The roots of the equation $a(b-c)x^{2}+b(c-a)x+c(a-b)=0$ are
- $1, \displaystyle \frac{c(a-b)}{a(b-c)}$
- $1, \displaystyle \frac{b(c-a)}{a(b-c)}$
- $\displaystyle \frac{c(a-b)}{a(b-c)}, \frac{b(c-a)}{a(b-c)}$
- $a, \displaystyle \frac{c(a-b)}{a(b-c)}$
The equation is a(b-c)x^2 + b(c-a)x + c(a-b) = 0. Notice that the sum of the coefficients is a(b-c) + b(c-a) + c(a-b) = ab - ac + bc - ba + ca - cb = 0. If the sum of coefficients is zero, then x = 1 is always a root. Using the product of roots formula (c/a), the other root is c(a-b) / a(b-c).
By direct substitution, if we place x = 1 into the expression a(b-c)x^2 + b(c-a)x + c(a-b), we get a(b-c) + b(c-a) + c(a-b). Expanding this gives ab - ac + bc - ab + ac - bc, which perfectly cancels out to 0, confirming 1 is a root. For a quadratic equation Ax^2 + Bx + C = 0 where A = a(b-c), B = b(c-a) and C = c(a-b), the product of the roots is C/A. Therefore, the second root is c(a-b) / a(b-c).