(x^2-20)(x^2-15)(x^2-10)(x^2-5) 0

Jasonbradley

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Jun 17, 2024

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Solving the Equation: (x^2 - 20)(x^2 - 15)(x^2 - 10)(x^2 - 5) = 0

This equation presents a unique challenge: it's a quartic equation (highest power of x is 4) but with a specific form that allows for a relatively straightforward solution. Let's break down the process:

Understanding the Equation

The equation consists of four factors, each representing a quadratic expression in the form (x^2 - a). For the entire product to equal zero, at least one of the factors must equal zero.

Solving for x

1. Set each factor equal to zero:

   • x^2 - 20 = 0
   • x^2 - 15 = 0
   • x^2 - 10 = 0
   • x^2 - 5 = 0

2. Solve for x^2 in each equation:

   • x^2 = 20
   • x^2 = 15
   • x^2 = 10
   • x^2 = 5

3. Take the square root of both sides of each equation (remembering both positive and negative roots):

   • x = ±√20
   • x = ±√15
   • x = ±√10
   • x = ±√5

The Solutions

Therefore, the solutions to the equation (x^2 - 20)(x^2 - 15)(x^2 - 10)(x^2 - 5) = 0 are:

• x = √20
• x = -√20
• x = √15
• x = -√15
• x = √10
• x = -√10
• x = √5
• x = -√5

Important Note: These solutions can be expressed in simplified radical form (e.g., √20 = 2√5) if desired.

Conclusion

By setting each factor of the equation to zero and solving the resulting quadratic equations, we successfully find all eight solutions to the equation. This process highlights the power of factoring and the principle that for a product to be zero, at least one of its factors must be zero.