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Question-198575

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Question Number 198575 by ajfour last updated on 22/Oct/23
Commented by ajfour last updated on 22/Oct/23
p^3 =p+c ∀ 0<c<(2/(3(√3))) . Find p in surd form.
$${p}^{\mathrm{3}} ={p}+{c}\:\:\:\:\:\forall\:\:\mathrm{0}<{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\:\:\:.\:{Find}\:{p}\:{in} \\ $$$${surd}\:{form}. \\ $$
Commented by mr W last updated on 22/Oct/23
i don′t think it is possible.
$${i}\:{don}'{t}\:{think}\:{it}\:{is}\:{possible}. \\ $$
Commented by ajfour last updated on 22/Oct/23
Earlier i did not understand why so is not possible. Now i do understand, but yet trying out of habit.��
Answered by ajfour last updated on 22/Oct/23
k=h^3 −h−c line equation: y=(((k−t)/h))x+t s^2 =(((((k−t)/h))p+t)/(1+(((k−t)/h))^2 )) p=s^2 ((h/(k−t))+((k−t)/h))−(((ht)/(k−t))) p=(((s^2 −t)h^2 +s^2 (h^3 −h−c−t)^2 )/(h(h^3 −h−c−t))) say p=(((S−t)h^2 +S(H−t)^2 )/(h(H−t))) p^3 =p+c (S−t)^3 h^6 +S^3 (H−t)^6 +3h^4 S(S−t)^2 (H−t)^2 +3h^2 S^2 (S−t)(H−t)^4 = h^3 (S−t)(H−t)^2 +h^2 S(H−t)^4 +ch^3 (H−t)^3 ⇒ S^3 {h^6 +(H−t)^6 +3h^4 (H−t)^2 +3h^2 (H−t)^4 } +S^2 {−3h^6 t−6h^4 t(H−t)^2 −3h^2 t(H−t)^4 } +S{3h^6 t^2 +3h^4 t^2 (H−t)^2 −h^3 (H−t)^2 −h^2 (H−t)^4 } −h^3 {h^3 t^3 +t(H−t)^2 −c(H−t)^3 }=0 ⇒ h^3 z^3 +((hz)/h)=c if z=(t/(H−t)) D=0 ⇒ (c^2 /4)+(1/(27h^3 ))=0 ⇒ h=−(((27c^2 )/4))^(1/3) & hz=2((c/2))^(1/3) ⇒ z^3 =4c×(−(4/(27c^2 )))=−((16)/(27c)) z=(t/(H−t)) ⇒ t=((Hz)/(1+z))=(((h^3 −h−c)z)/(1+z)) t=((−((27c^2 )/4)+(((27c^2 )/4))^(1/3) −c)/(1−(((27c)/(16)))^(1/3) )) t, h obtained; S =0, S_1 , S_2 p=(((S−t)h^2 +S(H−t)^2 )/(h(H−t))) havnt checked though!
$${k}={h}^{\mathrm{3}} −{h}−{c} \\ $$$${line}\:{equation}:\:\:{y}=\left(\frac{{k}−{t}}{{h}}\right){x}+{t} \\ $$$${s}^{\mathrm{2}} =\frac{\left(\frac{{k}−{t}}{{h}}\right){p}+{t}}{\mathrm{1}+\left(\frac{{k}−{t}}{{h}}\right)^{\mathrm{2}} } \\ $$$${p}={s}^{\mathrm{2}} \left(\frac{{h}}{{k}−{t}}+\frac{{k}−{t}}{{h}}\right)−\left(\frac{{ht}}{{k}−{t}}\right) \\ $$$$\:\:{p}=\frac{\left({s}^{\mathrm{2}} −{t}\right){h}^{\mathrm{2}} +{s}^{\mathrm{2}} \left({h}^{\mathrm{3}} −{h}−{c}−{t}\right)^{\mathrm{2}} }{{h}\left({h}^{\mathrm{3}} −{h}−{c}−{t}\right)} \\ $$$${say}\:\:{p}=\frac{\left({S}−{t}\right){h}^{\mathrm{2}} +{S}\left({H}−{t}\right)^{\mathrm{2}} }{{h}\left({H}−{t}\right)} \\ $$$${p}^{\mathrm{3}} ={p}+{c} \\ $$$$\left({S}−{t}\right)^{\mathrm{3}} {h}^{\mathrm{6}} +{S}^{\mathrm{3}} \left({H}−{t}\right)^{\mathrm{6}} \\ $$$$\:\:\:\:\:\:+\mathrm{3}{h}^{\mathrm{4}} {S}\left({S}−{t}\right)^{\mathrm{2}} \left({H}−{t}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:+\mathrm{3}{h}^{\mathrm{2}} {S}^{\mathrm{2}} \left({S}−{t}\right)\left({H}−{t}\right)^{\mathrm{4}} \\ $$$$\:\:\:\:=\:{h}^{\mathrm{3}} \left({S}−{t}\right)\left({H}−{t}\right)^{\mathrm{2}} +{h}^{\mathrm{2}} {S}\left({H}−{t}\right)^{\mathrm{4}} \\ $$$$\:\:\:\:\:+{ch}^{\mathrm{3}} \left({H}−{t}\right)^{\mathrm{3}} \\ $$$$\Rightarrow \\ $$$${S}^{\mathrm{3}} \left\{{h}^{\mathrm{6}} +\left({H}−{t}\right)^{\mathrm{6}} +\mathrm{3}{h}^{\mathrm{4}} \left({H}−{t}\right)^{\mathrm{2}} +\mathrm{3}{h}^{\mathrm{2}} \left({H}−{t}\right)^{\mathrm{4}} \right\} \\ $$$$+{S}^{\mathrm{2}} \left\{−\mathrm{3}{h}^{\mathrm{6}} {t}−\mathrm{6}{h}^{\mathrm{4}} {t}\left({H}−{t}\right)^{\mathrm{2}} −\mathrm{3}{h}^{\mathrm{2}} {t}\left({H}−{t}\right)^{\mathrm{4}} \right\} \\ $$$$+{S}\left\{\mathrm{3}{h}^{\mathrm{6}} {t}^{\mathrm{2}} +\mathrm{3}{h}^{\mathrm{4}} {t}^{\mathrm{2}} \left({H}−{t}\right)^{\mathrm{2}} −{h}^{\mathrm{3}} \left({H}−{t}\right)^{\mathrm{2}} \right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:−{h}^{\mathrm{2}} \left({H}−{t}\right)^{\mathrm{4}} \right\} \\ $$$$−{h}^{\mathrm{3}} \left\{{h}^{\mathrm{3}} {t}^{\mathrm{3}} +{t}\left({H}−{t}\right)^{\mathrm{2}} −{c}\left({H}−{t}\right)^{\mathrm{3}} \right\}=\mathrm{0} \\ $$$$\Rightarrow\:\:{h}^{\mathrm{3}} {z}^{\mathrm{3}} +\frac{{hz}}{{h}}={c}\:\:\:\:{if}\:\:\:{z}=\frac{{t}}{{H}−{t}} \\ $$$${D}=\mathrm{0}\:\:\:\Rightarrow\:\:\:\frac{{c}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{27}{h}^{\mathrm{3}} }=\mathrm{0} \\ $$$$\Rightarrow\:\:\:{h}=−\left(\frac{\mathrm{27}{c}^{\mathrm{2}} }{\mathrm{4}}\right)^{\mathrm{1}/\mathrm{3}} \:\:\:\&\:\:{hz}=\mathrm{2}\left(\frac{{c}}{\mathrm{2}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\Rightarrow\:\:{z}^{\mathrm{3}} =\mathrm{4}{c}×\left(−\frac{\mathrm{4}}{\mathrm{27}{c}^{\mathrm{2}} }\right)=−\frac{\mathrm{16}}{\mathrm{27}{c}} \\ $$$${z}=\frac{{t}}{{H}−{t}}\:\:\:\Rightarrow\:\:\:{t}=\frac{{Hz}}{\mathrm{1}+{z}}=\frac{\left({h}^{\mathrm{3}} −{h}−{c}\right){z}}{\mathrm{1}+{z}} \\ $$$$\:\:\:{t}=\frac{−\frac{\mathrm{27}{c}^{\mathrm{2}} }{\mathrm{4}}+\left(\frac{\mathrm{27}{c}^{\mathrm{2}} }{\mathrm{4}}\right)^{\mathrm{1}/\mathrm{3}} −{c}}{\mathrm{1}−\left(\frac{\mathrm{27}{c}}{\mathrm{16}}\right)^{\mathrm{1}/\mathrm{3}} } \\ $$$${t},\:{h}\:{obtained};\:{S}\:=\mathrm{0},\:{S}_{\mathrm{1}} ,\:{S}_{\mathrm{2}} \\ $$$${p}=\frac{\left({S}−{t}\right){h}^{\mathrm{2}} +{S}\left({H}−{t}\right)^{\mathrm{2}} }{{h}\left({H}−{t}\right)} \\ $$$$\:\:{havnt}\:{checked}\:{though}! \\ $$

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