Solving Equations With Tables: A Step-by-Step Guide

Hey everyone! Today, we're diving into a cool math problem where we'll approximate the solution to an equation using a table of values. Sounds fun, right? The equation we're tackling is: $2 \sqrt{x-1}+2=\frac{3 x}{\pi-1}$. We'll go through the process step-by-step, making sure it's easy to follow. Our goal is to find the approximate solution to the nearest quarter of a unit, just like the multiple-choice options we're given. So, let's break it down and get started! This is a great skill to have, as it helps visualize and understand how equations behave. Let's start with the basics.

Understanding the Problem and Setting Up the Table

First off, let's understand what the question is asking. We're given an equation, and we need to find the value of x that makes the equation true. But, we're not going to solve it algebraically (you know, isolating x). Instead, we'll use a table of values to estimate the solution. Think of it like this: we'll plug in different x values and see which one gets us closest to making the equation balance. The options we are given are: A. $x \approx 2.5$, B. $x \approx 3$, C. $x \approx 4.75$, and D. $x \approx 2.75$. Now we have a clear idea. First, we need to pick x values to put into our table. Since we're approximating to the nearest quarter of a unit, it's smart to start with the provided options. Then, we calculate the left-hand side (LHS) and the right-hand side (RHS) of the equation for each x value. The solution is where the LHS and RHS are approximately equal. Let's start building our table. We'll include the x values from the multiple-choice answers in our table to make things easy. The LHS of the equation is $2 \sqrt{x-1}+2$, and the RHS is $\frac{3 x}{\pi-1}$. We are looking for an x value where the LHS equals the RHS.

Now, let's populate the table. We'll start with the values from the answer choices: 2.5, 3, 4.75, and 2.75. For each x value, we'll calculate the LHS and RHS separately and fill in the table. This is the heart of the method. Guys, this is where we actually do the math! For x = 2.5, we'll calculate the LHS: $2 \sqrt{2.5-1} + 2 = 2 \sqrt{1.5} + 2 \approx 4.45$. For the RHS: $\frac{3 * 2.5}{\pi-1} \approx 2.62$. For x = 3, we calculate the LHS: $2 \sqrt{3-1} + 2 = 2 \sqrt{2} + 2 \approx 4.83$. For the RHS: $\frac{3 * 3}{\pi-1} \approx 3.94$. Next, for x = 4.75, LHS: $2 \sqrt{4.75-1} + 2 = 2 \sqrt{3.75} + 2 \approx 5.87$. For the RHS: $\frac{3 * 4.75}{\pi-1} \approx 6.26$. Finally, for x = 2.75, LHS: $2 \sqrt{2.75-1} + 2 = 2 \sqrt{1.75} + 2 \approx 4.64$. RHS: $\frac{3 * 2.75}{\pi-1} \approx 3.61$. The table will help us see the trend and make an informed decision.

Performing the Calculations and Filling in the Table

Alright, let's get those calculators fired up and compute the values for each side of the equation. This is where precision matters, so double-check your calculations! Remember, the goal is to get the values for both sides of the equation as close as possible. This involves substituting the x value into both sides of the original equation and simplifying. Accuracy is key. When calculating the LHS, be extra careful with the square root. For x = 2.5, the LHS is $2\sqrt{2.5-1} + 2 \approx 4.45$, and the RHS is $\frac{3 * 2.5}{\pi-1} \approx 2.62$. Notice that the LHS and RHS are not equal. Now, for x = 3, LHS is $2\sqrt{3-1} + 2 \approx 4.83$, and RHS is $\frac{3 * 3}{\pi-1} \approx 3.94$. These values are also not equal. The next one is x = 4.75, where the LHS is $2\sqrt{4.75-1} + 2 \approx 5.87$, and the RHS is $\frac{3 * 4.75}{\pi-1} \approx 6.26$. These are also not equal, but are closer to each other than the previous examples. Finally, for x = 2.75, we have LHS is $2\sqrt{2.75-1} + 2 \approx 4.64$, and the RHS is $\frac{3 * 2.75}{\pi-1} \approx 3.61$. This isn't equal either. The table will clearly display the results. We will have four rows, one for each x value, showing the LHS and RHS results side by side. By comparing the two values for each x, we can identify which one brings us closest to a solution.

Here’s what our table might look like:

Analyzing the Results and Approximating the Solution

Now, let's analyze the table. We're looking for an x value where the LHS and RHS are closest to each other. Comparing the results in the table, we'll see which x value gives us the closest match between the LHS and RHS. The goal is to find the x value where the difference between the LHS and RHS is the smallest. Look at each row, and compare the two values. For x = 2.5, the difference is about 1.83. For x = 3, the difference is about 0.89. For x = 4.75, the difference is about 0.39. For x = 2.75, the difference is about 1.03. We're aiming for the point where the LHS and RHS are as close as possible. It is pretty clear from the table that the x value that makes the LHS and RHS closest is x = 4.75. So, looking at the table, we can see that when x = 4.75, the LHS and RHS are the closest, with values of approximately 5.87 and 6.26, respectively. Therefore, we can say that the solution to the equation, approximated to the nearest quarter of a unit, is x ≈ 4.75. Guys, the best part of this method is that it visually represents the equation. Understanding how to use a table of values is a valuable tool for solving and understanding equations, especially when dealing with square roots and other more complex functions. In this case, option C, x ≈ 4.75, looks like the most accurate approximation. We have now successfully solved the problem using a table of values! We've systematically gone through each step, making sure to show every calculation and explaining why each step is important. By comparing the LHS and RHS values for each x, we could identify the closest match and find our approximate solution.

So, the answer is C. $x \approx 4.75$. Congratulations, you have successfully used a table of values to approximate the solution to the equation! This method is a great way to visualize and understand how equations work. Keep practicing, and you'll become a pro at this. Remember, the key is to be organized, methodical, and pay close attention to your calculations. With a little practice, you'll be able to solve these types of problems with confidence.

Why this method is useful

Using a table of values is super useful for several reasons. Firstly, it provides a visual representation of the equation. You can actually see how the LHS and RHS change as x changes. This is incredibly helpful in understanding the behavior of the equation, especially when it involves more complex functions. Secondly, it's a great approximation technique. Sometimes, you might not be able to solve an equation directly, but using a table, you can get a pretty close estimate of the solution. Thirdly, it builds your problem-solving skills. It encourages you to think systematically, break down the problem into smaller parts, and use logical reasoning. Finally, it's a good way to check your answers. If you've solved an equation algebraically, you can use a table to verify your solution. So, knowing how to create and use a table of values is a valuable skill in your math toolkit.