(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0

3 min read Jun 17, 2024

Solving the Equation: (x-a)(x-b) + (x-b)(x-c) + (x-c)(x-a) = 0

This equation might look intimidating at first glance, but it can be solved using simple algebraic manipulations. Let's break it down step-by-step:

Expanding the Equation

First, we expand the equation by multiplying out the terms:

(x-a)(x-b) + (x-b)(x-c) + (x-c)(x-a) = 0
x² - ax - bx + ab + x² - bx - cx + bc + x² - cx - ax + ac = 0

Simplifying the Equation

Now, we combine like terms:

3x² - 2(a+b+c)x + (ab + bc + ac) = 0

Solving for x

The equation is now in a quadratic form (ax² + bx + c = 0). We can solve for x using the quadratic formula:

x = (-b ± √(b² - 4ac)) / 2a

In this case:

a = 3
b = -2(a+b+c)
c = ab + bc + ac

Substituting these values into the quadratic formula gives us the solutions for x.

Interpretation of the Solution

The solutions for x represent the points where the expression (x-a)(x-b) + (x-b)(x-c) + (x-c)(x-a) equals zero. This could be interpreted graphically as the x-intercepts of the function represented by this expression.

Conclusion

Solving the equation (x-a)(x-b) + (x-b)(x-c) + (x-c)(x-a) = 0 involves expanding, simplifying, and then applying the quadratic formula. This process demonstrates the power of basic algebraic techniques in handling complex-looking equations. The solutions for x provide valuable information about the behavior of the function represented by the expression.