ax-3-bx-2-c-0-x-1-x-2-x-3- Tinku Tara June 4, 2023 None 0 Comments FacebookTweetPin Question Number 183202 by liuxinnan last updated on 23/Dec/22 ax3+bx2+c=0x1=x2=x3= Answered by Frix last updated on 23/Dec/22 ax3+bx2+c=0m=ba∧n=cax3+mx2+n=0x=t−m3t3+m23t+2m327+n=0p=m23∧q=2m327+nt3+pt+q=0DependingonD=p327+q24=><0youcanuseCartano′sFormulaortheTrigonometricFormula:D>0⇒u=−q2−p327+q243∧v=−q2+p327+q243[youmusttaketherealroots:−r3=−r3]t1=u+vt2=(−12+32i)u+(−12−32i)vt3=(−12−32i)u+(−12+32i)vD=0∧p≠0∧q≠0⇔p327=−q24⇒u=v=−q23t1=2−q23=3qpt2=t3=−−q23=−3q2pD<0⇒tk=−4p33×cos(2πk+cos−1(−q2×−27p3)3)withk=1,2,3Youmustinsertbackwardsp=m23∧q=2m327+nxk=tk−m3m=ba∧n=catogetthedemandedformula Answered by mr W last updated on 23/Dec/22 wecanalsosolvefor1xinsteadofforx.1x3+bcx+ac=0Δ=14(ac)2+127(bc)3ifΔ>0:―1x1=Δ−a2c3−Δ+a2c31x2,3=−12(Δ−a2c3−Δ+a2c3)±(Δ−a2c3+Δ+a2c3)iifΔ=0:―1x1=−2a2c31x2=1x3=a2c3ifΔ<0:―1x1,2,3=2−b3csin[13sin−1(3a−2b−3cb)+2kπ3](k=0,1,2) Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Find-lim-x-sin-x-x-Next Next post: Question-52129 Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.