Question Number 129832 by ajfour last updated on 19/Jan/21
Commented by ajfour last updated on 19/Jan/21
$${If}\:{p},\:{q},\:{r}\:{are}\:{roots}\:{of}\:{y}={x}^{\mathrm{3}} −{x}−{c} \\ $$$${then}\:{find}\:{p}\:{or}\:{q}\:{or}\:{r},\:{with}\:{the} \\ $$$${help}\:{of}\:{s}. \\ $$
Commented by MJS_new last updated on 20/Jan/21
![my opinion: x^4 +ax^2 +bx+c=0 can be solved exactly if we can find exact factors α, β, γ with (x^2 −αx−β)(x^2 +αx−γ)=x^4 +ax^2 +bx+c ⇔ we must find at least one useable exact solution for y^3 +Py+Q=0 with P=−((a^2 +12c)/3)∧Q=−((2a^3 −72ac+27b^2 )/(27)) now what you seem to want is finding s in order to solve the following: (X^3 −X−C)(X−s)=0 ⇔ X^4 −sX^3 −X^2 −(C−s)X+Cs=0 ⇔ [X=x+(s/4)] ⇔ x^4 −((3s^2 +8)/8)x^2 −((s^3 −4s+8C)/8)x−((3s(s^3 −16s−64C))/(256))=0 ⇒ we must be able to find a solution of y^3 +Py+Q=0 with P=−((3s^2 +9Cs+1)/3) Q=−((27Cs^3 +18s^2 +27Cs+37C^2 −2)/(27)) try with C=(1/3) I don′t think you can find a fitting s following this path without solving another 3^(rd) degree polynome](https://www.tinkutara.com/question/Q129861.png)
$$\mathrm{my}\:\mathrm{opinion}: \\ $$$${x}^{\mathrm{4}} +{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$$\mathrm{can}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{exactly}\:\mathrm{if}\:\mathrm{we}\:\mathrm{can}\:\mathrm{find}\:\mathrm{exact} \\ $$$$\mathrm{factors}\:\alpha,\:\beta,\:\gamma\:\mathrm{with} \\ $$$$\left({x}^{\mathrm{2}} −\alpha{x}−\beta\right)\left({x}^{\mathrm{2}} +\alpha{x}−\gamma\right)={x}^{\mathrm{4}} +{ax}^{\mathrm{2}} +{bx}+{c} \\ $$$$\Leftrightarrow \\ $$$$\mathrm{we}\:\mathrm{must}\:\mathrm{find}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:{useable}\:\mathrm{exact} \\ $$$$\mathrm{solution}\:\mathrm{for} \\ $$$${y}^{\mathrm{3}} +{Py}+{Q}=\mathrm{0}\:\mathrm{with}\:{P}=−\frac{{a}^{\mathrm{2}} +\mathrm{12}{c}}{\mathrm{3}}\wedge{Q}=−\frac{\mathrm{2}{a}^{\mathrm{3}} −\mathrm{72}{ac}+\mathrm{27}{b}^{\mathrm{2}} }{\mathrm{27}} \\ $$$$\mathrm{now}\:\mathrm{what}\:\mathrm{you}\:\mathrm{seem}\:\mathrm{to}\:\mathrm{want}\:\mathrm{is}\:\mathrm{finding}\:{s}\:\mathrm{in} \\ $$$$\mathrm{order}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{following}: \\ $$$$\left({X}^{\mathrm{3}} −{X}−{C}\right)\left({X}−{s}\right)=\mathrm{0} \\ $$$$\Leftrightarrow \\ $$$${X}^{\mathrm{4}} −{sX}^{\mathrm{3}} −{X}^{\mathrm{2}} −\left({C}−{s}\right){X}+{Cs}=\mathrm{0}\:\Leftrightarrow \\ $$$$\:\:\:\:\:\left[{X}={x}+\frac{{s}}{\mathrm{4}}\right] \\ $$$$\Leftrightarrow\:{x}^{\mathrm{4}} −\frac{\mathrm{3}{s}^{\mathrm{2}} +\mathrm{8}}{\mathrm{8}}{x}^{\mathrm{2}} −\frac{{s}^{\mathrm{3}} −\mathrm{4}{s}+\mathrm{8}{C}}{\mathrm{8}}{x}−\frac{\mathrm{3}{s}\left({s}^{\mathrm{3}} −\mathrm{16}{s}−\mathrm{64}{C}\right)}{\mathrm{256}}=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{we}\:\mathrm{must}\:\mathrm{be}\:\mathrm{able}\:\mathrm{to}\:\mathrm{find}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{of} \\ $$$${y}^{\mathrm{3}} +{Py}+{Q}=\mathrm{0} \\ $$$$\mathrm{with} \\ $$$${P}=−\frac{\mathrm{3}{s}^{\mathrm{2}} +\mathrm{9}{Cs}+\mathrm{1}}{\mathrm{3}} \\ $$$${Q}=−\frac{\mathrm{27}{Cs}^{\mathrm{3}} +\mathrm{18}{s}^{\mathrm{2}} +\mathrm{27}{Cs}+\mathrm{37}{C}^{\mathrm{2}} −\mathrm{2}}{\mathrm{27}} \\ $$$$\mathrm{try}\:\mathrm{with} \\ $$$${C}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{you}\:\mathrm{can}\:\mathrm{find}\:\mathrm{a}\:\mathrm{fitting}\:{s}\:\mathrm{following} \\ $$$$\mathrm{this}\:\mathrm{path}\:{without}\:\mathrm{solving}\:\mathrm{another}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{degree} \\ $$$$\mathrm{polynome} \\ $$
Answered by ajfour last updated on 20/Jan/21
$$\:\:{y}={x}^{\mathrm{4}} −{sx}^{\mathrm{3}} −{x}^{\mathrm{2}} +\left({s}−{c}\right){x}+{cs} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{4}{x}^{\mathrm{3}} −\mathrm{3}{sx}^{\mathrm{2}} −\mathrm{2}{x}+{s}−{c} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{12}{x}^{\mathrm{2}} −\mathrm{6}{sx}−\mathrm{2} \\ $$$${let}\:\:\mathrm{4}{x}^{\mathrm{3}} −\mathrm{3}{sx}^{\mathrm{2}} −\mathrm{2}{x}+{s}−{c} \\ $$$$\:\:\:\:\:\:\:=\:{s}−{c} \\ $$$$\Rightarrow\:\:\:\mathrm{4}{x}^{\mathrm{2}} −\mathrm{3}{sx}−\mathrm{2}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\mathrm{3}{sx}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{4}{c}=\mathrm{0} \\ $$$$\Rightarrow\:\:\left(\mathrm{8}{c}−\mathrm{3}{s}\right){x}=\mathrm{6}{cs}−\mathrm{2} \\ $$$$\Rightarrow\:\:{x}=\frac{\mathrm{6}{cs}−\mathrm{2}}{\mathrm{8}{c}−\mathrm{3}{s}} \\ $$$$\:\mathrm{4}\left(\mathrm{6}{cs}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{3}{s}\left(\mathrm{6}{cs}−\mathrm{2}\right)\left(\mathrm{8}{c}−\mathrm{3}{s}\right) \\ $$$$\:\:−\mathrm{2}\left(\mathrm{8}{c}−\mathrm{3}{s}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$… \\ $$