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

2 min read Jun 17, 2024

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

This equation, although it looks complex, can be solved using the quadratic formula. Let's break down the steps:

First, we need to expand the given equation:

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

Expanding the products, we get:

x² - ax - bx + ab + x² - bx - cx + bc + x² - cx - ax + ac = 0

Next, we combine like terms:

3x² - 2(a+b+c)x + (ab + ac + bc) = 0

This equation now has the form of a quadratic equation: ax² + bx + c = 0, where:

The quadratic formula is used to solve for x in the equation ax² + bx + c = 0:

x = (-b ± √(b² - 4ac)) / 2a

Substituting the values of a, b, and c from our equation, we get:

x = (2(a+b+c) ± √((-2(a+b+c))² - 4 * 3 * (ab + ac + bc))) / (2 * 3)

After simplifying the expression, we arrive at the solution for x:

x = (a+b+c ± √((a+b+c)² - 3(ab + ac + bc))) / 3

This is the solution for the equation (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0.

Note: The solution for x will depend on the specific values of a, b, and c.