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Question Number 63373 by MJS last updated on 31/Jul/19
just found this on the web I thought it might help in some cases where quartics appear i.e. Sir Aifour′s geometric questions. sometimes we know the nature of the roots, but how to use this information? ax^4 +bx^3 +cx^2 +dx+e=0 1. divide by a 2. x=z−(b/(4a)) this leads to the reduced z^4 +pz^2 +qz+r=0 now we find the nature of the roots: T_1 =16p^4 r−4p^3 q^2 −128p^2 r^2 +144pq^2 r−27q^4 +256r^3 T_2 =p^2 +12r T_3 =−p^2 +4r T_1 <0 ⇒ 2 distinct real and 2 conjugated complex roots T_1 >0∧(p<0∧T_3 <0) ⇒ 4 distinct real roots T_1 >0∧(p>0∨T_3 >0) ⇒ 2 pairs of conjugated complex roots T_1 =0∧(p<0∧T_3 <0∧T_2 ≠0) ⇒ 1 real double and 2 real simple roots T_1 =0∧(T_3 >0∨(p>0∧(T_3 ≠0∨q≠0))) ⇒ 1 real double and 2 conjugated complex roots T_1 =0∧(T_2 =0∧T_3 ≠0) ⇒ 1 real triple and 1 real simple roots T_1 =0∧(T_3 =0∧p<0) ⇒ 2 real double roots T_1 =0∧(T_3 =0∧p>0∧q=0) ⇒ 2 conjugated complex double roots T_1 =0∧T_2 =0 ⇒ all roots are equal
$$\mathrm{just}\:\mathrm{found}\:\mathrm{this}\:\mathrm{on}\:\mathrm{the}\:\mathrm{web} \\ $$$$\mathrm{I}\:\mathrm{thought}\:\mathrm{it}\:\mathrm{might}\:\mathrm{help}\:\mathrm{in}\:\mathrm{some}\:\mathrm{cases}\:\mathrm{where} \\ $$$$\mathrm{quartics}\:\mathrm{appear}\:\mathrm{i}.\mathrm{e}.\:\mathrm{Sir}\:\mathrm{Aifour}'\mathrm{s}\:\mathrm{geometric} \\ $$$$\mathrm{questions}.\:\mathrm{sometimes}\:\mathrm{we}\:\mathrm{know}\:\mathrm{the}\:\mathrm{nature}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{roots},\:\mathrm{but}\:\mathrm{how}\:\mathrm{to}\:\mathrm{use}\:\mathrm{this}\:\mathrm{information}? \\ $$$$ \\ $$$${ax}^{\mathrm{4}} +{bx}^{\mathrm{3}} +{cx}^{\mathrm{2}} +{dx}+{e}=\mathrm{0} \\ $$$$\mathrm{1}.\:\mathrm{divide}\:\mathrm{by}\:{a} \\ $$$$\mathrm{2}.\:{x}={z}−\frac{{b}}{\mathrm{4}{a}} \\ $$$$\mathrm{this}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{the}\:\mathrm{reduced} \\ $$$$ \\ $$$${z}^{\mathrm{4}} +{pz}^{\mathrm{2}} +{qz}+{r}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{now}\:\mathrm{we}\:\mathrm{find}\:\mathrm{the}\:\mathrm{nature}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}: \\ $$$${T}_{\mathrm{1}} =\mathrm{16}{p}^{\mathrm{4}} {r}−\mathrm{4}{p}^{\mathrm{3}} {q}^{\mathrm{2}} −\mathrm{128}{p}^{\mathrm{2}} {r}^{\mathrm{2}} +\mathrm{144}{pq}^{\mathrm{2}} {r}−\mathrm{27}{q}^{\mathrm{4}} +\mathrm{256}{r}^{\mathrm{3}} \\ $$$${T}_{\mathrm{2}} ={p}^{\mathrm{2}} +\mathrm{12}{r} \\ $$$${T}_{\mathrm{3}} =−{p}^{\mathrm{2}} +\mathrm{4}{r} \\ $$$${T}_{\mathrm{1}} <\mathrm{0}\:\Rightarrow\:\mathrm{2}\:\mathrm{distinct}\:\mathrm{real}\:\mathrm{and}\:\mathrm{2}\:\mathrm{conjugated}\:\mathrm{complex}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} >\mathrm{0}\wedge\left({p}<\mathrm{0}\wedge{T}_{\mathrm{3}} <\mathrm{0}\right)\:\Rightarrow\:\mathrm{4}\:\mathrm{distinct}\:\mathrm{real}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} >\mathrm{0}\wedge\left({p}>\mathrm{0}\vee{T}_{\mathrm{3}} >\mathrm{0}\right)\:\Rightarrow\:\mathrm{2}\:\mathrm{pairs}\:\mathrm{of}\:\mathrm{conjugated}\:\mathrm{complex}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} =\mathrm{0}\wedge\left({p}<\mathrm{0}\wedge{T}_{\mathrm{3}} <\mathrm{0}\wedge{T}_{\mathrm{2}} \neq\mathrm{0}\right)\:\Rightarrow\:\mathrm{1}\:\mathrm{real}\:\mathrm{double}\:\mathrm{and}\:\mathrm{2}\:\mathrm{real}\:\mathrm{simple}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} =\mathrm{0}\wedge\left({T}_{\mathrm{3}} >\mathrm{0}\vee\left({p}>\mathrm{0}\wedge\left({T}_{\mathrm{3}} \neq\mathrm{0}\vee{q}\neq\mathrm{0}\right)\right)\right)\:\Rightarrow\:\mathrm{1}\:\mathrm{real}\:\mathrm{double}\:\mathrm{and}\:\mathrm{2}\:\mathrm{conjugated}\:\mathrm{complex}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} =\mathrm{0}\wedge\left({T}_{\mathrm{2}} =\mathrm{0}\wedge{T}_{\mathrm{3}} \neq\mathrm{0}\right)\:\Rightarrow\:\mathrm{1}\:\mathrm{real}\:\mathrm{triple}\:\mathrm{and}\:\mathrm{1}\:\mathrm{real}\:\mathrm{simple}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} =\mathrm{0}\wedge\left({T}_{\mathrm{3}} =\mathrm{0}\wedge{p}<\mathrm{0}\right)\:\Rightarrow\:\mathrm{2}\:\mathrm{real}\:\mathrm{double}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} =\mathrm{0}\wedge\left({T}_{\mathrm{3}} =\mathrm{0}\wedge{p}>\mathrm{0}\wedge{q}=\mathrm{0}\right)\:\Rightarrow\:\mathrm{2}\:\mathrm{conjugated}\:\mathrm{complex}\:\mathrm{double}\:\mathrm{roots} \\ $$$${T}_{\mathrm{1}} =\mathrm{0}\wedge{T}_{\mathrm{2}} =\mathrm{0}\:\Rightarrow\:\mathrm{all}\:\mathrm{roots}\:\mathrm{are}\:\mathrm{equal} \\ $$
Commented by mr W last updated on 03/Jul/19
good information, thanks sir!
$${good}\:{information},\:{thanks}\:{sir}! \\ $$
Commented by MJS last updated on 31/Jul/19
I just corrected 2 typos it′s not r but q in 2 cases...
$$\mathrm{I}\:\mathrm{just}\:\mathrm{corrected}\:\mathrm{2}\:\mathrm{typos} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{not}\:{r}\:\mathrm{but}\:{q}\:\mathrm{in}\:\mathrm{2}\:\mathrm{cases}… \\ $$

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