Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 123210 by Khalmohmmad last updated on 24/Nov/20

Answered by Lordose last updated on 24/Nov/20

$$$$$$\sqrt{2+\sqrt{2-\sqrt{2+\mathrm{x}}}}=\mathrm{x}$$$$2+\sqrt{2-\sqrt{2+\mathrm{x}}}={\mathrm{x}}^2$$$$2-\sqrt{2+\mathrm{x}}={({\mathrm{x}}^2-2)}^2$$$${(2-{({\mathrm{x}}^2-2)}^2)}^2=2+\mathrm{x}$$$${\mathrm{x}}^8-{8\mathrm{x}}^6+{20\mathrm{x}}^4-{16\mathrm{x}}^2+4-2-\mathrm{x}=0$$$${\mathrm{x}}^8-{8\mathrm{x}}^6+{20\mathrm{x}}^4-{16\mathrm{x}}^2-\mathrm{x}+2=0$$$$(\mathrm{x}+1)(\mathrm{x}-2)({\mathrm{x}}^6+{\mathrm{x}}^5-{5\mathrm{x}}^4-{3\mathrm{x}}^3+{7\mathrm{x}}^2+\mathrm{x}-1)=0$$$$\mathrm{x}=-1\mathrm{or}\mathrm{x}=2\mathrm{\forall }\{\mathrm{x}\in \mathbb{Z}\}$$

Commented by MJS_new last updated on 24/Nov/20

$$x=-1\mathrm{must}\mathrm{be}\mathrm{wrong}\mathrm{because}\sqrt{r}=-1\mathrm{has}$$$$\mathrm{no}\mathrm{solution}\mathrm{in}\mathbb{C}$$$$x=2\mathrm{leads}\mathrm{to}\sqrt{2+\sqrt{2-\sqrt{2+2}}}=\sqrt{2+\sqrt{2-2}}=\sqrt{2}\ne 2$$$$\mathrm{as}\mathrm{you}\mathrm{should}\mathrm{know}\mathrm{squaring}\mathrm{introduces}$$$$\mathrm{wrong}\mathrm{solutions}$$

Commented by 676597498 last updated on 24/Nov/20

$$true$$

Commented by MJS_new last updated on 24/Nov/20

$$x^6+x^5-5x^4-3x^3+7x^2+x-1=0$$$$(x^3-3x+1)(x^3+x^2-2x-1)=0$$$$\mathrm{solving}\mathrm{both}\mathrm{using}\mathrm{the}\mathrm{trigonometric}\mathrm{method}$$$$\mathrm{and}\mathrm{testing}\mathrm{the}\mathrm{solutions}\mathrm{we}\mathrm{get}$$$$x_1=-2\cos \frac{\pi }{9}\mathrm{false}$$$$x_2=2\sin \frac{\pi }{18}\mathrm{false}$$$$x_3=2\cos \frac{2\pi }{9}\mathrm{true}$$$$x_4=-\frac{1}{3}-\frac{2\sqrt{7}}{3}\cos (\frac{\pi }{6}+\frac{1}{3}\arcsin \frac{\sqrt{7}}{14})\mathrm{false}$$$$x_5=-\frac{1}{3}-\frac{2\sqrt{7}}{3}\sin \left(\frac{1}{3}\arcsin \frac{\sqrt{7}}{14}\right)\mathrm{false}$$$$x_6=-\frac{1}{3}+\frac{2\sqrt{7}}{3}\sin (\frac{\pi }{3}+\frac{1}{3}\arcsin \frac{\sqrt{7}}{14})\mathrm{false}$$$$\Rightarrow x=2\cos \frac{2\pi }{9}$$

Commented by Lordose last updated on 24/Nov/20

$$\mathrm{True}$$$${\mathrm{Didn}}^{\prime }\mathrm{t}\mathrm{thought}\mathrm{that}$$

Commented by Lordose last updated on 24/Nov/20

$$\mathrm{Thanks}\mathrm{sir}$$

Commented by Khalmohmmad last updated on 24/Nov/20

$$Thankssir$$

Answered by Dwaipayan Shikari last updated on 24/Nov/20

$$\sqrt{2+\sqrt{2-\sqrt{2+2cos\theta }}}=2cos\theta x=2cos\theta$$$$\sqrt{2+\sqrt{2-2cos\frac{\theta }{2}}}=2cos\theta$$$$\sqrt{2+2sin\frac{\theta }{4}}=2cos\theta$$$$\sqrt{2}(sin\frac{\theta }{8}+cos\frac{\theta }{8})=2cos\theta$$$$\left(cos(\frac{\pi }{4}-\frac{\theta }{8})\right)=cos\theta \Rightarrow \theta =2k\pi +\frac{2\pi }{9}$$$$x=2cos\frac{2\pi }{9}As\left(xispositive\right)$$

Commented by Khalmohmmad last updated on 24/Nov/20

$$Thankssir$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com