Question Number 64758 by Tawa1 last updated on 21/Jul/19
$$\mathrm{Solve}:\:\:\:\:\:\mathrm{x}^{\mathrm{4}} \:+\:\mathrm{5x}^{\mathrm{3}} \:−\:\mathrm{4x}^{\mathrm{2}} \:+\:\mathrm{7x}\:−\:\mathrm{1}\:\:=\:\:\mathrm{0} \\ $$
Answered by ajfour last updated on 21/Jul/19
![x=t−(5/4) say we then get t^4 −at^2 +bt−c=0 ⇒ (t^2 −pt+q)(t^2 +pt+r)=0 ⇒ t^4 +(r+q−p^2 )t^2 +(−pr+pq)t+qr=0 ⇒ p^2 −(q+r)=a p(q−r)=b qr=−c ⇒ (p^2 −a)^2 −(b^2 /p^2 )=−4c let p^2 =s ⇒s^3 −2as^2 +(a^2 +4c)s−b^2 =0 let s=k+((2a)/3) ⇒ k^3 +((4a^2 k)/3)+((8a^3 )/(27))−((8a^2 k)/3)−((8a^3 )/9) +(a^2 +4c)k+((2a(a^2 +4c))/3)−b^2 =0 say k^3 +Pk+Q=0 k=[−(Q/2)+(√((Q^2 /4)+(P^( 3) /(27))))]^(1/3) +[−(Q/2)−(√((Q^2 /4)+(P^( 3) /(27))))]^(1/3) P =4c−(a^2 /3) ; Q= ((2a^3 )/(27))+((8ac)/3)−b^2 ■](Q64761.png)
$${x}={t}−\frac{\mathrm{5}}{\mathrm{4}} \\ $$$${say}\:\:{we}\:{then}\:{get} \\ $$$$\:\:\:{t}^{\mathrm{4}} −{at}^{\mathrm{2}} +{bt}−{c}=\mathrm{0} \\ $$$$\Rightarrow\:\left({t}^{\mathrm{2}} −{pt}+{q}\right)\left({t}^{\mathrm{2}} +{pt}+{r}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:{t}^{\mathrm{4}} +\left({r}+{q}−{p}^{\mathrm{2}} \right){t}^{\mathrm{2}} +\left(−{pr}+{pq}\right){t}+{qr}=\mathrm{0} \\ $$$$\Rightarrow\:\:{p}^{\mathrm{2}} −\left({q}+{r}\right)={a} \\ $$$$\:\:\:\:\:\:\:{p}\left({q}−{r}\right)={b} \\ $$$$\:\:\:\:\:\:\:{qr}=−{c} \\ $$$$\Rightarrow\:\:\left({p}^{\mathrm{2}} −{a}\right)^{\mathrm{2}} −\frac{{b}^{\mathrm{2}} }{{p}^{\mathrm{2}} }=−\mathrm{4}{c} \\ $$$${let}\:\:\:\:{p}^{\mathrm{2}} ={s} \\ $$$$\Rightarrow{s}^{\mathrm{3}} −\mathrm{2}{as}^{\mathrm{2}} +\left({a}^{\mathrm{2}} +\mathrm{4}{c}\right){s}−{b}^{\mathrm{2}} =\mathrm{0} \\ $$$${let}\:\:\:\:{s}={k}+\frac{\mathrm{2}{a}}{\mathrm{3}} \\ $$$$\Rightarrow\:\:{k}^{\mathrm{3}} +\frac{\mathrm{4}{a}^{\mathrm{2}} {k}}{\mathrm{3}}+\frac{\mathrm{8}{a}^{\mathrm{3}} }{\mathrm{27}}−\frac{\mathrm{8}{a}^{\mathrm{2}} {k}}{\mathrm{3}}−\frac{\mathrm{8}{a}^{\mathrm{3}} }{\mathrm{9}} \\ $$$$\:\:\:\:\:+\left({a}^{\mathrm{2}} +\mathrm{4}{c}\right){k}+\frac{\mathrm{2}{a}\left({a}^{\mathrm{2}} +\mathrm{4}{c}\right)}{\mathrm{3}}−{b}^{\mathrm{2}} =\mathrm{0} \\ $$$${say} \\ $$$$\:\:\:\:\:\:\:\:\:\:{k}^{\mathrm{3}} +{Pk}+{Q}=\mathrm{0} \\ $$$${k}=\left[−\frac{{Q}}{\mathrm{2}}+\sqrt{\frac{{Q}^{\mathrm{2}} }{\mathrm{4}}+\frac{{P}^{\:\mathrm{3}} }{\mathrm{27}}}\right]^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:+\left[−\frac{{Q}}{\mathrm{2}}−\sqrt{\frac{{Q}^{\mathrm{2}} }{\mathrm{4}}+\frac{{P}^{\:\mathrm{3}} }{\mathrm{27}}}\right]^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:{P}\:=\mathrm{4}{c}−\frac{{a}^{\mathrm{2}} }{\mathrm{3}}\:\:\:;\: \\ $$$$\:\:\:{Q}=\:\frac{\mathrm{2}{a}^{\mathrm{3}} }{\mathrm{27}}+\frac{\mathrm{8}{ac}}{\mathrm{3}}−{b}^{\mathrm{2}} \: \\ $$$$\blacksquare \\ $$
Commented by Tawa1 last updated on 21/Jul/19
$$\mathrm{Wow},\:\:\mathrm{waiting}\:\mathrm{for}\:\mathrm{the}\:\mathrm{rest}\:\mathrm{sir} \\ $$
Answered by MJS last updated on 21/Jul/19
$$\mathrm{no}\:``\mathrm{beautiful}''\:\mathrm{solution} \\ $$$${x}_{\mathrm{1}} \approx−\mathrm{5}.\mathrm{88645} \\ $$$${x}_{\mathrm{2}} \approx.\mathrm{153681} \\ $$$${x}_{\mathrm{3},\:\mathrm{4}} \approx.\mathrm{366384}\pm.\mathrm{985486i} \\ $$
Commented by ajfour last updated on 21/Jul/19
$${Thanks}\:{sir},\:{i}\:{had}\:{made}\:{blunder}, \\ $$$${had}\:{practiced}\:{quintic}\:{excessively} \\ $$$${before},\:{thats}\:{why}.. \\ $$