(x-2)^x-3(x+1)(x-1)+6x^2=5

3 min read Jun 17, 2024

Solving the Equation: (x-2)^x - 3(x+1)(x-1) + 6x^2 = 5

This equation presents a unique challenge due to the presence of a variable exponent in the first term. Let's break down the steps to solve it:

Simplifying the Equation

Expand the products:
- (x+1)(x-1) = x^2 - 1
- 3(x+1)(x-1) = 3(x^2 - 1) = 3x^2 - 3
Combine terms:
- The equation now becomes: (x-2)^x - 3x^2 + 3 + 6x^2 = 5
- Simplifying further: (x-2)^x + 3x^2 + 3 = 5
Isolate the exponential term:
- (x-2)^x = 2 - 3x^2

Addressing the Variable Exponent

Unfortunately, there's no direct algebraic method to solve equations with variable exponents. Here are some approaches:

1. Numerical Methods:

Graphing: Plot the left-hand side and right-hand side of the equation as separate functions. The points where the graphs intersect represent the solutions.
Iteration Methods: Techniques like Newton-Raphson can be used to iteratively approximate the solution.

2. Approximations:

Trial and Error: Substitute different values of x and observe if the equation holds true. This is not very efficient but can provide a general idea of the solution range.
Linearization: For specific values of x, the exponential term can be approximated using Taylor series expansion, allowing for a more manageable equation.

Example: Using Graphing

We can visualize the equation by plotting the functions:

f(x) = (x-2)^x
g(x) = 2 - 3x^2

The intersection points of these graphs will represent the solutions to the original equation.

Note:

It's important to remember that the above approaches will provide approximate solutions. The accuracy of the solution depends on the chosen method and the desired level of precision.

Additionally, the equation might have multiple solutions, depending on the behavior of the functions. Therefore, exploring different methods and using a combination of techniques can be helpful in identifying all possible solutions.