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Question Number 88141 by A8;15: last updated on 08/Apr/20

Answered by MJS last updated on 08/Apr/20

$$\mathrm{squaring}\mathrm{and}\mathrm{transforming}3\mathrm{times}\mathrm{leads}\mathrm{to}$$$$x^8-28x^6+238x^4-588x^2+128x-7=0$$$$\mathrm{this}\mathrm{can}\mathrm{be}\mathrm{factorized}$$$$(x^2+2x-7){(x^3-x^2-9x+1)}^2=0$$$$x_{1,2}=-1\pm 2\sqrt{2}$$$$x_3=\frac{1}{3}(1-4\sqrt{7}\sin \frac{\arcsin \frac{\sqrt{7}}{14}}{3})$$$$x_4=\frac{1}{3}(1+4\sqrt{7}\sin \frac{\pi +\arcsin \frac{\sqrt{7}}{14}}{3})$$$$x_5=\frac{1}{3}(1-4\sqrt{7}\sin \frac{2\pi +\arcsin \frac{\sqrt{7}}{14}}{3})$$$$\mathrm{testing}\mathrm{all}\mathrm{solutions}\mathrm{leads}\mathrm{to}x=x_4$$

Commented by ajfour last updated on 08/Apr/20

$$thatitcouldbefactorised,how$$$$canoneguess,Sir?$$

Commented by MJS last updated on 08/Apr/20

$$\mathrm{I}\mathrm{tried}\mathrm{to}\mathrm{approximate}\mathrm{and}\mathrm{found}$$$$x_1\approx -3.82843\mathrm{and}{x}_2\approx 1.82843$$$$\mathrm{knowing}\mathrm{that}\sqrt{2}\approx 1.41421\mathrm{it}\mathrm{was}\mathrm{easy}\mathrm{to}$$$$\mathrm{guess}{x}_1\stackrel{?}{=}-1-2\sqrt{2}\mathrm{and}{x}_2\stackrel{?}{=}-1+2\sqrt{2}$$$$\mathrm{it}\mathrm{was}\mathrm{just}\mathrm{good}\mathrm{luck}$$

Commented by ajfour last updated on 08/Apr/20

$$isee,thankuSir.$$

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