-If `a,b,c in R` then prove that the roots of the equation `1/(x-a)+1/(x-b)+1/(x-c)=0` are always real cannot have roots if `a=b=c`

Options

A. One roots lies in (a,b)
B. One root lies in (b,c)
C. One root must be non-real
D. Three roots are real.

Answer

A, B, D

Explanation

Simplifying
`x(x-b)(x-c)+x(x-c)(x-a)+x(x-a)(x-b)-(x-a)(x-b)(x-c)=0`
let `f(x)=x(x-b)(x-c)+x(x-c)(x-a)+x(x-a)(x-b)-(x-a)(x-b)(x-c)`
`becausef(a)=a(a-b)(a-c)gt0`
`f(b)=b(b-c)(b-a)gt0`
`f(x)=c(c-a)(c-b)gt0`
(assuming `altbltc)`
If two roots are real, then the polynomial of degree three has the third root which must be real.