If $x^2+ax+bc=0$ and $x^2+bx+ca=0$ $(c\neq 0)$ have a common root. Then other root satisfies the equation
- $x^2+cx+ab=0$
- $x^2+ax+bc=0$
- $x^2+bc+ac=0$
- $x^2+abx+c=0$
Let alpha be the common root of x^2 + ax + bc = 0 and x^2 + bx + ca = 0, and let beta be the other root of the first equation while gamma is the other root of the second equation. By Vieta's formulas for the first equation, alpha + beta = -a and alpha times beta = bc. For the second equation, alpha + gamma = -b and alpha times gamma = ca. Subtracting the two equations gives (alpha + beta) - (alpha + gamma) = -a - (-b), which simplifies to beta - gamma = b - a. From the product of the roots, beta = bc divided by alpha and gamma = ca divided by alpha. Substituting these into the difference yields (bc - ca) divided by alpha = b - a. Factoring out c gives c(b - a) divided by alpha = b - a. Since a and b are distinct, b - a is not equal to zero, leaving alpha = c. Replacing alpha with c in the sum of the roots for the first equation gives beta = -a - c. Using Vieta's formulas, the sum of the new roots for the required quadratic is beta + gamma = -c and the product is beta times gamma = ab. Therefore, the required equation is x^2 + cx + ab = 0.