(x^2-x-20)(x^2+2x-3)=(x^2+2x-8)(x^2-8x+15)

Jasonbradley

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Jun 17, 2024

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Solving the Equation: (x^2-x-20)(x^2+2x-3)=(x^2+2x-8)(x^2-8x+15)

This equation involves factoring quadratic expressions on both sides. Let's break down the steps to solve it:

1. Factor each quadratic expression

• Left side:

  • (x^2 - x - 20) factors into (x - 5)(x + 4)
  • (x^2 + 2x - 3) factors into (x + 3)(x - 1)

• Right side:

  • (x^2 + 2x - 8) factors into (x + 4)(x - 2)
  • (x^2 - 8x + 15) factors into (x - 5)(x - 3)

2. Rewrite the equation with factored expressions

The equation now becomes: (x - 5)(x + 4)(x + 3)(x - 1) = (x + 4)(x - 2)(x - 5)(x - 3)

3. Simplify by canceling common factors

Notice that both sides share the factors (x - 5) and (x + 4). Cancelling these out, we get:

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

4. Expand and solve for x

• Expanding both sides: x² + 2x - 3 = x² - 5x + 6
• Combining like terms: 7x = 9
• Solving for x: x = 9/7

Conclusion

Therefore, the solution to the equation (x^2-x-20)(x^2+2x-3)=(x^2+2x-8)(x^2-8x+15) is x = 9/7.