Question Number 114263 by Khalmohmmad last updated on 18/Sep/20

3x^3 −3x−(1/2)=0  x_1 ,x_2 ,x_3  ?

Answered by MJS_new last updated on 18/Sep/20

x^3 −x−(1/6)=0  p=−1∧q=−(1/6)  D=(p^3 /(27))+(q^2 /4)=−((13)/(432))<0 ⇒ trigonometric solution  x_(j+1) =((2(√(−3p)))/3)sin (((2πj)/3)+(1/3)arcsin ((3(√3)q)/(2(√(−p^3 )))))) with j=0, 1, 2  in this case  x_1 =−((2(√3))/3)sin ((1/3)arcsin ((√3)/4)) ≈−.171731  x_2 =((2(√3))/3)sin ((π/3)+(1/3)arcsin ((√3)/4)) ≈1.07474  x_3 =−((2(√3))/3)cos ((π/6)+(1/3)arcsin ((√3)/4)) ≈−.903013

Commented byKhalmohmmad last updated on 19/Sep/20

you can explain the formula  please

Commented byMJS_new last updated on 19/Sep/20

(1)  x^3 +px+q=0  D=(p^3 /(27))+(q^2 /4)  if D≥0 ⇒ Cardano′s solution  if D<0:  (p^3 /(27))+(q^2 /4)<0 ⇔ −1<((27q^2 )/(4p^3 ))<1  ⇒ we can put ((27q^2 )/(4p^3 ))=sin θ but it′s convenient  to let θ=3α  ((27q^2 )/(4p^3 ))=sin 3α  now let x=az and a=2(√((−p)/3)) ⇒ equation (1)  becomes  a^3 z^3 +az+q=0  ⇒  4z^3 −3z−((3(√3)q)/( 2(√(−p^3 ))))=0  now (−((3(√3)q)/( 2(√(−p^3 )))))^2 =∣((27q^2 )/(4p^3 ))∣=sin 3α  4z^3 −3z+sin 3α =0  z^3 −(3/4)z+((sin 3α)/4)=0  we know that sin 3α =3sin α −4sin^3  α ⇔  ⇔ sin^3  α −(3/4)sin α +((sin 3α)/4)=0  we can put z=sin α and get the solution

Answered by Olaf last updated on 19/Sep/20

  x = u+v  3(u^3 +v^3 )+9uv(u+v)−3(u+v)−(1/2) = 0  3(u^3 +v^3 )+3(u+v)(3uv−1)−(1/2) = 0  we choose u and v such as uv = (1/3)  ⇒ u^3 v^3  = (1/(27))  and 3(u^3 +v^3 )−(1/2) = 0  u^3 +v^3  = (1/6)  Let U = u^3 , V = v^3   and S = U+V = (1/6), P = UV = (1/(27))  We know S and P :  z^2 −Sz+P = 0  Δ = S^2 −4P = (1/(36))−(4/(27)) = −((13)/(108))  U = ((S−(√Δ))/2) = (1/2)((1/6)−(1/6)i(√((13)/3)))  U = ((S−(√Δ))/2) = (1/(12))(1−i(√((13)/3)))  V = ((S+(√Δ))/2) = (1/(12))(1+i(√((13)/3)))  ∣U∣ = ∣V∣ = (1/(12))×(√((16)/3)) = (1/(3(√3)))  Let θ = ArgU = −ArgV = arctan(√((13)/3))  U = (1/(3(√3)))e^(iθ)  and V = (1/(3(√3)))e^(−iθ)   u = ^3 (√U) = (1/( (√3)))e^(i((θ/3)+((2kπ)/3)))  k = 0,1,2  v = ^3 (√V) = (1/( (√3)))e^(i(−(θ/3)+((2kπ)/3)))  k = 0,1,2  for k = 0, u = (1/( (√3)))e^(i(θ/3))  and v = (1/( (√3)))e^(−i(θ/3))   x = u+v = (1/( (√3)))(e^(i(θ/3)) +e^(−i(θ/3)) )  x = u+v = (2/( (√3)))cos(θ/3)  x = u+v = (2/( (√3)))cos((1/3)arctan(√((13)/3)))  we have the first root...

Commented byMJS_new last updated on 19/Sep/20

yes but the formula I posted is standard...