Question Number 130089 by ajfour last updated on 22/Jan/21
![To solve x^3 =x+c Let y=(x−p)(x^3 −x−c) (dy/dx)=(x^3 −x−c)+(3x^2 −1)(x−p) let (dy/dx)=mx ⇒ (3x^2 −1)(x−p)=mx 3x^3 −3px^2 −(m+1)x+p=0 3(x+c)−3px^2 −(m+1)x+p=0 3px^2 +(m−2)x−(3c+p)=0 and since x^3 =x+c (m−2)x^2 +(2p−3c)x+3cp=0 [9cp^2 +(3c+p)(m−2)]x +[3cp(m−2)+(3c+p)(2p−3c)] =0 x=−(((3c+p)(2p−3c)+3cp(m−2))/(9cp^2 +(3c+p)(m−2))) (m−2)x^2 +(2p−3c)x+(3cp)=0 Now choosing suitable ′p′ value we find ′m′; and then x=f(p,m) ★](Q130089.png)
$${To}\:{solve}\:\:\:{x}^{\mathrm{3}} ={x}+{c} \\ $$$${Let}\:\:{y}=\left({x}−{p}\right)\left({x}^{\mathrm{3}} −{x}−{c}\right) \\ $$$$\:\:\frac{{dy}}{{dx}}=\left({x}^{\mathrm{3}} −{x}−{c}\right)+\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}−{p}\right) \\ $$$$\:\:{let}\:\:\:\frac{{dy}}{{dx}}={mx} \\ $$$$\Rightarrow\:\:\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}−{p}\right)={mx} \\ $$$$\mathrm{3}{x}^{\mathrm{3}} −\mathrm{3}{px}^{\mathrm{2}} −\left({m}+\mathrm{1}\right){x}+{p}=\mathrm{0} \\ $$$$\mathrm{3}\left({x}+{c}\right)−\mathrm{3}{px}^{\mathrm{2}} −\left({m}+\mathrm{1}\right){x}+{p}=\mathrm{0} \\ $$$$\mathrm{3}{px}^{\mathrm{2}} +\left({m}−\mathrm{2}\right){x}−\left(\mathrm{3}{c}+{p}\right)=\mathrm{0} \\ $$$${and}\:{since}\:{x}^{\mathrm{3}} ={x}+{c} \\ $$$$\left({m}−\mathrm{2}\right){x}^{\mathrm{2}} +\left(\mathrm{2}{p}−\mathrm{3}{c}\right){x}+\mathrm{3}{cp}=\mathrm{0} \\ $$$$\left[\mathrm{9}{cp}^{\mathrm{2}} +\left(\mathrm{3}{c}+{p}\right)\left({m}−\mathrm{2}\right)\right]{x} \\ $$$$+\left[\mathrm{3}{cp}\left({m}−\mathrm{2}\right)+\left(\mathrm{3}{c}+{p}\right)\left(\mathrm{2}{p}−\mathrm{3}{c}\right)\right] \\ $$$$=\mathrm{0} \\ $$$${x}=−\frac{\left(\mathrm{3}{c}+{p}\right)\left(\mathrm{2}{p}−\mathrm{3}{c}\right)+\mathrm{3}{cp}\left({m}−\mathrm{2}\right)}{\mathrm{9}{cp}^{\mathrm{2}} +\left(\mathrm{3}{c}+{p}\right)\left({m}−\mathrm{2}\right)} \\ $$$$\left({m}−\mathrm{2}\right){x}^{\mathrm{2}} +\left(\mathrm{2}{p}−\mathrm{3}{c}\right){x}+\left(\mathrm{3}{cp}\right)=\mathrm{0} \\ $$$${Now}\:{choosing}\:{suitable}\:'{p}'\:{value} \\ $$$${we}\:{find}\:'{m}'; \\ $$$${and}\:{then}\:{x}={f}\left({p},{m}\right)\:\bigstar \\ $$
Commented by bobhans last updated on 22/Jan/21
$${example}\::\:{x}^{\mathrm{3}} ={x}+\mathrm{6} \\ $$$${y}=\left({x}−{p}\right)\left({x}^{\mathrm{3}} −{x}−\mathrm{6}\right) \\ $$$$\frac{{dy}}{{dx}}\:=\:{x}^{\mathrm{3}} −{x}+\mathrm{6}+\left({x}−{p}\right)\left(\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$\frac{{dy}}{{dx}}=\:\mathrm{4}{x}^{\mathrm{3}} −\mathrm{4}{px}^{\mathrm{2}} −\mathrm{2}{x}+{p}+\mathrm{6} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{4}{x}+\mathrm{24}−\mathrm{4}{px}^{\mathrm{2}} −\mathrm{2}{x}+{p}+\mathrm{6}=\mathrm{2}{x}−\mathrm{4}{px}^{\mathrm{2}} +\mathrm{30}+{p} \\ $$$${let}\:\frac{{dy}}{{dx}}\:=\:{mx}\:\Rightarrow\mathrm{4}{px}^{\mathrm{2}} +\left({m}−\mathrm{2}\right){x}−\left({p}+\mathrm{30}\right)=\mathrm{0} \\ $$$${x}=\frac{\mathrm{2}−{m}\pm\sqrt{\left({m}−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{16}{p}\left({p}+\mathrm{30}\right)}}{\mathrm{8}{p}} \\ $$$${now}\:\begin{cases}{{p}=?}\\{{m}=?}\end{cases} \\ $$
Commented by ajfour last updated on 23/Jan/21
Answered by ajfour last updated on 23/Jan/21
![x=−(((3c+p)(2p−3c)+3cp(m−2))/(9cp^2 +(3c+p)(m−2))) (m−2)x^2 +(2p−3c)x+(3cp)=0 say m−2=M x=−((3cpM+(3c+p)(2p−3c))/((3c+p)M+9cp^2 )) let (M/p)=s , (1/p)=q x=−((3cs+(3cq+1)(2−3cq))/((3cq+1)s+9c)) s[3cs+(3cq+1)(2−3cq)]^2 −(2−3cq)[3cs+(3cq+1)(2−3cq)][(3cq+1)s+9c] +3c[(3cq+1)s+9c]^2 =0 9c^2 s^3 +6c(3cq+1)(2−3cq)s^2 +(3cq+1)^2 (2−3cq)^2 s_(−) −6c(3cq+1)s^2 −54c^2 s +9c^2 q(3cq+1)s^2 +81c^3 qs −(3cq+1)^2 (2−3cq)^2 s_(−) −9c(3cq+1)(2−3cq)^2 +3c(3cq+1)^2 s^2 +54c(3cq+1)s +243c^3 =0 ⇒ (9c^2 )s^3 +[6c(3cq+1)(2−3cq) −6c(3cq+1)+9c^2 q(3cq+1) +3c(3cq+1)^2 ]s^2 + [54c^2 +81c^3 q+54c(3cq+1)]s +[−9c(3cq+1)(2−3cq)^2 +243c^3 ]=0 let q=0 ⇒ s^3 +(1/c)s^2 +6((1/c)+1)s −((4/c)−27c)=0 s=−2.54409 , −0.95044 for c=((30)/(19(√(19)))) x=−((3cs+(3cq+1)(2−3cq))/((3cq+1)s+9c)) and for q=0 x=−((3cs+2)/(s+9c)) = right answer is x=1.14708](Q130106.png)
$${x}=−\frac{\left(\mathrm{3}{c}+{p}\right)\left(\mathrm{2}{p}−\mathrm{3}{c}\right)+\mathrm{3}{cp}\left({m}−\mathrm{2}\right)}{\mathrm{9}{cp}^{\mathrm{2}} +\left(\mathrm{3}{c}+{p}\right)\left({m}−\mathrm{2}\right)} \\ $$$$\left({m}−\mathrm{2}\right){x}^{\mathrm{2}} +\left(\mathrm{2}{p}−\mathrm{3}{c}\right){x}+\left(\mathrm{3}{cp}\right)=\mathrm{0} \\ $$$${say}\:\:{m}−\mathrm{2}={M} \\ $$$$\boldsymbol{{x}}=−\frac{\mathrm{3}\boldsymbol{{cpM}}+\left(\mathrm{3}\boldsymbol{{c}}+\boldsymbol{{p}}\right)\left(\mathrm{2}\boldsymbol{{p}}−\mathrm{3}\boldsymbol{{c}}\right)}{\left(\mathrm{3}\boldsymbol{{c}}+\boldsymbol{{p}}\right)\boldsymbol{{M}}+\mathrm{9}{cp}^{\mathrm{2}} } \\ $$$${let}\:\:\frac{{M}}{{p}}={s}\:\:,\:\:\frac{\mathrm{1}}{{p}}={q} \\ $$$$\:\:\:\boldsymbol{{x}}=−\frac{\mathrm{3}{c}\boldsymbol{{s}}+\left(\mathrm{3}{cq}+\mathrm{1}\right)\left(\mathrm{2}−\mathrm{3}{cq}\right)}{\left(\mathrm{3}{cq}+\mathrm{1}\right)\boldsymbol{{s}}+\mathrm{9}{c}} \\ $$$$\boldsymbol{{s}}\left[\mathrm{3}\boldsymbol{{cs}}+\left(\mathrm{3}\boldsymbol{{cq}}+\mathrm{1}\right)\left(\mathrm{2}−\mathrm{3}\boldsymbol{{cq}}\right)\right]^{\mathrm{2}} \\ $$$$−\left(\mathrm{2}−\mathrm{3}\boldsymbol{{cq}}\right)\left[\mathrm{3}\boldsymbol{{cs}}+\left(\mathrm{3}\boldsymbol{{cq}}+\mathrm{1}\right)\left(\mathrm{2}−\mathrm{3}\boldsymbol{{cq}}\right)\right]\left[\left(\mathrm{3}\boldsymbol{{cq}}+\mathrm{1}\right)\boldsymbol{{s}}+\mathrm{9}\boldsymbol{{c}}\right] \\ $$$$+\mathrm{3}\boldsymbol{{c}}\left[\left(\mathrm{3}\boldsymbol{{cq}}+\mathrm{1}\right)\boldsymbol{{s}}+\mathrm{9}\boldsymbol{{c}}\right]^{\mathrm{2}} =\mathrm{0} \\ $$$$ \\ $$$$\mathrm{9}\boldsymbol{{c}}^{\mathrm{2}} \boldsymbol{{s}}^{\mathrm{3}} +\mathrm{6}\boldsymbol{{c}}\left(\mathrm{3}\boldsymbol{{cq}}+\mathrm{1}\right)\left(\mathrm{2}−\mathrm{3}\boldsymbol{{cq}}\right)\boldsymbol{{s}}^{\mathrm{2}} \\ $$$$+\underset{−} {\left(\mathrm{3}\boldsymbol{{cq}}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{2}−\mathrm{3}\boldsymbol{{cq}}\right)^{\mathrm{2}} \boldsymbol{{s}}} \\ $$$$−\mathrm{6}\boldsymbol{{c}}\left(\mathrm{3}\boldsymbol{{cq}}+\mathrm{1}\right)\boldsymbol{{s}}^{\mathrm{2}} −\mathrm{54}\boldsymbol{{c}}^{\mathrm{2}} \boldsymbol{{s}} \\ $$$$+\mathrm{9}\boldsymbol{{c}}^{\mathrm{2}} \boldsymbol{{q}}\left(\mathrm{3}\boldsymbol{{cq}}+\mathrm{1}\right)\boldsymbol{{s}}^{\mathrm{2}} +\mathrm{81}\boldsymbol{{c}}^{\mathrm{3}} \boldsymbol{{qs}} \\ $$$$−\underset{−} {\left(\mathrm{3}\boldsymbol{{cq}}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{2}−\mathrm{3}\boldsymbol{{cq}}\right)^{\mathrm{2}} \boldsymbol{{s}}} \\ $$$$−\mathrm{9}\boldsymbol{{c}}\left(\mathrm{3}\boldsymbol{{cq}}+\mathrm{1}\right)\left(\mathrm{2}−\mathrm{3}\boldsymbol{{cq}}\right)^{\mathrm{2}} \\ $$$$+\mathrm{3}\boldsymbol{{c}}\left(\mathrm{3}\boldsymbol{{cq}}+\mathrm{1}\right)^{\mathrm{2}} \boldsymbol{{s}}^{\mathrm{2}} +\mathrm{54}\boldsymbol{{c}}\left(\mathrm{3}\boldsymbol{{cq}}+\mathrm{1}\right)\boldsymbol{{s}} \\ $$$$+\mathrm{243}\boldsymbol{{c}}^{\mathrm{3}} =\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{9}{c}^{\mathrm{2}} \right)\boldsymbol{{s}}^{\mathrm{3}} +\left[\mathrm{6}{c}\left(\mathrm{3}{cq}+\mathrm{1}\right)\left(\mathrm{2}−\mathrm{3}{cq}\right)\right. \\ $$$$\:\:−\mathrm{6}{c}\left(\mathrm{3}{cq}+\mathrm{1}\right)+\mathrm{9}{c}^{\mathrm{2}} {q}\left(\mathrm{3}{cq}+\mathrm{1}\right) \\ $$$$\left.\:\:+\mathrm{3}{c}\left(\mathrm{3}{cq}+\mathrm{1}\right)^{\mathrm{2}} \right]\boldsymbol{{s}}^{\mathrm{2}} + \\ $$$$\:\:\:\left[\mathrm{54}{c}^{\mathrm{2}} +\mathrm{81}{c}^{\mathrm{3}} {q}+\mathrm{54}{c}\left(\mathrm{3}{cq}+\mathrm{1}\right)\right]\boldsymbol{{s}} \\ $$$$+\left[−\mathrm{9}{c}\left(\mathrm{3}{cq}+\mathrm{1}\right)\left(\mathrm{2}−\mathrm{3}{cq}\right)^{\mathrm{2}} +\mathrm{243}{c}^{\mathrm{3}} \right]=\mathrm{0} \\ $$$${let}\:\:{q}=\mathrm{0} \\ $$$$\Rightarrow\:\:{s}^{\mathrm{3}} +\frac{\mathrm{1}}{{c}}{s}^{\mathrm{2}} +\mathrm{6}\left(\frac{\mathrm{1}}{{c}}+\mathrm{1}\right){s} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:−\left(\frac{\mathrm{4}}{{c}}−\mathrm{27}{c}\right)=\mathrm{0} \\ $$$${s}=−\mathrm{2}.\mathrm{54409}\:,\:−\mathrm{0}.\mathrm{95044} \\ $$$${for}\:{c}=\frac{\mathrm{30}}{\mathrm{19}\sqrt{\mathrm{19}}} \\ $$$$\:\:\:\boldsymbol{{x}}=−\frac{\mathrm{3}{c}\boldsymbol{{s}}+\left(\mathrm{3}{cq}+\mathrm{1}\right)\left(\mathrm{2}−\mathrm{3}{cq}\right)}{\left(\mathrm{3}{cq}+\mathrm{1}\right)\boldsymbol{{s}}+\mathrm{9}{c}} \\ $$$${and}\:{for}\:\:{q}=\mathrm{0} \\ $$$$\:\:{x}=−\frac{\mathrm{3}{cs}+\mathrm{2}}{{s}+\mathrm{9}{c}}\:=\: \\ $$$${right}\:{answer}\:{is}\:{x}=\mathrm{1}.\mathrm{14708} \\ $$$$ \\ $$