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Question Number 123451 by Fikret last updated on 25/Nov/20

a,b,c∈R ax^4 +bx^3 +cx^2 +bx+a=0 sulation set of the equation?

$${a},{b},{c}\in\mathbb{R} \\ $$$${ax}^{\mathrm{4}} +{bx}^{\mathrm{3}} +{cx}^{\mathrm{2}} +{bx}+{a}=\mathrm{0} \\ $$$${sulation}\:{set}\:{of}\:{the}\:{equation}? \\ $$$$ \\ $$

Answered by TANMAY PANACEA last updated on 25/Nov/20

ax^2 +(a/x^2 )+bx+(b/x)+c=0 a(x^2 +(1/x^2 ))+b(x+(1/x))+c=0 x+(1/x)=y a(y^2 −2)+b(y)+c=0 ay^2 +by+c−2a=0 y=((−b±(√(b^2 −4a(c−2a))) )/(2a))=m and n m=considering +ve sign before (√ ) sign n=considering −ve sign before (√ ) sign x+(1/x)=m x^2 −mx+1=0 x=((m±(√(m^2 −4)))/2) x=((n±(√(n^2 −4)))/2)

$${ax}^{\mathrm{2}} +\frac{{a}}{{x}^{\mathrm{2}} }+{bx}+\frac{{b}}{{x}}+{c}=\mathrm{0} \\ $$$${a}\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)+{b}\left({x}+\frac{\mathrm{1}}{{x}}\right)+{c}=\mathrm{0} \\ $$$${x}+\frac{\mathrm{1}}{{x}}={y} \\ $$$${a}\left({y}^{\mathrm{2}} −\mathrm{2}\right)+{b}\left({y}\right)+{c}=\mathrm{0} \\ $$$${ay}^{\mathrm{2}} +{by}+{c}−\mathrm{2}{a}=\mathrm{0} \\ $$$${y}=\frac{−{b}\pm\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{a}\left({c}−\mathrm{2}{a}\right)}\:}{\mathrm{2}{a}}={m}\:{and}\:\:{n} \\ $$$${m}={considering}\:+{ve}\:{sign}\:{before}\:\sqrt{\:}\:{sign} \\ $$$${n}={considering}\:−{ve}\:{sign}\:{before}\:\sqrt{\:}\:{sign} \\ $$$${x}+\frac{\mathrm{1}}{{x}}={m} \\ $$$${x}^{\mathrm{2}} −{mx}+\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{{m}\pm\sqrt{{m}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}} \\ $$$${x}=\frac{{n}\pm\sqrt{{n}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}} \\ $$

Commented by MJS_new last updated on 25/Nov/20

good!

$$\mathrm{good}! \\ $$

Commented by TANMAY PANACEA last updated on 25/Nov/20

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Answered by Bird last updated on 25/Nov/20

e ⇒x^(4 ) +(b/a)x^3 +(c/a)x^2 +(b/a)x +1=0 (a≠0) equation at form x^4 +αx^3 +βx^(2 ) +αx +1 =0⇒ x^2 +αx +β +(α/x) +(1/x^2 )=0 ⇒ (x+(1/x))^2 −2+α(x+(1/x))+β=0 we put x+(1/x)=z ⇒ z^2 +αz+β−2=0 Δ=α^2 −4(β−2) =α^2 −4β+8 Δ>0 ⇒z_1 =((−α+(√(α^2 −4β+8)))/2) z_2 =((−α−(√(α^2 −4β+8)))/2) after we solve x+(1/x)=z_i if Δ<0⇒z_1 =((−α+i(√(−Δ)))/2) z_2 =((−α−i(√(−Δ)))/2)....

$${e}\:\Rightarrow{x}^{\mathrm{4}\:} \:+\frac{{b}}{{a}}{x}^{\mathrm{3}} \:+\frac{{c}}{{a}}{x}^{\mathrm{2}} \:+\frac{{b}}{{a}}{x}\:+\mathrm{1}=\mathrm{0} \\ $$$$\left({a}\neq\mathrm{0}\right)\:{equation}\:{at}\:{form} \\ $$$${x}^{\mathrm{4}} \:+\alpha{x}^{\mathrm{3}} \:+\beta{x}^{\mathrm{2}\:} +\alpha{x}\:+\mathrm{1}\:=\mathrm{0}\Rightarrow \\ $$$${x}^{\mathrm{2}} \:+\alpha{x}\:+\beta\:+\frac{\alpha}{{x}}\:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{0}\:\Rightarrow \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}+\alpha\left({x}+\frac{\mathrm{1}}{{x}}\right)+\beta=\mathrm{0} \\ $$$${we}\:{put}\:{x}+\frac{\mathrm{1}}{{x}}={z}\:\Rightarrow \\ $$$${z}^{\mathrm{2}} +\alpha{z}+\beta−\mathrm{2}=\mathrm{0} \\ $$$$\Delta=\alpha^{\mathrm{2}} −\mathrm{4}\left(\beta−\mathrm{2}\right)\:=\alpha^{\mathrm{2}} −\mathrm{4}\beta+\mathrm{8} \\ $$$$\Delta>\mathrm{0}\:\Rightarrow{z}_{\mathrm{1}} =\frac{−\alpha+\sqrt{\alpha^{\mathrm{2}} −\mathrm{4}\beta+\mathrm{8}}}{\mathrm{2}} \\ $$$${z}_{\mathrm{2}} =\frac{−\alpha−\sqrt{\alpha^{\mathrm{2}} −\mathrm{4}\beta+\mathrm{8}}}{\mathrm{2}} \\ $$$${after}\:{we}\:{solve}\:{x}+\frac{\mathrm{1}}{{x}}={z}_{{i}} \\ $$$${if}\:\Delta<\mathrm{0}\Rightarrow{z}_{\mathrm{1}} =\frac{−\alpha+{i}\sqrt{−\Delta}}{\mathrm{2}} \\ $$$${z}_{\mathrm{2}} =\frac{−\alpha−{i}\sqrt{−\Delta}}{\mathrm{2}}.... \\ $$

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