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Algebra

Hari Shankar's Avatar
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Joined: 28 Feb 2007
Post: 2173
19 Jul 2008 11:25:32 IST
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Yet another polynomial prob
None

f(z) is a non-constant polynomial with real coefficients such that for some real number a, we have


f(a) \ne 0, f


Prove that f(z) cannot have all roots real.


I happen to know the official solution. So, only original looking solutions get rated.


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abhishek sinha's Avatar

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Joined: 18 Dec 2007
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19 Jul 2008 18:09:12 IST
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write f(x)=c\prod(x-\alpha_{i})} ,where \alpha_{i} s are the roots of the polynomial


given that f(a) !=0 whch means none of the roots are a .


Now we get f'(a)=0


so taking logarithm of both sides and then taking derivative at x=a , we obtain


\sum{\frac{1}{a-\alpha_{i}}=0   ......(2)


again taking the derivative of f'(x ) at x=a and imposing the condition that f''(a)=0 , we get using (2)


\sum{\frac{1}{(a-\alpha_{i})^2}=0  ....(3)


but it is given that a is real . So if all the roots are real then surely (3)can't hold true .


Hence proved .




 


 

Hari Shankar's Avatar

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Joined: 28 Feb 2007
Posts: 2173
20 Jul 2008 07:37:12 IST
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thats the solution i was looking for!


A bit of clarification may be required here:


We have for a polynomial f(x) with roots \alpha_i:


\frac{f


Since f'(a) = f"(a) = 0, we have


\sum_{i=1}^n \frac{1}{(x-\alpha_i)^2} = 0




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