(x-b)%2B(x-b)(x-c)%2B(x-c)(x-a)%3D0-solve-by-quadratic-formula.jpeg "(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 Solve By Quadratic Formula")
Solving the Equation (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 using the Quadratic Formula
This article will guide you through solving the equation (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 using the quadratic formula.
Expanding the Equation
First, we need to expand the equation to obtain a standard quadratic form:
(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0
Expanding each product:
- x² - ax - bx + ab
- x² - bx - cx + bc
- x² - cx - ax + ac
Combining like terms:
3x² - 2(a+b+c)x + (ab+bc+ac) = 0
Applying the Quadratic Formula
Now, we have a quadratic equation in the form ax² + bx + c = 0. We can apply the quadratic formula to solve for x:
x = (-b ± √(b² - 4ac)) / 2a
In our equation:
- a = 3
- b = -2(a+b+c)
- c = ab+bc+ac
Substitute these values into the quadratic formula:
x = (2(a+b+c) ± √((-2(a+b+c))² - 4 * 3 * (ab+bc+ac))) / (2 * 3)
Simplifying the Solution
After simplification, we arrive at the solution:
x = (a+b+c ± √((a+b+c)² - 3(ab+bc+ac))) / 3
Conclusion
By expanding the equation and applying the quadratic formula, we have successfully solved the equation (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0 for x. The solution provides us with the two possible values of x that satisfy the given equation.