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Question Number 126329 by mnjuly1970 last updated on 19/Dec/20

Answered by Olaf last updated on 19/Dec/20

$$x^5-5x^4-10x^3+10x^2+5x-1=0$$$$(x^5-1)-5(x^4-x)-10(x^3-x^2)=0$$$$x=1\mathrm{or}:$$$$(x^4+x^3+x^2+x+1)-5x(x^2+x+1)-10x^2=0$$$$x^4-4x^3-14x^2-4x+1=0$$$$x^2-4x-14-\frac{4}{x}+\frac{1}{x^2}=0$$$$(x^2+\frac{1}{x^2})-4(x+\frac{1}{x})-14=0$$$${(x+\frac{1}{x})}^2-2-4(x+\frac{1}{x})-14=0$$$${(x+\frac{1}{x})}^2-4(x+\frac{1}{x})-16=0$$$$\mathrm{Let}u=x+\frac{1}{x}$$$$u^2-4u-16=0$$$$u=\frac{4\pm \sqrt{16-4\left(1\right)(-16)}}{2}=2(1\pm \sqrt{5})$$$$x+\frac{1}{x}=u$$$$x^2-ux+1=0$$$$x=\frac{u\pm \sqrt{u^2-4}}{2}$$$$\mathrm{If}u=2(1-\sqrt{5}),u^2-4=4(5-2\sqrt{5})$$$$x=1-\sqrt{5}\pm 2\sqrt{5-2\sqrt{5}}$$$$\mathrm{If}u=2(1+\sqrt{5}),u^2-4=4(5+2\sqrt{5})$$$$x=1+\sqrt{5}\pm 2\sqrt{5+2\sqrt{5})}$$

Commented by mnjuly1970 last updated on 19/Dec/20

$$merceymrolaf...thankyou...$$

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