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Question Number 12797 by tawa last updated on 01/May/17

$$1+\mathrm{x}+{\mathrm{x}}^2+...{\mathrm{x}}^{49}=\frac{1}{2}({\mathrm{x}}^{49}-\frac{1}{\mathrm{x}})$$$$\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{x}$$

Commented by prakash jain last updated on 01/May/17

$$\frac{x^{50}-1}{x-1}=\frac{1}{2}\left(\frac{x^{50}-1}{x}\right)$$$$x\ne 1,x\ne 0$$$$(x^{50}-1)2x=(x-1)(x^{50}-1)$$$$(x^{50}-1)(x+1)=0$$$$x=-1$$$$x^{50}=1\Rightarrow x=\cos \frac{2k\pi }{50}+i\sin \frac{2k\pi }{50}$$$$k=1,2,...,49$$$$k=0\Rightarrow x=1,invalidsincex\ne 1$$

Commented by tawa last updated on 01/May/17

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

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 01/May/17

$$2(x^{50}-1)=(x-1)(x^{50}-1)/x\Rightarrow$$$$2x=x-1\Rightarrow x=-1.◼$$$$if:x^{50}-1=0\Rightarrow thereisnosolotionforx.$$

Commented by tawa last updated on 01/May/17

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

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