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(a-5)^2+(b-c)^2+(c-d)^2+(b+c+d-9)^2=0

3 min read Jun 16, 2024
(a-5)^2+(b-c)^2+(c-d)^2+(b+c+d-9)^2=0

Exploring the Equation (a-5)^2 + (b-c)^2 + (c-d)^2 + (b+c+d-9)^2 = 0

This equation represents a unique mathematical problem where we are asked to find the values of a, b, c, and d that satisfy the given condition. Let's break down why this is interesting and how we can solve it.

Understanding the Equation

The equation involves the sum of four squared terms. The key observation is that the square of any real number is always greater than or equal to zero. This means each individual term in the equation:

  • (a-5)^2
  • (b-c)^2
  • (c-d)^2
  • (b+c+d-9)^2

must be greater than or equal to zero.

Solving for the Variables

For the entire equation to equal zero, each individual term must be equal to zero. This is because the sum of non-negative numbers can only be zero if each of those numbers is zero.

Therefore, we can set each term equal to zero and solve:

  1. (a-5)^2 = 0 => a - 5 = 0 => a = 5

  2. (b-c)^2 = 0 => b - c = 0 => b = c

  3. (c-d)^2 = 0 => c - d = 0 => c = d

  4. (b+c+d-9)^2 = 0 => b + c + d - 9 = 0

Since we know b = c and c = d, we can substitute these into the fourth equation:

b + b + b - 9 = 0 => 3b - 9 = 0 => 3b = 9 => b = 3

Therefore, we have:

  • a = 5
  • b = 3
  • c = 3
  • d = 3

Conclusion

The equation (a-5)^2 + (b-c)^2 + (c-d)^2 + (b+c+d-9)^2 = 0 has only one solution, which is a = 5, b = c = d = 3. This solution arises from the fundamental property of squares being non-negative, leading us to solve for each term individually.

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