rownania wielomianowe

[adamos64]

Ustal krotnosci pierwiastkow rownania \(\displaystyle{ (2x^2+px+1)^2}\)=0 w zaleznosci od parametru p.

[frej]

\(\displaystyle{ \Delta 0}\) dwa pierwiastki 2-krotne: \(\displaystyle{ \frac{-p-\sqrt{p^2-8}}{4}}\), \(\displaystyle{ \frac{-p+\sqrt{p^2-8}}{4}}\)

[ Dodano: 22 Sierpnia 2008, 23:00 ]
\(\displaystyle{ \Delta=p^2-8}\)
\(\displaystyle{ \Delta=0 \Leftrightarrow p^2-8=0 \Leftrightarrow p^2=8 \Leftrightarrow p=2\sqrt{2} \vee p=-2\sqrt{2}}\)
\(\displaystyle{ \Delta >0 \Leftrightarrow p^2-8>0 \Leftrightarrow p^2>8 p\in (-\infty,-2\sqrt{2}) \cup (2\sqrt{2},\infty)}\)