If each pair of the three equations `x^(2)+ax+b=0, x^(2)+cx+d=0` and `x^(2)+ex+f=0` has exactly one root in common then show that `(a+c+e)^(2)=4(ac+ce+ea-b-d-f)`

Explanation

Given equations are
`x^(2)+ax+b=0`……….i
`x^(2)+cx+d=0`…………..ii
`x^(2)+ex+f=0`………….iii
Let `alpha, beta` be the roots of Eq. (i) `beta, gamma` be the roots of Eq. ii and `gamma, delta` be the roots of Eq. iii then
`alpha+beta=-a, alpha beta=b`.........iv ltbr `beta+gamma-c, beta gamma=d`.........v
`gamma+alpha=-e,gamma alpha=f`...........vi
`:.LHS=(a+c+e)^(2)=(-alpha-beta-beta-gamma-gamma-alpha)^(2)`
[from eqs iv , v and vi]
`=4(alpha+beta+gamma)^(2)`........vii
`RHS=4(ac+ce+ea-b-d-f)`
`=4{(alpha+beta)(beta+gamma)+beta+gamma)(gamma+alpha)+(gamma+alpha)`
`(alpha+beta)-alpha beta-beta gamma-gamma alpha)}`
[from eqs iv, v and vi]
`=4(alpha^(2)+beta^(2)+gamma^(2)+2alpha beta+2beta tamma+2 gamma alpha)`
`=4(alpha+beta+gamma)^(2)` ...........iii
From Eq vii and viii then we get
`(a+c+e)^(2)=4(ac+ce+cea-b-d-f)`