1Ans i think is
3) no complex roots
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The equation A2/(x-a) + B2/(x-b) + C2/(x-c) + ..... + H2/(x-h) = k has
1) No real root
2) At most one real root
3) No complex root
4) At most two complex roots
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8 Answers
21DanTe^ ,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)
341I 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
62one more proof could be by cross multiplying...
We see that the polynomial we have is of n degree...
Moreover the original question has one root between xi and xi+1 this is simply done by using limits because the right limit at xi is +ve infinity and left limit at xi+1 is - infinity
Also by saying that the function is continuous in between the limits...
Hence the polynomial has atleast n-1 roots...
There cant be 1 single complex root..
Hence all roots are Real.
1got it nishant bhaiya has used the asymptotes ,this gives the same resullt as that of shubodip
