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Radacinile unei ecuatii
Posted: Tue Jan 20, 2009 12:25 pm
by Beniamin Bogosel
Fie \( n>3 \) natural si \( a \in \mathbb{R}, a > 2 \) si fie \( \alpha \) radacina ecuatiei \( x^n-ax^{n-1}+1=0 \) ce satisface \( 0< \alpha < 1 \). Aratati ca orice alta radacina nereala \( z \) satisface \( |z| <\alpha \).
Posted: Tue Jan 20, 2009 6:49 pm
by mihai++
\( n\in? \), ca daca \( n\in\mathbb{N} \) atunci ecuatia nu are radacini nereale.
Posted: Wed Jan 21, 2009 3:34 pm
by Beniamin Bogosel
\( f(1)=2-a<0,\ f(0)=1>0 \) deci ecuatia are o radacina reala in (0,1).
Posted: Wed Jan 21, 2009 5:22 pm
by mihai++
Nu la radacinile reale ma refeream, ci la cele nereale.
Din cate vad eu ecuatia este polinomiala, si cum \( n<3 \), este ori de gradul 1 ori de gradul 2. Nu inteleg unde privesc gresit.
Posted: Wed Jan 21, 2009 6:06 pm
by mychrom
Cred ca n>3.
Posted: Wed Jan 21, 2009 7:55 pm
by Beniamin Bogosel
Imi pare rau... numai acum am vazut... era \( n>3 \)