(x^2-25)^2+(x^2+3x-10)^2=0

3 min read Jun 17, 2024

Solving the Equation: (x^2-25)^2 + (x^2+3x-10)^2 = 0

This equation presents a unique challenge due to its structure involving squares. Let's break down the steps to solve it:

Understanding the Properties of Squares

The key observation here is that squares of real numbers are always non-negative. This means:

(x^2 - 25)^2 ≥ 0 for all real values of x
(x^2 + 3x - 10)^2 ≥ 0 for all real values of x

Analyzing the Equation

The equation (x^2 - 25)^2 + (x^2 + 3x - 10)^2 = 0 states that the sum of two non-negative squares is zero. This can only be true if both squares are simultaneously equal to zero.

Solving for x

Therefore, we have two separate equations to solve:

(x^2 - 25)^2 = 0
- Taking the square root of both sides gives us x^2 - 25 = 0
- Factoring the difference of squares yields (x + 5)(x - 5) = 0
- This leads to solutions: x = -5 and x = 5
(x^2 + 3x - 10)^2 = 0
- Similar to the first equation, taking the square root gives us x^2 + 3x - 10 = 0
- Factoring the quadratic equation yields (x + 5)(x - 2) = 0
- This leads to solutions: x = -5 and x = 2

The Final Solution

We have found that the solutions for x are -5, 5, and 2. However, x = -5 is the only solution that satisfies both equations simultaneously. Therefore, the solution to the original equation is x = -5.

Conclusion

The equation (x^2 - 25)^2 + (x^2 + 3x - 10)^2 = 0 has a single solution, x = -5. This solution is obtained by understanding the non-negative nature of squares and applying the principle that the sum of two non-negative squares can only be zero if both squares are individually zero.