If `a b+b c+c a=0,` then solve `a(b-2c)x^2+b(c-2a)x+c(a-2b)=0.`

Explanation

As it is given that`ab + bc + ca = 0`, Putting x =1 in the equation, we get
`f(1) = a(b - 2c) + b(c - 2a)+ c(a - 2b)`
`= - (ab + bc + ca)`
=0
So x = 1 is a root of the equation.
Let the other root be `alpha`
Now product of the roots is
`1xxalpha = (c(a - 2b))/(a(b - 2c))`
or `alpha =(c)/(a) ((a - 2b)/(b - 2c))`