Question

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

Answer 1

Given: The equation  (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0

To find: Solve the given equation.

Solution:

• Now the equation given is (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0

• Expanding it, we get:

                    x² - (a + b)x + ab + x² - (b + c)x + bc + x² - (c + a)x + ca

                    3x² - 2(a + b + c)x + (ab + bc + ca)

• Now discriminant will be:

                    D = b^2 - 4ac = 0

                    D = {2(a + b + c)}² - 4(ab + bc + ca)(3) = 0

                    4{a² + b² + c² + 2(ab + bc + ca)} -12(ab + bc + ca) = 0

                    a² + b² + c² - ab - bc - ca = 0

                    2a² + 2b² + 2c² - 2ab - 2bc - 2ca = 0

                    (a - b)² + (b - c)² + (c - a)² = 0

                    If a = b = c then roots are equal.

Answer:

            a = b = c if roots of given equation are equal.