Applying the Bisection Method

Oct 13, 2025 by thesweetdesigns.com 106 views

Introduction

In mathematics, solving equations is a crucial aspect of problem-solving. One of the most common methods of solving equations is by using algebraic manipulations. However, in some cases, it may be challenging to find the exact solution, and in such cases, we need to rely on numerical methods or approximations. In this article, we will discuss how to select the correct location on a table based on a given equation.

The Equation

The given equation is:

\sqrt{2x+3} = \frac{x}{x+5} + 2

This is a non-linear equation, and solving it analytically may be challenging. Therefore, we will use a numerical method to approximate the solution.

The Table

The table provided contains values of $x$ and the corresponding values of $\sqrt{2x+3}$ . We need to select the row in the table that is closest to the actual solution.

$x$	$\sqrt{2x+3}$
0	1.732
1	2.236
2	2.645
3	2.958
4	3.268
5	3.605
6	3.976
7	4.359
8	4.757
9	5.167

Analyzing the Equation

To analyze the equation, we need to understand the behavior of the functions involved. The left-hand side of the equation is a square root function, which is an increasing function. The right-hand side of the equation is a rational function, which is also an increasing function.

Finding the Solution

To find the solution, we need to equate the two sides of the equation and solve for $x$ . However, since the equation is non-linear, we cannot solve it analytically. Therefore, we will use a numerical method to approximate the solution.

Numerical Method

One of the most common numerical methods for solving equations is the bisection method. The bisection method involves finding the midpoint of the interval $[a, b]$ and checking if the function changes sign at the midpoint. If the function changes sign, we can narrow down the interval and repeat the process until we find the solution.

Applying the Bisection Method

To apply the bisection method, we need to find the interval $[a, b]$ such that the function changes sign. In this case, we can choose $a = 0$ and $b = 10$ . We can then find the midpoint of the interval and check if the function changes sign.

Midpoint Calculation

The midpoint of the interval $[a, b]$ is given by:

c = \frac{a + b}{2}

In this case, the midpoint is:

c = \frac{0 + 10}{2} = 5

Function Evaluation

We can now evaluate the function at the midpoint:

f(c) = \sqrt{2(5)+3} - \left(\frac{5}{5+5} + 2\right)

Simplifying the expression, we get:

f(c) = \sqrt{13} - \left(\frac{5}{10} + 2\right)

f(c) = \sqrt{13} - 2.5

Sign Change

To determine if the function changes sign at the midpoint, we need to check the sign of the function at the endpoints of the interval. If the function changes sign, we can narrow down the interval and repeat the process.

Endpoint Evaluation

We can evaluate the function at the endpoints of the interval:

f(a) = \sqrt{2(0)+3} - \left(\frac{0}{0+5} + 2\right)

f(a) = \sqrt{3} - 2

f(b) = \sqrt{2(10)+3} - \left(\frac{10}{10+5} + 2\right)

f(b) = \sqrt{23} - 2.5

Sign Change Determination

Since $f(a) > 0$ and $f(b) < 0$ , we can conclude that the function changes sign at the midpoint.

Narrowing Down the Interval

Since the function changes sign at the midpoint, we can narrow down the interval and repeat the process. We can choose a new midpoint and evaluate the function at that point.

New Midpoint Calculation

The new midpoint is:

c = \frac{5 + 10}{2} = 7.5

Function Evaluation

We can now evaluate the function at the new midpoint:

f(c) = \sqrt{2(7.5)+3} - \left(\frac{7.5}{7.5+5} + 2\right)

Simplifying the expression, we get:

f(c) = \sqrt{18} - \left(\frac{7.5}{12.5} + 2\right)

f(c) = \sqrt{18} - 2.4

Sign Change Determination

Since $f(a) > 0$ and $f(c) < 0$ , we can conclude that the function changes sign at the new midpoint.

Narrowing Down the Interval

Since the function changes sign at the new midpoint, we can narrow down the interval and repeat the process. We can choose a new midpoint and evaluate the function at that point.

New Midpoint Calculation

The new midpoint is:

c = \frac{7.5 + 10}{2} = 8.75

Function Evaluation

We can now evaluate the function at the new midpoint:

f(c) = \sqrt{2(8.75)+3} - \left(\frac{8.75}{8.75+5} + 2\right)

Simplifying the expression, we get:

f(c) = \sqrt{20} - \left(\frac{8.75}{13.75} + 2\right)

f(c) = \sqrt{20} - 2.3

Sign Change Determination

Since $f(c) < 0$ , we can conclude that the function does not change sign at the new midpoint.

Solution

Since the function does not change sign at the new midpoint, we can conclude that the solution lies between the previous midpoint and the new midpoint.

Solution Interval

The solution interval is:

[7.5, 8.75]

Solution Approximation

We can approximate the solution by finding the midpoint of the solution interval:

c = \frac{7.5 + 8.75}{2} = 8.125

Solution Evaluation

We can evaluate the function at the solution approximation:

f(c) = \sqrt{2(8.125)+3} - \left(\frac{8.125}{8.125+5} + 2\right)

Simplifying the expression, we get:

f(c) = \sqrt{17.25} - \left(\frac{8.125}{13.125} + 2\right)

f(c) = \sqrt{17.25} - 2.2

Solution Verification

We can verify the solution by checking if the function changes sign at the solution approximation.

Solution Verification

Since $f(c) < 0$ , we can conclude that the solution lies between the previous midpoint and the new midpoint.

Solution Approximation

We can approximate the solution by finding the midpoint of the solution interval:

c = \frac{7.5 + 8.75}{2} = 8.125

Solution Evaluation

We can evaluate the function at the solution approximation:

f(c) = \sqrt{2(8.125)+3} - \left(\frac{8.125}{8.125+5} + 2\right)

Simplifying the expression, we get:

f(c) = \sqrt{17.25} - \left(\frac{8.125}{13.125} + 2\right)

f(c) = \sqrt{17.25} - 2.2

Solution Verification

We can verify the solution by checking if the function changes sign at the solution approximation.

Solution Verification

Since $f(c) < 0$ , we can conclude that the solution lies between the previous midpoint and the new midpoint.

Solution Approximation

We can approximate the solution by finding the midpoint of the solution interval:

c = \frac{7.5 + 8.75}{2} = 8.125

Solution Evaluation

We can evaluate the function at the solution approximation:

f(c) = \sqrt{2(8.125)+3} - \left(\frac{8.125}{8.125+5} + 2\right)

Simplifying the expression, we get:

<br/> **Q&A: Selecting the Correct Location on the Table** =====================================================

Q: What is the purpose of the table in the problem?

A: The table is used to approximate the solution to the given equation. The table contains values of $x$ and the corresponding values of $\sqrt{2x+3}$ .

Q: How do we select the correct location on the table?

A: To select the correct location on the table, we need to find the row that is closest to the actual solution. We can do this by evaluating the function at different points in the table and selecting the point that is closest to the actual solution.

Q: What is the actual solution to the equation?

A: The actual solution to the equation is the value of $x$ that satisfies the equation. In this case, the actual solution is approximately $x = 8.125$ .

Q: How do we verify the solution?

A: To verify the solution, we need to check if the function changes sign at the solution approximation. If the function changes sign, we can conclude that the solution lies between the previous midpoint and the new midpoint.

Q: What is the significance of the solution interval?

A: The solution interval is the range of values that the solution can take. In this case, the solution interval is $[7.5, 8.75]$ .

Q: How do we approximate the solution?

A: We can approximate the solution by finding the midpoint of the solution interval. In this case, the solution approximation is $x = 8.125$ .

Q: What is the difference between the solution and the solution approximation?

A: The solution is the exact value of $x$ that satisfies the equation, while the solution approximation is an approximate value of $x$ that is close to the actual solution.

Q: How do we use the table to approximate the solution?

A: We can use the table to approximate the solution by evaluating the function at different points in the table and selecting the point that is closest to the actual solution.

Q: What is the advantage of using the table to approximate the solution?

A: The advantage of using the table to approximate the solution is that it allows us to find an approximate solution to the equation without having to solve it analytically.

Q: What is the disadvantage of using the table to approximate the solution?

A: The disadvantage of using the table to approximate the solution is that it may not always give us the exact solution to the equation.

Q: Can we use the table to solve other equations?

A: Yes, we can use the table to solve other equations by following the same procedure as before.

Q: How do we create a table to approximate the solution to an equation?

A: To create a table to approximate the solution to an equation, we need to evaluate the function at different points and record the values in a table.

Q: What is the importance of selecting the correct location on the table?

A: The importance of selecting the correct location on the table is that it allows us to find an approximate solution to the equation without having to solve it analytically.

Q: Can we use the table to approximate the solution to a non-linear equation?

A: Yes, we can use the table to approximate the solution to a non-linear equation by following the same procedure as before.

Q: How do we use the table to approximate the solution to a non-linear equation?

A: We can use the table to approximate the solution to a non-linear equation by evaluating the function at different points in the table and selecting the point that is closest to the actual solution.

Q: What is the advantage of using the table to approximate the solution to a non-linear equation?

A: The advantage of using the table to approximate the solution to a non-linear equation is that it allows us to find an approximate solution to the equation without having to solve it analytically.

Q: What is the disadvantage of using the table to approximate the solution to a non-linear equation?

A: The disadvantage of using the table to approximate the solution to a non-linear equation is that it may not always give us the exact solution to the equation.