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Question Number 66197 by Rasheed.Sindhi last updated on 11/Aug/19

If x+(1/x)=1,prove that: x^n +x^(n−2) +x^(n−4) =0

$$\mathrm{If}\:\:\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}=\mathrm{1},\mathrm{prove}\:\mathrm{that}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{n}} +\mathrm{x}^{\mathrm{n}−\mathrm{2}} +\mathrm{x}^{\mathrm{n}−\mathrm{4}} =\mathrm{0} \\ $$

Commented by mr W last updated on 10/Aug/19

VERY NICE!

$$\mathcal{VERY}\:\mathcal{NICE}! \\ $$

Commented by Rasheed.Sindhi last updated on 11/Aug/19

x^n +x^(n−2) +x^(n−4) =x^(n−2) (x^2 +1+(1/x^2 )) =x^(n−2) (x^2 +2+(1/x^2 )−1) =x^(n−2) ( (x+(1/x))^2 −1) =x^(n−2) ({1}^2 −1) =x^(n−2) .0 =0

$$\:\:\:\:\:\:\:\:\:\:\mathrm{x}^{\mathrm{n}} +\mathrm{x}^{\mathrm{n}−\mathrm{2}} +\mathrm{x}^{\mathrm{n}−\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{x}^{\mathrm{n}−\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{x}^{\mathrm{n}−\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{2}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }−\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{x}^{\mathrm{n}−\mathrm{2}} \left(\:\left(\mathrm{x}+\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{x}^{\mathrm{n}−\mathrm{2}} \left(\left\{\mathrm{1}\right\}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{x}^{\mathrm{n}−\mathrm{2}} .\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:=\mathrm{0} \\ $$

Commented by Rasheed.Sindhi last updated on 11/Aug/19

ThankSs for Encouraging Sir!

$$\mathcal{T}{hank}\mathcal{S}{s}\:{for}\:\mathcal{E}{ncouraging}\:\mathcal{S}{ir}! \\ $$

Answered by som(math1967) last updated on 10/Aug/19

x+(1/x)=1 ((x^2 +1)/x)=1 x^2 −x+1=0⇒(x+1)(x^2 −x+1)=0 ⇒x^3 +1=0∴x^3 =−1 ⇒x^3 .x=−x ∴x^4 =−x Now x^n +x^(n−2) +x^(n−4) =x^n (1+x^(−2) +x^(−4) ) =x^n (1+(1/x^2 )+(1/x^4 ))=x^n (((x^4 +x^2 +1)/x^4 )) =x^n (((x^2 −x+1)/x^4 )) ★ =x^n ((0/x^4 ))=0 ★[★ ★x^4 =−x ★★x^2 −x+1=0

$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{1} \\ $$$$\frac{{x}^{\mathrm{2}} +\mathrm{1}}{{x}}=\mathrm{1} \\ $$$${x}^{\mathrm{2}} −{x}+\mathrm{1}=\mathrm{0}\Rightarrow\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{3}} +\mathrm{1}=\mathrm{0}\therefore{x}^{\mathrm{3}} =−\mathrm{1}\:\Rightarrow{x}^{\mathrm{3}} .{x}=−{x} \\ $$$$\therefore{x}^{\mathrm{4}} =−{x} \\ $$$${Now}\:{x}^{{n}} +{x}^{{n}−\mathrm{2}} +{x}^{{n}−\mathrm{4}} \\ $$$$={x}^{{n}} \left(\mathrm{1}+{x}^{−\mathrm{2}} +{x}^{−\mathrm{4}} \right) \\ $$$$={x}^{{n}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right)={x}^{{n}} \left(\frac{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{4}} }\right) \\ $$$$={x}^{{n}} \left(\frac{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{{x}^{\mathrm{4}} }\right)\:\:\bigstar \\ $$$$={x}^{{n}} \left(\frac{\mathrm{0}}{{x}^{\mathrm{4}} }\right)=\mathrm{0}\:\:\bigstar\left[\bigstar\right. \\ $$$$\bigstar{x}^{\mathrm{4}} =−{x}\:\:\:\:\:\: \\ $$$$\bigstar\bigstar{x}^{\mathrm{2}} −{x}+\mathrm{1}=\mathrm{0} \\ $$

Commented by Rasheed.Sindhi last updated on 11/Aug/19

THanks Sir!

$$\mathcal{TH}{anks}\:\mathcal{S}{ir}! \\ $$

Answered by mr W last updated on 10/Aug/19

x≠0 x+(1/x)=1 ⇒x^2 −x+1=0 ⇒x^2 =x−1 ⇒x^4 =(x−1)^2 =x^2 −2x+1 ⇒x^4 +x^2 +1=2x^2 −2x+2 ⇒x^4 +x^2 +1=2(x^2 −x+1) ⇒x^4 +x^2 +1=0 ⇒x^(n−4) (x^4 +x^2 +1)=0 ⇒x^n +x^(n−2) +x^(n−4) =0

$${x}\neq\mathrm{0} \\ $$$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −{x}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} ={x}−\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{4}} =\left({x}−\mathrm{1}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}=\mathrm{2}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2} \\ $$$$\Rightarrow{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}=\mathrm{2}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right) \\ $$$$\Rightarrow{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{x}^{{n}−\mathrm{4}} \left({x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}^{{n}} +{x}^{{n}−\mathrm{2}} +{x}^{{n}−\mathrm{4}} =\mathrm{0} \\ $$

Commented by Rasheed.Sindhi last updated on 11/Aug/19

Thαnks Sir!

$$\mathbb{T}\mathrm{h}\alpha\mathrm{n}\Bbbk\mathrm{s}\:\mathrm{Sir}! \\ $$

Answered by gunawan last updated on 11/Aug/19

x^2 +x^(−2) +2=1 x^2 +x^(−2) +1=0 x^n +x^(n−2) +x^(n−4) =0 x^(n−2) (1+x^2 +x^(−2) ) =x^(n−2) ×0=0

$${x}^{\mathrm{2}} +{x}^{−\mathrm{2}} +\mathrm{2}=\mathrm{1} \\ $$$${x}^{\mathrm{2}} +{x}^{−\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$ \\ $$$${x}^{{n}} +{x}^{{n}−\mathrm{2}} +{x}^{{n}−\mathrm{4}} =\mathrm{0} \\ $$$${x}^{{n}−\mathrm{2}} \left(\mathrm{1}+{x}^{\mathrm{2}} +{x}^{−\mathrm{2}} \right) \\ $$$$={x}^{{n}−\mathrm{2}} ×\mathrm{0}=\mathrm{0} \\ $$

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