Radacinile unei ecuatii

Post by **Beniamin Bogosel** » Tue Jan 20, 2009 12:25 pm

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 \( |z| <\alpha \).

mihai++ · Post by **mihai++** » Tue Jan 20, 2009 6:49 pm

\( n\in? \), ca daca \( n\in\mathbb{N} \) atunci ecuatia nu are radacini nereale.

Post by **Beniamin Bogosel** » Wed Jan 21, 2009 3:34 pm

\( f(1)=2-a<0,\ f(0)=1>0 \) deci ecuatia are o radacina reala in (0,1).

mihai++ · Post by **mihai++** » Wed Jan 21, 2009 5:22 pm

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.

mychrom · Post by **mychrom** » Wed Jan 21, 2009 6:06 pm

Cred ca n>3.

Post by **Beniamin Bogosel** » Wed Jan 21, 2009 7:55 pm

Imi pare rau... numai acum am vazut... era \( n>3 \)