The roots of the equation `a(b-2c)x^(2)+b(c-2a)x+c(a-2b)=0` are, when `ab+bc+ca=0`

Options

A. 1, `(c(a-2b))/(a(b-2c))`
B. `(c )/(a)`, `(a-2b)/(b-2c)`
C. `(a-2b)/(a-2c)`, `(a-2b)/(b-2c)`
D. none of these

Answer

A

Explanation

`(a)` As it is given that `ab+bc+ca=0`, so putting `x=1` in the equation, we get
`f(1)=a(b-2c)+b(c-2a)+c(a-2b)`
`implies f(1)=-(ab+bc+ac)=0`
So, `1` is a root of the equation,
Now product of the roots be `1`
`:. 1*alpha=(c(a-2b))/(a(b-2c))impliesalpha=(c )/(a)((a-2b)/(b-2c))`