Question

If $a,b,c$ are distinct positive numbers, then the nature of roots of the equation $\frac{1}{(x-a)}+\frac{1}{(x-b)}+\frac{1}{(x-c)}=\frac{1}{x}$ is

• A. all real and distinct
• B. all real and at least two are distinct
• C. at least two real
• D. all non-real

Answer 1

The correct option is A all real and distinct

$\frac{1}{(x-a)}+\frac{1}{(x-b)}+\frac{1}{(x-c)}=\frac{1}{x}$

The equation on simplifying gives

$x(x-b)(x-c)+x(x-c)(x-a)+x(x-a)(x-b)-(x-a)(x-b)(x-c)=0$

Let,

$f\left(x\right)=x(x-b)(x-c)+x(x-c)(x-a)+x(x-a)(x-b)-(x-a)(x-b)(x-c)$

We can assume without loss of generality that $a<b<c.$ Now,

$f\left(a\right)=a(a-b)(a-c)>0$

$f\left(b\right)=b(b-c)(b-a)<0$

$f\left(c\right)=c(c-a)(c-b)>0$

So, one root of $\left(1\right)$ lies in $(a,b)$ and one root in $(b,c).$ Obviously the third root must also be real.