we will have a new equation that is simpler and easier to solve. Remember, the goal is to isolate x, and each step we take is a step closer to achieving that. Let's proceed with adding 16 to both sides and see how the equation simplifies further.

4. Adding 16 to Both Sides

Adding 16 to both sides, we get:

  • 4xβˆ’16+16=1296+164x - 16 + 16 = 1296 + 16

  • This simplifies to 4x=13124x = 1312.

Adding 16 to both sides of the equation is a fundamental algebraic operation. It's based on the principle of maintaining balance in the equation. By adding the same value to both sides, we ensure that the equality remains true. In this case, adding 16 to the left side cancels out the -16, effectively isolating the term with x. On the right side, we add 16 to 1296, resulting in 1312. This step has simplified the equation significantly. We've gone from having a square root and a constant term on the left side to having just a term with x multiplied by a coefficient. This brings us closer to our ultimate goal of solving for x. Now, we have a much simpler equation, 4x=13124x = 1312, which can be easily solved by dividing both sides by the coefficient of x. Remember, each step we take is a deliberate effort to isolate x and find its value.

5. Dividing by the Coefficient of x

To finally isolate x, we divide both sides of the equation by 4:

  • rac{4x}{4} = \frac{1312}{4}

  • This gives us x=328x = 328.

Dividing both sides of the equation by the coefficient of x is the final step in isolating x and finding its value. In our case, the coefficient of x is 4. By dividing both sides of the equation 4x=13124x = 1312 by 4, we are essentially undoing the multiplication of x by 4. This leaves us with x on its own on the left side of the equation. On the right side, we perform the division 1312Γ·41312 \div 4, which results in 328. Therefore, we have found that x equals 328. This is the solution to the equation (4xβˆ’16)12=36(4x - 16)^{\frac{1}{2}} = 36. However, it's always a good practice to check our solution by substituting it back into the original equation to ensure it's a valid solution. This helps us avoid extraneous solutions that might arise from squaring both sides of the equation. So, let's verify our solution to be absolutely sure.

Verifying the Solution

To ensure our solution is correct, we substitute x=328x = 328 back into the original equation:

  • (4(328)βˆ’16)12=36(4(328) - 16)^{\frac{1}{2}} = 36

  • (1312βˆ’16)12=36(1312 - 16)^{\frac{1}{2}} = 36

  • (1296)12=36(1296)^{\frac{1}{2}} = 36

  • 36=3636 = 36

The substitution of the calculated value of x back into the original equation is a crucial step in verifying the solution. This process helps us confirm that the value we found for x indeed satisfies the equation. By replacing x with 328 in the original equation, we can see if both sides of the equation are equal. This step is particularly important when dealing with equations involving square roots, as squaring both sides can sometimes introduce extraneous solutions. Extraneous solutions are values that satisfy the transformed equation but not the original one. By substituting the value back, we can rule out any such extraneous solutions. In this case, when we substitute x = 328, we find that the equation holds true, meaning that 328 is indeed the correct solution. This verification step gives us confidence in our answer and ensures that we have solved the equation accurately. It's always a good practice to include this step in your problem-solving process.

Conclusion

Therefore, the solution to the equation (4xβˆ’16)12=36(4x - 16)^{\frac{1}{2}} = 36 is x=328x = 328. The correct answer is D. Understanding how to solve equations like this is crucial for mastering algebra. Remember to always verify your solutions, especially when dealing with square roots.

Throughout this guide, we've covered several key concepts that are fundamental to solving algebraic equations. Firstly, we emphasized the importance of isolating the variable. This involves performing operations on both sides of the equation to get the variable by itself. Secondly, we discussed the concept of inverse operations. This is crucial for undoing operations and simplifying the equation. For example, we used squaring to undo the square root. Thirdly, we highlighted the significance of maintaining equality. Any operation performed on one side of the equation must also be performed on the other side to keep the equation balanced. Fourthly, we stressed the importance of verification. Substituting the solution back into the original equation helps ensure that it's a valid solution and not an extraneous one. These key concepts are not only applicable to this specific problem but also to a wide range of algebraic equations. By mastering these concepts, you'll be well-equipped to tackle various mathematical challenges. Remember, practice is key to success in mathematics. The more you practice, the more comfortable you'll become with these concepts, and the better you'll be at solving equations.

Practice Problems

To further solidify your understanding of solving equations with square roots, try these practice problems:

  1. Solve for x: (2x+5)12=7(2x + 5)^{\frac{1}{2}} = 7
  2. Solve for y: (3yβˆ’2)12=4(3y - 2)^{\frac{1}{2}} = 4
  3. Find the value of z: (5z+1)12=9(5z + 1)^{\frac{1}{2}} = 9

Working through these problems will give you valuable practice and help you build confidence in your problem-solving abilities. Remember to follow the steps outlined in this guide: isolate the square root, square both sides, isolate the variable, and verify your solution. If you encounter any difficulties, review the steps and explanations provided in this article. Don't be afraid to make mistakes; mistakes are opportunities for learning. The key is to understand where you went wrong and how to correct it. With consistent practice and a clear understanding of the concepts, you'll be able to solve equations with square roots effectively. Mathematics is a journey, and each problem you solve is a step forward. So, keep practicing and keep learning.