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Question Number 174092 by behi834171 last updated on 24/Jul/22

x^4 +ax^2 +14x−210=0 x_1 .x_2 =22, a=?

$$\:\:\:\:\:\:\boldsymbol{{x}}^{\mathrm{4}} +\boldsymbol{{ax}}^{\mathrm{2}} +\mathrm{14}\boldsymbol{{x}}−\mathrm{210}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{{x}}_{\mathrm{1}} .\boldsymbol{{x}}_{\mathrm{2}} =\mathrm{22},\:\:\:\:\:\:\:\boldsymbol{{a}}=? \\ $$

Answered by Rasheed.Sindhi last updated on 24/Jul/22

x_2 =((22)/x_1 ) x_1 ^4 +ax_1 ^2 +14x_1 −210=(((22)/x_1 ))^4 +a(((22)/x_1 ))^2 +14(((22)/x_1 ))−210=0 x_1 ^4 +ax_1 ^2 +14x_1 −210=((22^4 )/x_1 ^4 )+a(((22^2 )/x_1 ^2 ))+14(((22)/x_1 ))−210=0 x_1 ^8 +ax_1 ^6 +14x_1 ^5 −210x_1 ^4 =22^4 +22^2 ax_1 ^2 +14(22)x_1 ^3 −210x_1 ^4 =0 x_1 ^8 +ax_1 ^6 +14x_1 ^5 =22^4 +22^2 ax_1 ^2 +14(22)x_1 ^3 =0 Continue

$${x}_{\mathrm{2}} =\frac{\mathrm{22}}{{x}_{\mathrm{1}} } \\ $$$${x}_{\mathrm{1}} ^{\mathrm{4}} +{ax}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{14}{x}_{\mathrm{1}} −\mathrm{210}=\left(\frac{\mathrm{22}}{{x}_{\mathrm{1}} }\right)^{\mathrm{4}} +{a}\left(\frac{\mathrm{22}}{{x}_{\mathrm{1}} }\right)^{\mathrm{2}} +\mathrm{14}\left(\frac{\mathrm{22}}{{x}_{\mathrm{1}} }\right)−\mathrm{210}=\mathrm{0} \\ $$$${x}_{\mathrm{1}} ^{\mathrm{4}} +{ax}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{14}{x}_{\mathrm{1}} −\mathrm{210}=\frac{\mathrm{22}^{\mathrm{4}} }{{x}_{\mathrm{1}} ^{\mathrm{4}} }+{a}\left(\frac{\mathrm{22}^{\mathrm{2}} }{{x}_{\mathrm{1}} ^{\mathrm{2}} }\right)+\mathrm{14}\left(\frac{\mathrm{22}}{{x}_{\mathrm{1}} }\right)−\mathrm{210}=\mathrm{0} \\ $$$${x}_{\mathrm{1}} ^{\mathrm{8}} +{ax}_{\mathrm{1}} ^{\mathrm{6}} +\mathrm{14}{x}_{\mathrm{1}} ^{\mathrm{5}} −\mathrm{210}{x}_{\mathrm{1}} ^{\mathrm{4}} =\mathrm{22}^{\mathrm{4}} +\mathrm{22}^{\mathrm{2}} {ax}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{14}\left(\mathrm{22}\right){x}_{\mathrm{1}} ^{\mathrm{3}} −\mathrm{210}{x}_{\mathrm{1}} ^{\mathrm{4}} =\mathrm{0} \\ $$$${x}_{\mathrm{1}} ^{\mathrm{8}} +{ax}_{\mathrm{1}} ^{\mathrm{6}} +\mathrm{14}{x}_{\mathrm{1}} ^{\mathrm{5}} =\mathrm{22}^{\mathrm{4}} +\mathrm{22}^{\mathrm{2}} {ax}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{14}\left(\mathrm{22}\right){x}_{\mathrm{1}} ^{\mathrm{3}} =\mathrm{0} \\ $$$$ \\ $$$${Continue} \\ $$

Commented by behi834171 last updated on 24/Jul/22

waiting for final resualts..... dear master

$${waiting}\:{for}\:{final}\:{resualts}.....\:{dear}\:{master} \\ $$

Commented by MJS_new last updated on 24/Jul/22

the key is x^4 +0x^3 +px^2 +qx+r always has two square factors (x^2 −αx−β)(x^2 +αx−γ)

$$\mathrm{the}\:\mathrm{key}\:\mathrm{is} \\ $$$${x}^{\mathrm{4}} +\mathrm{0}{x}^{\mathrm{3}} +{px}^{\mathrm{2}} +{qx}+{r} \\ $$$$\mathrm{always}\:\mathrm{has}\:\mathrm{two}\:\mathrm{square}\:\mathrm{factors} \\ $$$$\left({x}^{\mathrm{2}} −\alpha{x}−\beta\right)\left({x}^{\mathrm{2}} +\alpha{x}−\gamma\right) \\ $$

Commented by Rasheed.Sindhi last updated on 25/Jul/22

Thanks for real help sir!

$$\mathbb{T}\boldsymbol{\mathrm{han}}\Bbbk\boldsymbol{\mathrm{s}}\:\mathrm{for}\:\mathrm{real}\:\mathrm{help}\:\boldsymbol{\mathrm{sir}}! \\ $$

Answered by MJS_new last updated on 24/Jul/22

let x_1 =p∧x_2 =((22)/p) ⇒ x^4 +ax^2 +14x−210=(x^2 −((p^2 +22)/p)x+22)(x^2 +((p^2 +22)/p)x−((105)/(11))) since p≠0 we get by matching constants { ((p^2 −((154)/(347))p+22=0)),((a=−((11p^4 +347p^2 +5324)/(11p^2 )))) :} ⇒ p=((77)/(347))±((√(2643069))/(347))i a=((16235157)/(1324499))

$$\mathrm{let}\:{x}_{\mathrm{1}} ={p}\wedge{x}_{\mathrm{2}} =\frac{\mathrm{22}}{{p}} \\ $$$$\Rightarrow \\ $$$${x}^{\mathrm{4}} +{ax}^{\mathrm{2}} +\mathrm{14}{x}−\mathrm{210}=\left({x}^{\mathrm{2}} −\frac{{p}^{\mathrm{2}} +\mathrm{22}}{{p}}{x}+\mathrm{22}\right)\left({x}^{\mathrm{2}} +\frac{{p}^{\mathrm{2}} +\mathrm{22}}{{p}}{x}−\frac{\mathrm{105}}{\mathrm{11}}\right) \\ $$$$\mathrm{since}\:{p}\neq\mathrm{0}\:\mathrm{we}\:\mathrm{get}\:\mathrm{by}\:\mathrm{matching}\:\mathrm{constants} \\ $$$$\begin{cases}{{p}^{\mathrm{2}} −\frac{\mathrm{154}}{\mathrm{347}}{p}+\mathrm{22}=\mathrm{0}}\\{{a}=−\frac{\mathrm{11}{p}^{\mathrm{4}} +\mathrm{347}{p}^{\mathrm{2}} +\mathrm{5324}}{\mathrm{11}{p}^{\mathrm{2}} }}\end{cases} \\ $$$$\Rightarrow \\ $$$${p}=\frac{\mathrm{77}}{\mathrm{347}}\pm\frac{\sqrt{\mathrm{2643069}}}{\mathrm{347}}\mathrm{i} \\ $$$${a}=\frac{\mathrm{16235157}}{\mathrm{1324499}} \\ $$

Commented by behi834171 last updated on 24/Jul/22

thank you very much dear master.

$${thank}\:{you}\:{very}\:{much}\:{dear}\:{master}. \\ $$

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