(x-1)(x-2)+(x+4)(x-4)+3x=0

2 min read Jun 17, 2024
(x-1)(x-2)+(x+4)(x-4)+3x=0

Solving the Quadratic Equation: (x-1)(x-2)+(x+4)(x-4)+3x=0

This article will guide you through solving the quadratic equation (x-1)(x-2)+(x+4)(x-4)+3x=0. We will use algebraic manipulation to simplify the equation and find the solutions for x.

Expanding and Simplifying the Equation

First, we need to expand the products in the equation:

  • (x-1)(x-2) = x² - 3x + 2
  • (x+4)(x-4) = x² - 16

Now, let's substitute these expansions back into the original equation:

x² - 3x + 2 + x² - 16 + 3x = 0

Combining like terms, we get:

2x² - 14 = 0

Solving for x

Now, we have a simplified quadratic equation. We can solve for x using a few methods:

1. Factoring:

  • Divide both sides by 2: x² - 7 = 0
  • Add 7 to both sides: x² = 7
  • Take the square root of both sides: x = ±√7

2. Quadratic Formula:

  • The quadratic formula is used to solve for x in any equation of the form ax² + bx + c = 0: x = (-b ± √(b² - 4ac)) / 2a
  • In our equation, a = 2, b = 0, and c = -14.
  • Substitute these values into the formula: x = (0 ± √(0² - 4 * 2 * -14)) / (2 * 2) x = ±√(112) / 4 x = ±√(16 * 7) / 4 x = ±4√7 / 4 x = ±√7

Solutions

Therefore, the solutions to the quadratic equation (x-1)(x-2)+(x+4)(x-4)+3x=0 are:

x = √7 and x = -√7

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