Question Number 131605 by physicstutes last updated on 06/Feb/21
$$\mathrm{solve}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\:\:{m}^{\mathrm{4}} −\mathrm{7}{m}^{\mathrm{3}} +\:\mathrm{14}{m}^{\mathrm{2}} −\mathrm{7}{m}\:+\:\mathrm{1}\:=\:\mathrm{0}\: \\ $$
Answered by malwan last updated on 06/Feb/21
$${m}^{\mathrm{4}} +\left(−\mathrm{3}{m}^{\mathrm{3}} −\mathrm{4}{m}^{\mathrm{3}} \right) \\ $$$$+\left({m}^{\mathrm{2}} +\mathrm{12}{m}^{\mathrm{2}} +{m}^{\mathrm{2}} \right) \\ $$$$+\left(−\mathrm{4}{m}−\mathrm{3}{m}\right)+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left({m}^{\mathrm{4}} −\mathrm{3}{m}^{\mathrm{2}} +{m}^{\mathrm{2}} \right) \\ $$$$+\left(−\mathrm{4}{m}^{\mathrm{3}} +\mathrm{12}{m}^{\mathrm{2}} −\mathrm{4}{m}\right) \\ $$$$+\left({m}^{\mathrm{2}} −\mathrm{3}{m}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{m}^{\mathrm{2}} \left({m}^{\mathrm{2}} −\mathrm{3}{m}+\mathrm{1}\right) \\ $$$$−\mathrm{4}{m}\left({m}^{\mathrm{2}} −\mathrm{3}{m}+\mathrm{1}\right) \\ $$$$+\left({m}^{\mathrm{2}} −\mathrm{3}{m}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({m}^{\mathrm{2}} −\mathrm{3}{m}+\mathrm{1}\right)\left({m}^{\mathrm{2}} −\mathrm{4}{m}+\mathrm{1}\right)=\mathrm{0} \\ $$$${m}^{\mathrm{2}} −\mathrm{3}{m}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{m}=\frac{\mathrm{3}\pm\sqrt{\mathrm{9}−\mathrm{4}}}{\mathrm{2}}\:=\:\frac{\mathrm{3}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${or}\:{m}^{\mathrm{2}} −\mathrm{4}{m}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{m}=\frac{\mathrm{4}\pm\sqrt{\mathrm{16}−\mathrm{4}}}{\mathrm{2}}\:=\:\mathrm{2}\pm\sqrt{\mathrm{3}} \\ $$
Commented by physicstutes last updated on 06/Feb/21
$$\mathrm{thats}\:\mathrm{awsome}\:\mathrm{sir} \\ $$
Commented by malwan last updated on 06/Feb/21
$${thanks} \\ $$
Answered by EDWIN88 last updated on 06/Feb/21
![(1)(m^2 +km+1)(m^2 +ℓm+1)=m^4 −7m^3 +14m^2 −7m+1 ⇒m^4 +(ℓ+k)m^3 +(2+kℓ)m^2 +(k+ℓ)m+1= m^4 −7m^3 +14m^2 −7m+1 ⇒ { ((k+ℓ=−7⇒k=−7−ℓ)),((2+kℓ=14;kℓ=12)) :} ⇒ℓ(−7−ℓ)=12; ℓ^2 +7ℓ+12=0 → { ((ℓ=−3)),((k=−4)) :} then ⇔ (m^2 −3m+1)(m^2 −4m+1)=0 ⇔[(m−(3/2))^2 −(5/4) ] [(m−2)^2 −3 ] = 0 ⇔ m = { (((3/2)±((√5)/2))),((2± (√3))) :} .](https://www.tinkutara.com/question/Q131628.png)
$$\left(\mathrm{1}\right)\left(\mathrm{m}^{\mathrm{2}} +\mathrm{km}+\mathrm{1}\right)\left(\mathrm{m}^{\mathrm{2}} +\ell\mathrm{m}+\mathrm{1}\right)=\mathrm{m}^{\mathrm{4}} −\mathrm{7m}^{\mathrm{3}} +\mathrm{14m}^{\mathrm{2}} −\mathrm{7m}+\mathrm{1} \\ $$$$\Rightarrow\mathrm{m}^{\mathrm{4}} +\left(\ell+\mathrm{k}\right)\mathrm{m}^{\mathrm{3}} +\left(\mathrm{2}+\mathrm{k}\ell\right)\mathrm{m}^{\mathrm{2}} +\left(\mathrm{k}+\ell\right)\mathrm{m}+\mathrm{1}= \\ $$$$\:\:\:\:\mathrm{m}^{\mathrm{4}} −\mathrm{7m}^{\mathrm{3}} +\mathrm{14m}^{\mathrm{2}} −\mathrm{7m}+\mathrm{1} \\ $$$$\Rightarrow\begin{cases}{\mathrm{k}+\ell=−\mathrm{7}\Rightarrow\mathrm{k}=−\mathrm{7}−\ell}\\{\mathrm{2}+\mathrm{k}\ell=\mathrm{14};\mathrm{k}\ell=\mathrm{12}}\end{cases} \\ $$$$\Rightarrow\ell\left(−\mathrm{7}−\ell\right)=\mathrm{12};\:\ell^{\mathrm{2}} +\mathrm{7}\ell+\mathrm{12}=\mathrm{0}\:\rightarrow\begin{cases}{\ell=−\mathrm{3}}\\{\mathrm{k}=−\mathrm{4}}\end{cases} \\ $$$$\mathrm{then}\:\Leftrightarrow\:\left(\mathrm{m}^{\mathrm{2}} −\mathrm{3m}+\mathrm{1}\right)\left(\mathrm{m}^{\mathrm{2}} −\mathrm{4m}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Leftrightarrow\left[\left(\mathrm{m}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{4}}\:\right]\:\left[\left(\mathrm{m}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{3}\:\right]\:=\:\mathrm{0} \\ $$$$\Leftrightarrow\:\mathrm{m}\:=\:\begin{cases}{\frac{\mathrm{3}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}}\\{\mathrm{2}\pm\:\sqrt{\mathrm{3}}}\end{cases}\:. \\ $$