(x^2+6x-7)·(2x^2-5x-3)=0

2 min read Jun 17, 2024
(x^2+6x-7)·(2x^2-5x-3)=0

Solving the Equation (x^2 + 6x - 7) · (2x^2 - 5x - 3) = 0

This equation involves a product of two quadratic expressions equaling zero. The key to solving this type of equation lies in the Zero Product Property, which states that if the product of two or more factors is zero, then at least one of the factors must be zero.

1. Factor each quadratic expression:

  • (x^2 + 6x - 7) This expression can be factored into: (x + 7)(x - 1)

  • (2x^2 - 5x - 3) This expression can be factored into: (2x + 1)(x - 3)

2. Apply the Zero Product Property:

Now we have the equation: (x + 7)(x - 1)(2x + 1)(x - 3) = 0

For this product to be zero, at least one of the factors must be zero. Therefore, we set each factor equal to zero and solve for x:

  • x + 7 = 0 => x = -7
  • x - 1 = 0 => x = 1
  • 2x + 1 = 0 => x = -1/2
  • x - 3 = 0 => x = 3

3. Solutions:

The solutions to the equation (x^2 + 6x - 7) · (2x^2 - 5x - 3) = 0 are:

  • x = -7
  • x = 1
  • x = -1/2
  • x = 3

These are the values of x that make the original equation true.

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