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Question Number 188073 by Shrinava last updated on 25/Feb/23

$$\mathrm{If}\mathrm{x}=\sqrt{\frac{1+\sqrt{5}}{\sqrt{5}-1}}\mathrm{find}{5\mathrm{x}}^2-5\mathrm{x}-1=?$$

Answered by som(math1967) last updated on 25/Feb/23

$$x=\sqrt{\frac{{(\sqrt{5}+1)}^2}{4}}=\frac{\sqrt{5}+1}{2}$$$$\frac{1}{x}=\frac{2}{\sqrt{5}+1}=\frac{2(\sqrt{5}-1)}{4}=\frac{\sqrt{5}-1}{2}$$$$x-\frac{1}{x}=1$$$$\frac{x^2-1}{x}=1\Rightarrow x^2-1=x$$$$5x^2-5-5x+4=5(x^2-1)-5x+4$$$$=5x-5x+4=4$$

Commented by Shrinava last updated on 26/Feb/23

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{professor}$$

Answered by Frix last updated on 25/Feb/23

$$\frac{1+\sqrt{5}}{\sqrt{5}-1}=\frac{{(1+\sqrt{5})}^2}{4}\Rightarrow x=\frac{1+\sqrt{5}}{2}=\phi$$$${\phi }^2=\phi +1\Rightarrow$$$$5(\phi +1)-5\phi -1=4$$

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