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2014-08-05T09:58:25+03:00
Ecuatia de gradul  3 are forma ax³+bx²+cx+d=0,  a,b,c,d -nr reale
Conventional se rezolva printr-un sir de substitutii :
1. inlocuim x cu (y-b)/3a si obtinem o alta ecuatie de forma
y³+py+q=0
p si q sunt coeficienti si se afla ci rapoartele
p= \frac{c}{a}- \frac{b^2}{3a^2}
q= \frac{2b^3}{27a^3} - \frac{bc}{3a^2} +  \frac{d}{a}
Aflam discriminantul
Δ=(p/3)^3+(q/2)^2
Notam P=∛-q/2+√Δ si Q=∛-q/2-√Δ
Si acum , daca nu ne doare mana, aflam radacinile
y1=P+Q
 y_{2} =- \frac{P+Q}{2}+i \frac{P-Q}{2}   \sqrt{3}
 y_{3} =- \frac{P+Q}{2}-i \frac{P-Q}{2}   \sqrt{3}