If each pair of the following three equations $ { x }^{ 2 }+ax+b=0$, ${ x }^{ 2 }+cx+d=0$, ${ x }^{ 2 }+ex+f=0$ has exactly one root in common, then
- $ \left( { a+c+e } \right) ^{ 2 }=3\left( ac+ce+ea-b-d-f \right) $
- $ \left( { a+c+e } \right) ^{ 2 }=8\left( ac+ce+ea-b-d-f \right) $
- $ \left( { a+c+e } \right) ^{ 3 }=4\left( ac+ce+ea-b-d-f \right) $
- $ \left( { a+c+e } \right) ^{ 2 }=4\left( ac+ce+ea-b-d-f \right) $
If each pair has a common root, let the roots be alpha, beta, gamma. The equations are (x-alpha)(x-beta)=0, (x-beta)(x-gamma)=0, (x-gamma)(x-alpha)=0. Expanding these gives coefficients related to the sums and products of roots. The condition leads to the identity (a+c+e)^2 = 4(ac+ce+ea-b-d-f).
Let alpha, beta, and gamma be the distinct common roots for the three pairs of equations. We can express the polynomials as (x - alpha)(x - beta) = 0, (x - beta)(x - gamma) = 0, and (x - gamma)(x - alpha) = 0. Expanding these gives x^2 - (alpha + beta)x + alpha*beta = 0, and similarly for the others. By matching coefficients, we find a = -(alpha + beta), c = -(beta + gamma), and e = -(gamma + alpha). We then evaluate (a + c + e)^2 as (-2(alpha + beta + gamma))^2, which equals 4 times (alpha^2 + beta^2 + gamma^2 + alpha*beta + beta*gamma + gamma*alpha). We also evaluate the right side expression ac + ce + ea - b - d - f, which similarly simplifies to exactly the same expression, proving the identity (a + c + e)^2 = 4(ac + ce + ea - b - d - f).