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ax-3-bx-2-c-0-x-1-x-2-x-3-




Question Number 183202 by liuxinnan last updated on 23/Dec/22
ax^3 +bx^2 +c=0  x_1 =  x_2 =  x_3 =
ax3+bx2+c=0x1=x2=x3=
Answered by Frix last updated on 23/Dec/22
ax^3 +bx^2 +c=0  m=(b/a)∧n=(c/a)  x^3 +mx^2 +n=0  x=t−(m/3)  t^3 +(m^2 /3)t+((2m^3 )/(27))+n=0  p=(m^2 /3)∧q=((2m^3 )/(27))+n  t^3 +pt+q=0  Depending on D=(p^3 /(27))+(q^2 /4) =_< ^>  0 you can use  Cartano′s Formula or the Trigonometric  Formula:  D>0 ⇒       u=((−(q/2)−(√((p^3 /(27))+(q^2 /4)))))^(1/3) ∧v=((−(q/2)+(√((p^3 /(27))+(q^2 /4)))))^(1/3)             [you must take the real roots: ((−r))^(1/3) =−(r)^(1/3) ]       t_1 =u+v       t_2 =(−(1/2)+((√3)/2)i)u+(−(1/2)−((√3)/2)i)v       t_3 =(−(1/2)−((√3)/2)i)u+(−(1/2)+((√3)/2)i)v  D=0∧p≠0∧q≠0 ⇔ (p^3 /(27))=−(q^2 /4) ⇒       u=v=((−(q/2)))^(1/3)        t_1 =2((−(q/2)))^(1/3) =((3q)/p)       t_2 =t_3 =−((−(q/2)))^(1/3) =−((3q)/(2p))  D<0 ⇒       t_k =((−((4p)/3)))^(1/3) ×cos (((2πk+cos^(−1)  (−(q/2)×(√(−((27)/p^3 )))))/3)) with k=1, 2, 3  You must insert backwards  p=(m^2 /3)∧q=((2m^3 )/(27))+n  x_k =t_k −(m/3)  m=(b/a)∧n=(c/a)  to get the demanded formula
ax3+bx2+c=0m=ban=cax3+mx2+n=0x=tm3t3+m23t+2m327+n=0p=m23q=2m327+nt3+pt+q=0DependingonD=p327+q24=><0youcanuseCartanosFormulaortheTrigonometricFormula:D>0u=q2p327+q243v=q2+p327+q243[youmusttaketherealroots:r3=r3]t1=u+vt2=(12+32i)u+(1232i)vt3=(1232i)u+(12+32i)vD=0p0q0p327=q24u=v=q23t1=2q23=3qpt2=t3=q23=3q2pD<0tk=4p33×cos(2πk+cos1(q2×27p3)3)withk=1,2,3Youmustinsertbackwardsp=m23q=2m327+nxk=tkm3m=ban=catogetthedemandedformula
Answered by mr W last updated on 23/Dec/22
we can also solve for (1/x) instead of   for x.  (1/x^3 )+(b/(cx))+(a/c)=0  Δ=(1/4)((a/c))^2 +(1/(27))((b/c))^3   if Δ>0:  (1/x_1 )=((Δ−(a/(2c))))^(1/3) −((Δ+(a/(2c))))^(1/3)   (1/x_(2,3) )=−(1/2)(((Δ−(a/(2c))))^(1/3) −((Δ+(a/(2c))))^(1/3) )±(((Δ−(a/(2c))))^(1/3) +((Δ+(a/(2c))))^(1/3) )i  if Δ=0:  (1/x_1 )=−2((a/(2c)))^(1/3)   (1/x_2 )=(1/x_3 )=((a/(2c)))^(1/3)   if Δ<0:  (1/x_(1,2,3) )=2(√(−(b/(3c)))) sin [(1/3)sin^(−1) (((3a)/(−2b))(√(−((3c)/b))))+((2kπ)/3)]  (k=0,1,2)
wecanalsosolvefor1xinsteadofforx.1x3+bcx+ac=0Δ=14(ac)2+127(bc)3ifΔ>0:1x1=Δa2c3Δ+a2c31x2,3=12(Δa2c3Δ+a2c3)±(Δa2c3+Δ+a2c3)iifΔ=0:1x1=2a2c31x2=1x3=a2c3ifΔ<0:1x1,2,3=2b3csin[13sin1(3a2b3cb)+2kπ3](k=0,1,2)

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