(2x-1)^2+(x+3)^2-5(x+7)(x-7)=0

2 min read Jun 16, 2024

Solving the Quadratic Equation: (2x-1)^2 + (x+3)^2 - 5(x+7)(x-7) = 0

This article will guide you through the steps of solving the quadratic equation (2x-1)^2 + (x+3)^2 - 5(x+7)(x-7) = 0. Let's break down the process:

1. Expanding the Equation:

First, we need to expand the equation to get rid of the parentheses.

(2x-1)^2: This expands to (2x-1)(2x-1) = 4x^2 - 4x + 1
(x+3)^2: This expands to (x+3)(x+3) = x^2 + 6x + 9
5(x+7)(x-7): This is a difference of squares pattern, resulting in 5(x^2 - 49) = 5x^2 - 245

Now, our equation becomes: 4x^2 - 4x + 1 + x^2 + 6x + 9 - 5x^2 + 245 = 0

2. Simplifying the Equation:

Combining like terms, we get:

2x + 255 = 0

3. Solving for x:

Subtract 255 from both sides: 2x = -255
Divide both sides by 2: x = -127.5

Therefore, the solution to the quadratic equation (2x-1)^2 + (x+3)^2 - 5(x+7)(x-7) = 0 is x = -127.5.

Conclusion:

By expanding, simplifying, and applying basic algebraic operations, we have successfully solved the given quadratic equation. This process demonstrates the power of algebraic manipulation in simplifying complex equations and finding their solutions.