(x^2-4)^2+(x^2-3x-10)^2=0

2 min read Jun 17, 2024

Solving the Equation: (x^2-4)^2 + (x^2-3x-10)^2 = 0

This equation presents an interesting challenge because it involves squares of expressions. Let's explore how to find its solutions.

Understanding the Equation

The equation (x^2-4)^2 + (x^2-3x-10)^2 = 0 has a key characteristic: The sum of two squares equals zero. Remember that the square of any real number is always non-negative (zero or positive). This leads us to a crucial observation:

For the sum of two squares to equal zero, both squares must be individually equal to zero.

Solving for x

Let's apply this logic to our equation:

Set each squared term to zero:
- (x^2 - 4)^2 = 0
- (x^2 - 3x - 10)^2 = 0
Solve each equation individually:
- (x^2 - 4) = 0 => x^2 = 4 => x = ±2
- (x^2 - 3x - 10) = 0 => (x - 5)(x + 2) = 0 => x = 5 or x = -2
Combine the solutions: The solutions to the original equation are x = 2, x = -2, and x = 5.

Verification

We can verify our solutions by plugging them back into the original equation. We'll find that all three solutions satisfy the equation.

Conclusion

By understanding the properties of squares and applying a systematic approach, we were able to solve the equation (x^2-4)^2 + (x^2-3x-10)^2 = 0 and find its three distinct solutions: x = 2, x = -2, and x = 5.