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Question Number 123025 by ajfour last updated on 21/Nov/20

Commented by ajfour last updated on 22/Nov/20

If α, β, γ are roots of cubic eq. x^3 −x−c =0 (0<c<(2/(3(√3)))) ; wirh the help of suitable p , find α.

$${If}\:\alpha,\:\beta,\:\gamma\:{are}\:{roots}\:{of}\:{cubic}\:{eq}. \\ $$$${x}^{\mathrm{3}} −{x}−{c}\:=\mathrm{0}\:\:\:\left(\mathrm{0}<{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\right)\:;\:{wirh}\:{the} \\ $$$${help}\:{of}\:{suitable}\:{p}\:,\:{find}\:\alpha. \\ $$

Commented by ajfour last updated on 22/Nov/20

value of p is to be chosen, so that the degree four can be solved if not the degree three.

$${value}\:{of}\:{p}\:{is}\:{to}\:{be}\:{chosen},\:{so} \\ $$$${that}\:{the}\:{degree}\:{four}\:{can}\:{be}\:{solved} \\ $$$${if}\:{not}\:{the}\:{degree}\:{three}. \\ $$

Commented by mr W last updated on 22/Nov/20

what should be p?

$${what}\:{should}\:{be}\:{p}? \\ $$

Commented by MJS_new last updated on 22/Nov/20

just a few thoughts... if α>0∧β<0∧γ<0∧β≠γ ⇒ α=−(β+γ) x^3 +px+q=0 with p=−1 ⇒ β=−(γ/2)±((√(4−3γ^2 ))/2) ⇒ c=γ(γ^2 −1) introducing p we get (x−p)(x−γ)(x−(γ/2)−((√(4−3γ^2 ))/2))(−(γ/2)+((√(4−3γ^2 ))/2))=0 and it would be nice if p=−γ so we have to solve c=γ(γ^2 −1) for γ but this is exactly the same problem as before: γ^3 −γ−c=0 I don′t think we can find any fitting p without solving this 3^(rd) degree polynome...

$$\mathrm{just}\:\mathrm{a}\:\mathrm{few}\:\mathrm{thoughts}... \\ $$$$\mathrm{if}\:\alpha>\mathrm{0}\wedge\beta<\mathrm{0}\wedge\gamma<\mathrm{0}\wedge\beta\neq\gamma\:\Rightarrow\:\alpha=−\left(\beta+\gamma\right) \\ $$$${x}^{\mathrm{3}} +{px}+{q}=\mathrm{0}\:\mathrm{with}\:{p}=−\mathrm{1}\:\Rightarrow\:\beta=−\frac{\gamma}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{4}−\mathrm{3}\gamma^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\Rightarrow\:{c}=\gamma\left(\gamma^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$\mathrm{introducing}\:{p}\:\mathrm{we}\:\mathrm{get} \\ $$$$\left({x}−{p}\right)\left({x}−\gamma\right)\left({x}−\frac{\gamma}{\mathrm{2}}−\frac{\sqrt{\mathrm{4}−\mathrm{3}\gamma^{\mathrm{2}} }}{\mathrm{2}}\right)\left(−\frac{\gamma}{\mathrm{2}}+\frac{\sqrt{\mathrm{4}−\mathrm{3}\gamma^{\mathrm{2}} }}{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\mathrm{and}\:\mathrm{it}\:\mathrm{would}\:\mathrm{be}\:\mathrm{nice}\:\mathrm{if}\:{p}=−\gamma \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{solve}\:{c}=\gamma\left(\gamma^{\mathrm{2}} −\mathrm{1}\right)\:\mathrm{for}\:\gamma\:\mathrm{but}\:\mathrm{this} \\ $$$$\mathrm{is}\:\mathrm{exactly}\:\mathrm{the}\:\mathrm{same}\:\mathrm{problem}\:\mathrm{as}\:\mathrm{before}: \\ $$$$\gamma^{\mathrm{3}} −\gamma−{c}=\mathrm{0} \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{we}\:\mathrm{can}\:\mathrm{find}\:\mathrm{any}\:\mathrm{fitting}\:{p}\:\mathrm{without} \\ $$$$\mathrm{solving}\:\mathrm{this}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{degree}\:\mathrm{polynome}... \\ $$

Answered by ajfour last updated on 26/Nov/20

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