Nature of roots

yes how did you get that?

DanTe^ ,sorry i made stupid mistake.

I am proving this for the equation A²/(x-a) + B²/(x-b) + C²/(x-c)+ D²/(x-d) = k (same arguments hold a,b,c,d,e,f,g,h )

Design a polynomial function f(x) = A²(x-a) B²(x-b)C²(x-c)D²(x-d)

Taking log and differentiating we can say f'(x)/f(x) = A²/(x-a) + B²/(x-b) + C²/(x-c)+ D²/(x-d)

So we have to solve f'(x)f(x) = k , or kf(x) - f'(x) = 0

Let P(x) = kf(x) - f'(x) ,Let k>0 (similar arguments hold for k<0)

From the graph of f(x)[drawn with help of Rolle's theorem] check the sign of p(x). from the figure given.

There is at least one root in between points of each sign change. Hence we clearly see 3 real roots of p(x). As p(x) is an polynomial equation with real co-efficients complex roots must occur in conjugate pairs. so there can't be a single complex root. so all roots are real.

(while checking the sign of p(x) , use the fact that when slope of tangent at any point is negative f'(x) at that point is negative)

I take it that A,B, ...,H, a,b,c...,k are all real

So, suppose z is a root, so is its conjugate

Hence

\frac{A^2}{z-a} + \frac{B^2}{z-b} +....+\frac{H^2}{z-h} =k

and

\frac{A^2}{\overline{z}-a} + \frac{B^2}{\overline{z}-b} +....+\frac{H^2}{\overline{z}-h} =k

Let z-\overline{z} = iR, R \in \mathbb{R}

Then subtracting and simplifying, we obtain

iR \left(\frac{A^2}{|z-a|^2} + \frac{B^2}{|z-b|^2} +....+\frac{H^2}{|z-h|^2} \right)=0

Since the bracketed term is positive, we must have R=0.

In other words roots are purely real

Thanks Shubhodip, hsbhatt Sir and Nishant Sir.