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Question Number 223374 by behi834171 last updated on 22/Jul/25
Answered by Ghisom last updated on 22/Jul/25
r^4 +4r+1=0 r^4 =−4r−1 r^4 −1=−4r−2 (r−r_1 )(r−r_2 )(r−r_3 )(r−r_4 )=r^4 +4r+1 ⇒ A=r_1 +r_2 +r_3 +r_4 =0 B=r_1 r_2 +r_1 r_3 +r_1 r_4 +r_2 r_3 +r_2 r_4 +r_3 r_4 =0 C=r_1 r_2 r_3 +r_1 r_2 r_4 +r_1 r_3 r_4 +r_2 r_3 r_4 =−4 D=r_1 r_2 r_3 r_4 =1 ⇒e Π_(i=1) ^4 (r_i ^4 −1)= =(−2)^4 (2r_1 −1)(2r_2 +1)(2r_3 −1)(2r_4 −1)= =32A+64B+128C+256D+16= =−240
$${r}^{\mathrm{4}} +\mathrm{4}{r}+\mathrm{1}=\mathrm{0} \\ $$$${r}^{\mathrm{4}} =−\mathrm{4}{r}−\mathrm{1} \\ $$$${r}^{\mathrm{4}} −\mathrm{1}=−\mathrm{4}{r}−\mathrm{2} \\ $$$$ \\ $$$$\left({r}−{r}_{\mathrm{1}} \right)\left({r}−{r}_{\mathrm{2}} \right)\left({r}−{r}_{\mathrm{3}} \right)\left({r}−{r}_{\mathrm{4}} \right)={r}^{\mathrm{4}} +\mathrm{4}{r}+\mathrm{1} \\ $$$$\Rightarrow \\ $$$${A}={r}_{\mathrm{1}} +{r}_{\mathrm{2}} +{r}_{\mathrm{3}} +{r}_{\mathrm{4}} =\mathrm{0} \\ $$$${B}={r}_{\mathrm{1}} {r}_{\mathrm{2}} +{r}_{\mathrm{1}} {r}_{\mathrm{3}} +{r}_{\mathrm{1}} {r}_{\mathrm{4}} +{r}_{\mathrm{2}} {r}_{\mathrm{3}} +{r}_{\mathrm{2}} {r}_{\mathrm{4}} +{r}_{\mathrm{3}} {r}_{\mathrm{4}} =\mathrm{0} \\ $$$${C}={r}_{\mathrm{1}} {r}_{\mathrm{2}} {r}_{\mathrm{3}} +{r}_{\mathrm{1}} {r}_{\mathrm{2}} {r}_{\mathrm{4}} +{r}_{\mathrm{1}} {r}_{\mathrm{3}} {r}_{\mathrm{4}} +{r}_{\mathrm{2}} {r}_{\mathrm{3}} {r}_{\mathrm{4}} =−\mathrm{4} \\ $$$${D}={r}_{\mathrm{1}} {r}_{\mathrm{2}} {r}_{\mathrm{3}} {r}_{\mathrm{4}} =\mathrm{1} \\ $$$$\Rightarrow{e} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\mathrm{4}} {\prod}}\left({r}_{{i}} ^{\mathrm{4}} −\mathrm{1}\right)= \\ $$$$=\left(−\mathrm{2}\right)^{\mathrm{4}} \left(\mathrm{2}{r}_{\mathrm{1}} −\mathrm{1}\right)\left(\mathrm{2}{r}_{\mathrm{2}} +\mathrm{1}\right)\left(\mathrm{2}{r}_{\mathrm{3}} −\mathrm{1}\right)\left(\mathrm{2}{r}_{\mathrm{4}} −\mathrm{1}\right)= \\ $$$$=\mathrm{32}{A}+\mathrm{64}{B}+\mathrm{128}{C}+\mathrm{256}{D}+\mathrm{16}= \\ $$$$=−\mathrm{240} \\ $$
Commented by behi834171 last updated on 23/Jul/25
thank you so much sir.
$${thank}\:{you}\:{so}\:{much}\:{sir}. \\ $$
Answered by mr W last updated on 23/Jul/25
r^4 −1=−4(r+(1/2))=−4p (r+(1/2)−(1/2))^4 +4(r+(1/2)−(1/2))+1=0 (p−(1/2))^4 +4(p−(1/2))+1=0 ...+(−(1/2))^4 +4(−(1/2))+1=0 ...−((15)/(16))=0 (we only need the constant term) ⇒Π_(i=1) ^4 p_i =−((15)/(16)) Π_(i=1) ^4 (r_i ^4 −1)=(−4)^4 Π_(i=1) ^4 p_i =256×(−((15)/(16)))=−240 ✓
$${r}^{\mathrm{4}} −\mathrm{1}=−\mathrm{4}\left({r}+\frac{\mathrm{1}}{\mathrm{2}}\right)=−\mathrm{4}{p} \\ $$$$\left({r}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}} +\mathrm{4}\left({r}+\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{1}=\mathrm{0} \\ $$$$\left({p}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}} +\mathrm{4}\left({p}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{1}=\mathrm{0} \\ $$$$…+\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{4}} +\mathrm{4}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{1}=\mathrm{0} \\ $$$$…−\frac{\mathrm{15}}{\mathrm{16}}=\mathrm{0}\:\:\:\:\:\left({we}\:{only}\:{need}\:{the}\:{constant}\:{term}\right) \\ $$$$\Rightarrow\underset{{i}=\mathrm{1}} {\overset{\mathrm{4}} {\prod}}{p}_{{i}} =−\frac{\mathrm{15}}{\mathrm{16}} \\ $$$$\underset{{i}=\mathrm{1}} {\overset{\mathrm{4}} {\prod}}\left({r}_{{i}} ^{\mathrm{4}} −\mathrm{1}\right)=\left(−\mathrm{4}\right)^{\mathrm{4}} \underset{{i}=\mathrm{1}} {\overset{\mathrm{4}} {\prod}}{p}_{{i}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{256}×\left(−\frac{\mathrm{15}}{\mathrm{16}}\right)=−\mathrm{240}\:\checkmark \\ $$
Commented by behi834171 last updated on 23/Jul/25
wow! very nice work.thank alot dear master.
$${wow}! \\ $$$${very}\:{nice}\:{work}.{thank}\:{alot}\:{dear}\:{master}. \\ $$
Answered by behi834171 last updated on 23/Jul/25
r^4 −1=(r+1)(r−1)(r+i)(r−i); i=(√(−1)) r^4 +4r+1=𝚷_1 ^4 (r−r_i ) ⇒ { ((1+4×1+1=Π(1−r_i ))),((1−4×1+1=𝚷(−1−r_i ))),((1+4i+1=𝚷(i−r_i ))),((1−4i+1=𝚷(−i−r_i ))) :} ⇒(6)(−2)(2+4i)(2−4i)=Π(r_i ^4 −1)⇒ Π(r_i ^4 −1)=−12(4−16i^2 )=−12×20=−240. ■
$$\boldsymbol{{r}}^{\mathrm{4}} −\mathrm{1}=\left(\boldsymbol{{r}}+\mathrm{1}\right)\left(\boldsymbol{{r}}−\mathrm{1}\right)\left(\boldsymbol{{r}}+\boldsymbol{{i}}\right)\left(\boldsymbol{{r}}−\boldsymbol{{i}}\right);\:\boldsymbol{{i}}=\sqrt{−\mathrm{1}} \\ $$$$\boldsymbol{{r}}^{\mathrm{4}} +\mathrm{4}\boldsymbol{{r}}+\mathrm{1}=\underset{\mathrm{1}} {\overset{\mathrm{4}} {\boldsymbol{\prod}}}\left(\boldsymbol{{r}}−\boldsymbol{{r}}_{\boldsymbol{{i}}} \right) \\ $$$$\Rightarrow\begin{cases}{\mathrm{1}+\mathrm{4}×\mathrm{1}+\mathrm{1}=\Pi\left(\mathrm{1}−\boldsymbol{{r}}_{\boldsymbol{{i}}} \right)}\\{\mathrm{1}−\mathrm{4}×\mathrm{1}+\mathrm{1}=\boldsymbol{\Pi}\left(−\mathrm{1}−\boldsymbol{{r}}_{\boldsymbol{{i}}} \right)}\\{\mathrm{1}+\mathrm{4}\boldsymbol{{i}}+\mathrm{1}=\boldsymbol{\Pi}\left(\boldsymbol{{i}}−\boldsymbol{{r}}_{\boldsymbol{{i}}} \right)}\\{\mathrm{1}−\mathrm{4}\boldsymbol{{i}}+\mathrm{1}=\boldsymbol{\Pi}\left(−\boldsymbol{{i}}−\boldsymbol{{r}}_{\boldsymbol{{i}}} \right)}\end{cases} \\ $$$$\Rightarrow\left(\mathrm{6}\right)\left(−\mathrm{2}\right)\left(\mathrm{2}+\mathrm{4}{i}\right)\left(\mathrm{2}−\mathrm{4}{i}\right)=\Pi\left(\boldsymbol{{r}}_{\boldsymbol{{i}}} ^{\mathrm{4}} −\mathrm{1}\right)\Rightarrow \\ $$$$\Pi\left(\boldsymbol{{r}}_{\boldsymbol{{i}}} ^{\mathrm{4}} −\mathrm{1}\right)=−\mathrm{12}\left(\mathrm{4}−\mathrm{16}\boldsymbol{{i}}^{\mathrm{2}} \right)=−\mathrm{12}×\mathrm{20}=−\mathrm{240}.\:\:\blacksquare \\ $$

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