Solving the Equation ln(x) - x^2 + 2x = 0 Using the False Position Method

In this article, we will explore the solution to the equation ln(x) - x^2 + 2x = 0 using the False Position Method, also known as the Regula Falsi method. This technique is a numerical method for finding the roots of a function, and it is particularly useful when the function is continuous and has at least one root within the given interval.

The False Position Method is an iterative process that uses two initial guesses, a and b, to find the root of the equation. The method is based on the assumption that the function changes sign between the two guesses, indicating that a root exists within the interval [a, b].

Let's apply the False Position Method to solve the equation ln(x) - x^2 + 2x = 0 for the smallest root, correct to at least two decimal places.

1. Choose the initial guesses: We will start with a = 0.5 and b = 1.5, as these values seem to bracket a root of the equation.

2. Evaluate the function at the initial guesses: Substituting the initial guesses into the equation, we get:

• f(a) = ln(0.5) - 0.5^2 + 2(0.5) = -0.6931 - 0.25 + 1 = -0.0431
• f(b) = ln(1.5) - 1.5^2 + 2(1.5) = 0.4055 - 2.25 + 3 = 1.1555

1. Calculate the new approximation: The False Position Method uses the following formula to calculate the new approximation: x = b - f(b) * (a - b) / (f(a) - f(b)) Substituting the values, we get: x = 1.5 - 1.1555 * (0.5 - 1.5) / (-0.0431 - 1.1555) = 0.8404

2. Check the accuracy: We want the solution to be accurate to at least two decimal places, so we need to check if the absolute value of the difference between the new approximation and the previous approximation is less than or equal to 0.01.

• |x - b| = |0.8404 - 1.5| = 0.6596 > 0.01, so we need to continue the iteration.

1. Update the interval: Since f(a) and f(b) have different signs, the root must lie within the interval [a, b]. We can update the interval as follows:

• a = 0.5
• b = 0.8404

1. Repeat the process: Repeat steps 2-4 with the updated interval.

• f(a) = ln(0.5) - 0.5^2 + 2(0.5) = -0.6931 - 0.25 + 1 = -0.0431
• f(b) = ln(0.8404) - 0.8404^2 + 2(0.8404) = -0.1741 - 0.7065 + 1.6808 = 0.8002
• x = 0.8404 - 0.8002 * (0.5 - 0.8404) / (-0.0431 - 0.8002) = 0.6677
• |x - b| = |0.6677 - 0.8404| = 0.1727 > 0.01, so we need to continue the iteration.

1. Update the interval: a = 0.5, b = 0.6677

2. Repeat the process: Repeat steps 2-4 with the updated interval.

• f(a) = ln(0.5) - 0.5^2 + 2(0.5) = -0.6931 - 0.25 + 1 = -0.0431
• f(b) = ln(0.6677) - 0.6677^2 + 2(0.6677) = -0.4027 - 0.4457 + 1.3354 = 0.4870
• x = 0.6677 - 0.4870 * (0.5 - 0.6677) / (-0.0431 - 0.4870) = 0.5795
• |x - b| = |0.5795 - 0.6677| = 0.0882 > 0.01, so we need to continue the iteration.

1. Update the interval: a = 0.5, b = 0.5795

2. Repeat the process: Repeat steps 2-4 with the updated interval.

   • f(a) = ln(0.5) - 0.5^2 + 2(0.5) = -0.6931 - 0.25 + 1 = -0.0431
   • f(b) = ln(0.5795) - 0.5795^2 + 2(0.5795) = -0.5448 - 0.3359 + 1.1590 = 0.2783
   • x = 0.5795 - 0.2783 * (0.5 - 0.5795) / (-0.0431 - 0.2783) = 0.5450
   • |x - b| = |0.5450 - 0.5795| = 0.0345 < 0.01, so we have found the solution accurate to at least two decimal places.

The smallest root of the equation ln(x) - x^2 + 2x = 0 using the False Position Method is approximately x = 0.55.