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Question Number 194637 by manxsol last updated on 12/Jul/23

$$$$$$x+y=1$$$$x^2+y^2=2$$$$x^{11}+y^{11}=?$$$$$$$$$$

Commented by Rasheed.Sindhi last updated on 14/Jul/23

$$\text{You can't use 'macro parameter character \#' in math mode}$$

Answered by Rasheed.Sindhi last updated on 12/Jul/23

$$y=1-x\Rightarrow y^2={(1-x)}^2$$$$y^2=2-x^2$$$$1-2x+x^2=2-x^2$$$$2x^2-2x-1=0$$$$x=\frac{2\pm \sqrt{4-8}}{4}=\frac{2\pm 2i}{4}=\frac{1\pm i}{2}$$$$y=1-x=1-\frac{1\pm i}{2}=\frac{2-1\mp i}{2}=\frac{1\mp i}{2}$$$$x+y=\frac{1\pm i}{2}+\frac{1\mp i}{2}=\frac{2}{2}=1$$$$satisfyx+y=1$$$$x^2+y^2={\left(\frac{1\pm i}{2}\right)}^2+{\left(\frac{1\mp i}{2}\right)}^2$$$$=\frac{2i}{4}+\frac{-2i}{4}=0$$$$Donotsatisfy{x}^2+y^2=2$$$$x^{11}+y^{11}={\left(\frac{1\pm i}{2}\right)}^{11}+{\left(\frac{1\mp i}{2}\right)}^{11}$$$$=\frac{-1}{32}???$$

Commented by Tinku Tara last updated on 13/Jul/23

$$2x^2-2x-1=0$$$$\Rightarrow x=\frac{2\pm \sqrt{4+8}}{4}=\frac{1\pm \sqrt{3}}{2}$$$$y=1-x=\frac{1\mp \sqrt{3}}{2}$$$$\mathrm{Mistake}\mathrm{in}\mathrm{your}\mathrm{solution}-8\mathrm{should}$$$$\mathrm{be}+8$$$${\left(\frac{1+\sqrt{3}}{2}\right)}^{11}+{\left(\frac{1-\sqrt{3}}{2}\right)}^{11}$$

Commented by Rasheed.Sindhi last updated on 13/Jul/23

$$\mathbb{T}\mathbf{han}\mathbb{k}\mathbf{s}\mathbf{sir}\mathrm{for}\mathrm{pointing}\mathrm{out}\mathrm{my}$$$$\mathrm{mistake}!$$

Answered by Rasheed.Sindhi last updated on 12/Jul/23

$$x+y=1,x^2+y^2=2,x^{11}+y^{11}=?$$$${(x+y)}^2={\left(1\right)}^2$$$$x^2+y^2+2xy=1$$$$2+2xy=1$$$$xy=-\frac{1}{2}$$$$(x+y)(x^2+y^2)=\left(1\right)\left(2\right)$$$$x^3+y^3+xy(x+y)=2$$$$x^3+y^3+(-\frac{1}{2})\left(1\right)=2$$$$x^3+y^3=2+\frac{1}{2}=\frac{5}{2}$$$$(x^3+y^3)(x+y)=\left(\frac{5}{2}\right)\left(1\right)$$$$x^4+y^4+xy(x^2+y^2)=\frac{5}{2}$$$$x^4+y^4+(-\frac{1}{2})\left(2\right)=\frac{5}{2}$$$$x^4+y^4=\frac{7}{2}$$$$(x^3+y^3)(x^2+y^2)=\left(\frac{5}{2}\right)\left(2\right)$$$$x^5+y^5+x^2{y}^2(x+y)=5$$$$x^5+y^5+{(-\frac{1}{2})}^2\left(1\right)=5$$$$x^5+y^5=5-\frac{1}{4}=\frac{19}{4}$$$${(x^4+y^4)}^2={\left(\frac{7}{2}\right)}^2$$$$x^8+y^8+2x^4{y}^4=\frac{49}{4}$$$$x^8+y^8+2{(-\frac{1}{2})}^4=\frac{49}{4}$$$$x^8+y^8=\frac{49}{4}-\frac{1}{8}=\frac{97}{8}$$$$(x^8+y^8)(x^3+y^3)=\left(\frac{97}{8}\right)\left(\frac{5}{2}\right)$$$$x^{11}+y^{11}+x^3{y}^3(x^5+y^5)=\frac{485}{16}$$$$x^{11}+y^{11}+{(-\frac{1}{2})}^3\left(\frac{19}{4}\right)=\frac{485}{16}$$$$x^{11}+y^{11}=\frac{485}{16}+\frac{19}{32}=\frac{970+19}{32}=\frac{989}{32}$$

Answered by deleteduser1 last updated on 13/Jul/23

$${(x+y)}^2-2xy=2\Rightarrow xy=\frac{-1}{2}$$$$x^{11}+y^{11}=(x^6+y^6)(x^5+y^5)-{\left(xy\right)}^5(x+y)$$$$(x^6+y^6)=[x^2+y^2][{(x^2+y^2)}^2-3{\left(xy\right)}^2]=2(4-\frac{3}{4})=\frac{26}{4}$$$$x^5+y^5=(x^2+y^2)\left[(x+y)\{{(x+y)}^2-3xy\}\right]-{\left(xy\right)}^2(x+y)$$$$=2[1+\frac{3}{2}]-\frac{1}{4}=\frac{19}{4}$$$$\Rightarrow x^{11}+y^{11}=\frac{26}{4}\times \frac{19}{4}+\frac{1}{32}=\frac{989}{32}$$

Answered by manxsol last updated on 15/Jul/23

$$thanksRasheed,AST,Tinkutara$$$$mysolution$$$$r(cos\theta +sin\theta )=1$$$$r=\sqrt{2}$$$$sin(\theta +45)=sin\left(150\right)$$$$\theta =105$$$$(cos{105}^{11}+sin{105}^{11})$$$$\sqrt{2}{}^{11}[{\left(\frac{\sqrt{6}+\sqrt{2}}{4}\right)}^{11}-{\left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)}^{11}]$$$$\sqrt{2}{}^{11}.\frac{\sqrt{2}{}^{11}}{2^{22}}[{(\sqrt{3}+1)}^{11}-{(\sqrt{3}-1)}^{11}]$$$$\frac{1}{2048}\left[2\underset{1}{\sum ^6}{C}_{2k-1}^{11}{3}^{6-k}\right]$$$$\frac{3^6}{1024}\underset{1}{\sum ^6}{C}_{2k-1}^{11}{3}^{-k}$$$$30.90625={\left(\frac{1+\sqrt{3}}{2}\right)}^{11}+{\left(\frac{1-\sqrt{3}}{2}\right)}^{11}$$$$$$$$$$

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