Matematické Fórum

stuart clark · 17. 05. 2011 04:31

Prove That the equation $\frac{A^2}{x-a}+\frac{B^2}{x-b}+\frac{C^2}{x-c}+...........................+\frac{H^2}{x-h}=k$ has all Real roots.

Pavel · 17. 05. 2011 12:38 — Editoval Pavel (17. 05. 2011 12:42)

↑ stuart clark:

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check_drummer · 17. 05. 2011 18:26

↑ stuart clark:
It has to be said from what number set the values a-h,k,A-H are, I suppose they are real numbers, aren't?

stuart clark · 23. 05. 2011 09:21

thanks Paul for Nice solution.

and check_drummer you are saying Right.

musixx · 23. 05. 2011 12:47 — Editoval musixx (23. 05. 2011 13:43)

Let's use $A_i$ instead of A,B,C, ..., and let's use $a_i$ instead of a,b,c, ... .

Then we are looking for zeroes of a function $f(x)=$\sum_{j=1}^?\frac{A_j^2}{x-a_j}$-k$ . As Pavel has remarked, all the possible roots are also roots of some polynomial with real coefficients, so all its roots are realized within complex numbers.

So what is $f(m+n{\rm i})$ ? The imaginary part clearly is ${\rm i}\cdot n\cdot\sum_{j=1}^{?}\frac{A_j^2}{(m-a_j)^2+n^2}$ , so the sum cannot be zero (EDIT: if at least one of A_i and one of a_i are nonzero which is a natural assumption), and since we are looking for $f(m+n{\rm i})=0$ , we have n=0, so therefore only real roots.

Matematické Fórum

#1 17. 05. 2011 04:31

real roots

#2 17. 05. 2011 12:38 — Editoval Pavel (17. 05. 2011 12:42)

Re: real roots

#3 17. 05. 2011 18:26

Re: real roots

#4 23. 05. 2011 09:21

Re: real roots

#5 23. 05. 2011 12:47 — Editoval musixx (23. 05. 2011 13:43)

Re: real roots

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