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Question Number 74742 by TawaTawa last updated on 30/Nov/19

$$\mathrm{If}\alpha \mathrm{and}\beta \mathrm{are}\mathrm{the}\mathrm{roots}\mathrm{of}{\mathrm{x}}^2-\mathrm{x}+1=0,$$$$\mathrm{Find}{\alpha }^{23}+{\beta }^{23}\mathrm{without}{\mathrm{demoivre}}^{\prime }\mathrm{s}\mathrm{theorem}.$$

Commented by abdomathmax last updated on 02/Dec/19

$$x^2-x+1=0\to \mathrm{\Delta }=1-4=-3\Rightarrow \alpha =\frac{1+i\sqrt{3}}{2}$$$$and\beta =\frac{1-i\sqrt{3}}{2}=\stackrel{-}{\alpha }wehave{x}^2-x+1=0\Rightarrow$$$$x^2=x-1\Rightarrow x^{2n}={(x-1)}^n\Rightarrow x^{2n+1}=x{(x-1)}^n\Rightarrow$$$${\alpha }^{23}={\alpha }^{2\times 11+1}=\alpha {(\alpha -1)}^{11}also$$$${\beta }^{23}={\left(\stackrel{-}{\alpha }\right)}^{2\times 11+1}=\stackrel{-}{\alpha }{(\stackrel{-}{\alpha }-1)}^{11}\Rightarrow$$$${\alpha }^{23}+{\beta }^{23}=\alpha {(\alpha -1)}^{11}+\stackrel{-}{\alpha }{(\stackrel{-}{\alpha }-1)}^{11}$$$$=2Re\left(\alpha {(\alpha -1)}^{11}\right)but\alpha {(\alpha -1)}^{11}$$$$=\frac{1+i\sqrt{3}}{2}{\left(\frac{i\sqrt{3}-1}{2}\right)}^{11}=\frac{1}{2^{12}}(1+i\sqrt{3})\sum _{k=0}^{11}{C}_{11}^k{\left(i\sqrt{3}\right)}^k{(-1)}^{11-k}$$$$=\frac{1}{2^{12}}(1+i\sqrt{3})\{\sum _{p=0}^5{C}_{11}^{2p}{(-1)}^p{3}^p{(-1)}^{11-2p}$$$$+\sum _{p=1}^5{C}_{11}^{2p+1}i{(-1)}^p\sqrt{3}\times 3^p{(-1)}^{10-2p}\}$$$$resttoextractReslofthisquantity....$$

Answered by Smail last updated on 30/Nov/19

$$\alpha =-jand\beta =-j^2$$$${(-j)}^{23}+{(-j^2)}^{23}=-(j^{23}+\left(j^{46}\right)$$$$=-(j^{21}{j}^2+{jj}^{45})=-({\left(j^3\right)}^7{j}^2+j{\left(j^3\right)}^{15})$$$$=-(j^2+j)=1$$$$Thus,{\alpha }^{23}+{\beta }^{23}=1withj=e^{2i\pi /3}$$

Commented by TawaTawa last updated on 30/Nov/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by TawaTawa last updated on 30/Nov/19

$$\mathrm{But}\mathrm{is}\mathrm{it}\mathrm{possible}\mathrm{without}\mathrm{complex}\mathrm{number}\mathrm{and}$$$$\mathrm{cube}\mathrm{root}\mathrm{of}\mathrm{unity}$$

Answered by mind is power last updated on 30/Nov/19

$$(\mathrm{x}+1)({\mathrm{x}}^2-\mathrm{x}+1)={\mathrm{x}}^3+1=0$$$$\Rightarrow {\alpha }^3={\beta }^3=-1$$$$\Rightarrow \mathrm{\forall }\mathrm{k}\in \mathbb{N}{\alpha }^3={\beta }^3={(-1)}^{\mathrm{k}}=1\mathrm{if}\mathrm{k}\equiv 0\left(2\right),-1\mathrm{else}$$$$23=3.7+2$$$$\Rightarrow {\alpha }^{3.7+2}+{\beta }^{3.7+2}$$$$=(-1).{\alpha }^2-{\left(\beta \right)}^2=-{\alpha }^2-{\beta }^2$$$${\alpha }^2=-1+\alpha ,\beta =-1+\beta$$$$=-(\alpha +\beta )+2$$$$\alpha +\beta =-\frac{-1}{1}=1,\mathrm{som}\mathrm{of}\mathrm{root}$$$$=-1+2=1$$$$$$

Commented by TawaTawa last updated on 30/Nov/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by TawaTawa last updated on 30/Nov/19

$$\mathrm{Can}\mathrm{we}\mathrm{use}\mathrm{it}\mathrm{for}:$$$${\mathrm{x}}^2-2\mathrm{x}+4=0$$$$\mathrm{Find}{\alpha }^{23}+{\beta }^{23}$$

Commented by mind is power last updated on 30/Nov/19

$$\iff \frac{{\mathrm{x}}^2}{4}-\frac{\mathrm{x}}{2}+1=0$$$$\mathrm{let}\mathrm{Y}=\frac{\mathrm{x}}{2}\iff {\mathrm{Y}}^2-\mathrm{Y}+1=0...\mathrm{E}$$$$\mathrm{root}\mathrm{of}\mathrm{E}\mathrm{are}\frac{\alpha }{2},\frac{\beta }{2}$$$$\mathrm{by}\mathrm{previous}\mathrm{Quation}$$$$\mathrm{we}\mathrm{have}{\left(\frac{\alpha }{2}\right)}^{23}+{\left(\frac{\beta }{2}\right)}^{23}=1$$$$\Rightarrow {\alpha }^{23}+{\beta }^{23}=2^{23}$$

Commented by TawaTawa last updated on 30/Nov/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by TawaTawa last updated on 30/Nov/19

$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}.$$

Commented by TawaTawa last updated on 30/Nov/19

$$\mathrm{Sir},\mathrm{is}\mathrm{there}\mathrm{any}\mathrm{way}\mathrm{this}\mathrm{method}\mathrm{cannot}\mathrm{be}\mathrm{used}\mathrm{appart}$$$$\mathrm{from}\mathrm{complex}\mathrm{number}?$$$$\mathrm{Or}\mathrm{the}\mathrm{method}\mathrm{can}\mathrm{work}\mathrm{for}\mathrm{any}\mathrm{question}??$$

Commented by TawaTawa last updated on 30/Nov/19

$$\mathrm{Something}\mathrm{like}:{3\mathrm{x}}^2+5\mathrm{x}+7=0$$$$\mathrm{Find}{\alpha }^{23}+{\beta }^{23}$$

Commented by mind is power last updated on 30/Nov/19

$$\mathrm{in}\mathrm{this}\mathrm{you}\mathrm{have}\mathrm{too}\mathrm{solve}\mathrm{Equation}\mathrm{and}$$$$\mathrm{use}\mathrm{complex}\mathrm{anslysis}$$$$\mathrm{previous}\mathrm{one}\mathrm{our}\mathrm{equation}\Rightarrow {\mathrm{X}}^3=-1\mathrm{witch}\mathrm{redious}\mathrm{calclus}$$$$\mathrm{ther}{3\mathrm{x}}^2+5\mathrm{x}+7=0$$$$\Rightarrow \mathrm{x}=\frac{-5-\mathrm{i}\sqrt{59}}{6},{\mathrm{x}}_2=\frac{-5+\mathrm{i}\sqrt{59}}{6},\mathrm{isee}\mathrm{no}\mathrm{other}\mathrm{methode}\mathrm{i}\mathrm{will}\mathrm{try}\mathrm{later}$$$$$$$$$$$$$$

Commented by TawaTawa last updated on 30/Nov/19

$$\mathrm{I}\mathrm{can}\mathrm{use}\mathrm{complex}\mathrm{number}$$$$\mathrm{Help}\mathrm{me}\mathrm{find}\mathrm{other}\mathrm{method}\mathrm{sir}$$

Commented by TawaTawa last updated on 30/Nov/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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