Question Number 102490 by ajfour last updated on 09/Jul/20

x^3 −bx−c=0 ; b, c >0 ; ((b/3))^3 >((c/2))^2 To find the three real roots without the use of trigonometric solution to cubic polynomial...

Answered by ajfour last updated on 10/Jul/20

let z=(x/(√b)) , a=(c/(b(√b))) As (c/2)<((b(√b))/(3(√3))) ⇒ a<(2/(3(√3))) ⇒ z^3 −z−a=0 Now let (z−1)(z^3 −z−a)=0 z^4 −z^3 −z^2 +(1−a)z+a=0 Now let z=((t+s)/(t+1)) ⇒ (t^4 +4st^3 +6s^2 t^2 +4s^3 t+s^4 ) −(t+1)(t^3 +3st^2 +3s^2 t+s^3 ) −(t^2 +2t+1)(t^2 +2st+s^2 ) +(1−a)(t+s)(t^3 +3t^2 +3t+1) +a(t^4 +4t^3 +6t^2 +4t+1)=0 ⇒ t^3 {4s−3s−1−2s−2+3(1−a) +s(1−a)+4a}+ t^2 {6s^2 −3s^2 −3s−s^2 −4s−1 +3(1−a)+3s(1−a)+6a}+ t{4s^3 −s^3 −3s^2 −2s^2 −2s +(1−a)+3s(1−a)+4a}+ {s^4 −s^3 −s^2 +s(1−a)+a} = 0 ⇒ t^3 {a(1−s)}+t^2 {2s^2 −s(4+3a)+3a+2} +t{s^3 −5s^2 −(3a−1)s+3a+1} +{s^4 −s^3 −s^2 +s(1−a)+a}=0 Now let 2s^2 −s(4+3a)+3a+2=0 ⇒ s=((3a+4)/2)±((√((3a+4)^2 −2(3a+2)))/2) = ((3a+4±(√((3a+4)^2 −2(3a+4)+4)))/2) =((3(a+1)+1±(√(9(a+1)^2 +3)))/2) Now t^3 +t{((s^3 −5s^2 −(3a−1)s+3a+1)/(a(1−s)))} +{((s^4 −s^3 −s^2 +s(1−a)+a)/(a(1−s)))}= 0 ⇒ t^3 +t{((s^3 −5s^2 −(3a−1)s+3a+1)/(a(1−s)))} −(1/a)(s^3 −s−a) = 0 ⇒ t^3 +(t/a){(s+3)^2 +3a+4(((2−s)/(s−1)))} −(1/a)(s^3 −s−a) = 0 lets say above expression is t^3 +((p/a))t−(q/a)=0 we can apply Cardano′s formula if p>0 ; say t=t_0 , then z_0 =((t_0 +s)/(t_0 +1)) and x=(√b)(z_0 ) ....... lets check the sign of p = (s+3)^2 +3a+4(((2−s)/(s−1))) where s =((3(a+1)+1±(√(9(a+1)^2 +3)))/2) and a<(2/(3(√3))) (someone please help here... can p be >0 for one of the two values of s ? )