Menu Close

x-3-px-q-0-If-equation-has-all-its-roots-real-find-them-

Tinku Tara 0 Comments
Pin




Question Number 43756 by ajfour last updated on 15/Sep/18
x^3 +px+q = 0 If equation has all its roots real, find them.
$$\:\:\boldsymbol{{x}}^{\mathrm{3}} +\boldsymbol{{px}}+\boldsymbol{{q}}\:=\:\mathrm{0} \\ $$$$\boldsymbol{{If}}\:\boldsymbol{{equation}}\:\boldsymbol{{has}}\:\boldsymbol{{all}}\:\boldsymbol{{its}}\:\boldsymbol{{roots}} \\ $$$$\boldsymbol{{real}},\:\boldsymbol{{find}}\:\boldsymbol{{them}}. \\ $$
Answered by ajfour last updated on 15/Sep/18
A poor self attempt : ________________________ 3sin ΞΈβˆ’4sin^3 ΞΈ = sin 3ΞΈ let sin 𝛉= mx k such that it makes this equation equivalent to that in question. β‡’ 3mxβˆ’4m^3 x^3 = sin 3𝛉 or x^3 βˆ’((3x)/(4m^2 )) βˆ’((sin 3𝛉)/(4m^3 )) = 0 β‡’ p = βˆ’(3/(4m^2 )) β‡’ m=Β±(√((βˆ’3)/(4p))) q = βˆ’((sin 3ΞΈ)/(4m^3 )) or sin (3ΞΈβˆ’2kΟ€) = βˆ’4m^3 q sin ΞΈ = sin[ ((2kΟ€)/3)+(1/3)sin^(βˆ’1) (Β±((3q)/p)(√((βˆ’3)/(4p))) )] β‡’ Β±(√((βˆ’3)/(4p))) x = sin[ ((2kΟ€)/3)+(1/3)sin^(βˆ’1) (Β±((3q)/p)(√((βˆ’3)/(4p))) )] β‡’ x = Β±(√((βˆ’4p)/3)) sin [((2kΟ€)/3)Β±(1/3)sin^(βˆ’1) (((3q)/p)(√((βˆ’3)/(4p))) )] unless i know how to choose the signs, i might obtain 12 roots. please help..
$$\:\:{A}\:{poor}\:{self}\:{attempt}\:: \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$$\:\:\mathrm{3sin}\:\thetaβˆ’\mathrm{4sin}\:^{\mathrm{3}} \theta\:=\:\mathrm{sin}\:\mathrm{3}\theta \\ $$$${let}\:\:\mathrm{sin}\:\boldsymbol{\theta}=\:\boldsymbol{{mx}}\: \\ $$$${k}\:{such}\:{that}\:{it}\:{makes}\:{this}\:{equation} \\ $$$${equivalent}\:{to}\:{that}\:{in}\:{question}. \\ $$$$\:\Rightarrow\:\:\mathrm{3}\boldsymbol{{mx}}βˆ’\mathrm{4}\boldsymbol{{m}}^{\mathrm{3}} \boldsymbol{{x}}^{\mathrm{3}} \:=\:\mathrm{sin}\:\mathrm{3}\boldsymbol{\theta} \\ $$$$\:\:\:\:{or}\:\:\:\boldsymbol{{x}}^{\mathrm{3}} βˆ’\frac{\mathrm{3}\boldsymbol{{x}}}{\mathrm{4}\boldsymbol{{m}}^{\mathrm{2}} }\:βˆ’\frac{\mathrm{sin}\:\mathrm{3}\boldsymbol{\theta}}{\mathrm{4}\boldsymbol{{m}}^{\mathrm{3}} }\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\:{p}\:=\:βˆ’\frac{\mathrm{3}}{\mathrm{4}{m}^{\mathrm{2}} }\:\:\Rightarrow\:{m}=\pm\sqrt{\frac{βˆ’\mathrm{3}}{\mathrm{4}{p}}} \\ $$$$\:\:\:\:\:\:\:{q}\:=\:βˆ’\frac{\mathrm{sin}\:\mathrm{3}\theta}{\mathrm{4}{m}^{\mathrm{3}} }\:\: \\ $$$${or}\:\:\:\mathrm{sin}\:\left(\mathrm{3}\thetaβˆ’\mathrm{2}{k}\pi\right)\:=\:βˆ’\mathrm{4}{m}^{\mathrm{3}} {q} \\ $$$$\:\mathrm{sin}\:\theta\:=\:\mathrm{sin}\left[\:\frac{\mathrm{2}{k}\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{βˆ’\mathrm{1}} \left(\pm\frac{\mathrm{3}{q}}{{p}}\sqrt{\frac{βˆ’\mathrm{3}}{\mathrm{4}{p}}}\:\right)\right] \\ $$$$\:\Rightarrow\:\pm\sqrt{\frac{βˆ’\mathrm{3}}{\mathrm{4}{p}}}\:\boldsymbol{{x}}\:=\:\mathrm{sin}\left[\:\frac{\mathrm{2}{k}\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{βˆ’\mathrm{1}} \left(\pm\frac{\mathrm{3}{q}}{{p}}\sqrt{\frac{βˆ’\mathrm{3}}{\mathrm{4}{p}}}\:\right)\right] \\ $$$$\Rightarrow\:{x}\:=\:\pm\sqrt{\frac{βˆ’\mathrm{4}{p}}{\mathrm{3}}}\:\mathrm{sin}\:\left[\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\pm\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{βˆ’\mathrm{1}} \left(\frac{\mathrm{3}{q}}{{p}}\sqrt{\frac{βˆ’\mathrm{3}}{\mathrm{4}{p}}}\:\right)\right] \\ $$$$\:\:{unless}\:{i}\:{know}\:{how}\:{to}\:{choose} \\ $$$${the}\:{signs},\:{i}\:{might}\:{obtain}\:\mathrm{12}\:{roots}. \\ $$$${please}\:{help}.. \\ $$
Commented by MJS last updated on 15/Sep/18
x^3 +px+q=0 3 real rootsβ‰ 0 ⇔ ((q/2))^2 +((p/3))^3 <0 β‡’ β‡’ p<0 ∧ 27q^2 +4p^3 <0 β‡’ ∣((27q^2 )/(4p^3 ))∣<1 we put x=az, a=2(√(βˆ’(p/3))) a^3 z^3 +apz+q=0 z^3 +(p/a^2 )z+(q/a^3 )=0 a^2 =βˆ’((4p)/3) a^3 =βˆ’((8p)/3)(√(βˆ’(p/3))) z^3 βˆ’(3/4)zβˆ’((3q)/(8p))(√(βˆ’(3/p)))=0 4z^3 βˆ’3zβˆ’((3q)/2)(√(βˆ’(3/p)))=0 (((3q)/2)(√(βˆ’(3/p))))^2 =∣((27q^2 )/(4p^3 ))∣<1 β‡’ βˆ’1<((3q)/2)(√(βˆ’(3/p)))<1 β‡’ β‡’ we put βˆ’((3q)/2)(√(βˆ’(3/p)))=sin 3Ξ± z^3 βˆ’(3/4)z+((sin 3Ξ±)/4)=0 if we put a=βˆ’2(√(βˆ’(p/3))) we get the same equation by putting ((3q)/2)(√(βˆ’(3/p)))=sin 3Ξ±
$${x}^{\mathrm{3}} +{px}+{q}=\mathrm{0} \\ $$$$\mathrm{3}\:\mathrm{real}\:\mathrm{roots}\neq\mathrm{0}\:\Leftrightarrow\:\left(\frac{{q}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{p}}{\mathrm{3}}\right)^{\mathrm{3}} <\mathrm{0}\:\Rightarrow \\ $$$$\Rightarrow\:{p}<\mathrm{0}\:\wedge\:\mathrm{27}{q}^{\mathrm{2}} +\mathrm{4}{p}^{\mathrm{3}} <\mathrm{0}\:\Rightarrow\:\mid\frac{\mathrm{27}{q}^{\mathrm{2}} }{\mathrm{4}{p}^{\mathrm{3}} }\mid<\mathrm{1} \\ $$$$\mathrm{we}\:\mathrm{put}\:{x}={az},\:{a}=\mathrm{2}\sqrt{βˆ’\frac{{p}}{\mathrm{3}}} \\ $$$${a}^{\mathrm{3}} {z}^{\mathrm{3}} +{apz}+{q}=\mathrm{0} \\ $$$${z}^{\mathrm{3}} +\frac{{p}}{{a}^{\mathrm{2}} }{z}+\frac{{q}}{{a}^{\mathrm{3}} }=\mathrm{0} \\ $$$${a}^{\mathrm{2}} =βˆ’\frac{\mathrm{4}{p}}{\mathrm{3}}\:\:\:\:\:{a}^{\mathrm{3}} =βˆ’\frac{\mathrm{8}{p}}{\mathrm{3}}\sqrt{βˆ’\frac{{p}}{\mathrm{3}}} \\ $$$${z}^{\mathrm{3}} βˆ’\frac{\mathrm{3}}{\mathrm{4}}{z}βˆ’\frac{\mathrm{3}{q}}{\mathrm{8}{p}}\sqrt{βˆ’\frac{\mathrm{3}}{{p}}}=\mathrm{0} \\ $$$$\mathrm{4}{z}^{\mathrm{3}} βˆ’\mathrm{3}{z}βˆ’\frac{\mathrm{3}{q}}{\mathrm{2}}\sqrt{βˆ’\frac{\mathrm{3}}{{p}}}=\mathrm{0} \\ $$$$\left(\frac{\mathrm{3}{q}}{\mathrm{2}}\sqrt{βˆ’\frac{\mathrm{3}}{{p}}}\right)^{\mathrm{2}} =\mid\frac{\mathrm{27}{q}^{\mathrm{2}} }{\mathrm{4}{p}^{\mathrm{3}} }\mid<\mathrm{1}\:\Rightarrow\:βˆ’\mathrm{1}<\frac{\mathrm{3}{q}}{\mathrm{2}}\sqrt{βˆ’\frac{\mathrm{3}}{{p}}}<\mathrm{1}\:\Rightarrow \\ $$$$\Rightarrow\:\mathrm{we}\:\mathrm{put}\:βˆ’\frac{\mathrm{3}{q}}{\mathrm{2}}\sqrt{βˆ’\frac{\mathrm{3}}{{p}}}=\mathrm{sin}\:\mathrm{3}\alpha \\ $$$${z}^{\mathrm{3}} βˆ’\frac{\mathrm{3}}{\mathrm{4}}{z}+\frac{\mathrm{sin}\:\mathrm{3}\alpha}{\mathrm{4}}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{put}\:{a}=βˆ’\mathrm{2}\sqrt{βˆ’\frac{{p}}{\mathrm{3}}}\:\mathrm{we}\:\mathrm{get}\:\mathrm{the}\:\mathrm{same}\:\mathrm{equation} \\ $$$$\mathrm{by}\:\mathrm{putting}\:\frac{\mathrm{3}{q}}{\mathrm{2}}\sqrt{βˆ’\frac{\mathrm{3}}{{p}}}=\mathrm{sin}\:\mathrm{3}\alpha \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *