Question Number 192370 by Tawa11 last updated on 15/May/23
$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}\:\:\:\mathrm{x}^{\mathrm{4}} \:\:−\:\:\mathrm{2x}^{\mathrm{3}} \:\:+\:\:\mathrm{4x}^{\mathrm{2}} \:\:+\:\:\mathrm{6x}\:\:\:−\:\:\mathrm{21}\:\:\:=\:\:\:\mathrm{0},\:\: \\ $$$$\mathrm{Given}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{two}\:\mathrm{of}\:\mathrm{its}\:\mathrm{roots}\:\mathrm{is}\:\mathrm{zero} \\ $$
Answered by Frix last updated on 16/May/23
![(x−a)(x+a)(x−b)(x−c)=0 x^4 −(b+c)x^3 −(a^2 −bc)x^3 +a^2 (b+c)x−a^2 bc=0 [1] b+c=2 [2] a^2 −bc=−4 [3] a^2 (b+c)=6 [4] a^2 bc=21 [1]&[3] ⇒ a^2 =3 ⇒ a=±(√3) ⇒ [2], [4] bc=7 ⇒ b+c=2∧bc=7 ⇒ b=1±(√6)i∧c=1∓(√6)i](https://www.tinkutara.com/question/Q192379.png)
$$\left({x}−{a}\right)\left({x}+{a}\right)\left({x}−{b}\right)\left({x}−{c}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{4}} −\left({b}+{c}\right){x}^{\mathrm{3}} −\left({a}^{\mathrm{2}} −{bc}\right){x}^{\mathrm{3}} +{a}^{\mathrm{2}} \left({b}+{c}\right){x}−{a}^{\mathrm{2}} {bc}=\mathrm{0} \\ $$$$\left[\mathrm{1}\right]\:\:\:\:\:{b}+{c}=\mathrm{2} \\ $$$$\left[\mathrm{2}\right]\:\:\:\:\:{a}^{\mathrm{2}} −{bc}=−\mathrm{4} \\ $$$$\left[\mathrm{3}\right]\:\:\:\:\:{a}^{\mathrm{2}} \left({b}+{c}\right)=\mathrm{6} \\ $$$$\left[\mathrm{4}\right]\:\:\:\:\:{a}^{\mathrm{2}} {bc}=\mathrm{21} \\ $$$$ \\ $$$$\left[\mathrm{1}\right]\&\left[\mathrm{3}\right]\:\Rightarrow\:{a}^{\mathrm{2}} =\mathrm{3}\:\Rightarrow\:{a}=\pm\sqrt{\mathrm{3}} \\ $$$$\Rightarrow \\ $$$$\left[\mathrm{2}\right],\:\left[\mathrm{4}\right]\:\:\:\:\:{bc}=\mathrm{7} \\ $$$$\Rightarrow \\ $$$${b}+{c}=\mathrm{2}\wedge{bc}=\mathrm{7} \\ $$$$\Rightarrow\:{b}=\mathrm{1}\pm\sqrt{\mathrm{6}}\mathrm{i}\wedge{c}=\mathrm{1}\mp\sqrt{\mathrm{6}}\mathrm{i} \\ $$
Commented by Tawa11 last updated on 16/May/23
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by a.lgnaoui last updated on 16/May/23
![l equatoon admetant deux racines opposees (α et −α verifie[alors l equation (x^2 −𝛂)(x^2 +ax+b) x^4 +ax^3 + (b− 𝛂)x^2 − aα x− bα=0 ⇒ a=−2 b−𝛂 =4 ∗ −a𝛂 =+6 ⇒ α=((+6)/2) = +3 −bα =−21 ⇒ b=((21)/3) = +7 ∗b−α=7−3=4 (verifie) donc l equation (E) (x^2 −3)(x^2 −2x+7) { ((x_1 =(√3) x_2 =−x_1 =−(√3) )),((x_3 =((2−2i(√6))/2) x_4 =((2+2i(√6))/2) (△=−24))) :} [ x_3 =1−i(√(6 )) x_4 =1+i(√6) ]](https://www.tinkutara.com/question/Q192380.png)
$$\mathrm{l}\:\mathrm{equatoon}\:\mathrm{admetant}\:\:\mathrm{deux}\:\mathrm{r}\boldsymbol{\mathrm{a}}\mathrm{cines}\:\mathrm{opposees} \\ $$$$\left(\alpha\:\mathrm{et}\:−\alpha\:\mathrm{verifie}\left[\mathrm{alors}\:\mathrm{l}\:\mathrm{equation}\right.\right. \\ $$$$\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\alpha}\right)\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{ax}}+\boldsymbol{\mathrm{b}}\right) \\ $$$$\:\:\boldsymbol{\mathrm{x}}^{\mathrm{4}} +\boldsymbol{\mathrm{ax}}^{\mathrm{3}} \:+\:\left(\boldsymbol{\mathrm{b}}−\:\boldsymbol{\alpha}\right)\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\:\:\mathrm{a}\alpha\:\boldsymbol{\mathrm{x}}−\:\mathrm{b}\alpha=\mathrm{0} \\ $$$$\Rightarrow\:\:\boldsymbol{\mathrm{a}}=−\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\mathrm{b}}−\boldsymbol{\alpha}\:=\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\:\:\ast \\ $$$$\:\:\:\:−\boldsymbol{\mathrm{a}\alpha}\:\:\:=+\mathrm{6}\:\:\:\:\:\:\:\Rightarrow\:\:\alpha=\frac{+\mathrm{6}}{\mathrm{2}}\:\:=\:+\mathrm{3} \\ $$$$\:\:\:\:\:\:\:−\mathrm{b}\alpha\:=−\mathrm{21}\:\:\:\Rightarrow\:\:\mathrm{b}=\frac{\mathrm{21}}{\mathrm{3}}\:\:\:=\:+\mathrm{7} \\ $$$$\:\:\:\ast\mathrm{b}−\alpha=\mathrm{7}−\mathrm{3}=\mathrm{4}\:\:\left(\mathrm{verifie}\right) \\ $$$$\:\:\mathrm{donc}\:\:\mathrm{l}\:\mathrm{equation}\:\left(\mathrm{E}\right) \\ $$$$\:\:\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{3}\right)\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{7}\right)\:\: \\ $$$$\:\:\:\begin{cases}{\boldsymbol{\mathrm{x}}_{\mathrm{1}} =\sqrt{\mathrm{3}}\:\:\:\:\boldsymbol{\mathrm{x}}_{\mathrm{2}} =−\boldsymbol{\mathrm{x}}_{\mathrm{1}} =−\sqrt{\mathrm{3}}\:}\\{\boldsymbol{\mathrm{x}}_{\mathrm{3}} =\frac{\mathrm{2}−\mathrm{2i}\sqrt{\mathrm{6}}}{\mathrm{2}}\:\:\:\:\:\:\:\:\mathrm{x}_{\mathrm{4}} =\frac{\mathrm{2}+\mathrm{2i}\sqrt{\mathrm{6}}}{\mathrm{2}}\:\:\:\left(\bigtriangleup=−\mathrm{24}\right)}\end{cases}\: \\ $$$$\:\:\:\:\:\left[\:\:\:\boldsymbol{\mathrm{x}}_{\mathrm{3}} =\mathrm{1}−\mathrm{i}\sqrt{\mathrm{6}\:}\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}_{\mathrm{4}} =\mathrm{1}+\mathrm{i}\sqrt{\mathrm{6}}\:\:\:\:\right] \\ $$$$ \\ $$
Commented by Tawa11 last updated on 16/May/23
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$