Solving the Quadratic Equation 2x 3^2 25: A Comprehensive Guide

Introduction to Quadratic Equations

A quadratic equation is any equation that can be expressed in the form (ax^2 bx c 0), where (a), (b), and (c) are constants and (a eq 0). This guide focuses on a specific example: 2x 3^2 25. We will explore different methods to solve this type of equation.

Step-by-Step Solution

Let's begin by solving the equation 2x 3^2 25. First, we recognize that this equation can be simplified into a standard quadratic form.

Square Root Method

One approach is to use the square root method. Here are the steps:

Square both sides of the equation 2x 3^2 25. 2x 3 ±√25 2x 3 ±5

Now, we solve for x in both cases.

Case 1

For 2x 3 5:

2x 8 x 4

Case 2

For 2x 3 -5:

2x -8 x -4

Therefore, the solutions are x 4 and x -4. However, upon re-evaluating the original equation, we realize that x 4 is not a valid solution. Let's correct this by revisiting the equation and ensuring each step is strictly followed.

Revisiting the Equation

The equation 2x 3^2 25 can be rewritten and solved step-by-step as follows:

2x 3^2 25

Take the square root of both sides:

(sqrt{2x cdot 3^2} sqrt{25}) 2x 3 ±5

Now, we solve for x in both cases.

Case 1

For 2x 3 5:

2x 2 x 1

Case 2

For 2x 3 -5:

2x -8 x -4

The solutions are x 1 and x -4. These are the correct and valid solutions.

Diagrammatic Representation

For a better understanding, a diagram can be drawn to illustrate the process:

Draw the initial equation 2x 3^2 25. Decompose the equation into 2x 3 ±5. Solve for x in both cases using the values ±5. Mark the final solutions on a number line: x 1 and x -4.

Conclusion and Final Answer

The final valid solutions for the equation 2x 3^2 25 are x 1 and x -4. This method demonstrates the importance of careful algebraic manipulation and revising intermediate steps to ensure accuracy.