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Question Number 172326 by nurtani last updated on 25/Jun/22

If (x^3 +x^2 +x)+((1/x^3 )+(1/x^2 )+(1/x)) = 70 x^4 +(1/x^4 ) = ?

$${If}\: \\ $$$$\:\:\:\:\:\:\:\:\:\left({x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}\right)+\left(\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}}\right)\:=\:\mathrm{70} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\:=\:\:? \\ $$

Commented by infinityaction last updated on 25/Jun/22

x^3 +(1/x^3 )+x^2 +(1/x^2 )+x+(1/x) = 70 (x+(1/x))^3 −3(x+(1/x))+(x+(1/x))^2 −2+x+(1/x) = 70 let x+(1/x) = a a^3 +a^2 −2a−72 = 0 (a−4)(a^2 +5a+18) = 0 a= 4 , −(5/2)±(((√(47))i)/2) a = 4 so x+(1/x) = 4 x^2 + (1/x^2 ) = 14 x^4 +(1/x^4 ) = 194

$$\:\:\:\:{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+{x}+\frac{\mathrm{1}}{{x}}\:=\:\mathrm{70} \\ $$$$\:\:\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} −\mathrm{3}\left({x}+\frac{\mathrm{1}}{{x}}\right)+\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}+{x}+\frac{\mathrm{1}}{{x}}\:=\:\mathrm{70} \\ $$$$\:\:\:\:\:{let}\:\:\:\:\:{x}+\frac{\mathrm{1}}{{x}}\:\:=\:\:{a} \\ $$$$\:\:\:\:\:{a}^{\mathrm{3}} +{a}^{\mathrm{2}} −\mathrm{2}{a}−\mathrm{72}\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\left({a}−\mathrm{4}\right)\left({a}^{\mathrm{2}} +\mathrm{5}{a}+\mathrm{18}\right)\:=\:\mathrm{0} \\ $$$$\:\:\:\:\:\:{a}=\:\mathrm{4}\:\:,\:\:\:−\frac{\mathrm{5}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{47}}{i}}{\mathrm{2}} \\ $$$$\:\:\:\:\:{a}\:\:=\:\:\mathrm{4} \\ $$$$\:\:\:{so} \\ $$$$\:\:\:\:\:\:\:\:\:{x}+\frac{\mathrm{1}}{{x}}\:\:=\:\:\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} +\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\:\:=\:\:\mathrm{14} \\ $$$$\:\:\:\:\:\:\:\:\:{x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\:\:=\:\:\mathrm{194} \\ $$

Commented by nurtani last updated on 25/Jun/22

Thank you sir !

$$\mathcal{T}\boldsymbol{{hank}}\:\boldsymbol{{you}}\:\boldsymbol{{sir}}\:! \\ $$

Commented by Tawa11 last updated on 25/Jun/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Commented by peter frank last updated on 27/Jun/22

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

Answered by greougoury555 last updated on 25/Jun/22

(x^2 +x+1)((1/x^2 )+(1/x)+1)=70 (x^2 +x+1)(((x^2 +x+1)/x^2 ))=70 (x^2 +x+1)^2 = 70x^2 (x^2 +x+1−x(√(70)))(x^2 +x+x(√(70))+1)=0 { ((x^2 +(1−(√(70)))x+1=0)),((x^2 +(1+(√(70)))x+1=0)) :}

$$\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}}+\mathrm{1}\right)=\mathrm{70} \\ $$$$\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\left(\frac{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{{x}^{\mathrm{2}} }\right)=\mathrm{70} \\ $$$$\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} \:=\:\mathrm{70}{x}^{\mathrm{2}} \\ $$$$\left({x}^{\mathrm{2}} +{x}+\mathrm{1}−{x}\sqrt{\mathrm{70}}\right)\left({x}^{\mathrm{2}} +{x}+{x}\sqrt{\mathrm{70}}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\begin{cases}{{x}^{\mathrm{2}} +\left(\mathrm{1}−\sqrt{\mathrm{70}}\right){x}+\mathrm{1}=\mathrm{0}}\\{{x}^{\mathrm{2}} +\left(\mathrm{1}+\sqrt{\mathrm{70}}\right){x}+\mathrm{1}=\mathrm{0}}\end{cases} \\ $$$$ \\ $$

Answered by Rasheed.Sindhi last updated on 26/Jun/22

x^2 (x+1+(1/x))+(1/x^2 )((1/x)+1+x)=70 (x+1+(1/x))(x^2 +(1/x^2 ))=70 (x+(1/x)+1)((x+(1/x))^2 −2)=70 (y+1)(y^2 −2)=70 ; x+(1/x)=y y^3 +y^2 −2y−72= (y−4)(y^2 +5y+18)=0 y=4 ∣ y=((−5±(√(25−72)))/2)=((−5±i(√(47)))/2) x+(1/x)=4 x^2 +(1/x^2 )=16−2=14 x^4 +(1/x^4 )=196−2=194

$${x}^{\mathrm{2}} \left({x}+\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left(\frac{\mathrm{1}}{{x}}+\mathrm{1}+{x}\right)=\mathrm{70} \\ $$$$\left({x}+\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)=\mathrm{70} \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}}+\mathrm{1}\right)\left(\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}\right)=\mathrm{70} \\ $$$$\left({y}+\mathrm{1}\right)\left({y}^{\mathrm{2}} −\mathrm{2}\right)=\mathrm{70}\:;\:\:\:{x}+\frac{\mathrm{1}}{{x}}={y} \\ $$$${y}^{\mathrm{3}} +{y}^{\mathrm{2}} −\mathrm{2}{y}−\mathrm{72}= \\ $$$$\left({y}−\mathrm{4}\right)\left({y}^{\mathrm{2}} +\mathrm{5}{y}+\mathrm{18}\right)=\mathrm{0} \\ $$$${y}=\mathrm{4}\:\mid\:{y}=\frac{−\mathrm{5}\pm\sqrt{\mathrm{25}−\mathrm{72}}}{\mathrm{2}}=\frac{−\mathrm{5}\pm{i}\sqrt{\mathrm{47}}}{\mathrm{2}} \\ $$$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{4} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{16}−\mathrm{2}=\mathrm{14} \\ $$$${x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }=\mathrm{196}−\mathrm{2}=\mathrm{194} \\ $$

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